### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# chem CHM1046

FSU

### View Full Document

## 33

## 1

1 review

## Popular in General Chemistry 2

## Popular in Chemistry

This page Bundle was uploaded by ff Notetaker on Thursday November 19, 2015. The Bundle belongs to CHM1046 at Florida State University taught by Kenneth Hanson in Fall 2015. Since its upload, it has received 33 views. For similar materials see General Chemistry 2 in Chemistry at Florida State University.

## Reviews for chem

Why didn't I know about this earlier? This notetaker is awesome, notes were really good and really detailed. Next time I really need help, I know where to turn!

-*Nathen Fadel*

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 11/19/15

CHAPTER 3 MATTER 8 ENERGY Problems 150 5758 6174 107 109115 32 What is Matter Matter Anything that has mass and occupies volume We study matter at different levels macroscopic the level that can be observed with the naked eye eg geologists study rocks and stone at the macroscopic level microscopic the level that can be observed with a microscope eg scientists study tiny animals plants or crystals at microscopic level particulate at the level of atoms and molecules also called atomic or molecularlevel cannot be observed directly even with the most powerful microscopes where the term nanotechnology comes from since many atoms and molecules are about a few nanometers in size lSpeciallzed techniques 39 are required to rosarye I IE Elli Thomson Higher Education Substances like water can be represented using different symbols eg H20 and models i ii H M Lewis Structure or Electronpt liagrtam Hl llazalla tlEtfiuk medal 5Faggsfil limg 15513 CHEM 121 TM Chapter 3 page 1 of 13 33 Classifying Matter According to Its State Solid Liquid and Gas Matter exists in one of three physical states solid liquid gas Copyright 21109 Pearson Prentice Hall inc solid Has definite shape and a fixed or constant and rigid volume Particles only vibrate in place liquid Has a fixed or constant volume but its shape can change Takes the shape of its container because particles are moving Particles are packed closely together but can move around each other gas Volume is variable and particles are far apart from one another Takes the shape of the container because particles are moving gt If container volume expands particles move apart to fill container gt If container volume decreases particles move closer together gt Gases are compressible ie can be forced to occupy a smaller volume Particles are in constant random motion CHEM 121 TM Chapter 3 page 2 of 13 34 Classifying Matter According to Its Composition Elements Compounds and Mixtures We can classify matter into pure substances and mixtures pure substance a single chemical consisting of only one kind of matter There are two types of pure substances elements and compounds In the figure below copper rods are an example of an element and sugar is an example of a compound mixture consists of two or more elements andor compounds Mixtures can be homogeneous or heterogeneous Homogeneous mixtures have a uniform appearance and composition because the particles in them mix uniformly eg solutions like sweetened tea below Heterogeneous mixtures do not have a uniform composition eg chocolate chip cookie water and C8H18 mixture below shown as separate layers Cepyright 2009 Peereen Prent iee Hell lhe CHEM 121 Tro Chapter 3 page 3 of 13 elements consist of only one type of atom atoms cannot be broken down into smaller components by chemical reaction eg copper wire Cu sulfur powder 88 Examples also include sodium Na barium Ba hydrogen gas H2 oxygen gas 02 and chlorine gas H2 compounds consist of more than one type of atom and have a specific chemical formula Examples include hydrogen chloride HCI water H20 sodium chloride NaCl which is table salt barium chloride BaClg Two or more pure substances combine to form mixtures mixtures consist of many compounds andor elements with no specific formula Matter having variable composition with definite or varying properties can be separated into component elements andor compounds eg any alloy like brass steel 10K to 18K gold sea water carbonated soda air consists of nitrogen oxygen and other trace gases The image at the right shows that air is a mixture of mostly nitrogen N2 in blue and some oxygen 02 in red while salt water consists of salt Na and Cl39 ions or charged particles dissolved in water Example ls salt water a homogeneous or heterogeneous mixture Explain ohlorido Each E 39IIzd 2111quot Thomoon Higll39ror Eduoation Copyright E g Pearson Prentice Hall lino V or Maiural Pairit iouilato Illustration Maoroooopio Brimstone Formula Form not to some Photograph Hoium Ha itom A colorless gas 7 Atom Eooium Hir Erostaiiino solid A romi Barium Ea Eruora llilrne 5mm Humorgan H Molecule e colorless goo Elixirgen D Moloouilra Fl milorloao giro Chlorino Cl Molooulo Hurlrogon A r r ohlorido Hm gleam Water Hg Moloou a Sodium CWEEEIEMB thiorido 3 339 mild Eloriusrn Ewstallir ro CHEM 121 TM Chapter 3 page 4 of 13 35 How We Tell Different Kinds of Matter Apart Physical and Chemical Properties The characteristics that distinguish one substance from another are called properties Physical Properties inherent characteristics of a substance independent of other substances physical state solid liquid gas electricalampheat conductivity color odor density hardness melting and boiling points solubility doesdoes not dissolve in water Chemical Properties how a substance reacts with other substances eg hydrogen reacts explosively with oxygen 36 How Matter Changes Physical and Chemical Changes physical change a process that does not alter the chemical makeup of the starting materials Note in the figure below that the H20 molecules remain H20 regardless of the physical state solid liquid or gas gt Changes in physical state are physical changes Other examples of physical changes include hammering gold into foil dry ice subliming Dissolving table salt or sugar in water is also a physical change A substance dissolved in water is the fourth physical state aqueous CHEM 121 TM Chapter 3 page 5 of 13 Know the terms for transitions from one physical state to another freezing liquid gt solid condensinggas gt liquid melting solid gt liquid evaporating or vaporizing liquid gt gas Two less common transitions sublimation solid gt gas eg dry ice sublimes deposition gas gt solid eg water vapor deposits on an icebox chemical change a process that does change the chemical makeup of the starting materials We can show H2 and 02 reacting to form water H20 below Since the H20 has a different chemical makeup than H2 and 02 this is a chemical change Other examples of chemical changes eg oxidation of matter burning or rusting release of gas bubbles fizzing mixing two solutions to form an insoluble solid precipitation and other evidence indicating the starting materials reactants were changed to a different substance The following examples are all chemical changes that convert the reactants to completely different compounds andor elements relea r gas bble formation ofinsoluble solid fizzing precipitation oxidaio buring or usting CHEM 121 TM Chapter 3 page 6 of 13 Example 1 Consider the following molecularlevel representations of different substances For each figure above indicate if it represents an element a compound or a mixture AND if it represents a solid liquid or gas A element compound mixture solid liquid gas B element compound mixture solid liquid gas C element compound mixture solid liquid gas D element compound mixture solid liquid gas E element compound mixture solid liquid gas F element compound mixture solid liquid gas Ex 2 Circle all of the following that are chemical changes burning condensing dissolving rusting vaporizing precipitating CHEM 121 TM Chapter 3 page 7 of 13 37 Conservation of Mass There is No New Matter Chemical Reaction REACTANTS gt PRODUCTS starting materials substances after reaction For the reaction C 02 gt C02 The reactants are carbon and oxygen gas and the product is carbon dioxide Antoine Lavoisier 17431794 a French chemist carried out experiments on combustion by burning different substances and measuring their masses before and after burning He found that there was no change in the overall mass of the sample and air around it gt Law of Conservation of Mass Matter is neither created nor destroyed in a chemical reaction so mass is conserved Mass of the products in a reaction must be equal to the mass of the reactants For example 112 g hydrogen 888 9 oxygen 1000 g water Ex 1 Methane burns by reacting with oxygen present in air to produce steam and carbon dioxide gas Calculate the mass of oxygen that reacts if burning 500 g of methane produces 1123 g of steam and 1371 g of carbon dioxide 38 Energy the capacity to do work potential energy PE energy due to position or its composition chemical bonds A 10Ib bowling ball has higher PE when it is 10 feet off the ground compared to 10 inches off the ground gt Greater damage on your foot after falling 10 feet compared to falling only 10 inches In terms of chemical bonds the stronger the bond gt more energy is required to break the bond gt the higher the potential energy of the bond CHEM 121 TM Chapter 3 page 8 of 13 kinetic energy KE energy associated with an object s motion eg a car moving at 75 mph has much greater KE than the same car moving at 15 mph gt Greater damage if the car crashes at 75 mph than at 15 mph Six Forms of Energy heat light chemical electrical mechanical and nuclear Each can be converted to another Example Identify at least two types of energy involved for each of the following 1 When you turn on a lamp 2 When using solar panels 3 At the Springfield Power Plant in The Simpsons Energy changes accompany physical and chemical changes due to changes in potential and kinetic energy Kinetic Energy and Physical States Solids have the lowest KE of the three physical states Highest attraction between particles gt particles are fixed Liquids have slightly higher KE than solids Particles are still attracted to each other but can move past one another gt particles are less restricted Gases have greatest KE compared to solids and liquids Attractive forces completely overcome so particles fly freely within container gt particles are completely unrestricted So d CHEM 121 Tro Chapter 3 page 9 of 13 310 Temperature Random Molecular and Atomic Motion Heat Energy that is transferred from a body at a higher temperature to one at a lower temperature gt heat always transfers from the hotter to the cooler object quotheat flowquot means heat transfer Heat Transfer and Temperature One becomes hotter by gaining heat One becomes colder by losing heat ie when you feel cold you are actually losing heat Ex 1 Fill in the blanks to indicate how heat is transferred a You burn your hand on a hot frying pan loses heat and gains the heat b Your tongue feels cold when you eat ice cream loses heat and gains the heat Ex 2 A small chunk of gold is heated in beaker 1 which contains boiling water The gold chunk is then transferred to beaker 2 which contains roomtemperature water a The temperature of the water in beaker 2 T L stays the same b Fill in the blanks loses heat and gains the heat Ex 3 Why do surfaces like a stone countertop or metal flatware forks knives etc feel cold in restaurant Aren t they at room temperature like you Explain 39 Energy and Chemical and Physical Change endothermic change a physical or chemical change that requires energy or heat to occur boiling water requires energy H20I heat energy gt H20g electrolysis of water requires energy 2 H20I electrical energy gt 2 H2g 029 exothermic change a physical or chemical change that releases energy or heat water condensing releases energy H20g gt H20I heat energy hydrogen burning releases energy 2H2g 029 gt 2H20g heat energy CHEM 121 TM Chapter 3 page 10 of 13 For physical changes consider whether the reactants or products have more kinetic energy If the reactants have greater kinetic energy than the products gt exothermic process If the products have greater kinetic energy than the reactants gt endothermic process a Melecules close teglether same 5 391le nearest neighbor 139 a 39 ever time b Melecules close rm tiugiethler but Liquid mnnvinlgexc hangiilng UH 9 9 If k Vquot 39J V nearest netgihbers J w c Meleeules widely separated moving it Gas rapidlyr i system that part of the universe being studied surroundings the rest of the universe outside the system For chemical changes observe if the surroundings including you feel hotter or colder after the reaction has occurred If the surroundings are hotter the reaction released heat gt exothermic reaction If the surroundings are colder the reaction absorbed heat gt endothermic reaction Ex 1 Circle all of the following changes that are exothermic freezing vaporizing sublimation melting deposition CHEM 121 TM Chapter 3 page 11 of 13 Ex 2 A student adds ammonium chloride NH4CI salt to a test tube containing water and notices that the test tube feels colder as the ammonium chloride dissolves Is this process exothermic or endothermic Explain Ex 3 A student mixes two solutions hydrochloric acid and sodium hydroxide and notices the beaker with the substances feels hotter as they mix Is this reaction exothermic or endothermic Explain Units of Energy calorie cal unit of energy used most often in the US amount of energy required to raise the temperature of 1 g of water by 1 C 1 cal 5 4184 J Note This is EXACT But a nutritional calorie abbreviated Cal is actually 1000 cal 1 Cal 1 kcal 4184 kJ joule J SI unit of energy To recognize the size of ajoule note that 1 watt 1 g gt So a 100watt light bulb uses 100 J every second Heat is also often reported in kilojoules kJ where 1 kJ 1000 J Electricity usage on our electricity bills are generally reported in kilowatthour kWh Example A 100 W lightbulb running for 100 hours requires 1 kWh of energy If the average Seattle home uses 25 kWhday this is equal to how many 100 W light bulbs running nonstop for one day CHEM 121 TM Chapter 3 page 12 of 13 Energy and Food Values food value The amount of heat released when food is burned completely usually reported in kJg food or Calg food Most of the energy needed by our bodies comes from carbohydrates and fats and the carbohydrates decompose in the intestines into glucose 06H1206 The combustion of glucose produces energy that is quickly supplied to the body C5H1205g 6 029 6 C02g 6 H20g heat energy The body also produces energy from proteins and fats which can be stored because fats are insoluble in water and produce more energy than proteins and carbohydrates The energy content reported on food labels is generally determined using a bomb calorimeter similar to that described in the previous section 311 Temperature Changes Heat Capacity heat capacity amount of heat necessary to raise the temperature of a given amount of any substance by 1 C in units of J C specific heat capacity amount of heat necessary to raise the temperature of 1 gram of any or specific heat substance by 1 C has units of Jg C Water has a relatively high specific heat 4184 Jg C compared to the specific heats of rocks and other solids 13 Jg C for dry Earth 09 Jg C for concrete 046 Jg C for iron Example Water covers most of the Earth Compare the specific heat of water with Earth and other solids given above then explain what would happen if the Earth were covered mostly in land not water CHEM 121 TM Chapter 3 page 13 of 13 3091 Introduction to Solid State Chemistry Lecture Notes No 2 CHEMICAL BONDING Sources for Further Reading 1 Pauling L General Chemistry WH Freeman and Co San Francisco 1970 Chapter 6 Gray HB Chemical Bonds Benjamin 1973 West AR Solid State Chemistry J Wiley 1984 Borg Rd and GJ Dienes Physical Chemistry of Solids Academic Press 1992 Parker SB Editor Physical Chemistry Source Book McGrawHill 1987 it 2 3 4 5 The electronic configurations of the elements as specified in the previous chapter apply in principle only to isolated atoms atoms separated by distances over which no mutual interactions of their electronic orbitals can occur infinite distance This condition is never met in condensed phases ie liquids and solids it is only encountered in high vacua where atoms move over long distances without mutual interaction Under normal conditions particularly in the mentioned condensed phases atoms are separated over distances controlled in essence by the dimension of their respective outermost occupied electronic orbitals Whenever the outer electron shells of two or more atoms come in contact with each other overlap to any extent the potential for interaction reaction exists Spontaneous reaction consisting of a rearrangement of the electronic orbitals andor actual transfer of electrons from one atom to another will take place whenever such a rearrangement results in a lower energy configuration This means that the driving force for mutual interaction and rearrangement of electronic configurations is in most instances but not always manifested through the release of heat a form of energy to the environment Typical case Mg 12 02 gt Reaction Product MgO light heat ie energy Since the same amount of energy must be supplied to the system if the original state ie LN 2 separation of the species is to be reestablished the reaction partners are bonded together by comparable energies The strength of the bonds can obviously vary from system to system with the nature of the electronic rearrangement Even the inert gases particularly the heavier ones like xenon are capable of forming associations with other atoms Sometimes we find that two atoms assume a more stable state by sharing electrons at other times an atom may transfer electrons to another atom in order to achieve a greater stability In still other instances the rearrangement may simply be an orbital distortion or an internal charge redistribution In either event the mutual benefit that accrues is the formation of what is commonly called a chemical bond Through these bonds atoms combine with each other to form very different kinds of particles referred to as molecules and ions 1 NATURE OF CHEMICAL BONDS In order for a chemical bond to be formed between two atoms there must be a net decrease in the energy of the system the two atoms the ions or molecules produced by electronic rearrangements must be in a lower energy state than the atoms were prior to interaction prior to bond formation Since atoms of each of the elements have different electronic structures the variety of possible chemical bonds differing from each other in at least some small way is considerable and is even further increased by the effects of neighboring atoms on the bond under consideration The modes of bond formation can be categorized into two basic types each representing a type of bonding The bonding types are called electrovalent or ionic bonding and covalent bonding Electrovalent bonding arises from complete transfer of one or more electrons from one atom to another covalent bonding arises from the sharing of two or more electrons between atoms Since these models represent the limiting cases we can anticipate that most real bonds will fall LN 2 between these two extremes Two additional types of bonding metallic bonding and Van der Waals bonding will be discussed later Before discussing these models in detail it is appropriate to consider the relationships between the electronic structures of atoms and their chemical reactivity The inert gases Group VIII are the most stable elements with regard to bond formation ie toward electronic rearrangements It is therefore useful to examine the reasons for their stability lnert gases all have electronic structures consisting of filled subshells For all but helium the outer or valence shell contains eight electrons with filled s and p sublevels ns2p6 The electronic structure of helium is 1s2 which is equivalent to the structure of the other inert gases since there is no 1p sublevel lnert gases have high ionization energies because each electron in the sublevel of highest energy is poorly screened from the nucleus by other electrons in its same sublevel Each electron sees relatively high positive charge on the nucleus and a large amount of energy is required to remove it from the atom lnert gases have very low electron affinity because any added electron must enter a significantly higher energy level We find therefore that the electronic structures of inert gases are particularly resistant to changes by either loss or gain of electrons and further that atoms of other elements with fewer or more electrons than inert gas configuration tend to gain or lose electrons respectively to achieve such inert gas structure 2 ELECTROVALENT IONIC BONDING An electrovalent bond is formed by the transfer of one or more electrons from one atom to another Consider first atoms that have electronic structures differing from an inert gas structure by only a few 1 2 or 3 electrons These include the representative elements of Groups I II and III in the Periodic Table which have respectively 1 2 and 3 electrons more than a neighboring inert gas and the representative elements of LN 2 Groups V VI and VII which have respectively 3 2 and 1 electrons less than a neighboring inert gas fig 1 Unstable G Valence shell 6 Q 9 configurations lt9 6 Q 0 G e O 6 G O 63 e e E Q G e 6 Q Valence shell 1 2 5 with stable n3 n3 Octet Configuration p Fig1 Stable and Unstable Valence Shell Configurations The elements of Groups I II and Ill can form the electronic structure of an inert gas by losing their outer 1 2 and 3 valence electrons The resulting species are positively charged ions In a similar electron transfer which however involves the acquisition of electrons in the outer valence levels elements of Groups V VI and VII form an inert gas electronic structure by formation of negatively charged ions It is through the electron transfer between an electronlosing element and an electrongaining element that compounds are formed which involve electrostatic attraction electrovalent bonds of oppositely charged species called ions 1e Na F a Na F 2e 0 Mg 2F gt Mg 2F In the notation Nao the dot indicates the outermost electron which is in excess of the rare gas configuration It is referred to as a valence electron LN 2 Elements immediately following the inert gases in the horizontal columns of the Periodic Table lose electrons and those immediately preceding the inert gases gain electrons on interaction The resulting compounds are called electrovalent the valence number charge on the ion of a particular element when it forms an electrovalent compound is given by the number of electrons lost or gained in changing from the atomic to the ionic state The stoichiometric formula of an electrovalent compound reflects the ratio usually very simple of positive to negative ions that gives a neutral aggregate Hence the ions Na and F form a compound whose formula is NaF because these ions are singly charged and are present in the compound in a onetoone ratio Magnesium nitride composed of Mg2 and N3 has the formula Mg3N2 because this composition represents electroneutrality The electrovalent bond is the result of electrostatic attraction between ions of opposite charge This attractive force accounts for the stability of these compounds typified by NaF LiCl CaO and KCI The ions individually possess the electronic structures of neighboring inert gases their residual charge arises from an imbalance in the number of electrons and protons in their structures Isolated ions and simple isolated pairs of ions as represented by the formula NaCl exist only in the gaseous state Their electrostatic forces are active in all directions they attract oppositely charged species and thus can form regular arrays resulting in ordered lattice structures ie the solid state fig 2 Even in the liquid state and in solutions where disruptive thermal forces reach values close to that of the attractive electrostatic bonding forces attraction between ions and with other species remains effective LN 2 Ionic interactions are The resulting low energy configuration omnidirectional and An ordered 3dimensional network nonsaturated a quotcrystallinequot solid Figure 2 Formation of a crystalline ordered body a SOLID a direct consequence of ion interaction in conjunction with energy minimization Energetics of Ionic Bonding Ionic bonding is the simplest type of chemical bonding to visualize since it is totally or almost totally electrostatic in nature The principle of the energetics of ionic bond formation is realized when considering the formation of NaCl our common salt from its constituents Na metal and Gig chlorine gas Formally this reaction is Na 3 12 CI2 g gt NaCl 3 AH 414 kJmol The equation as written indicates that 1 mole sodium reacts with 12 mole chlorine CI2 under formation of 1 mole ionically bonded sodium chloride this reaction is accompanied by the release of 414 kJ of energy AH referred to as the heat of reaction From earlier considerations it is clear that electronic rearrangement reaction or bond formation takes place because the resulting solid product NaCl is at a lower energy state than the sum total of the energies of the original components LN 2 The energetics associated with ionic bond formation may be determined quantitatively by considering the energy changes associated with the individual steps leading from the starting materials to the final product HaberBorn cycle The bond formation in NaCl may be formally presented as an electrontransfer reaction 1e Na Cl gt Na Cl The reactions involved in this process which result in the formation of 1 mole of solid salt are 1 Ionization of Na Na gas gt Na 1e El 497 kJmole The energy change associated with this step energy of ionization El is 497 kJmole 2 Acquisition of one electron by Cl Cl gas 1e gt Cl EA 364 kJmole The energy change associated with this step electron affinity EA is 364 kJmole The minus sign reflects an energy release a lowering of the energy state associated with the achievement of stable rare gas configuration by chlorine So far the energy balance appears positive AE 133 kJ this means the reaction is not favored since the final products are at a higher energy state than the starting products However there are additional steps involved since 3 Vaporization of Na Na metal gt Na gas AHV 109 kJmole The energy required to transform Na metal into Na gas the latent heat of vaporization AHV is 109 kJmole now reaction appears even less favorable LN 2 4 Dissociation of Gig CI2 gt 2Cl ED 242 kJ The energy associated with breaking up the stable chlorine molecule into two reactive chlorine atoms dissociation energy ED is 242 kJmole Since the formation of one mole NaCl involves only 1 mole Cl and since 2 moles are formed from one mole Cl2 the energy required in this step is 12 ED or 121 kJ The total energy change associated with reactions 1 through 4 AHT 364 kJ This indicates that one or more basic reactions which lead to an overall decrease in energy are still unconsidered 5 Unconsidered as yet is the coulombic attraction of the reaction products which are of opposite charge and an energy term associated with the formation of the solid state product NaCl The two ionic species originally at zero energy infinite distance of separation attract each other with an accompanying energy change decrease The energetics associated with the approach of the reaction partners fig 3 is best considered by fixing the position of one for example Na and letting the other Cl approach as Cl approaches Na E decreases according to Coulomb s law with 92411280l With the approach of the two oppositely charged ions the outermost electronic shells will come into contact and a repulsive force will become active as the shells interpenetrate The repulsive force Erep increases with br12 and thus is only active in the immediate vicinity of the sodium ion but at that stage increases rapidly The two ions at close proximity are under the influence of both attractive and repulsive forces and will assume a distance of separation at which the two forces balance each other a distance which is referred to as the equilibrium separation re and which LN 2 Etotal Eattr Erep quotBond equilibrium distance of separation r0 at Energy minimum Bond energy Figure 3 Energetics of ionic bonding corresponds to the energy minimum for the NaCl molecules COU39 4nsoro r 2 where e electronic charge so permittivity of free space 885 x 10 12 Fm and b is a constant r0 for gaseous molecules may be obtained from physical measurements It the presently considered coulombic attractive energy term is taken into consideration as reaction 5 it will decrease the overall positive energy change AHT it will however not make it change sign from to Thus NaCl molecules in gaseous form are not the final reaction product nor would reaction occur if AHT remained positive Considering an ionically bonded gaseous NaCl molecule it is clear that its electrostatic forces and are not saturated they remain active in all possible directions This means that Cl will attract Na ions from other directions as will the Na LN 2 atoms attract additional Cl ions The result of these attractive forces in all directions is the formation of a giant size ionic body a solid body of macroscopic dimensions From the preceding it is recognized that the minimum energy configuration is given by a body in which Na and Cl are arranged with extreme periodicity and order since any ion located outside of its equilibrium position will be in a higher energy configuration such an ordered body is referred to as a crystalline solid or frequently just called a solid fig 2 The total energy change associated with the formation of one mole of crystalline ionic solid from its ionic constituents is given as E M NAe2Q1Qz11 n where M 1747 Madelung constant for NaCl reflecting multiple interactions for the particular geometric arrangement of ions in the solid NA Avogadro s number Q number of charges per ion 1 for Na and Cl r0 equilibrium distance of separation of ions and n repulsive exponent n 12 for NaCl The relationship for the crystal energy AEcryst is readily obtained from equilibrium energy considerations 2 b E e COUI 4nsor 4rcsorn At equilibrium distance of ion separation r0 dEdr 0 Thus dE 92 nb 0 r0 a 4nsor 4nsor31 and b ezrg 1 n and 10 LN 2 2 2 e e EoCoul 43580 n 435130rO E 92 lt 1 oCou 4rc80ro For a molar crystal with ionic charges 01 and 02 the molar AE0Coul is thus given as MNAQ10292 1 AE Here M and NA stand as indicated above for the conventional terms the Madelung constant and Avogadro s number For the present system NaCl Now considering reactions 1 through 5 which is identical with the value experimentally determined and given for the reaction Na metal Cl2 gas NaCl solid In the HaberBorn cycle the reaction energy AH associated with the formation of NaCl from Na Cl2 may be summarized as AH E EA AHV 12 ED AHcryst AH the heat of reaction may thus be obtained from the energetics of the steps leading to the end product In most instances however the reaction energy AH is determined experimentally in a calorimeter and the HaberBorn cycle is used to obtain the value of AEcryst or EA which are both extremely hard to come by 11 LN 2 Conclusions A primary drive for atomic interactions leading to bonding is the achievement of valence shell octets which exhibit a high degree of stability lf atoms on the left and right side of the Periodic Table interact ie atoms with a large difference in electron affinity AEA stabilization is achieved by electron octet formation through charge transfer The reaction products exhibit opposite charges EB cations e anions and are subject to Coulombic attraction ionic bonds are formed Since the electrostatic forces are nondirectional and nonsaturated energy minimization will result in the formation of macroscopic bodies that are highly ordered on the atomic scale crystalline ionic solids lonic solids have mostly predictable basic properties 0 atomic arrangements are a function of the ion size ratio the charge ratio of ions and their electronic structure 0 electrical and thermal conductivities are expected to be low because the high stability of the octets formed results in bound electrons which do not contribute to conduction there is a large energy gap to be crossed for electrons to move into a higher energy state optically ionic solids are mostly transparent or translucent reflecting octet stability of the electrons and macro or microcrystallinity o melting points are high increasing with the electronic charges on the cations and anions o ionic solids are normally hard and brittle 3 COVALENT BONDING Wave Mechanical Concepts and Conclusions In 1924 L DeBroglie advanced the hypothesis that all matter in motion possesses wave properties and can be attributed a particle wavelength KP hmv where h is the Planck constant m is the mass of the matter and v is its velocity 12 LN 2 The credibility of this hypothesis was established by Davisson and Germer in 1927 when they demonstrated that electrons like electromagnetic radiation are diffracted by crystal lattices An important consequence of the dual nature of matter it exhibits both particle and wave properties is the uncertainty principle established in 1927 by W Heisenberg It states that it is impossible to simultaneously know with certainty both the momentum and position of a moving particle Apx AX 2 h We can paraphrase the uncertainty principle in the following manner If the energy of a particle is known measured with high precision its location is associated with a high degree of uncertainty lf electrons occupied simple orbits as postulated by BohrSommerfeld their momentum and position could be determined exactly at any moment in violation of the uncertainty principle According to Heisenberg if the energy of an electron is specified with precision sharpness of spectral lines its location can only be specified in terms of the probability of finding this electron in a certain location volume element These arguments give rise to the concepts of probability density and electron cloud which are inherent to the wavemechanical electron concept emanating from the solutions of Schrodinger s wave equation which relates the energy of an orbiting electron to its wave properties When solving exactly the Schrodinger wave equation for an electron in a hydrogen atom a quantization results according to which electrons can only assume certain energies which are in quantitative agreement with those obtained from the Bohr theory In comparison to the Bohr theory significant differences are observed for the orbital quantization orbital quantum number I which specifies the orbital shape For n 1 I 0 1s orbital wave mechanics predicts a spherical electron density distribution with a maximum probability density W2 at a distance of 0529 A a0 from the nucleus For 13 LN 2 l 1 p orbitals however it is found that the orbitals three form lobes aligned with rectangular coordinates see LN1 fig 3 For I 2 d orbitals five complex orbital configurations are obtained Previous considerations suggest that all elements attempt to assume a stable octet configuration with eight electrons in the valence shell For hydrogen with only one shell occupied the stable configuration consists of two electrons in the K shell which is the maximum number of electrons that can be accommodated Stable octet formation is possible through electron transfer and ionic bond formation when for example elements in columns IA IIA and IIIA react with elements in columns VA VIA and VIIA respectively In these instances the reaction partners exhibit rather pronounced differences in electron affinity and upon reaction one assumes octet configuration by losing one or more electrons while the other does so by acquiring the missing number of electrons This mechanism is clearly not possible if H reacts with H to form an H2 molecule where two atoms are bonded together The same argument holds for the formation of N2 CI2 and 02 molecules Inert gas configuration octet configuration in such elements is achieved by a mechanism called orbital sharing and the resulting bond is called covalent or electron pair bond A covalent bond is somewhat more difficult to visualize than an ionic or electrovalent bond because it involves the sharing of a pair of electrons between atoms The stability of this bond can be attributed to the complex mutual attraction of two positively charged nuclei by the shared pair of electrons In principle the bond can be understood if it is recognized that both electrons in the bonding orbital spend more time between the two nuclei than around them and thus must exercise attractive forces which constitute the bond In this arrangement it is clear that each electron regardless of its source exerts an attractive force on each of the bonded nuclei The pair of electrons 14 LN 2 in a covalent bond is unique to the extent that the Pauli exclusion principle precludes the presence of additional electrons in the same orbital Furthermore the pairing phenomenon neutralizes the separate electronic spins of the single electrons and the resulting electron pair with its zero spin momentum interacts less strongly with its surroundings than do two independent electrons Covalent bonds are conveniently symbolized through the dot notation introduced first by Lewis fig 4 LEWIS NOTATION ln LEWIS notation the covalent molecular bond is indicated as a BAR or as two DOTS standing for the paired electrons HH or H H H 39 I I CC1 or ICIIClIorCICl CH4 H H I Formal valence shell octet stabilization can be achieved by electron sharing H whereby one electron from each reaction partner share spin paired the molecular bonding orbital Figure 4 Lewis notation Quantum mechanics makes it possible to rigorously describe these bonds for very simple cases such as the hydrogen molecule which is composed of two protons and two electrons It can thus be shown that the potential energy for the system reaches a minimum for a certain equilibrium distance between the nuclei with increased electron density between the nuclei At shorter distances between the nuclei repulsive forces are found to increase very rapidly In the hydrogen molecule H2 the two hydrogen atoms are effectively linked together by one molecular electron orbital termed a o orbital which comprises both atoms and contains two electrons fig 5 Each of these two electrons can be considered to have originated from one of the two atoms they were originally both 1s electrons with the same spin value s 12 In the molecular orbit comprising both atoms the spins of 15 LN 2 quotsingly occupied atomic orbitals can on overlap and spin pairing form molecular orbitals sigma 6 bonds in which the electron density between the rection partner is increased along the connecting axis SS Sp pp7 sp3 sp2 sp s or p see later G bond Figure 5 Covalent bond formation the two electrons must align antiparallel opposite spin This spinpairing process results in a considerable release of energy and thus contributes significantly to the stability strength of the covalent bond formed It is interesting that according to Newtonian mechanics no stable configuration can arise from the placement of two electrons into the same region orbit wave mechanics see below however predicts increased stability from such configurations Similar spinpairing occurs in the filling of atomic orbitals In the molecular case spinpairing has the consequence that both the probability distributions W2 and spatial distribution of electrons are such that maximum overlap of orbitals of combining atoms occurs The bond formed in hydrogen is a single bond containing two electrons with paired spins This formulation predicts that there will be a fixed internuclear separation which for the hydrogen molecule has the value of 074 A 5 bonds are formed not only by s orbital overlap but also by ps and pp orbital overlap as well as by the overlap of s or p orbitals with hybridized orbitals to be discussed below see fig 5 16 LN 2 It is important to recognize that octet stabilization by electron orbital sharing results in bond properties which differ fundamentally from those encountered in ionic electrovalent bonding with the formation of the covalent bond between the hydrogen atoms H2 molecule formation the bond forming capabilities of the two hydrogen atoms are saturated the final product is a distinct molecule H2 rather than a giantsized solid body which is obtained as the final product with ionic bond formation Covalent solid bodies however also do exist they are formed if the elements involved have the capability of forming more than one bond For example carbon will form four covalent bonds in tetrahedral configuration the result is diamond a covalently bonded three dimensional network Diatomic Molecules Involving Dissimilar Atoms Consider the compound hydrogen chloride in the gaseous liquid or solid state This compound is not ionic because the energy state on complete ionization would be higher Instead bonding between hydrogen and chlorine atoms is accomplished by the sharing of electrons in a molecular orbital thus forming a single covalent bond fig 6 l IDy asymmetric charge distribution 90395quot COValenCy With asymmetric electron distribution simplified presentation l 5 5 l Figure 6 17 LN 2 The electrons involved are the 1s electron of the hydrogen atom and the unpaired 3pz electron of the chlorine atom After spinpairing a typical 6 bond results in which each atom attains a noble gas structure in HCI the hydrogen atom is involved with two electrons as in helium and the chlorine atom with 18 electrons as in argon The other inner electrons of chlorine do not participate in the bonding they are termed nonbonding According to the above considerations the bonding of HCI may be taken as very similar to that of H2 However in the hydrogen molecule the electrons participating in the 6 bond formation do not favor the proximity of either of the hydrogen nuclei instead they are equally shared On the other hand a preference for one nucleus chlorine is shown by the electrons of the 6 bond in hydrogen chloride This is because electrons for energetic reasons favor the environment of the more electronegative atom in hydrogen chloride this is the chlorine atom and consequently the 6 bond electrons spend more time in the vicinity of the chlorine atom This situation results in the chlorine end of the molecule being fractionally negatively charged 5 and the hydrogen end being fractionally positively charged 5 We denote such an internal charge redistribution by the symbolism H8 Cl5 where the 5 sign indicates a partial electronic charge the bond is said to be polar The hydrogen molecule does not a priori exhibit such an asymmetric charge distribution it may however acquire a temporary polarization Molecules with asymmetric electronic charge distribution have a permanent dipole moment the value of which is given by the product of the fractional charge 5 must be equal to 5 and the distance of charge separation L Although dipole moments can be measured their values do not allow us to directly calculate the polarity of bonds between atoms since the detailed internal charge distribution is unknown The direction of electron drift and to some extent the magnitude of its effect can be estimated from the magnitude of the difference in electron affinity between the 18 LN 2 reaction partners Because of a limited data base on electron affinity L Pauling introduced a related term the relative electronegafivify see below From the above considerations it should be clear that no sharp dividing line exists between ionic and covalent bonding We might consider a completely ionic bond to result in cases where the electron drift is such that one atom the cation becomes entirely deficient in one or more electrons and the other atom the anion becomes correspondingly electronrich the bonding electrons being entirely under the influence of the latter Hydrogen halides are 6 bonded and have dipole moments the bonds are referred to as polar covalency The homonuclear diatomic molecules formed by the halogens viz F2 Cl2 Br2 and I2 are all obonded systems involving spinpairing of the various pz electrons 2pz 3pz 4pz and 5pz respectively They have no permanent dipole moments Energetics of Covalent Bonding Pauling extensively treated the energetics of polar covalencies encountered in all heteronuclear systems such as H Cl which have permanent dipole moments H8 Cl5 His approach visualizes the bonding to consist of two components a pure covalency and an ionic bonding component with attraction resulting from the interaction of the fractional charges on the nuclei involved Pauling determines the basic covalent bonding component for example in the formation of HCI from the experimentally obtained bond energies associated with the molecular species of the components H2 and Gig For this purpose he made the basic assumption that the covalent bond component between the dissimilar atoms is given by the geometric mean of the pure covalent bond energies associated with these 19 LN 2 molecular species Thus Pure Covalent Bond Energy of HCI BEHCI BEH2 x BECIZ H2 H H Bond Energy BEH2 430 kJmole Cl2 Cl Cl Bond Energy BECI 238 kJmole BEHCI l43o x 238 320 kJmole The ionic contribution A to the polar covalent bonding is then obtained from the difference between the experimentally determined bond energy for HCI for example which obviously must contain both components and the calculated pure covalent bond energy A BEHCI experimental BEHCI theoretical covalent The experimentally determined BEHCI 426 kJmole D II 426 320 106 kJmole In connection with the presently discussed work Linus Pauling established the now generally used electronegativity scale or better the scale of relative electronegativities This scale included in the Periodic Table of the Elements is extremely helpful since values of the electron affinity are known thus far for only very few elements The electronegativity x scale lists the relative tendency of the neutral elements to attract an additional electron The values of x are conventionally listed in electron Volts The scale listed was obtained by arbitrarily fixing the value of xH 22 Pauling obtained the values for the other elements by relating differences in electronegativity of reaction partners to the fractional ionic character ionic bonding component of the bond 20 LN 2 established between them experimental BEAB BEAA x BEBB kxA xB2 or A 963 xA xB2 kJ According to Pauling the bonding character between two different elements may be defined as ionic bonding for Ax gt 17 covalent bonding for Ax lt 17 The fractional ionicity of polar covalent bonding as listed in the PT is obtained by the relationship quot0 ionic bonding 1 e 025XA XB2 x 100 Bonding In Polyatomic Molecules In the formation of covalent bonds between atoms in polyatomic molecules the conditions that atomic orbitals distort so that maximum overlap may be achieved when bonding occurs produces more extensive changes of orbital geometry than is the case with diatomic molecules It will be remembered that the changes which result in a 6 bond formation in diatomic molecules are a distortion of the s orbitals or p lobes so that strongest bond formation results from the maximum overlap between the two joined nuclei For polyatomic molecules extensive alterations in the spatial disposition of atomic orbitals do occur and very often the unique orbital geometry of the original atomic orbitals is completely lost fig 7 It is sometimes convenient to view these alterations as occurring in each atom prior to bonding by a process involving the mixing or hybridization of atomic orbitals 21 LN 2 8 difference in electron affinity between 0 and H leads to fractional charges on the hydrogen atoms and to bond angle distortion because of electrostatic repulsion 5H 5 q 31gt 1070 5 H saturated nonbonding pZorbital a 5 The water molecule has a permanent dipole moment Figure 7 Bonding Involving Carbon ln methane CH4 the four outer or valence electrons of carbon are shared with the electrons of hydrogen there is spinpairing resulting in bond formation between each individual hydrogen electron and one of the carbon valence electrons The noble gas structure is thus attained by each nucleus the carbon nucleus sees eight outer electrons and each H nucleus sees two electrons Consider now the orbitals which are involved in more detail Each hydrogen has one spherical valence orbital the 1s orbital containing one electron and covalent bond formation results from its distortion overlap with the valence orbitals of carbon and spinpairing According to the Aufbau principle and Hund s rule in its ground state carbon has two electrons in the filled K shell and four electrons in the L shell two 2s electrons and two 2p electrons in singly occupied orbitals which are capable of covalent bond formation This configuration provides in principle only two orbitals 2px 2py for covalent bond 22 LN 2 formation However with two covalencies carbon will not yield the desirable octet configuration Such a configuration can be obtained if one of the two 2s electrons is promoted into the empty 2pz orbital since this process results in four singly occupied orbitals all of which being singly occupied are capable of bond formation The dissimilar singly occupied orbitals can assume and therefore m assume upon bond formation a lower energy configuration involving a process called hybridization whereby the four orbitals of two different types hybridize into four identical orbitals of maximum equal spacing from each other fig 8 Thus the hybridized orbitals sp3 hybrids are lobes emanating from the carbon atom into the corners of a tetrahedron forming bond angles of 109 28 sp3 hybridization is characteristic for carbon however other types of hybridization sp2 and sp are encountered in other elements as well Boron for example will tend to promote one of its two 2s electrons into a 2p state and by hybridization form three equivalent sp3 orbitals which assume planar orientation with band angles of 120 Beryllium forms sp hybrid orbitals of linear orientation the bond angle is 180 All hybrid orbitals are capable of 6 bond formation The great diversity of carbon compounds it forms more compounds than all the rest of the elements in the Periodic Table can be attributed to the hybridization capability of the carbon orbitals sp3 sp2 and sp As a consequence carbon forms not only axisymmetric 6 bonds through axial orbital overlap but also it bonds through lateral orbital overlap In the compound ethane H2O CH2 the interactive carbon atoms undergo sp2 hybridization for a 6 bond by overlap of two sp2 hybrid orbitals and form in addition a it bond by lateral overlap of the remaining nonhybridized p orbitals Double bonds involve one o and one it bond In acetylene HC E CH we find sp hybridization and by axial overlap 6 bond formation as well as lateral overlap of the remaining nonhybridized px and py orbitals which form two it bonds fig 9 23 LN 2 CH4 formed by 4 molecular orbitals obonds based on overlap of sp3 and s atomic orbitals BA10902839 Octet formation in the carbon valence shell can be achieved by sp3 hybridization which leads to 4 equivalent bonding orbitals spaced tetrahedrally at 10902839 4 sp3 hybrid orbitals A A A energy level 4 4 l V 293 60 1292 A A 232 N gt 2314 hybrid bonds 3 sp2 hybrid p orbitals orbital a partial hybridization A A 3 hybrid bonds 2 sp hybrid Hybrid bonding orbital configurations 2 pf orbitals orbitals A A sp3 tetrahedral EA 1090 C02 BeH2 f 4 sp2 planar EA 1200 s03 BF3 2 hybrid bonds sp linear EA 1800 BeH2 002 Figure 8 Orbital Hybridization 24 LN 2 7 T lateral overlap A of porbitals leads to at bond form he overlapping Ethylene has a double bond the C atoms are bonded orbital C to each other with a o bond sp2sp2 overlap and with etweef t e b d a ar bond lateral porbital overlap atoms 0quot a 039 0 Acetylene has a triple bond C atoms are bonded by overlap of sphybrid orbitals obond and by lateral overlap of 2 sets of porbitals 2 at bonds configured at right angles to each other Figure 9 Double and Triple Bonds 3 METALLIC BONDING The stability of both covalent and metallic bonds may be regarded as arising from the potential energy lowering experienced by valence electrons under the influence of more than one nucleus In metals where the valence electrons are not as tightly bound to their ion cores we cannot expect the formation of strong electronpair bonds The bond energies of known diatomic molecules of metallic elements are in fact smaller than those consisting of nonmetallic elements 104 kJmole for Hg2 Only diatomic molecules of the semi metals have relatively high binding energies 385 kJmole for As2 293 kJmole for Sb2 and 163 kJmole for Big These values reflect multiple bonding Much greater stability is possible in larger aggregates of atoms such as bulk metals The known properties of metals such as low electrical resistance and malleability support the conceptual view that the valence electrons in metals never remain near any 25 LN 2 particular atom very long but drift in a random manner through the lattice of ion cores We may therefore visualize metals as a lattice of ion cores being held together by a gas of free electrons Bonding in metallic systems is discussed in more detail in LN3 4 SECONDARY van der WAALS BONDING Primary bonding ionic covalent and metallic is strong and the energies involved range from about 100 to 1000 kJmole In contrast secondary bonding is weak involving energies ranging from about 01 to 10 kJmole While this type of bonding also referred to as residual is weak it is essential in the functioning of our environment Coke would likely be gaseous and not a bubbly refreshing brew were it not for secondary bonding nor would catalytic converters function The energy difference between the liquid and vapor states of a given system is given by the heat of vaporization ie the heat required to convert a given liquid into a vapor normally at the boiling point temperature at 1 atm pressure The energy difference is due to intermolecular attraction between molecules at close distance of separation This phenomenon of attraction through secondary bonding can best be considered between a single pair of molecules but recognizing that the forces are of longer range Four types of intermolecular forces can be identified 1 DipoleDipole Interaction Molecules with permanent dipoles such as water alcohol and other organic compounds with functional groups exert a net attractive force on each other as a result of varying degrees of alignment of oppositely charged portions of the molecules fig 10 For two polar molecules with a dipole moment of u separated by a distance of r the energy of attraction can be quantified as 26 LN 2 Schematic presentation of a dipole exhibiting a Dipolemoment p which is given by uL q where L is the distance of separation of the fractional electronic charges aq and aq in abbreviated form given as 5 and 5 ln molecules the dipole moment is given by the vector sum of the polar bonds Thus because of molecular geometry symmetry molecules such as CCI4 exhibit no dipolemoment DipoleDipole Interaction Figure 10 EDipole Dipole where u is the dipole moment r is the distance of approach of the oppositely charged molecular portions k is the Boltzmann constant see PT and T is the absolute temperature in K The molar energies of attraction associated with dipoledipole interaction range from 0 to about 10 kJmole These forces are primarily responsible for the liquid state at room temperature of most polar organic molecules they are a contributing factor for H20 to be liquid at room temperature and are responsible for alcohol being a liquid 2 DipoleInduced Dipole Interaction A dipole in one molecule can interact with and polarize the electrons of a neighboring nonpolar molecule thus generating an induced dipole which will experience an attractive force with the polarizing polar dipole fig 11 P Debye showed that in a molecule with a polarizability of or the attractive potential arising from dipoleinduced dipole interaction is given as E 20cu2 Dipole lnduceddipole Dr6 27 LN 2 DipoleInduced Dipole Interaction Figure 11 Induced dipole interaction is important in aqueous solutions and very effective during adsorption of inert molecules on active solid substrates 3 London Dispersion Forces It is a wellknown fact that all substances including rare gases and hydrogen assume liquid state at finite temperature an indication of the existence of attractive interatomic and intermolecular forces even in the absence of permanent dipole systems The origin of this force has been proposed by F London in 1930 Accordingly orbiting electrons will at any instance generate a temporary dipole the configuration of which changes as the electrons move Since all atoms of a given system similarly experience temporary instantaneous dipole moments their effect is expected to be cancelled because of the statistically random orientation of dipoles It is evident that should the dipoles be synchronized in a given assembly of atoms then a net attractive force would result fig 12 But since such an attractive force constitutes a lowering of the energy of a given system synchronization can and will take place because all systems will attempt to assume minimum energy configuration The London dispersion force can be formulated as Ka2 ELondon r6 The attractive London forces are small as manifested by the very low boiling points of the smaller rare gases of hydrogen and nitrogen 4 Hydrogen Bonding The short and long range dipole interactions calculated from molecular dipole moments are inadequate in explaining a multitude of phenomena in organic as well as some inorganic systems L Pauling studied such 28 LN 2 London Dispersion Fluctuating Dipole Interaction Electron motion in atoms generates fluctuating dipoles Since the polarization is random attractions are cancelled by repulsions London Upon synchronization of electron motion inter atomic attraction will be established in an assembly of atoms Since this attraction constitutes a lowering of the energy of the system attraction by synchronization will take place as all systems attempt to assume lowest energy configuration No net attraction Net attraction Figure 12 phenomena and concluded the existence of highly specific attractive interaction between hydrogen that is acidic carries a fractional positive charge and the elements 0 F N and to a lesser extent 8 in both organic and inorganic molecules This interaction which can as yet not be formulated is referred to as hydrogen bonding its magnitude ranging up to 40 kJmole is significantly larger than that of any other secondary bonding type Hydrogen bonding is considered instrumental in controlling most properties of water is a key element in the structure of nucleic acid and thought to be an essential component in memory functions of the human brain 29 MIT OpenCourseWare httpocwmitedu 309180 Introduction to Solid State Chemistry FaH2009 For information about citing these materials or our Terms of Use visit httpocwmiteduterms CHAPTER 3 MATTER 8 ENERGY Problems 150 5758 6174 107 109115 32 What is Matter Matter Anything that has mass and occupies volume We study matter at different levels macroscopic the level that can be observed with the naked eye eg geologists study rocks and stone at the macroscopic level microscopic the level that can be observed with a microscope eg scientists study tiny animals plants or crystals at microscopic level particulate at the level of atoms and molecules also called atomic or molecularlevel cannot be observed directly even with the most powerful microscopes where the term nanotechnology comes from since many atoms and molecules are about a few nanometers in size lSpeciallzed techniques 39 are required to rosarye I IE Elli Thomson Higher Education Substances like water can be represented using different symbols eg H20 and models i ii H M Lewis Structure or Electronpt liagrtam Hl llazalla tlEtfiuk medal 5Faggsfil limg 15513 CHEM 121 TM Chapter 3 page 1 of 13 33 Classifying Matter According to Its State Solid Liquid and Gas Matter exists in one of three physical states solid liquid gas Copyright 21109 Pearson Prentice Hall inc solid Has definite shape and a fixed or constant and rigid volume Particles only vibrate in place liquid Has a fixed or constant volume but its shape can change Takes the shape of its container because particles are moving Particles are packed closely together but can move around each other gas Volume is variable and particles are far apart from one another Takes the shape of the container because particles are moving gt If container volume expands particles move apart to fill container gt If container volume decreases particles move closer together gt Gases are compressible ie can be forced to occupy a smaller volume Particles are in constant random motion CHEM 121 TM Chapter 3 page 2 of 13 34 Classifying Matter According to Its Composition Elements Compounds and Mixtures We can classify matter into pure substances and mixtures pure substance a single chemical consisting of only one kind of matter There are two types of pure substances elements and compounds In the figure below copper rods are an example of an element and sugar is an example of a compound mixture consists of two or more elements andor compounds Mixtures can be homogeneous or heterogeneous Homogeneous mixtures have a uniform appearance and composition because the particles in them mix uniformly eg solutions like sweetened tea below Heterogeneous mixtures do not have a uniform composition eg chocolate chip cookie water and C8H18 mixture below shown as separate layers Cepyright 2009 Peereen Prent iee Hell lhe CHEM 121 Tro Chapter 3 page 3 of 13 elements consist of only one type of atom atoms cannot be broken down into smaller components by chemical reaction eg copper wire Cu sulfur powder 88 Examples also include sodium Na barium Ba hydrogen gas H2 oxygen gas 02 and chlorine gas H2 compounds consist of more than one type of atom and have a specific chemical formula Examples include hydrogen chloride HCI water H20 sodium chloride NaCl which is table salt barium chloride BaClg Two or more pure substances combine to form mixtures mixtures consist of many compounds andor elements with no specific formula Matter having variable composition with definite or varying properties can be separated into component elements andor compounds eg any alloy like brass steel 10K to 18K gold sea water carbonated soda air consists of nitrogen oxygen and other trace gases The image at the right shows that air is a mixture of mostly nitrogen N2 in blue and some oxygen 02 in red while salt water consists of salt Na and Cl39 ions or charged particles dissolved in water Example ls salt water a homogeneous or heterogeneous mixture Explain ohlorido Each E 39IIzd 2111quot Thomoon Higll39ror Eduoation Copyright E g Pearson Prentice Hall lino V or Maiural Pairit iouilato Illustration Maoroooopio Brimstone Formula Form not to some Photograph Hoium Ha itom A colorless gas 7 Atom Eooium Hir Erostaiiino solid A romi Barium Ea Eruora llilrne 5mm Humorgan H Molecule e colorless goo Elixirgen D Moloouilra Fl milorloao giro Chlorino Cl Molooulo Hurlrogon A r r ohlorido Hm gleam Water Hg Moloou a Sodium CWEEEIEMB thiorido 3 339 mild Eloriusrn Ewstallir ro CHEM 121 TM Chapter 3 page 4 of 13 35 How We Tell Different Kinds of Matter Apart Physical and Chemical Properties The characteristics that distinguish one substance from another are called properties Physical Properties inherent characteristics of a substance independent of other substances physical state solid liquid gas electricalampheat conductivity color odor density hardness melting and boiling points solubility doesdoes not dissolve in water Chemical Properties how a substance reacts with other substances eg hydrogen reacts explosively with oxygen 36 How Matter Changes Physical and Chemical Changes physical change a process that does not alter the chemical makeup of the starting materials Note in the figure below that the H20 molecules remain H20 regardless of the physical state solid liquid or gas gt Changes in physical state are physical changes Other examples of physical changes include hammering gold into foil dry ice subliming Dissolving table salt or sugar in water is also a physical change A substance dissolved in water is the fourth physical state aqueous CHEM 121 TM Chapter 3 page 5 of 13 Know the terms for transitions from one physical state to another freezing liquid gt solid condensinggas gt liquid melting solid gt liquid evaporating or vaporizing liquid gt gas Two less common transitions sublimation solid gt gas eg dry ice sublimes deposition gas gt solid eg water vapor deposits on an icebox chemical change a process that does change the chemical makeup of the starting materials We can show H2 and 02 reacting to form water H20 below Since the H20 has a different chemical makeup than H2 and 02 this is a chemical change Other examples of chemical changes eg oxidation of matter burning or rusting release of gas bubbles fizzing mixing two solutions to form an insoluble solid precipitation and other evidence indicating the starting materials reactants were changed to a different substance The following examples are all chemical changes that convert the reactants to completely different compounds andor elements relea r gas bble formation ofinsoluble solid fizzing precipitation oxidaio buring or usting CHEM 121 TM Chapter 3 page 6 of 13 Example 1 Consider the following molecularlevel representations of different substances For each figure above indicate if it represents an element a compound or a mixture AND if it represents a solid liquid or gas A element compound mixture solid liquid gas B element compound mixture solid liquid gas C element compound mixture solid liquid gas D element compound mixture solid liquid gas E element compound mixture solid liquid gas F element compound mixture solid liquid gas Ex 2 Circle all of the following that are chemical changes burning condensing dissolving rusting vaporizing precipitating CHEM 121 TM Chapter 3 page 7 of 13 37 Conservation of Mass There is No New Matter Chemical Reaction REACTANTS gt PRODUCTS starting materials substances after reaction For the reaction C 02 gt C02 The reactants are carbon and oxygen gas and the product is carbon dioxide Antoine Lavoisier 17431794 a French chemist carried out experiments on combustion by burning different substances and measuring their masses before and after burning He found that there was no change in the overall mass of the sample and air around it gt Law of Conservation of Mass Matter is neither created nor destroyed in a chemical reaction so mass is conserved Mass of the products in a reaction must be equal to the mass of the reactants For example 112 g hydrogen 888 9 oxygen 1000 g water Ex 1 Methane burns by reacting with oxygen present in air to produce steam and carbon dioxide gas Calculate the mass of oxygen that reacts if burning 500 g of methane produces 1123 g of steam and 1371 g of carbon dioxide 38 Energy the capacity to do work potential energy PE energy due to position or its composition chemical bonds A 10Ib bowling ball has higher PE when it is 10 feet off the ground compared to 10 inches off the ground gt Greater damage on your foot after falling 10 feet compared to falling only 10 inches In terms of chemical bonds the stronger the bond gt more energy is required to break the bond gt the higher the potential energy of the bond CHEM 121 TM Chapter 3 page 8 of 13 kinetic energy KE energy associated with an object s motion eg a car moving at 75 mph has much greater KE than the same car moving at 15 mph gt Greater damage if the car crashes at 75 mph than at 15 mph Six Forms of Energy heat light chemical electrical mechanical and nuclear Each can be converted to another Example Identify at least two types of energy involved for each of the following 1 When you turn on a lamp 2 When using solar panels 3 At the Springfield Power Plant in The Simpsons Energy changes accompany physical and chemical changes due to changes in potential and kinetic energy Kinetic Energy and Physical States Solids have the lowest KE of the three physical states Highest attraction between particles gt particles are fixed Liquids have slightly higher KE than solids Particles are still attracted to each other but can move past one another gt particles are less restricted Gases have greatest KE compared to solids and liquids Attractive forces completely overcome so particles fly freely within container gt particles are completely unrestricted So d CHEM 121 Tro Chapter 3 page 9 of 13 310 Temperature Random Molecular and Atomic Motion Heat Energy that is transferred from a body at a higher temperature to one at a lower temperature gt heat always transfers from the hotter to the cooler object quotheat flowquot means heat transfer Heat Transfer and Temperature One becomes hotter by gaining heat One becomes colder by losing heat ie when you feel cold you are actually losing heat Ex 1 Fill in the blanks to indicate how heat is transferred a You burn your hand on a hot frying pan loses heat and gains the heat b Your tongue feels cold when you eat ice cream loses heat and gains the heat Ex 2 A small chunk of gold is heated in beaker 1 which contains boiling water The gold chunk is then transferred to beaker 2 which contains roomtemperature water a The temperature of the water in beaker 2 T L stays the same b Fill in the blanks loses heat and gains the heat Ex 3 Why do surfaces like a stone countertop or metal flatware forks knives etc feel cold in restaurant Aren t they at room temperature like you Explain 39 Energy and Chemical and Physical Change endothermic change a physical or chemical change that requires energy or heat to occur boiling water requires energy H20I heat energy gt H20g electrolysis of water requires energy 2 H20I electrical energy gt 2 H2g 029 exothermic change a physical or chemical change that releases energy or heat water condensing releases energy H20g gt H20I heat energy hydrogen burning releases energy 2H2g 029 gt 2H20g heat energy CHEM 121 TM Chapter 3 page 10 of 13 For physical changes consider whether the reactants or products have more kinetic energy If the reactants have greater kinetic energy than the products gt exothermic process If the products have greater kinetic energy than the reactants gt endothermic process a Melecules close teglether same 5 391le nearest neighbor 139 a 39 ever time b Melecules close rm tiugiethler but Liquid mnnvinlgexc hangiilng UH 9 9 If k Vquot 39J V nearest netgihbers J w c Meleeules widely separated moving it Gas rapidlyr i system that part of the universe being studied surroundings the rest of the universe outside the system For chemical changes observe if the surroundings including you feel hotter or colder after the reaction has occurred If the surroundings are hotter the reaction released heat gt exothermic reaction If the surroundings are colder the reaction absorbed heat gt endothermic reaction Ex 1 Circle all of the following changes that are exothermic freezing vaporizing sublimation melting deposition CHEM 121 TM Chapter 3 page 11 of 13 Ex 2 A student adds ammonium chloride NH4CI salt to a test tube containing water and notices that the test tube feels colder as the ammonium chloride dissolves Is this process exothermic or endothermic Explain Ex 3 A student mixes two solutions hydrochloric acid and sodium hydroxide and notices the beaker with the substances feels hotter as they mix Is this reaction exothermic or endothermic Explain Units of Energy calorie cal unit of energy used most often in the US amount of energy required to raise the temperature of 1 g of water by 1 C 1 cal 5 4184 J Note This is EXACT But a nutritional calorie abbreviated Cal is actually 1000 cal 1 Cal 1 kcal 4184 kJ joule J SI unit of energy To recognize the size of ajoule note that 1 watt 1 g gt So a 100watt light bulb uses 100 J every second Heat is also often reported in kilojoules kJ where 1 kJ 1000 J Electricity usage on our electricity bills are generally reported in kilowatthour kWh Example A 100 W lightbulb running for 100 hours requires 1 kWh of energy If the average Seattle home uses 25 kWhday this is equal to how many 100 W light bulbs running nonstop for one day CHEM 121 TM Chapter 3 page 12 of 13 Energy and Food Values food value The amount of heat released when food is burned completely usually reported in kJg food or Calg food Most of the energy needed by our bodies comes from carbohydrates and fats and the carbohydrates decompose in the intestines into glucose 06H1206 The combustion of glucose produces energy that is quickly supplied to the body C5H1205g 6 029 6 C02g 6 H20g heat energy The body also produces energy from proteins and fats which can be stored because fats are insoluble in water and produce more energy than proteins and carbohydrates The energy content reported on food labels is generally determined using a bomb calorimeter similar to that described in the previous section 311 Temperature Changes Heat Capacity heat capacity amount of heat necessary to raise the temperature of a given amount of any substance by 1 C in units of J C specific heat capacity amount of heat necessary to raise the temperature of 1 gram of any or specific heat substance by 1 C has units of Jg C Water has a relatively high specific heat 4184 Jg C compared to the specific heats of rocks and other solids 13 Jg C for dry Earth 09 Jg C for concrete 046 Jg C for iron Example Water covers most of the Earth Compare the specific heat of water with Earth and other solids given above then explain what would happen if the Earth were covered mostly in land not water CHEM 121 TM Chapter 3 page 13 of 13 Introductory Quantum Chemistry Chem 570a Lecture Notes Prof Victor S Batista Room Sterling Chemistry Laboratories SCL 18 Tuesdays and Thursdays 900 1015 am Yale University Department of Chemistry Contents 1 1 N h 10 11 12 SyHabus The Fundamental Postulates of Quantum Mechanics Continuous Representations Vector Space 41 Exercisel 42 Exercise2 Stationary States 51 Exercise 3 52 Exercise 4 53 Exercise 5 Particle in the Box 61 Exercise 6 Commutator 71 Exercise 7 Uncertainty Relations 81 Exercise 8 82 EPR Paradox Variational Theorem Digital GridBased Representations 101 Computational Probleml 102 Computational Problem 2 103 Computational Problem 3 104 Computational Problem 4 SOFT Method 111 Computational Problem 5 112 Computational Problem 6 113 Computational Problem 7 114 Computational Problem 8 115 Computational Problem 9 Time Independent Perturbation Theory 121 Exercise 9 7 10 10 13 13 14 14 14 15 15 16 17 17 17 18 18 19 20 20 21 22 22 23 23 24 24 25 25 25 26 13 14 15 16 17 18 19 20 21 Time Dependent Perturbation Theory 131 Exercise 10 132 Exercise 11 Problem Set 141 Exercise 11 142 Exercise 12 143 Exercise 13 144 Exercise 14 Exercises Suggested in Class 151 Copenhagen Interpretation 152 Feynman Interview 153 Momentum Operator 154 EPR Paradox 155 Schrodinger s cat paradox 156 Time Evolution Operator 1561 Evolution in the basis of eigenstates 1562 Trotter expansion of the time evolution operator 1563 Numerical Comparison Adiabatic Approximation Heisenberg Representation TwoLevel Systems Harmonic Oscillator 191 Exercise 15 Problem Set 201 Exercise 16 202 Exercise 17 203 Exercise 18 204 Exercise 19 Angular Momentum 211 Exercise 20 212 Exercise 21 213 Exercise 22 214 Exercise 23 215 Exercise 24 27 30 30 32 32 33 33 33 34 34 34 34 34 35 35 35 35 36 36 39 40 44 44 22 23 24 25 26 27 28 29 30 31 32 Spin Angular Momentum 221 Exercise 25 222 Exercise 26 223 Exercise 27 224 Exercise 28 Central Potential 231 Exercise 29 TwoParticle RigidRotor 241 Exercise 30 Problem Set 251 Exercise 31 252 Exercise 32 253 Exercise 33 254 Exercise 34 255 Exercise 35 256 Exercise 36 257 Exercise 37 Hydrogen Atom 261 Exercise 38 262 Exercise 39 263 Exercise 40 264 Exercise 41 265 Exercise 42 Helium Atom SpinAtom Wavefunctions Pauli Exclusion Principle Lithium Atom 301 Exercise 44 SpinOrbit Interaction 311 Exercise 45 Periodic Table 321 Exercise 46 322 Exercise 47 57 59 59 6O 61 61 63 64 64 64 64 64 65 65 65 65 65 66 67 68 68 68 7O 70 72 72 73 74 74 76 Problem Set 78 36 37 38 39 40 331 Exercise 48 78 332 Exercise 49 79 333 Exercise 50 79 334 Exercise 51 79 LCAO Method H Molecule 79 341 Exercise 52 81 342 Exercise 53 83 H2 Molecule 83 351 HeitlerLondonHL Method 85 352 Exercise 54 85 Homonuclear Diatomic Molecules 86 361 Exercise 55 89 Conjugated Systems Organic Molecules 90 SelfConsistent Field HartreeFock Method 92 381 Restricted Closed Shell Hartree Fock 96 Quantum MechanicsMolecular Mechanics Methods 101 Empirical Parametrization of Diatomic Molecules 102 401 Exercise 56 105 402 Exercise 57 106 Tunneling Current Landauer Formula 108 411 WKB Transmission 111 Solutions to Computational Assignments 114 421 Problem 1 114 422 Problem 2 117 423 Problem 3 122 424 Problem 4 125 425 Problem 5 131 426 Problem 6 132 427 Problem 7 143 428 Problem 8 161 429 Problem 9 178 1 SyHabus The goal of this course is to introduce fundamental concepts of Quantum Mechanics with emphasis on Quantum Dynamics and its applications to the description of molecular systems and their inter actions with electromagnetic radiation Quantum Mechanics involves a mathematical formulation and a physical interpretation establishing the correspondence between the mathematical elements of the theory e g functions and operators and the elements of reality e g the observable proper ties of real systems The presentation of the theory will be mostly based on the so called Orthodox Interpretation developed in Copenhagen during the rst three decades of the 20th century How ever other interpretations will be discussed including the pilot wave theory rst suggested by Pierre De Broglie in 1927 and independently rediscovered by David Bohm in the early 1950 s Textbooks The of cial textbook for this class is R1 Levine Ira N Quantum Chemistry 5th Edition Pearsor Prentice Hall 2009 However the lectures will be heavily complemented with material from other textbooks including R2 Quantum Theory by David Bohm Dover R3 Quantum Physics by Stephen Gasiorowicz Wiley R4 Quantum Mechanics by Claude Cohen Tannoudji Wiley Interscience R5 Quantum Mechanics by E Merzbacher Wiley R6 Modern Quantum Mechanics by J J Sakurai Addison Wesley All these references are on reserve at the Kline science library References to speci c pages of the textbooks listed above are indicated in the notes as follows R1190 indicates for more information see Reference 1 Page 190 Furthermore a useful mathematical reference is R Shankar Basic Training in Mathematics A Fitness Program for Science Students Plenum Press New York 1995 Useful search engines for mathematical and physical concepts can be found at httpscienceworldwolframcomphysics and httpmathworldwolframcom The lecture notes are posted online at httpwwwchemyaleedubatistavvvv570pdf Grading Grading and evaluation is the same for both undergraduate and graduate students The mid terms will be on 103 and 1121 The date for the Final Exam is determined by Yale s calendar of nal exams Homework includes exercises and computational assignments with due dates indicated in class Contact Information and Of ce Hours Prof Batista will be glad to meet with students at SCL 251 as requested by the students via email to victorbatistayaleedu or by phone at 203 432 6672 2 The Fundamental Postulates of Quantum Mechanics Quantum Mechanics can be formulated in terms of a few postulates ie theoretical principles based on experimental observations The goal of this section is to introduce such principles to gether with some mathematical concepts that are necessary for that purpose To keep the notation as simple as possible expressions are written for a 1 dimensional system The generalization to many dimensions is usually straightforward Postulate 1 Any system in a pure state can be described by a wavefunction rbt at where t is a parameter representing the time and at represents the coordinates of the system Such a function rbt 15 must be continuous single valued and square integrable Note 1 As a consequence of Postulate 4 we will see that P t at rbt crbt at dx represents the probability of nding the system between at and at dx at time t Postulate 2 Any observable ie any measurable property of the system can be described by an operator The operator must be linear and hermitian What is an operator 7 What is a linear operator 7 What is a hermitian operator De nition 1 An operator 0 is a mathematical entity that transforms a function f into another function gc as follows R496 where f and g are functions of 10 De nition 2 An operator 0 that represents an observable O is obtained by rst writing the clas sical expression of such observable in Cartesian coordinates eg O 01 p and then substi tuting the coordinate at in such expression by the coordinate operator E as well as the momentum p by the momentum operator p ihol 015 De nition 3 An operator 0 is linear if and only if iif 0afv b9 a l 5091 where a and b are constants De nition 4 An operator 0 is hermitian i f Manama mammal where the asterisk represents the complex conjugate De nition 5 A function ammo is an eigenfunction of 0 i f Origbig where On is a number called eigenvalue Property 1 The eigenvalues of a hermitian operator are real Proof Using De nition 4 we obtain minnow Manama 0 therefore on 02 manna 0 Since nl are square integrable functions then on 0 Property 2 Different eigenfunctions of a hermitian operator ie eigenfunctions with different eigenvalues are orthogonal ie the scalarAproduct of two different eigenfunctions is equal to zero Mathematically if 0ng Ono and Ogbm Omgbm with 0 7E Om then f dzcgbj bgbm 0 Proof lu 0 n lu uiulo on om dam521 0 Since On 7E Om then f dzcgbjfngbn 0 and Postulate 3 The only possible experimental results of a measurement of an observable are the eigenvalues of the operator that corresponds to such observable The average value of many measurements of an observable 0 when the system is described by rbc as equal to the expectation value 0 which is de ned as follows azimunpun f meme Postulate 5 The evolution of rb1 t in time is described by the timedependent Schrodinger equauon 01b1 t A h H t Z wltaj7 7 where H 2 is the operator associated with the total energy of the system E g m Vc Echansion Postulate R515 R497 The eigenfunctions of a linear and hermitian operator form a complete basis set Therefore any function rpm that is continuous single valued and square integrable can be expanded as a linear combination of eigenfunctions mm of a linear and hermitian operator fl as follows CijjW where Cj are numbers e g complex numbers called expansion coe icients Note that fl Z CjCjaj when M113 2 Z ngbj1 A jltgtzaj jlta and dv jv kx it This is because the eigenvalues CLj are the only possible experimental results of measurements of fl according to Postulate 3 and the expectation value fl is the average value of many measurements of fl when the system is described by the expansion rate 2 Ci gbj Postulate 4 Therefore the product Oj C can be interpreted as the probability weight associated with eigenvalue CLj ie the probability that the outcome of an observation of fl will be aj HilbertSpace According to the Expansion Postulate together with Postulate 1 the state of a system described by the function can be expanded as a linear combination of eigenfunctions gbj1 of a linear and hermitian operator eg 01gb11 02ng Usually the space de ned by these eigenfunctions ie functions that are continuous single valued and square integrable has an in nite number of dimensions Such space is called H ilbertSpace in honor to the mathematician Hilbert who did pioneer work in spaces of in nite dimensionalityR494 A representation of in such space of functions corresponds to a vector function C2 where 01 and 02 are the projections of along gbl1 and zc respectively All other components are omitted from the representation because they are orthogonal to the plane de ned by 1 and 2 3 Continuous Representations Certain operators have a continuous spectrum of eigenvalues For example the coordinate operator is one such operator since it satis es the equation i3 6 130 13 130 6 130 13 where the eigenvalues 130 de ne a continuum Delta functions 6130 13 thus de ne a continuous representation the so called coordinate representation for which W dwootoam as where Cm0 1b130 since d13613 61b13d13d010a613 66oz 13 1MB When combined with postulates 3 and 4 the de nition of the expansion coef cients 0st 1b130 implies that the probability of observing the system with coordinate eigenvalues between 130 and 130 d130 is P130 CmOCgodmo 1b1301b130d130 see Note 1 In general eigenstates gb01 13 with a continuum spectrum of eigenvalues 01 de ne continuous representations W momma with CO f d13gboz Delta functions and the plane waves are simply two particular examples of basis sets with continuum spectra Note 2 According to the Expansion Postulate a function 1b13 is uniquely and completely de ned by the coef cients Oj associated with its expansion in a complete set of eigenfunctions gbj However the coef cients of such expansion would be different if the same basis functions gbj depended on different coordinates e g gbj13 with 13 7E 13 In order to eliminate such ambiguity in the description it is necessary to introduce the concept of vectorket spaceR4108 4 Vector Space VectorKet Space 5 The vector ket space is introduced to represent states in a convenient space of vectors M gt instead of working in the space of functions gbj13 The main difference is that the coordinate dependence does not need to be speci ed when working in the vector ket space According to such representation function 1b13 is the component of vector W gt associated with index 13 vide infra Therefore for any function 1b13 Z Cj gbj13 we can de ne a ket vector W gt such that W gt 2leij gt The representation of l to gt in space a is 10 ng gt Ket Space e 02 l gt C1 lgb1 gt Note that the expansion coef cients Cj depend only on the kets l a gt and not on any speci c vector component Therefore the ambiguity mentioned above is removed In order to learn how to operate with kets we need to introduce the bra Space and the concept of linear functional After doing so this section will be concluded with the description of Postulate 5 and the Continuity Equation Linear functionals A functional X is a mathematical operation that transforms a function M113 into a number This concept is extended to the vector ket space 5 as an operation that transforms a vector ket into a number as follows X iv n or XW gt n where n is a number A linear functional satis es the following equation XltCLWgt bf WWW lad1 where a and b are constants Example The scalar productR4110 n manta is an example of a linear functional since such an operation transforms a function gbz into a number n In order to introduce the scalar product of kets we need to introduce the braSpace Bra Space 5 For every ket W gt we de ne a linear functional lt in l called bravector as follows lt WW gt fd iv Note that functional lt M is linear because the scalar product is a linear functional Therefore lt WWW gt blf gt a lt WW gt b lt Wlf gt 11 Note For convenience we will omit parenthesis so that the notation lt 1b M gt will be equivalent to lt w M gt Furthermore whenever we nd two bars next to each other we can merge them into a single one without changing the meaning of the expression Therefore lt W5 gtlt W5 gt The space of bra vectors is called dual space 5 simply because given a ket W gt2 Z Cj gbj gt the corresponding bra vector is lt M Z 05 lt gbjl In analogy to the ket space a bra vector lt w is represented in space 5 according to the following diagram lt gb Dual Space 5 A Cf 39lt 1 where C is the projection of lt 1b along lt gbj Projection Operator and Closure Relation Given a ket 1b gt in a certain basis set gbj gt w gt 20 gt 1 j where lt q klgbj gt2 Ski Oj lt ijllb gt 2 Substituting Eq 2 into Eql we obtain w gt Z M gtlt aw gt 3 From Eq3 it is obvious that Z ij gtlt qul 1 Closure Relation j where l is the identity operator that transforms any ket or function into itself 12 Note that E jgbj gtlt gbjj is an operator that transforms any vector W gt into a vector pointing in the direction of jgbj gt with magnitude lt gbjjrb gt The operator B7 is called the Projection Operator It projects jgbj gt according to pjlib gtlt W19 gt W7 gt Note that P P where P Pjpj This is true simply because lt gbjjgbj gt 1 41 Exercise 1 Prove that A 0P A rha Z m a A A A A A where 191 Pj HPj PjH Continuity Equation 42 Exercise 2 Prove that 3W aggro 75 m t 0 where joys me W W39 In general for higher dimensional problems the change in time of probability density oX t 77DX t x t is equal to minus the divergence of the probability uX j 9pm 75 at Vj This is the so called Continuity Equation Note Remember that given a vector eldj eg j 13 y z j11y Z 2j2113 y zjj31y z the divergence of j is de ned as the dot product of the del operator V 8y 32 and vector j as follows 13 5 Stationary States Stationary States are states for which the probability density 01 t 2 W133 trb1 t is constant at all times ie states for which W 0 and therefore V j 0 In this section we will show that if 1 t is factorizable according to 1 t gb1ft then 1 t is a stationary state Substituting 1t in the time dependent Schrodinger equation we obtain 8 h2 02 xih Q 400 5 ftVvltbv and dividing both sides by f tgb1 we obtain ih 8ft h2 033 m 075 2mgb1 8132 4 Since the right hand side rhs of Eq 4 can only be a function of 1 and the lhs can only be a function of t for any 1 and t and both functions have to be equal to each other then such function must be equal to a constant E Mathematically agt E ft f0eXp Et E2 82pm A 2m 0x2 vm E gt Ham E o The boxed equation is called the time independent Schrodinger equation Furthermore since f 0 is a constant function gbz f 0gbz also satis es the time independent Schrodinger equation as follows 5 m E 31 DU E 31 and was t Mam 39 Eq 5 indicates that E is the eigenvalue of H associated with the eigenfunction 51 Exercise 3 Prove that H is a Hermitian operator 52 Exercise 4 Prove that dim02 is a Hermitian operator 14 53 Exercise 5 Prove that if two hermitian operators Q and E satisfy the equation QE 2 E62 ie if P and Q commute vide infra the product operator QP is also hermitian Since H is hermitian E is a real number gt E E see Property 1 of Hermitian operators then MW 1375 3lt5 8 1D Since depends only on 1 g5 0 then Q 1 t1b1 t 0 This demonstration t 1 8t 1 8t proves that if 1b1 t gb1 f t then 1b1 is a stationary function 6 Particle in the Box The particle in the box can be represented by the following diagramR122 V1 00 Box V V 0 oo 1 39 1 0 a Particle The goal of this section is to show that a particle with energy E and mass m in the box potential Vx de ned as V 0 when ngga a oo otherwise has Stationary States and a discrete absorption Spectrum ie the particle absorbs only certain discrete values of energy called quanta To that end we rst solve the equation H gb1 Egb1 and then we obtain the stationary states 1b1 t gb1exp Et Since gb1 has to be continuous single valued and square integrable see Postulate l gb0 and gba must satisfy the appropriate boundary conditions both inside and outside the box The bound ary conditions inside the box lead to f 29ltIgtltgtEltIgtltgt ltIgtltgtAs39ltKgt lt6 1 1 1 1n 1 2771 612 7 Functions lt1gt1 determine the stationary states inside the box The boundary conditions outside the box are ltIgt1ooltlgt1EltIgt1 ltIgt10 15 and determine the energy associated with m inside the box as follows From Eq 6 we obtain flK2 EA and CMa ASinK a 0 gt Kazmr with n12 gt Note that the number of nodes of I ie the number of coordinates where m 0 is equal to n 1 for a given energy and the energy levels are E2 2 2 Ez n with n1 2 2m a2 eg h2 7T2 E 1 n 2m a27 h2 4712 E 2 Conclusion The energy of the particle in the box is quantized ie the absorption spectrum of the particle in the box is not continuous but discrete 61 Exercise 6 i Using the particle in the box model for an electron in a quantum dot e g a nanometer size silicon material explain why larger dots emit in the red end of the spectrum and smaller dots emit blue or ultraviolet Mighty Small Mulls mrlnzrmrimimr andminutesFirming i 1 will change Him 5 ll er hunmmsmmha39 ame in the mint El t fjm The5 rterrxgeazney ViiFarming nmpquot quot l mu Warfarechimi gj Imier 31539 l ii Consider the molecule hexatriene CH2 2 CH CH 2 CH CH 2 CH2 and assume that the 6 7r electrons move freely along the molecule Approximate the energy levels using the particle in the box model The length of the box is the sum of bond lengths with C C 154 A CC 135 A and an extra 154 A due to the ends of the molecule Assume that only 2 electrons can occupy each electronic state and compute A The energy of the highest occupied energy level B The energy of the lowest unoccupied energy level 16 C The energy difference between the highest and the lowest energy levels and compare such energy difference with the energy of the peak in the absorption spectrum at AMAX268nm D Predict whether the peak of the absorption spectrum for C H 2 C H C H C H n C H C H 2 would be red or blue shifted relative to the absorption spectrum of hexatriene 7 Commutator The commutator 121 B is de ned as followsR497 21 B AB BA Two operators fl and B are said to commute when 121 B O 71 Exercise 7 Prove that 5 ih 2 iii Hint Prove that f iharbc ih c where any is a function of 13 8 Uncertainty Relations The goal of this section is to show that the uncertainties AA lt 1 lt fl gt2 gt and AB lt B lt B gt2 gt of any pair of hermitian operators fl and B satisfy the uncertainty rela tionR3437 1 AAgt2ltABgt2 2 1 lt m B gt2 7 In particular when fl 1 and B 13 we obtain the Heisenberg uncertainty relation h A51 o Ap Z 8 Proof2A A A q E til lt A gt gbc E U iAVltIgtc V E B lt B gt A E fdzcgbcgbc Z 0 A manil lt A gtltIgtc wB lt B gtlt1gta1 lt A gtltIgtc wB lt B gtltIgtc A lt UCDlUCD gt A2 lt VimCI gt ilt VCDlUCD gt lt UCDlVCD gt 17 HMlt W hwmde Wgt lt WV VW gt2Q The minimum value of I A as a function of A is reached when 01 0A 2 818Xquot 0 This condition implies that 39lt A B gt 2AB AB L 7 Substituting this expression for A into Eq 9 we obtain i2ltABgt2 i2ltABgt2 AA 2 gt lt l39 MAB mABV Q AA2AB2 gt 4 81 Exercise 8 Compute lt X gt lt P gt AX and AP for the particle in the box in its minimum energy state and verify that AX and AP satisfy the uncertainty relation given by Eq 8 With the exception of a few concepts e g the Exclusion Principle that is introduced later in these lectures the previous sections have already introduced most of Quantum Theory Furthermore we have shown how to solve the equations introduced by Quantum Theory for the simplest possible problem which is the particle in the box There are a few other problems that can also be solved analytically eg the harmonicoscillator and the rigidrotor described later in these lectures However most of the problems of interest in Chemistry have equations that are too complicated to be solved analytically This observation has been stated by Paul Dirac as follows The underlying physical laws necessary for the mathematical theory of a large part of Physics and the whole of Chemistry are thus completed and the di iculty is only that exact application of these laws leads to the equations much too complicated to be soluble It is therefore essential to introduce approximate methods e g perturbation methods and variational methods 82 EPR Paradox Gedankenexperiments i e thought experiments have been proposed to determine hidden vari ables The most famous of these proposals has been the Einstein Podolski Rosen EPR gedanken experiment Phys Rev 1935 47 2777 780 where a system of 2 particles is initially prepared with total momentum pt At a later time when the two particles are far apart from each other the position 5151 is measured on particle l and the momentum p2 is measured on particle 2 The paradox is that the momentum of particle 1 could be obtained from the difference p1 2 pt p2 Therefore 18 the coordinate 521 and momentum p1 of particle 1 could be determined with more precision than established as possible by the uncertainty principle so long as the separation between the two par ticles could prevent any kind of interaction or disturbance of one particule due to a measurement on the other The origin of the paradox is the erroneous assumption that particles that are far apart from each other cannot maintain instantaneous correlations However quantum correlations between the properties of distant noninteracting systems can be maintained as described by Bohm and Aharonov Phys Rev 1957 1081070 1076 for the state of polarization of pairs of correlated photons Within the Bohmian picture of quantum mechanics these quantum correlations are estab lished by the quantum potential VQ q even when the particles are noninteracting i e Vq 0 Quantum correlations between distant noninteracting photons were observed for the rst time by Aspect and co workers in 1982 Phys Rev Lett 1982 4991 94 47 years after the EPR paradox was presented These quantum correlations constitute the fundamental physics eXploited by teleportation ie the transmission and reconstruction of quantum states over arbitrary large distances Nature 1997 390575 579 and ghost imaging ie a technique where the object and the image system are on separate optical paths Am J Phys 2007 75 2343 351 9 Variational Theorem The expectation value of the Hamiltonian computed with any trial wave function is always higher or equal than the energy of the ground state Mathematically lt gt2 E0 where Z Proof A in Z Oj oj where q j is a basis set of orthonormal eigenfunctions of the Hamiltonian H lt WW gt 22020 lt 1191qu gt j k j k Zcchj 2 E0 Egg 7 j where Z 00j 1 Variational Approach Starting with an initial trial wave function 1b de ned by the eXpansion coef cients 070 the optimum solution of an arbitrary problem described by the Hamiltonian H can be obtained by minimizing the eXpectation value lt rblH W gt with respect to the eXpansion coef cients 19 10 Digital GridBased Representations The standard formulation of quantum mechanics presented in previous sections relies upon the tools of calculus eg derivatives integrals etc and involves equations and operations with in nitesimal quantities as well as states in Hilbert space the in nite dimensional space of functions L2 The equations however seldom can be solved analytically Therefore computational solu tions are necessary However computers can not handle in nite spaces since they have only limited memory In fact all they can do is to store and manipulate discrete arrays of numbers Therefore the question is how can we represent continuum states and operators in the space of memory of digital computers In order to introduce the concept of a gridrepresentation we consider the state 14 a move 2 O2WO O 10 which can be expanded in the in nite basis set of delta functions 61 atquot as follows IJOU dz cv 6v 523 11 where Czv E 55 MPG IJOLquot All expressions are written in atomic units so h 1 A grid based representation of 110 can be obtained in the coordinate range 2 zjmin aim by discretizing Eq 11 as follows IJOI A Z cj6b 13 12 j1 where the array of numbers cj E aij lo represent the state 110 on a grid of equally spaced coordi nates go 2 Umm j 1A with nite resolution A aimC n 1 Note that the grid based representation introduced by Eq 12 can be trivially generalized to a grid based representation in the multidimensional space of parameters eg L39j pj yj etc when expanding the target state 110 as a linear combination of basis functions 351 pj 7 with expansion coef cients as cj E mjpjyj110 101 Computational Problem 1 Write a computer program to represent the wave packet introduced by Eq 10 on a grid of equally spaced coordinates 21 2 mm j 1A with nite resolution A glamC n 1 and visualize the output Choose x0 0 and p0 0 in the range x 2020 with oz 2 mm where m 1 and w 1 Next we consider grid based representations in momentum space Pop 091 13 20 Inserting the closure relation 1 f dxlx in Eq 13 we obtain that ltpwogt fdxltplwgtltxwogt lt2wgt12dme Wltxwogt lt14 is the Fourier transform of the initial state The second equality in Eq 14 was obtained by using 1131 2713912 ip 15 which is the eigenstate of the momentum operator p 239V with eigenvalue p since M23119 2 pWW The Fourier transform can be computationally implemented in ONlogN steps by using the Fast Fourier Transform FFT algorithm see Ch 12 of Numerical Recipes by WH Press BP Flannery SA Teukolsky and WT Vetterling Cambridge University Press Cambridge 1986 f12 2pdf when zjlkllo is represented on a grid with N 2 points where n is an integer In contrast the implementation of the Fourier transform by quadrature integration would require 0N2 steps 102 Computational Problem 2 Write a computer program to represent the initial state introduced by Eq 10 in the momentum space by applying the FFT algorithm to the grid based representation generated in Problem 1 and visualize the output Represent the wave packet amplitudes and phases in the range p 44 and compare your output with the corresponding values obtained from the analytic Fourier transform obtained by using dz eXp a212 111 a0 x7ra2 eXpa0 ai4a2 Next we consider the grid based representation of operators 6 g i3 13 Vi and T 132 2m and learn how these operators act on states represented on grids in coordinate and momentum spaces For simplicity we assume that the potential is Harmonic We 2 gmc 3 16 Consider rst applying the potential energy operator to the initial state as follows V x110a V1IJO1 E flow 17 Since i10lt l gt is just another function Eq 17 indicates that Vf can be represented on the same grid of coordinates as before ie equally spaced coordinates 19 xmm j 1A with nite resolution A aimC 1 Since for each 13 Omj V1jll1j the operator Vf can be represented just as an array of numbers V1j associated with the grid points 13 and its operation on a state is represented on such a grid as a simple multiplication 21 103 Computational Problem 3 Write a computer program to compute the expectation values of the position 5130 IIO IIO and the potential energy V WON5 KIIO where Vz is de ned according to Eq 16 for the initial wave packet introduced by Eq 10 with various possible values of 510 and p0 with oz 2 com where m 1 andw 1 Now consider applying the momentum operator 13 239V to the initial state IIO2 as follows 027 ltl lqjogt 5V 1 097 18 One simple way of implementing this operation when IIO1 is represented on a grid of equally spaced points 19 2 mm j 1A is by computing nite increment derivatives as follows I 0j1 WOW 1 G1j 2 2A 19 However for a more general operator eg T 132 2771 this nite increment derivative procedure becomes complicated In order to avoid such procedures one can represent the initial state in momentum space by Fourier transform of the initial state apply the operator by simple multiplication in momentum space and then transform the resulting product back to the coordinate representation by inverse Fourier transform This method can be derived by inserting the closure relation 1 f dplp p in Eq 18 GW Jil l l d dplt pgtltp 1 ogt 27TY12dpeipxmpl 1 oh 20 since phllo is de ned according to Eq 14 as the Fourier transform of the initial state Note that the second equality of Eq 20 is obtained by introducing the substitution mm 2 1262 21 While Eq 20 illustrates the method for the speci c operator 13 one immediately sees that any operator which is a function of 13 eg T 132 can be computed analogously according to the Fourier transform procedure 104 Computational Problem 4 Write a computer program to compute the expectation values of the initial momentum 190 110 13 110 and the kinetic energy T IIO 1322771 KIIO by using the Fourier transform procedure where IIO is the initial wave packet introduced by Eq 10 with 510 2 0 p0 0 and oz 2 com where m 1 and w 1 Compute the expectation value of the energy E 0119 KIIO where H 1322771 Vi with V1 de ned according to Eq 16 and compare your result with the zero point energy E0 w 2 22 11 SOFT Method The Split Operator Fourier Transform SOFT method is a numerical approach for solving the time dependent Schrodinger equation by using grid based representations of the time evolving states and operators It relies on the Fourier transform procedure to apply operators that are func tions of p by simple multiplication of array elements As an example we will illustrate the SOFT algorithm as applied to the propagation of the harmonic oscillator which can also be described analytically as follows Ilt1 d lemtl a lllo 22 where the Kernel 316 th 1513 is the quantum propagator ifIt gt mw mw we 2 l2 Q sinhmwfx 28inhmt 17 Coshwzt 21313 23 The essence of the method is to discretize the propagation time on a grid 75 k 1739 with k 1 n and time resolution 739 t n 1 and obtain the wave packet at the intermediate times 75 by recursively applying Eq 22 as follows imnmszmaawwmw m If 739 is a suf ciently small time increment ie n is large the time evolution operator can be approximated according to the Trotter expansion to second order accuracy 6 7117 iVE72 ip272m iVE72 0737 25 which separates the propagator into a product of three operators each of them depending either on i3 or p 111 Computational Problem 5 Expand the exponential operators in both sides of Eq 40 and show that the Trotter expansion is accurate to second order in powers of 739 Substituting Eq 40 into Eq 24 and inserting the closure relation 1 f dplp pl gives IjtkHH Z dle iV 72 lt pgt ip27 2m ltp lgt iVw 72Ijtl By substituting plx and p according to Eqs 15 and 21 respectively we obtain A 1 2 1 I 1V E7392 11 1T2m L E LVZE7392 Iltk11 e 27Tdpe p6 p 27Td1 e p e Iltka 27 According to Eq 27 then the computational task necessary to propagate Ilt1 for a time increment 739 involves the following steps 23 7LVI 7392 7VIj7392 1 Represent Iltk2 and e as arrays of numbers Iltk and 6 associated with a grid of equally spaced coordinates 29 3ij j 1A with nite resolution A aimw 1 2 Apply the potential energy part of the Trotter eXpansion 6 iV T2 to Iltk1 by simple multiplication of array elements 132 7 69 ij W 2 15 7 3 Fourier transform tk to obtain 13 pj and represent the kinetic energy part of the Trotter eXpansion e pgfQm as an array of numbers 6 7319372 7 associated with a grid of equally spaced momenta pj j 3ij 4 pply the kinetic energy part of the Trotter eXpansion 6 7492727 1 to the Fourier transform Iltk p by simple multiplication of array elements N i 2739 m Ijtkpj e p 2 qjtkpj 5 Inverse Fourier transform Ivltk pj to obtain Ivltk on the grid of equally spaced coordinates 13 j 6 Apply the potential energy part of the Trotter eXpansion 6 iV T2 to Ivltk1 by simple multiplication of array elements qjtk1 6 iVj72Ivjtk 112 Computational Problem 6 Write a computer program that propagates the initial state IIO1 for a single time increment 739 01 au Use 510 25 p0 0 and oz 2 mm where m 1 and w 1 Implement the SOFT method for the Hamiltonian H 1322771 Vi where V1 is de ned according to Eq 16 Compare the resulting propagated state with the analytic solution obtained by substituting Eq 23 into Eq 22 113 Computational Problem 7 Loop the computer program developed in Problem 5 with 510 2 25 and p0 0 for 100 steps with 739 01 au For each step compute the eXpectation values of coordinates 1t and momenta pt as done in Problems 3 and 4 respectively Compare your calculations with the analytic solutions obtained by substituting Eq 23 into Eq 22 Verify that these correspond to the classical trajec tories 1t E 30 Ecoswt and pt 2 p0 30 Ewm sinwt which can be computed according to the Velocity Verlet algorithm Pj1 27 FW Fj1T2 28 17j12j 24 114 Computational Problem 8 Change the potential to that of a Morse oscillator V De1 exp ai39 xe2 with we 2 0 De 2 8 and a kQDE where k mwQ Recompute the wave packet propagation with x0 05 and p0 0 for 100 steps with 739 01 au and compare the expectation values 51575 and pt with the corresponding classical trajectories obtained by recursively applying the Velocity Verlet algorithm 115 Computational Problem 9 Simulate the propagation of a wave packet with 520 2 55 and initial momentum p0 2 colliding with a barrier potential Vz 3 if absx lt 05 and Vx 0 otherwise Hint In order to avoid arti cial recurrences you might need to add an absorbing imaginary potential Vax iabsx 104 if absx gt 10 and Vax 0 otherwise 12 Time Independent Perturbation Theory Consider the time independent Schrodinger equationR2453 Wm 13mm 29 for a system described by the Hamiltonian H 132 2m V and assume that all the eigenfunctions q nx are known The goal of this section is to show that these eigenfunctions q nx can be used to solve the time independent Schrodinger equation of a slightly different problem a problem described by the Hamiltonian H H Ad This is accomplished by implementing the equations of Perturbation Theory derived in this section Consider the equation A H AcDltIgtnA x EnAltIgtnA 10 30 where A is a small parameter so that both EMA and EMA are well approximated by rapidly convergent cxpansions in powers of A ie expansions where only the rst few terms are important Expanding 1A we obtain ROWE ZCjn j 7 Substituting this expression in the time independent Schrodinger equation we obtain 2 CjnA 19 1 AWAM EM 2 CknA k therefore ClnAEl A Z and lt Wows gt Exclx 31 7 Expanding ij and E in powers of A we obtain 25 0 1 2 ijA Ck Ck A Ck A2 and EnA E55 ESSA EEW Substituting these expansions into Eq 31 we obtain 059E E 0O 2 MOZ BEZ 27 0 lt abllwlqu gt E 0C 139 X A2ClltjEl 27 179 lt allows gt E ECl Elmo Elncf 0 This equation must be valid for any A Therefore each of the terms in between parenthesis must be equal to zero 012El 19790 0 Zeroth order in A if 1 7E n then Cl 2 0 ifl n then 07892 1 and El ESP 1 0 A Cfanl E79 E 1Cl2 Z 0 lt gbllwlgbj gt First order in A ifl 7E n then ClEl ES 2 CTL92 lt gblldjlgbn gt ifl n then E90 0922 lt anlalan gt Note that 07 is not speci ed by the equations listed above 07 is obtained by normalizing the wave function writt2en to rst order in A 1 2 0 1 1 OlnEl En Zj Gin lt Cblld lgbj gt E78 Oln E78 Oln7 2 1 A lt nlwl gtlt Ial ngt 1fl n then En Z n 0 lt n w j gt Z n EE L ifl 7E n then Cl5El ES Zj jg lt gbl a gbj gt lt n ngtlt z ngt El ES lt jlal ngtlt zlal jgt lt nlal ngtlt zlal ngt t 3 37 139 El ES 39 2nd order in A 4 121 Exercise 9 Calculate the energy shifts to rst order in A for all excited states of the perturbed particle in the box described by the following potential 00 DO A A 26 Assume that the potential is described by perturbation Sim to the particle in the box 13 Time Dependent Perturbation Theory Given an arbitary stateR2410 15W 75 Z Ojj Ejt7 for the initially unperturbed system described by the Hamiltonian H for which Hi Ej lt1 and ihg f 2 H15 let us obtain the solution of the time dependent Schrodinger equation 373 H Maw 32 assuming that such solution can be written as a rapidly convergent eXpansion in powers of A 00 i am 75 Z Z C enlcpjpae 33 0 j l Substituting Eq 33 into Eq 32 we obtain 3973 C tAl C tAl EE JEN CtAlltltIgt ltIgtgtE AltltIgt AltIgtgt 43239 Mo mo lt h Jzltgtlt a 3 kiwi gteh Terms with A0 Zero order time dependent perturbation theory mama15E Ok0lttgtlt EkgteEktl Z Oj0lttgt6kjEjeE Gimme337a j Since 0760 t Z 07 gt Okio t Okio Therefore the unperturbed wave function is correct to zeroth order in A Terms with A First order time dependent perturbation theory ihlc klw iEktOk1tlt Eke iEktl Z 031t6kjEje Ejtcjot lt klalcbj gt et Ejt J out OJ00 lt chum gt e tlEj E 27 Therefore A 2A ethwe thlCIDj gt Okla 3 2 6900 lt cm gt2 3 2 6300 lt lt1 h j h j A 34 Eq 34 was obtained by making the substitution eHtltIgtj gt eEjtltIgtj gt which is justi ed in the note that follows this derivation Integrating Eq 34 we obtain 239 t A A Ckl t 2 dt ZCj00 lt 13k 6Ht COB EH15 gt 00 j which can also be written as follows t A A N 01105 dt lt Pk e m 675 MO gt 00 This expression gives the correction of the expansion coef cients to rst order in A Note The substitution made in Eq 34 can be justi ed as follows The exponential function is de ned in powers series as follows 00 n A 1 A E n20 In particular when A z39Hth Furthermore since gt2 Ej CIDj gt and A A A Hchbj gt Ejchbj gt Eflqu gt we obtain zc i 1 i 1 6 WI gt 1 15745 wEWEE HltIgtj gt 6 71374 gt which is the substitution implemented in Eq 34 Terms with A2 Second order time dependent perturbation theory i z A i zhlokxt Ok2t gEkl Eloymama 63105 lt ltIgtklwllt1gtj gt1e j 28 139 6Ht hmij gt 27175 239 0205 7 Z lt lt1gt j OO 2 t 25 0205 Z dt dt lt lt1gt j 00 00 Since 1 Z ij gtlt IDA 239 2 t t 02t dt dt ltlt1gt This expression gives the correction of the expansion coef cients to second order in A t A A 01205 dt Z lt ltIgt th toe th to gt Cj1t j 1AA1 A 1AIIA1AH ethwe hmij gtlt CIDj eth we hm rb0gt 1AA 139A A 1A ethwe hHt made th WOgt Limiting Cases 1 Impulsive Perturbation The perturbation is abruptly switched on R2412 0 According to the equations for rst order time dependent perturbation theory t Claw Z lt kiaiq j gt OjoOO dt EjEkt 7 j therefore i lt gt 1 C t Z Jo J HE Ekt 1 k1 EjEk 6 Assuming that initially Cj 6 gt Ojo Slj Therefore lt I a I gt 1 1105 6 El Em when k 7E Z Note that Cl1t must be obtained from the normalization of the wave function eXpanded to rst order in A 29 131 Exercise 10 Compare this expression of the rst order correction to the expansion coef cients due to an im pulsive perturbation with the expression obtained according to the time independent perturbation theory 2 Adiabatic limit The perturbation is switched on very slowly ltlt 6 with e arbitrarily smallR2448 6005 m m t i t z I 01105 dt lt Pklw qh gt e ElEkt Integrating by parts we obtain t zt 239 Ez Ekt t Ez Ekt 6w thZ i ltltIgtkwt CD1gt dt i ltDk Dlgt and since lt Pklw ooclgtl gt 0 Ckl t Z lt Dk wt ql gt 1 LEl Ekt7 El Etc when k 7E Z Note that Cl1t must be obtained from the normalization of the wave function expanded to rst order in A 132 Exercise 11 Compare this expression for the rst order correction to the expansion coef cients due to an adi abatic perturbation with the expression obtained according to the time independent perturbation theory 30 3 Sinusoidal Perturbation The sinusoidal perturbation is de ned as follows 205113 2 Sin 2t when t Z 0 and t 13 0 otherwise w1 t A Vt It is however more conveniently de ned in terms of exponentials a 7915 6 1915 22 e Therefore t Csz id lt h 0 with MO gt2 Z lecbj gt and HCIDj EjCIDj Substituting these expressions into Eq 16 we obtain eet39mtveet39no gt 35 1 t i i Ck1t 2 h Z Oj lt Maw gt 0 dt eaKEk Enwu e KEk E hmt 7 j and therefore 1 Ek Ej719t 1 Ek Ej 59t Ek Ej Ek Ej h Q h 9 1 016105 Z 3ij 7 Without lost of generality let us assume that Oj 6m ie initially only state n is occupied For k 2 n we obtain Ek En Ek En 2 wkn 2 1 z h Qt 1 1 h Qt 2 E Engt E Engt 9 9 Factor lekn determines the intensity of the transition eg the selection rules The rst term called antiresonant is responsible for emission The second term is called resonant and is re sponsible for absorption 31 For k 7E n Pk1t A2le1tl2 is the probability of nding the system in state k at time t to rst order in A lakn 4E2 EkEn h It is important to note that PM ltlt 1 indicates that the system has been slightly perturbed Such condition is satis ed only when 75 ltlt 2 h Therefore the theory is useful only at suf ciently wkn 39 short times 14 Problem Set 141 Exercise 11 Consider a distribution of charges Qi with coordinates 72 interacting with plane polarized radia tion Assume that the system is initially in the eigenstate CIDj of the unperturbed charge distribution A Write the eXpression of the sinusoidal perturbation in terms of Qi 73 and the radiation fre quency w and amplitude 60 B EXpand the time dependent wave function 1b of the charge distribution in terms of the eigen functions CIDk of the unperturbed charge distribution C Find the eXpansion coef cients according to rst order time dependent perturbation theory D What physical information is given by the square of the eXpansion coef cients E What frequency would be optimum to populate state k Assume Ek 2 E7 F Which other state could be populated with radiation of the optimum frequency found in term E G When would the transition j e k be forbidden 32 142 Exercise 12 A particle in the ground state of a square box of length lal is subject to a perturbation wt t2739 cube A What is the probability that the particle ends up in the rst excited state after a long time 75 gtgt 7 B How does that probability depend on 739 143 Exercise 13 V0 Figure l a Compute the minimum energy stationary state for a particle in the square well See Fig1 by solving the time independent Schrodinger equation b What would be the minimum energy absorbed by a particle in the potential well of Fig1 0 What would be the minimum energy of the particle in the potential well of Fig l d What would be the minimum energy absorbed by a particle in the potential well shown in Fig2 Assume that A is a small parameter give the answer to rst order in A A V0 0 05 1 144 Exercise 14 gure 2 a Prove that F 6 H is a hermitian operator b Prove that P CosH is a hermitian operator 33 15 Exercises Suggested in Class 151 Copenhagen Interpretation Describe the Copenhagen probabilistic formulation of Quantum Mechanics and show that a con sequence of Postulates 1 and 4 is that Pt x 2 W175 x t xdx represents the probability of observing the system described by M1375 between 15 and x dx at time t 152 Feynman Interview Watch Feynman talking about Quantum Mechanics at the Interview and comment on his obser vations in the context of the postulates of Quantum Mechanics 153 Momentum Operator Show that the momentum operator must be de ned as A 3 p Zh7 and the eigenfunction of the momentum operator with eigenvalue pj as a plane wave lt l gt le 37 15 2 p V 27Th since 135p pj pj5p pj 38 Hint Use the integral form of Dirac s delta function 6 p pj lh ff dxeppj 154 EPR Paradox In 1935 Einstein Podolsky and Rosen proposed a thought experiment where two systems that interact with each other are then separated so that they presumably interact no longer Then the position or momentum of one of the systems is measured and due to the known relationship be tween the measured value of the rst particle and the value of the second particle the observer is aware of the value in the second particle A measurement of the second value is made on the second particle and again due to the relationship between the two particles this value can then be known in the rst particle This outcome seems to violate the uncertainty principle since both the position and momentum of a single particle would be known with certainty Explain what is wrong with this paradox 34 155 Schrodinger s cat paradox A cat is placed in a steel box along with a Geiger counter a vial of poison a hammer and a radioactive substance When the radioactive substance decays the Geiger detects it and triggers the hammer to release the poison which subsequently kills the cat The radioactive decay is a random process and there is no way to predict when it will happen The atom exists in a state known as a superpositionboth decayed and not decayed at the same time Until the box is opened an observer doesn t know whether the cat is alive or deadbecause the cat s fate is intrinsically tied to whether or not the atom has decayed and the cat would as Schrodinger put it be living and dead in equal parts until it is observed In other words until the box was opened the cat s state is completely unknown and therefore the cat is considered to be both alive and dead at the same time until it is observed A The obvious contradiction is that the cat can not be both dead and alive so there must be a fundamental aw of the paradox or of the Copenhagen interpretation Explain what aspect of the Copenhagen interpretation of quantum mechanics is questioned by this gedanken experiment and what is wrong with the paradox 156 Time Evolution Operator 1561 Evolution in the basis of eigenstates Show that A A A A Htl jgt Ejtl jgt 39 when M is an eigenstate of H with eigenvalue Ej 1562 Trotter expansion of the time evolution operator Show that A 6 7117 Z iVE72 ip272m 7 V 7 2 C73 4O Hint Expand the exponential operators in both sides of Eq 40 and show that the Trotter expan sion is accurate to second order in powers of 739 35 1563 Numerical Comparison Consider a particle of unit mass m1 in a box of unit length L1 initially prepared in the superposition state 1 where gbl1 and gbg1 are the eigenstates with eigenvalues E1 and E2 obtained by solving the time independent Schrodinger equation 19rqu as Ej jv Compute the time evolved wavefunction at time t 107 with 739 1 by analytically applying the time evolution operator 6 th to 1M3 0 term by term Compare 1 t to the resulting wave function obtained by numerically applying the Trotter expansion of 6 iHT 10 times to the initial superposition state 70 1 2 16 Adiabatic Approximation The goal of this section is to solve the time dependent Schrodinger equation 01D A 2h at H w 41 for a time dependent Hamiltonian H V2 Vaz t where the potential Vx t undergoes signi cant changes but in a very large time scale eg a time scale much larger than the time associated with state transitionsR2496 Since Vxt changes very slowly we can solve the time independent Schrodinger equation at a speci c time t Ht ltlgtn 75 Ent ltIgtn 75 Assuming that 8 m 0 since Vxt changes very slowly we nd that the function Wat t mag mfg i I Enmdt39j is a good approximate solution to Eq 41 In fact it satis es Eq 41 exactly when 83 0 Expanding the general solution 1M3 t in the basis set Dn1t we obtain WW 2 ZCntlt1gtnatefJEnltt39gtdt and substituting this expression into Eq 41 we obtain ihZltqugtn 073 En0nlt1gtne f Enltt gtdt Z In owfig JENmg 36 where Note that a a H E at at then a a H E lt Mylo gt lt came gt 76 E lt Mch gt since lt cumin gt lt 3anka gt Furthermore if k 7E n then lt kl fln gt E Ek Substituting this expression into Eq 42 we obtain lt child gt2 lt Fi 13MB gt63 lt f dt Ent gtEkt gt En Ek Ck 0k lt gt ZCn n7 k Let us suppose that the system starts with Cn0 67 then solving by successive approximations we obtain that for k 7E j lt 1 29 I cbj gt z t I I I Ck 6 5 f0 dt EN Ekt Ek Ej Assuming that E7 75 and Ekt are slowly varying functions in time N lt gt 0 N e Ej Ekt EjEkto 7 Ej Ek2 since le iw M Ej Ekto S 2 Therefore a W a 4m lt kla flaj gt P Ej Ek4 The system remains in the initially populated state at all times whenever 29 is suf ciently small 8H E E 2 even when such state undergoes signi cant changes This is the so called adiabatic approximation It breaks down when Ej Ek because the inequality introduced by Eq 43 can not be satis ed Mathematically the condition that validates the adiabatic approximation can also be expressed in terms of the frequency 1 de ned by the equation Ej E 2 1w g or the time period 739 of the light emitted with frequency 1 as follows 37 lt gt ltlt 38 17 Heisenberg Representation Consider the eigenvalue problemR4124 R3240 HMO 2 Ella 44 for an arbitrary state Ml of a system eg an atom or molecule expanded in a basis set gbj as follows w Z 0 was 45 j where Cl 2 rm and gbjlgbk Sjk Substituting Eq 45 into Eq 44 we obtain 2 WM 2 MW j j Applying functional ltng to both sides of this equation we obtain ZWHHWNOZU Z Ellt kl jgt01ja 46 J where oklgbj 6M and k 12 n Introducing the notation H kj le gbj we obtain If I Hllcl l H12Cl2 HBOZB l H1n0ln ElCl1 OOZQ OCle7 k H210l1 l H22Cl2 H230l3 l H2n0ln OCZU ElOl2 OCle7 k n a qum Hmofm Hugo HMO 00 00 Ed 47 that can be conveniently written in terms of matrices and vectors as follows H11 H12 H1 Cl El 0 o of H21 H22 H2 of Z 0 El 0 of 48 H1 H2 Hm do 0 0 El Maj This is the Heisenberg representation of the eigenvalue problem introduced by Eq 44 According to the Heisenberg representation also called matrix representation the ket MM is represented by the vector 0 with components Cl 2 gbj W1 with j 1 n and the operator H is represented by the matrix H with elements ij gbj The expectation value of the Hamiltonian MHW Z ZOlklt iH jgtC k j 39 can be written in the matrix representation as follows H11 H12 Hln WWW CJHol gm Ola O gt 22 2 Z Note 1 It is important to note that according to the matrix representation the ketvector MM is repre sented by a column vector with components Cl 2 gbj MM and the bravector W is represented by a row vector with components Clj 2 If an operator is hermitian eg H it is represented by a hermitian matrix ie a matrix where any two elements which are symmetric with respect to the principal diagonal are complex conjugates of each other The diagonal elements of a hermitian matrix are real numbers therefore its eigenvalues are real 3 The eigenvalue problem has a non trivial solution only when the determinant detH 1E vanishes detH 1E 0 where i is the unity matrix This equation has n roots which are the eigenvalues of H 3 Finally we note that the matrix of column eigenvectors C satisfy the equation HC 2 EC where E is the diagonal matrix of eigenvalues H11 H12 Hut l 01 02 051 E1 0 o 01 02 021 IH21 H22 Hgnl 012 022 052 0 E2 0 012 022 072 49 18 TwoLevel Systems There are many problems in Quantum Chemistry that can be modeled in terms of the two level Hamiltonian ie a state space with only two dimensions Examples include electron transfer proton transfer and isomerization reactions Consider two states gbl gt and log gt of a system Assume that these states have similar energies E1 and E2 both of them well separated from all of the other energy levels of the system H0l 1 gt E1l 1 gt7 Hol 2 gt2 E2l 2 gt 0 A WA 0 40 In the presence of a perturbation the total Hamiltonian becomes H 2 H0 W Therefore states gbl gt and ng gt are no longer eigenstates of the system The goal of this section is to compute the eigenstates of the system in the presence of the perturbation W The eigenvalue problem H11 H12 0 2 El 0 0 H 21 H 22 Ila 0 El CF 7 is solved by nding the roots of the characteristic equation H 11 ElH22 El H 1211721 2 0 The values of E that satisfy such equation are 2 Elii W These eigenvalues Eli can be represented as a function of the energy difference E1 E2 accord ing to the following diagram El Eg Note that E1 and E2 cross each other but E and E repel each other Having found the eigen values Ei we can obtain the eigenstates Wt gt2 OS M1 If 2gt by solving for CE and If from the following equations 0907111 Ei Jme 0 231 GENO We see that in the presence of the perturbation the minimum energy state ML gt is always more stable than the minimum energy state of the unperturbed system Example 1 Resonance Structure 41 lt51 C52 The coupling between the two states makes the linear combination of the two more stable than the minimum energy state of the unperturbed system Example 2 Chemical Bond a 5 3 W e M51 gt C H H W2 gt The state of the system that involves a linear combination of these two states is more stable than Em because lt gbllHle2 gt7 0 Time Evolution Consider a two level system described by the Hamiltonian H 2 H0 W with H0 l gbl gt E1 l gbl gt Assume that the system is initially prepared in state l 1M0 gtl 1 gt Due to the presence of the perturbation W state gbl gt is not a stationary state Therefore the initial state evolves in time according to the time dependent Schrodinger Equation 3W gt at Z gt and becomes a linear superposition of states gbl gt and ng gt Mt gt 0mm gt 02t 2 gt 42 State 05 gt can be expanded in terms of the eigenstates Mi gt as follows W75 gt Clt0l gt C tl gt where the expansion coef cients Ci t evolve in time according to the following equations 80 75 2h EClttgt7 acne M at ECt Therefore state W05 gt can be written in terms of Mi gt as follows wt gt Cameterm gt 0ltogteEtw gt The probability amplitude of nding the system in state ng gt at time t is P1205 l lt MIN gt 2 0275C275a which can also be written as follows P1205 loan10 2340 2ReC C002C0eltEEgtt where 02 lt gbg Ili gt The following diagram represents P1205 as a function of time P 12 75 Rabi Oscillations 0 7th t E E The frequency 1 E E7rh is called Rabi Frequency It is observed eg in the absorption spectrum of H see Example 2 It corresponds to the frequency of the oscillating dipole moment which uctuates according to the electronic con gurations of W1 gt and W2 gt respectively The oscillating dipole moment exchanges energy with an external electromagnetic eld of its own characteristic frequency and therefore it is observed in the absorption spectrum of the system 43 19 Harmonic Oscillator Many physical systems including molecules with con gurations near their equilibrium positions can be described at least approximately by the Hamiltonian of the harmonic oscillatorR4483 R162 click A 2 H P 11711112532 2m 2 In order to nd the eigenfunctions of H we introduce two operators called creation 61 and annihi lation a which are de ned as follows 61 E 23915 and a E 115 where if 5 andp 2 Tim Using these de nitions of 61 and a we can write H as follows H 51a Introducing the number operator N de ned in terms of 61 and a as follows N E 51 we obtain that the Hamiltonian of the Harmonic Oscillator can be written as follows H 17 12hw 191 Exercise 15 Show that if 1 is an eigenfunction of H with eigenvalue EV then 1 is an eigenfunction of N with eigenvalue 1 5 Mathematically if H CIDV gt EVICIDV gt then NICIDV gt 1ltIgtV gt with 71w Eu 1 1 m 2 Theorem I The eigenvalues of N are greater or equal to zero ie 1 Z 0 Proof fdajl lt IZWCIDV gt 2 Z 0 lt Mame gt2 0 1 lt ILICE gt2 0 As a consequence alCIDO gt 0 eCIDO1 0 01nltIgt0 113823 mw DOW Z A can E322 where A 4 The wave function 130213 is the eigenfunction of N with 1 0 ie the ground State wave function because 1 Z 0 44 Theorem II If I gt 0 State dlCIDV gt is an eigenstate ofN with eigenvalue equal to V I Proof In order to prove this theorem we need to show that Nate gt u match gt 50 We rst observe that A N 6L amp Therefore A 6L ampampamp ampampamp A CAL CAL lb 646 N 6L 1amp because L 6L 2 1 amt 2 gym mp 2135 1313 ii mp 213i 1313 L L g 1 since 343 2 iii Applying the operator amp to state ME gt we obtain Na amp gt we gt and therefore NdlCIDV gt wlltlgt gt LlltIgtV gt which proves the theorem A natural consequence of theorems I and II is that I is an integer number greater or equal to zero The spectrum of N is therefore discrete and consists of integer numbers that are 2 0 In order to demonstrate such consequence we rst prove that Name gt2 u pampplltIgtV gt 51 In order to prove Eq 51 we apply 6L to both sides of Eq 50 Watch gt2 u 1amp2ltIgt gt and since A7 6L amp we obtain a Name gt2 u 1amp2ltIgt gt and A N cby gt V 2a2lt1gt gt 52 Applying EL to Eq 52 we obtain Maw gt2 u 2a3lt1gt gt and substituting amp7 by 6L Nd we obtain Na o gt V 3a3lt1gt gt Repeating this procedure p times we obtain Eq 51 45 Having proved Eq 51 we now realize that if 1 n with n an integer number we gt2 0 when p gt n This is because state dnlcbn gt is the eigenstate of N with eigenvalue equal to zero ie dnlcbn gt CIDO gt Therefore dlcbo gt dplcbn gt 0 whenp gt 71 Note that Eq 52 would contradict Theorem I if 1 was not an integer because starting with a nonzero function ME gt it would be possible to obtain a function dp ME gt different from zero with a negative eigenvalue Eigenfunctions of N In order to obtain eigenfunctions of N consider that Mo gt we gt and A NCAL CDV1 gt VCAL CDV1 gt Therefore dick1 gt is proportional to ME gt CAL CDV1 gt CV1 CDV gt Applying amp to Eq 53 we obtain N CIDMH gt2 CV1CAL CIV gt C 1 A V 11 02 A lt CDV1lltIgtV1 gt 1 12 lt cple no gt 11 CV1 I IDV1 gt DV gt7 Therefore MIL1 gt2 LlltIgtV gt2 1 z 1 The eigenfunctions of N can be generated from CIDO gt as follows lCIDV gtiim i p VlCIDO gt x E W h V 1 mac 0 cum 2 W a 7 mg 1301 46 For example h mu 2 in z 1311 25L 4 m i 6 3 5552 7T The pre exponential factor is the Hermite polynomial for 1 1 Time Evolution of Expectation Values In order to compute a time dependent expectation value 1212 lt lbtlAWt gt7 it is necessary to compute W gt by solving the time dependent Schrodinger equation mow gt 0t 2 H W gt This can be accomplished by rst nding all eigenstates of H En with eigenvalues EL and then computing W gt as follows lt gulp gt2 ZCne l Ent lt gulch gt TL where the expansion coef cients On are determined by the initial state lt xltbo gt The time depen dent expectation value lt wtlAWt gt is therefore 21 Z cgcnet Wm lt ctmmtctn gt nm Note that this approach might give you the wrong impression that the computational task necessary to solve the time dependent Schrodinger equation can always be reduced to nding the eigenstates and eigenvalues of H by solving the time independent Schrodinger equation While this is pos sible in principle it can only be implemented in practice for very simple problems e g systems with very few degrees of freedom Most of the problems of interest in Chemical Dynamics how ever require solving the time dependent Schrodinger equation explicitly by implementing other numerical techniques For animations see for example the following references 20 Problem Set 201 Exercise 16 A Show that lt Enlxlcbn gt n lamM1 an7H ShOW that lt En gt2 i mehn l 16ni n1 n7n1 C Show that n cby gt2 V 1H gt ch gt2 lCIDV1 gt 47 D Compute the ratio between the minimum Vibrational energies for bonds C H and C D assuming that the force constant k 2 77 is the same for both bonds E Estimate the energy of the rst excited Vibrational state for a Morse oscillator de ned as follows VR De1 exp aR Req2 202 Exercise 17 Prove that lt eagle gt En E lt ought gt when n 7A k and lt Pklcbn gt2 5m with A Htqgtjlj t Ejtlt1gtjt t 203 Exercise 18 Prove that V j 0 wherej E g Z wag andtb Re Et 204 Exercise 19 Consider a harmonic oscillator described by the following Hamiltonian A 1 1 H0 2 p2 imwQQZQ Consider that the system is initially in the ground state 130 with A 1 Compute the probability of nding the system in state 132 at time t after suddenly changing the frequency of the oscillator to w 21 Angular Momentum The angular momentum operator L is obtained by substituting 7 and p by their corresponding quantum mechanical operators f and 239th in the classical expression of the angular momentum L 39r X p The Cartesian components of L are a a L3 zhy 25y ypz zpy 48 01 0 0 0 L Z z zhmay 6 1 ypx These components satisfy the following commutation relations an Ly ypz 2193sz 113122 21pm zpml me vpz 21931721935 21911 122 y 2 2ac I z zlpy ihypx my 2 ihLz 211 Exercise 20 Show that L X L 2 ML Hint Show that ML 2 Lw L2 Note that this expression corresponds to the cyclic permutation where y is substituted by z 1 by y and 2 by 13 in the commutation relation ML 2 Ly L2 Cyclic permutations can be represented by the following diagram 97 Ly L2 239th gt f Lg L3 ihLy HaVing obtained the commutation relations we can show that L2 commutes with the Cartesian components of L e g L2 L3 0 We consider that L2 L3 L521C Li L3 L5 L27 Z L113 49 L2 L5 LyLy L3 Lw LLy LZLZ L50 Lu LLz and since Lw L3 ihLz Lw L3 ihLz L5 L3 ihLy then L2 L3 0 Due to the cyclic permutations we can also conclude that L2 Ly 0 and L2 L2 0 According to these equations both the magnitude of the angular momentum and one any of its components can be simultaneously determined since there is always a set of eigenfunctions that is common to L2 and any of the three Cartesian components Remember however that none of the individual components commute with each other Therefore if one component is determined the other two are completely undetermined Eigenvalues of L2 and L2 Ladder Operators In order to nd eigenfunctions Y that are common to L2 and L2 LQY aY 54 and LZY bY 55 we de ne the ladder operators L L5 iLy L 2 L5 iLy where L is the raising operator and L is the lowering operator In order to show the origin of these names we operate Eq 55 with LL and we obtain LLZY bLY Then we substitute LLz by L Lz LZL where L L2 L56 iLy Lz Lam L iLy Lz Since Lam Lz ihLy and Ly Lz 239th then LLz LZL ihLy iLm hL Consequently hL LZLY bLY and LzLY b hLY Thus the ladder operator L generates a new eigenfunction of Lz eg LY with eigenvalue 1 h when such operator is applied to the eigenfunction of Lz with eigenvalue b eg Y The operator L is therefore called the raising operator Applying p times the raising operation to Y we obtain LZLiY b hpLiY 50 212 Exercise 21 Show that LleiY b hpL1Y Therefore LL and L generate the following ladder of eigenvalues b 3h b 2h b h b bh b2h b3h Note that all functions LiY generated by the ladder operators are eigenfunctions of L2 with eigen value equal to a see Eq 54 Proof LQLPiY LpiLQY LiaY since L2 L36 L2 Ly L2 Li 2 0 and therefore L2 L1 0 Note that the ladder of eigenvalues must be bound LzYk kakn with Yk LiY and bk 2 b i Mi Therefore Lilk biYk L2lkj Z CLYk L3 L Y a b Yk qP non negative physical quantity gt a bi has to be positive a Z b gt a Z lbkl In order to avoid contradictions LYmaaC 0 and LYmm 0 L I L Ymin 0 LL L zLyL iLy LL L3 zLLy LyLw LE P ihLz LL L3 Li M L2 Li hLz Therefore a bin hbmm 0 56 because 51 Analogously LLme 0 U L2 Lg hLYm 0 and a 52 hbm 0 57 man Eqs 56 and 57 provide the following result 772716113 bmaaj 0 gt bmac Furthermore we know that bmaC bmm nh because all eigenvalues of L are separated by units of h Therefore 2bmw 2 Mi gt bmm h jh where j g a b727 hbmm j2h2 3 h2jj 1 and b jh j 1h j 2h Note that these quantization rules do not rule out the possibility that j might have half integer values In the next section we will see that such possibility is however ruled out by the requirement that the eigenfunctions of L2 must be 27r periodic Spherical Coordinates Spherical coordinates are de ned as follows 2 rCosQ y rSinQSingb 1 rSinQCosgb where 6 and q5 are de ned by the following diagram 52 213 Exercise 22 Write the Cartesian components of the linear momentum operator p 1333 g and 132 in spherical coordinates Hint af 0gb 0f 07 0f a1 y a W 06 W a W a art a W 5 M7 where g 913y72 and f f7 vy 2 Way 2 M1331 2 7 W tangb C080 2 E Z 7quot w2y2z2 0Cos6 Sine 1 2 21 gt r2Cos68in6Cosgb 01 01 2 332 y2 g 01 T331716 yaz yaz yaz 0tangb 1 i gt rSinQSingbCos2gb 01 Cos2gb 01 12 01 TQSiIl20COS2Q5 7 yaz yaz yaz gt E rSinQCosgb 01 2 7 01 7 yaZ yz 214 Exercise 23 Show that 0 Cos6 0 L3C 2h Sln Sine Cosgba gb 0 Cos6 0 Ly zh Cosgb Sine Sln a and a L 39 Z zha Squaring L56 Ly and Lz we obtain 02 C036 0 1 02 L2 2 2 o h 002 Sin6 06 311126 aw 53 Eigenfunctions of L2 Since L2 does not depend on r gt Y Y6 gb Furthermore if Y is an eigenfunction of L then LZY bY 0Y dlnY 1 dY b 39h bY gt Z 0gb 8gb Y 8gb iii ibgb Y A aph Since Ygb 2n Ygb we must have ail b 67 h 1 gt 27T 2am With m 0i1i2 Therefore where m is an integer In order to nd eigenfunctions that are common to Lz and L2 we assume A to be a function of theta A A6 02A Cose an 1 b2 ibgb ibgb LQY 2 A z A 9 hoe2smeaesm2e r12 eXph eXph 2A A b2 422 71262 SinQCosQaa Q n aA6Sin26 58 Making the substitution 1 Cos6 we obtain d2A dA a m2 2 1 1 x2 4039 59 215 Exercise 24 Obtain Eq 59 from Eq 5 8 Eq 59 is the associated Legendre equation whose solutions eXist only for a 2 WW 1 and b lh l 1h lh ie the quantum number 1 is an integer greater or equal to zero with g l The solutions of the associated Legendre equations are the associated Legendre polynomials Al m a39m39c0se For example the normalized polynomials for various values of l and m are A0 0 1 A1 0 MCOSQ A1 i1 M31116 54 The eigenstates that are common to L2 and Lz are called Spherical harmonics and are de ned as follows W qb Pl39m39mosmeimt The spherical harmonics are normalized as follows 27r 1 dgb dCosQ ZW 6 YZW6 5u5mm 0 1 Rotations and Angular Momentum A coordinate transformation that corresponds to a rotation can be represented by the following diagram This diagram shows that vector 77 can be speci ed either relative to the axes X y z or relative to the axes X y z where these two sets of coordinates are de ned relative to each other as follows 77 Roz z77 6O where 77 is the same vector 77 but with components eXpressed in the primed coordinate system oz Angle z Rotation aXis 1 rCosgb y rSingb 13 rCosgb oz 2 rCosgbCosoz Sinnginoz y rSingb oz 2 rSinq Cosoz Cosnginoz z z 13 szosoz ySinoz y yCosoz vSinoz 55 Therefore the coordinate transformation can be written in matrix representation as follows I 1 Cosoz Sinoz 0 1 y Sinoz Cosoz 0 y z 0 0 1 z The operator associated with the coordinate transformation is PRoz de ned as follows 15Rozazf7 flR 1OzaZ7 l Cosoz Sinoz 0 where R 1 is the transpose of R ie R 1 Sinoz Cosoz 0 0 0 1 Therefore PRM 2 f 77 f1Cosoz ySinoz xSinoz yCosoz An in nitesimal rotation is de ned as follows pr57 Zf7 at y5 25 y 2 rswam ea ay o ygaag 15R57 Zf7 at y Z Wag y m y 2 recall that ihza y 2 L2 therefore PR6 zf77 1 6Lzfrf A nite rotation through an angle oz can be de ned according to n in nitesimal rotations after subdividing oz into n angle increments oz 2 m5 and taking the limit n e co and 6 e 0 A 6 n i PRozz lim l l i Lz ze O Lz neoo eO h In general a nite rotation through an angle oz around an arbitrary aXis speci ed by a unit vector fl is de ned as follows ppoz efio 39L This equation establishes the connection between the operator associated with a coordinate trans formation and the angular momentum operator Note It is important to note that if coordinates are transformed according to 77 2 R77 the Hamiltonian is transformed according to a Similarity transformation which is de ned as follows E Proof Eonsider fgr E Hymn Egbr PRf7 Z PRHTPEE1PR T Z E R1T PRH71511 R17 E R1r HR17 R17 Therefore HUB 17 HRH7 P11 It is therefore evident that the Hamiltonian is an invariant operator ie H 7 H R 1r un der a coordinate transformation 77 2 R77 whenever the operator associated with the coordinate transformation commutes with the Hamiltonian 1513 H 0 56 22 Spin Angular Momentum The goal of this section is to introduce the Spin angular momentum S as a generalized angular momentum operator that satis es the general commutation relations The main dif ference between the angular momenta S and L is that S can have half integer quantum numbers Note Remember that the quantization rules established by the commutation relations did not rule out the possibility of half integer values for j However such possibility was ruled out by the periodicity requirement Y6 27139 Y6 associated with the eigenfunctions of L and L2 Since the spin eigenfunctions ie the Spinors do not depend on spatial coordinates they do not have to satisfy any periodicity condition and therefore their eigenvalues can be half integer Electron Spin A particular case of half integer spin is the spin angular momentum of an electron with l 12 see Goudsmit s historical recount of the discovery of the electron spin In discussing the spin properties of a particle we adopt the notation l S and m 2 ms The spin functions oz and 6 are eigenfunctions of 3 with eigenvalues h and h respectively These eigenfunctions are normalized according to 12 12 Z 104mg 1 Z 16072 1 61 ms12 mS 12 since m can be either or Also since the eigenfunctions oz and 6 correspond to different eigenvalues of Sz they must be orthogonal 12 Z altms6ltms 0 62 m5 12 In order to satisfy the conditions imposed by Eqs 61 and 62 05m8 Z SmS lg and Z 6m5712 It is useful to de ne the spin angular momentum ladder operators 3 2 ST iSy and S 2 ST iSy Here we prove that the raising operator 3 satis es the following equation Proof Using the normalization condition introduced by Eq 61 we obtain 12 12 Z altmsgtaltmsgt Z 3 lt3 1 ms12 mS 12 and 1012 2 331mm iSy ms 57 Now using the hermitian property of ST and Sy Z szg ZgSf m5 we Obtain we 2 2m 632S6 z s sw where lo 6SS6 i6SyS6 0 2 Ems 6S Sl67 lo 532 S ESZW 2 2 W Em 6 if 7 7W M2 7 Since the phase of C is arbitrary we can choose ch Similarly we Obtain Soz 2 EB Since oz is the eigenfunction with highest eigenvalue the Operator 3 acting on it must annihilate it as follows Soz 0 and SL5 0 1 smzrasg gx lgozjw L edl m z lg z Emo Similarly we nd 3556 iiioz and Syoz iih lt WA gt oz 6 lt Syl gt oz 6 lt lszl gt oz 6 oz 0 h2 oz 0 z39h2 oz h2 0 6 h2 0 6 zh2 0 6 0 h2 Therefore S 2 ho where 0 are the Pauli matrices de ned as follows 01 04 10 10 zo 0 1 where a a a 1 58 221 Exercise 25 Prove that the Pauli matrices anti commute with each other ie 039in 01707 0 wherez39 7E j andz39j 513y In order to nd the eigenfunctions of 3 called eigenspinors consider the following eigenvalue problem S i 2i i z vi 2 vi 7 1 0 161 161 o lviiviv ewe we Ul 711 U 71 Similarly we obtain and Therefore electron eigenspinors satisfy the eigen value problem h Sin iiXia HQ MIME Any spinor can be expanded in the complete set of eigenspinors as follows 3ia a v where lozl2 and Oz 2 are the probabilities that a measurement of 3 yields the value h and with h respectively when the system is described by state 3 222 Exercise 26 Prove that 32X 1X 59 223 Exercise 27 Consider an electron localized at a crystal site Assume that the spin is the only degree of freedom of the system and that due to the spin the electron has a magnetic moment M 3 2mc 7 where g m 2 m is the electron mass 6 is the electric Charge and c is the speed of light Therefore in the presence of an external magnetic eld B the Hamiltonian of the system is Hz MB Assume that B points in the z direction and that the state of the system is W em if Consider that initially ie at time t 0 the spin points in the 1 direction ie the spinor is an eigenstate of 03C with eigenvalue h Compute the expectation values of SC and Sy at time 75 Addition of Angular Momenta Since L depends on spatial coordinates and S does not then the two operators commute ie L S 0 It is therefore evident that the components of the total angular momentum JLamp satisfy the commutation relations J X J z hJ Eigenfunctions of J2 and J are obtained from the individual eigenfunctions of two angular mo mentum operators L1 and L2 with quantum numbers 1 m1 and 2 mg respectively as follows 37 2 gumzlmlzgmggag r l1m1l2m2 Clebsch Gordan Coef cients where J2 jm 52717 1ija szjm 6O 224 Exercise 28 Show that 771 2 11me OQYm1X is a common ei enfunction of J2 and J when J l l g 01 Z V lm1 and C2 3 lm or when Cl km and 02 Z ll mll 2l1 2l 1 2l1 39 Hint Analyze the particular case j l 12 and j l 12 Note that J2 L2 2 2L3 L2 2 2LZSZ LS LS JZ L 3 23 Central Potential Consider a two particle system represented by the following diagramR1123 R3168 Ngt Z1 t 39y1 A gt where 13 y and 2 represent distances between the two particles along the three Cartesian axes where 77 13 y z 772 771 with 771 and 772 the position vectors of particles 1 and 2 respectively The central potential V1 y z is a function of W V1132 y2 22 rather than a function of the individual Cartesian components Assuming that such function de nes the interaction between the two particles the Hamiltonian of the system has the form where T lf 1l2 l772l2 with l771l2 771 771 Changing variables 71 and 772 by the center of mass coordinates R and the relative coordinates 77 772 771 where m1771m2772 R TZTQ 1 7721 1 7722 weobtain m2 m1 rlzR r r2R r m1m2 m1m2 61 Therefore sz L L 1 LL 2 m1m2 m1m2 2 m1m2 m1m2 or m1m2 39 2 1 mlmg 2 1 92 1 2 T R r MR r 2 l l2m1mg 2 H 2 H Where M 7721 m2 is the total mass of the system and u E M is the reduced mass of the m1m2 two particle system Therefore the total Hamiltonian of the system can be written as follows H 1M 1 112 vm 13 133 vm 7 7 r 2 2 2M 2M Where the rst term corresponds to the kinetic energy of a particle of mass M and the second and third terms constitute the Hamiltonian of a single particle with coordinates 7 Therefore the time independent Schrodinger equation for the system is Trying a factorizable solution by separation of variables we I WWW we obtain h2 uvR2 M h2 MV72 u 1le M M2M J M M2M 1le depends on R depends on r Therefore each one of the parts of the Hamiltonian have to be equal to a constant 1le Vlrlj E w wM W 1 V 2w E 63 2 M wM R M M 7 722 1 2M 1 Eq 64 is the Schrodinger equation for a free particle with mass M The solution of such equation is V311 vq E with EM E E 64 I 2E2 WAR 27Th32 7 kR where 121M 2 EM 65 According to Eq 65 the energy EM is found by solVing the equation E2 vr2wu 1 WWW Z EMDM 66 62 Equations 65 and 66 have separated the problem of two particles interacting according to a central potential V fg 771 into two separate one particle problems that include 1 The translational motion of the entire system of mass M 2 The relative e g internal motion These results apply to any problem described by a central potential e g the hydrogen atom the two particle rigid rotor and the isotropic multidimensional harmonic oscillator Consider Eq 66 with V2 E 88 88 g g and V f a spherically symmetric potential ie a function of the distance 7 It is natural to work in spherical coordinates 231 Exercise 29 Prove that the Laplacian V2 can written in spherical coordinates as follows 82 2 a 1 A A a2 Cos6 8 1 82 L2 h L2 h2 w ere 992 Sing 06 sin26 6W v2 WJF f8 39r27i2 It is important to note that the commutator 2 V2L2 a 2 8 L2 LLQ L2 0 w ZE because 2 does not involve 7 but only 6 and gb Also since 2 does not involve 7 and V is a function of 7 ML zo Consequently H L2 0 whenever the potential energy of the system is de ned by a central potential Furthermore H Lz 0 because L ih Conclusion A system described by a central potential has eigenfunctions that are common to the operators H L2 and L Z ku Z lbw 21 2 mg 1 l 0 1 2 up hmzb m z z 1 1 Substituting these results into Eq 66 we obtain E2 a2 2 a h2 h2 1 wla WW VOTDwu E ww Since the eigenfunctions of 2 are spherical harmonics Ylmw gb we consider the solution 1 RWMmW 5 63 and we nd that RU must satisfy the equation J 3283 W 2M 87 2 7 87 2W210 1R Vl R EMR 67 24 TwoParticle RigidRotor The rigidrotor is a system of two particles for which the distance between them 77 l d is constant The Hamiltonian of the system is described by Eq 67 where the rst two terms are equal to zero 2 and EN 710 1 Vd w1th 1b Ylmw Q5 The moment of inertia of a system of particles is I C E 211 771732 where m is the mass of particle 239 and 7 is the particle distance to the C axis 241 Exercise 30 Prove that I 2 ud2 for the two particle rigid rotor where u m1m2 d 2 7 2 T1 and C is an axis with the center of mass of the system and is perpendicular to the axis that has the center of mass of both particles Assume that the center of mass lies at the origin of coordinates and that the 2 axis has the center of mass of both particles in the system The rotational energy levels of the rigid rotor are E2 EM ll1withl012 68 These energy levels usually give a good approximation of the rotational energy levels of di atomic molecules e g the HCl molecule 25 Problem Set 251 Exercise 31 Solve problems 65 and 66 of reference 1 252 Exercise 32 Prove that the angular momentum operator L 39r X p is hermitian 64 253 Exercise 33 Prove that W117 a W h p 1 where p ih8 813 and a is a nite displacement 254 Exercise 34 Let H be the Hamiltonian operator of a system Denote wk the eigenfunctions of H with eigenval ues Ek Prove that lt rbnl 62 H er gt 0 for any arbitrary operator Q when n k 255 Exercise 35 Prove that 113 H ipm where H p22m 256 Exercise 36 Prove that L YZ quot WU mltl m DEW 1 where LzYlm 2 thlm and L2Ylm 2 WW 257 Exercise 37 Consider a system described by the Hamiltonian matrix E0 A H A E a where the matrix elements H j lt rbj lH Wk gt Consider that the system is initially prepared in the ground state and is then in uenced by the perturbation Wt de ned as follows 0 6 1527392 7 wt W05 2 e tgTgliwt 0 Calculate the probability of nding the system in the excited state at time 75 gtgt 739 65 26 Hydrogen Atom Consider the hydrogen atom or hydrogen like ions eg He Li2 etc with nuclear charge 26 and mass M and the electron with charge e and mass m The potential energy of the system is a central potential eg the Coulombic potential 262k V 7a 1 in au where r is the electron nucleus distance and k 14713960 in SI units The total Hamiltonian is A E2 2h2 2 2me mm VB VT7 memn mn l39me the second and third terms of H is represented by the symbol H d and is called the electronic Hamiltonian because it depends only on the electronic coordinate r In order to nd the electronic eigenvalues we must solve the equation where a 2 Note that a me since me ltlt mm The Hamiltonian that includes only 19 Emil 69 Eq 69 is the eigenvalue problem of a one particle central potential We consider the factorizable solution we RrYlm6 a with l 0 1 2 lt l where Rr satis es the equation h2 02R 20R h2 Z 2R l l 1 2a 8r2 r 8r h2r2 R r ER 70 This equation could be solved by rst transforming it into the associated Lagaerre equation for which solutions are well known Here however we limit the presentation to note that Eq 70 has solutions that are nite single valued and square integrable only when Z 2 ae4 Z 262 E 71 27i2n27 or 2an27 where n 1 2 3 and a 3 is the Bohr radius These are the bound state energy levels of hydrogen like atoms responsible for the discrete nature of the absorption spectrum In particular the wavenumbers of the spectral lines are E2 E1 Z2ae4 1 1 w he hc2h2 n3 n 66 The eigenvalues can be represented by the following diagram E A Degeneracy Since the energy E depends only on the principal quantum number n and the wave function we depends on n l and m there are n2 possible states with the same energy States with the same energy are called degenerate states The number of states with the same energy is the degeneracy of the energy level n1 2 3 10 1 2 n l these are 11 states m l ll 0 12 l these are 2 l 1 states 261 Exercise 38 Prove that the degeneracy of the energy level E is n2 The complete hydrogen like bound state wave functions with quantum numbers n l and m are mm 6 as Emma Where are the associated Legendre polynomials and Rum are the Lagaerre associated polynomials TL l l 2 2r 39 h o RnlT rlena E bjrj where a E E 0529177A j0 and 22 jl1 n Ej1j212j39 j1 67 Example 1 Consider the ground state wave function of the H atom with n 1 l 0771 0 R100 2 6 57790 where b8 2 1 f000 ma a 2 and b0 2 23 Therefore 1 1 3 110W as air2727 Note An alternative notation for wave functions with orbital quantum number 1 0 1 2 is Example 2 The possible wave functions with n 2 are 25 2190 2191 219 17 19200 210 211 1521 17 262 Exercise 39 Show that i321 2a 27 2a 1 z S a 1 28 e zr2a7 2p1 52Te zr2asin66 iqu 1 1 z 72190 Z 2a 52reZT2 Cos6 21 352reZT2asin66i a 263 Exercise 40 Compute the ionization energy of He 264 Exercise 41 Use perturbation theory to rst order to compute the energies of states 210 211 and 214 when a hydrogen atom is perturbed by a magnetic eld B 2 B2 according to w 6LB where 6 220 The splitting of spectroscopic lines due to the perturbation of a magnetic eld is called Zeeman e ect 68 Radial Distribution Functions The probability of nding the electron in the region of space where 39r is between 7 to 7 d7 6 between 6 to 6 old and gb between gb and gb dgb is P 13mac mm6nm67231n6d7d6d Therefore the total probability of nding the electron with 7 between 7 and 7 dr is 7r 27r PTO dQ dnglm6Ylm8sin6 RTRrr2dr 0 0 where f0 d6 f027r dnglm6Ylm6sin6 1 For example the radial probability for mO and 11 can be visualized as follows I wm HELD1 U kerrHHHtrmtro nj Pictures of atomic orbitals with n g 10 are available here Real Hydrogenlike Functions Any linear combination of degenerate eigenfunctions of energy E is also an eigenfunction of the Hamiltonian with the same eigenvalue E Certain linear combinations of hydrogen like wave functions generate real eigenfunctions For example when l 1 mm n11 Rn1rsin6Cosgb E 132 wnll n11 Z Rn1TSiHQSiH E P2y7 210 3 1921327 are real and mutually orthogonal eigenfunctions Function 1bng is zero in the my plane positive above such plane and negative below it Func tions 21290 and 1amp2 py are zero at the 2y and 32 planes respectively 2111 and 2121 are eigenfunctions 69 of 2 with eigenvalue 27 However since 2111 and 2121 are eigenfunctions of L with different eigenvalues eg with eigenvalues h and h respectively linear combinations 21296 and 1amp2 py are eigenfunctions of 2 but not eigenfunctions of Lz 265 Exercise 42 A What is the most probable value of r for the ground state of a hydrogen atom Such value is represented by TM B What is the total probability of nding the electron at a distance 7 3 TM C Verify the orthogonality of functions 2P 2Py and 2B D Verify that the ground state of the hydrogen atom is an eigenstate of H but that such state is not an eigenstate of T or 27 Helium Atom The helium atom is represented by the following diagram e 7 12 e 7 2 7 1 2 This diagram represents two electrons with charge e and a nucleus with charge 2 The Hamiltonian of the Helium atom is A E2 2 262 E2 2 262 62 Note that the term couples two one electron hydrogenlike Hamiltonians In order to nd a solution to the eigenvalue problem A H w EID we implement an approximate method We rst solve the problem by neglecting the coupling term Then we consider such term to be a small perturbation and we correct the initially zeroth order eigenfunctions and eigenvalues by using perturbation theory Neglecting the coupling term the Hamiltonian becomes the sum of two independent one electron Hamiltonians The eigenfunctions of such Hamiltonian are 1 V 1 eimqban 7 Pquot 6 llt 2 l 1m 2 7 lb 2 Rnlquot 1sz91 and the eigenvalues are 0 z2ue4 221164 quot1 2 272271 2722712 Exercise 43 Prove that 2 e 5 z lt 100l W100 gt 2 T12 8 CL In order to illustrate how to correct the zeroth order solutions by implementing perturbation theory we compute the rst order correction to the ground state energy as follows 2 2 4 e 2 Me 5 2 z EEOlt gt e 11 19100 T12 W100 W 8 a Alternatively the variational method could be implemented to obtain better results with simple functions 1b e g products of hydrogenlike orbitals with an effective nuclear charge 2 15 A2e r1r2 According to the variational theorem the expectation value lt gt is always higher than the ground state energy Therefore the optimum coef cient 2 minimizes the eXpectation value Ez lt gt where E2 2 62 E2 2 62 2 z e2 2 z e2 62 V2 2 T1 T1 T2 T2 T1 T2 T12 Computing the eXpectation value of H analytically we obtain 2 2 Z T 2 2TZ 7172 2 2 2 7 a 71 72 2262 2z 2 2 62 356 2 61 8 61 Therefore the optimum parameter 2 is obtained as follows Ez aEz 5 5 2e 5 e2 5 5 2e 0 2 E 2 2 2 2 2 az TZOPt 16 T 201 16 a 16 a8 16 a 71 28 SpinAtom Wavefunctions The description of atoms can be formulated to a very good approximation under the assumption that the total Hamiltonian depends only on spatial coordinates and derivatives with respect to spatial coordinates but not on spin variables We can therefore separate the stationary state wave function according to a product of spatial and spin wavefunctions Example 1 The spin atom wavefunction of the hydrogen atom can be approximated as follows 77bel 1M1 ya Zgm87 where gms 2 oz 6 when 7725 12 12 respectively Since the Hamiltonian operator is assumed to be independent of spin variables it does not affect the spin function and the eigenvalues of the system are the same as the energies found with a wave function that did not involve spin coordinates Mathematically Hm ya z9msl 9msH a ya z E9ms i ya z The only consequence of modeling the hydrogen atom according to a spinatom wavefunction is that the degeneracy of the energy levels is increased Example 2 The ground electronic state energy of the helium atom has been modeled according to the zeroth order wave function 181 182 In order to take spin into account we must multiply such spatial wavefunction by a spin eigenfunction Since each electron has two possible spin states there are in principle four possible spin functions 0410627 06162a 61OZ27 and MUM Functions oz162 and 61oz2 however are not invariant under an electron permutation ie these functions make a distinction between electron 1 and electron 2 Therefore such functions are inadequate to describe the state of a system of indistinguishable quantum particles such as electrons Instead of working with functions 051 6 2 and 61oz2 it is necessary to construct linear combinations of such functions e g 1 EO 162i 61oz2l with correct exchange properties associated with indistinguishable particles p12 12 imam The two linear combinations together with functions oz1oz2 and 6 1 6 2 form the basis of four normalized two electron spin eigenfunctions of the helium atom 29 Pauli Exclusion Principle Pauli observed that relativistic quantum eld theory requires that particles with halfinteger spin s12 32 must have antisymmetric wave functions and particles with integer spin sO 1 72 must have symmetric wave functions Such observation is usually introduced as an additional postulate of quantum mechanics The wave function of a system of electrons must be antisymmetric with respect to interchange of any two electrons As a consequence of such principle is that two electrons with the same spin cannot have the same coordinates since the wavefunction must satisfy the following condition 7061132 77bv217 and therefore balm 0 For this reason the principle is known as the Pauli Exclusion Principle Another consequence of the Pauli Principle is that since the ground state wave function of the He atom must also be anti symmetric and since the spatial part of the zeroth order wave function is symmetric II 13 113 2 then the spin wave function X must be anti symmetric and the overall zeroth order wave function becomes 1 1b 1311S2 Oz162 61Oz2l 72 i Note that this anti symmetric spin atom wave function can be written in the form of the Slater determinant w i 131oz1 13161 n 18lt2gtalt2gt 18mm 30 Lithium Atom The spin factor affects primarily the degeneracy of the energy levels associated with the hydrogen and helium atoms To a good approximation the spin factors do not affect the energy levels of such atoms The lithium atom however has three electrons An antisymmetric spin wave function of three electrons could in principle be written as the Slater determinant a8 83 E53 73 X Z C14 C14 V5 03 63 043 Such Slater determinant however is equal to zero because two of the columns are equal to each other This fact rules out the possibility of having a zero order wave function that is the Fock product of three hydrogenlike functions w 2 131 132 133 74 73 Only if the construction of an antisymmetric spin wave function was possible we could proceed in analogy to the Helium atom and compute the perturbation due to repulsive coupling terms as follows 2 2 e2 e e E lt zbl lzb gt lt zbl lzb gt lt zbl lzb gt 7 12 7 23 7 13 where 1b is the product of hydrogenlike functions of Eq 48 Having ruled out such possibility we construct the zeroth order ground state wave function for lithium in terms of a determinant similar to Eq 47 but where each element is a spin orbital ie a product of a one electron spatial orbital and one electron spin function 131oz1 13161 231oz1 lt0gt 18lt2gtalt2gt 13262 2slt2gtalt2gt lt75 133oz3 13353 2S3oz3 where the third column includes the spatial orbital 23 instead of the orbital 13 because the Pauli exclusion principle rules out the possibility of having two electrons in the same spin orbital It is important to note that Eq 49 is not simply a product of spatial and spin parts as for the H and He atoms In contrast the wave function of Li involves a linear combination of terms which are products of non factorizable spatial and spin wavefunctions 301 Exercise 44 Show that for the lithium atom treating the electron electron repulsion interaction film as a per turbation 0 0 0 0 E E15 E1S E2sa and E 2 lt 1312S2 21S12S2 gt lt 1311S2 21S11S2 gt 7 12 7 12 lt 1312S2 2311S2 gt 7 12 31 SpinOrbit Interaction Although neglected up to this lecture the interaction between the electron spin and the orbital angular momentum must also be included in the atomic Hamiltonian Such interaction is described according to the Spinorbit Hamiltonian de ned as follows A 1 1 A A A A HSO ltWgtLsgLs 76 2771602 7 87 74 where V is the Coulombic potential of the electron in the eld of the atom Note that the spin orbit interaction is proportional to L S A proper derivation of Eq 76 requires a relativistic treatment of the electron which is beyond the scope of these lectures Note A classical description of such interaction also gives a perturbation proportional to i 3 This is because from the reference frame of the electron the nucleus is a moving charge that gen erates a magnetic eld B proportional to i Such magnetic eld interacts with the spin magnetic moment m8 eme Therefore the interaction between B and ms is proportional to i 3 Unfortunately however the proportionality constant predicted by such classical model is incorrect and a proper derivation requires a relativistic treatment of the electron as mentioned earlier in this section In order to compute the spin orbit Hamiltonian of a many electron atom it is necessary to compute rst an approximate effective potential V7 for each electron 739 in the total electric eld of electrons and nuclear charges Then we can compute the sum over all electrons as follows A 1 1 8 A A A A H m LS39 39L39oS39 77 80 2771602 Ti an 7 7 7 7 The correction of eigenfunctions and eigenvalues due to the spin orbit coupling is usually computed according to perturbation theory after solving the atomic eigenvalue problem in the ab sence of the spin orbit interaction For example the spin orbit correction to the eigenvalue of state l KI for a one electron atom is ESE 011 1 Si S l 11gt 78 Note that the L S product can be written in terms of J 2 L2 and 32 as followsL S J2 L2 SQ because J2 J J L SL S L2 32 2L S and since the unperturbed wave function is an eigenfunction of L2 32 and J2 L 31 gt2 72m 1 LL 1 slts 1111 gt Therefore 1 E30 i712 lt 5 gt JJ 1 LL 1 SS 1 It is important to note that due to the spin orbit coupling the total energy of a state depends on the value of the total angular momentum quantum number J Furthermore each of the energy levels is 2J 1 times degenerate as determined by the possible values of M J For example when Ll and Sl2 then the possible values of J are 12 and 32 since JLS LS l L S The spin orbit interaction is therefore responsible for the splitting of spectroscopic lines in atomic spectra 75 It is possible to remove the degeneracy of energy levels by applying an external magnetic eld that perturbs the system as follows H3 2 m B where m 2 mL ms with mL 2 L and m5 e S The external perturbation is therefore described by the following Hamiltonian me e e H L 23 B B 2me JSB 27716 The energy correction according to rst order perturbation theory is EB 2 BhMJ lt s gt ABMJ me where lt 3 gt hMJ JJ1 2LJLJ11SS1 and A is a proportionality constant Therefore the perturbation of an external magnetic eld splits the energy level characterized by quantum number J into 2Jl energy sub levels These sub levels correspond to different possible values of M J as described by the following diagram 23 1 J Levels States M J 1P 1P1 1 0 O 1 0 1 2 1 0 1 2 H0 H0 Hrep H0 HrepHso H0 HrepHsoHB 311 Exercise 45 A Calculate the energy of the spectroscopic lines associated with transitions 38 3P for Na in the absence of an external magnetic eld B Calculate the spectroscopic lines associated with transitions 38 3P for Na atoms perturbed by an external magnetic eld B as follows H3 m B gem 1a 5 A S S and EB lt mHBw gt eBMJg w1th g 1 W 55 1 76 32 Periodic Table Previous sections of these lectures have discussed the electronic structure of H He and Li atoms The general approach implemented in those sections is summarized as follows First we neglect the repulsive interaction between electrons and write the zeroth order ground state wave functions as antisymmetrized products of spin orbitals Slater determinants eg 9 i 131oz1 131e1 La a He A 1 131oz1 18161 2310z1 gr 132a2 13mm 232oz2 Li 5 133oz3 133e3 2S3oz3 with zeroth order energies Egg 2E1S and ES 2E1S E2S represented by the following diagram Energy Energy A 23 2S we 13 Helium Lithium It is important to note that these approximate wave functions are found by assuming that the elec trons do not interact with each other This is of course a very crude approximation It is nonethe less very useful because it is the underlying approximation for the construction of the periodic table Approximate zeroth order wave functions can be systematically constructed for all atoms in the periodic table by considering the energy order of hydrogenlike atomic orbitals in conjunction with Hand s Rules Hund s First Rule Other things being equal the state of highest multiplicity is the most stable Hund s Second Rule Among levels of equal electronic con guration and spin multiplicity the most stable level is the one with the largest angular momentum 77 These rules establish a distinction between the zeroth order wave functions of ground and excited electronic state con gurations For example according to Hund s rules the lithium ground state wave function is 131oz1 13161 231a1 W 132Oz2 13262 232Oz2 7 79 V5 133oz3 13363 2S3oz3 and the rst excited state wave function is 131oz1 13161 2P1oz1 Wazi 132oz2 13262 2P2a2 V5 133oz3 13363 2P3oz3 Note that the energy order of hydrogenlike atomic orbitals E is not suf cient to distin guish between the two electronic con gurations According to such expression orbitals 2p and 2s have the same energy E2 However Hund s second rule distinguishes the ground electronic state as the one with higher angular momentum This is veri ed by rst order perturbation theory since the perturbation energy of we is higher than the perturbation energy computed with 97quot 321 Exercise 46 Prove that according to rst order perturbation theory the energy difference AE between the two states is AEWW 6 2J1s2p J1s2s K1S2P K1828 ii 5 ogbgj gt2 Coulomb Intergral where J b 2 lt gb and K 17 2 lt dandy qbg kbgj gt2 Exchange Integral i 322 Exercise 47 Use Hund s Rules to predict that the ground states of nitrogen oxygen and uorine atoms are 4S 3 P and 2P respectively 33 Problem Set 331 Exercise 48 Use the variational approach to compute the ground state energy of a particle of mass m in the potential energy surface de ned as follows V1 AX4 Hint Use a Gaussian trial wave function 78 From tables 00 00 OO 2 3 7T 2 7T 2 1 7T d4 0 2 dare 0 dxzv2 0 00 405 oz 00 oz 00 2oz oz 332 Exercise 49 Compute the eigenvalues and normalized eigenvectors of 0 2 0y O39Z where 0 0 z39 1 0 y z 0 OZ 0 1 333 Exercise 50 Construct two excited state wavefunctions of He that obey the Pauli Exclusion principle with one electron in a 18 orbital and the other electron in the ZS orbital Explain the symmetry of spin and orbital wave functions 334 Exercise 51 Consider a spin 12 represented by the spinor Cosoz X sinoz 6738 39 What is the probability that a measurement of Sy would yield the value 3 when the spin is de scribed by X 34 LCAO Method H Molecule The H molecule can be represented by the following diagram 79 TB MBquot 7 A RAB RB MA RA where A and B represent two hydrogen nuclei and 6 represents the electron The Hamiltonian of the system isR1376 A W 2 W 2 H He 80 QMA VRA 2MB VRB T l where 2 2 2 2 A h Hel V2 3 81 2m TA TB RAB This is another three body Hamiltonian similar to the Helium atom Hamiltonian where instead of having two electrons and one nucleus we have two nuclei and one electron In order to compute the eigenstates we assume that the kinetic energy of the nuclei can be neglected when compared to the other terms in the Hamiltonian Bom Oppenheimer approximation The electronic energy is computed at various internuclear distances RAB by considering that the term 133 in Eq 81 is a constant factor parametrized by RAB In practice this constant factor is ignored when solving the eigenvalue problem since it can be added at the end of the calculation According to the linear combination of atomic orbitals LCAO method a convenient trial state for H can be written as follows qjgtOA 5AgtCB Bgt7 where gbA gt and gbB gt are 18 atomic orbitals of atoms A and B respectively According to the variational theorem the optimum coef cients CA and CB can be found by minimizing the expectation value of the energy lt WH elW gt Oil AA 2CACBHAB 012311733 ltEgt lt gt CESAA 2OAOBSAB l 0123333 7 with respect to CA and 03 Here ij lt gbjllil ellgbk gt Sjk lt jl k gt and 2 2 2 gel hvg 3 2m TA TB 80 341 Exercise 52 Show that the condition 855 0 implies HAA lt E gt SAACA HAB lt E gt SABCB 07 CB and aalt gt 0 HAB lt E gt SABOA l HBB lt E gt SBBCB 0 CA With 331739 lt jl jgt 1 In matrix form these equations called secular equations can be written in compact form as follows H ESC 0 82 where H is the Hamiltonian matrix C is the matrix of column eigenvectors E is the diagonal matrix of eigenvalues and S is the overlap matrix The secular equations have a nontrivial solution ie a solution different from the trivial solu tion CA 2 0 CB 2 0 when the determinant of H ES ie the so called secular determinant vanishes HAA ltEgt HAB SABltEgt HBA SBAltEgt HBB ltEgt 0 Since l gbA gt and l gbB gt are IS orbitals HAA H33 and SAB SBA S Therefore HAA lt E gt2 HA3 s lt E gt2 0 and E HAA i HAB i 1 i S Substituting lt E gt in the secular equations we obtain CA3 IIZOBZ Z Therefore Z CAlt A lt53 Where OA 2 1L 2 CAgbA gbB where CA 2 The strategy followed in this section for solving the eigenvalue problem of H 2 can be summarized as follows 1 Expand the solution l II gt according to a linear combination of atomic orbitals LCAO 2 Obtain a set of n secular equations according to the variational approach 3 Solve the secular determinant by nding the roots of the characteristic equation a polynomial of degree n in E 81 4 Substitute each root into the secular equations and nd the eigenvectors e g the expansion coef cients in the LCAO that correspond to such root The energies lt E gti are functions of H A A H A B and S The integral H A A is de ned as the sum of the energy of an electron in a 18 orbital and the attractive energy of the other nucleus E2 2 2 2 HAA d7 1 vg CbA E18H d7 bl A 83 TA TB TB As the nuclei A and B are brought closer together the second term in Eq 83 ie the term f drgbj 3 tends to make the energy of H if more negative increasing the stability of the molecule 2 Re is respons1ble for the repuls1on between nuclei and increases monotonically as the two nuclei get closer together counteracting the stabilization caused by 3 Therefore the sum 2 HAA R together The integral H A 3 de ned as follows The term B is not responsible for the stabilization of the system as the nuclei are brought closer E2 62 e2 HA3 2 drq51 V2 lt53 84 2m TA TB is called resonance integral and takes into account the fact that the electron is not restricted to any of the two 13 atomic orbitals but it can rather be exchanged between the two orbitals At large values of RAB the resonance integral H A 3 goes to zero Decreasing RAB H A 3 becomes more negative and stabilizes the molecule relative to the asymptotically separated atoms The eigenvalues lt E gti can be represented as a function of RAB by the following diagram Energy EM EB t RAB lt E gt is always larger or equal than E0 Exact answer for E0 e2 RAB At short distances RAB the internuclear repulsion dominates Note that lt E gt is lower than lt E gt because H A A and H A B are negative 82 In analogy to the variational approach implemented to study the Helium atom one could further im prove the variational solution of H J by optimizing the eXponents 5 e g effective nuclear charges in the functions that represent q A and Q53 gr2 Z a 2a AB 6 Such variational correction of the effective nuclear charge is known as scaling 342 Exercise 53 According to the quantum mechanical description of H J explain 1 Why do molecules form What is a chemical bond 2 Consider state rm 2 2 2S12XA X3 where nucleus A is at RA 2 g 0 0 and nucleus B is at RB g 0 0 Compute WED at the coordinate 000 and compare such probability density to the sum of probability amplitudes due to q A and Q53 35 H2 Molecule The H 2 molecule can be represented by the following diagram TA2 ETBl Us RAB The diagram includes two electrons represented by 61 and 62 and two protons A and B The 83 Hamiltonian of the system is H W V W V2 H 2MA RA 2MB RB 6 Where E2 2 2 E2 2 2 2 2 A e e e e e e Hel V2 V2 2m 1 TA1 7 31 2m 2 7712 7 32 7 12 RAB In analogy to the He atom it is possible to identify one electron Hamiltonians ie associated with electrons l and 2 E2 2 2 H2lt1gt V 6 2 TA1 7 31 and 2 2 2 h e e H22 v3 7712 7 32 62 A zeroth order solution is obtained by neglecting the repulsion between electrons Since R B contributes only with a constant value to the energy eg a constant parametrized by RAB we can make use of the theorem of separation of variables and obtain the solution of the eigenvalue problem HWgtEWgt as the product WgtAllt1gt1gtllt192gt7 86 where l 131 gt and l gbg gt are eigenstates of the H Hamiltonian and A is the anti symmetrizing spin wave function 1 A N la162 610z2l Note that the hydrogen molecule occupies the same place in the theory of molecular electronic structure as the helium atom in the theory of atomic electronic structure Therefore the correction due to electronic repulsion can be calculated according to rst order perturbation theory as follows 2 2 6 6 E 2EH2RAB lt turf 12w gt RAB 87 Note that the last term discounts the repulsion between nuclei that has been over counted The equilibrium distance RS3 is obtained by minimizing E with respect to RAB Substituting such value into Eq 87 we obtain the minimum energy of the H 2 molecule The complete ground state of H 2 is described as follows ID 04162 61042l 13A113A2 13A11SB2 133113A2 13311SB2l7 88 84 where N is a normalization factor obtained by substituting 131 gt and l 132 gt in Eq 88 by the ground state wave function of 1172 lt1 VLN 1840 1331 lt89 According to Eq 89 the probability of nding both electrons close to nucleus A ie the proba bility of nding the electronic con guration H g H g is determined by the square of the eXpansion coef cient associated with the term 13 A11S A2 Analogously the probability of nding both electrons close to nucleus B is proportional to the square of the eXpansion coef cient associated with the term 13311SB2 Therefore terms lSA11SA2 13311SB2 describe ionic con gurations while terms 13 A11S 32 and 13311SA2 describe covalent structures Unfortunately the LCAO wavefunction introduced by Eq 89 predicts the same probability for ionic and covalent con gurations Hng H Hg and H AH B respectively This is quite un satisfactory since it is contrary to the chemical eXperience The LCAO model predicts that upon dissociation half of the H 2 molecules break into ions H and H Contrary to such prediction the H 2 molecule dissociates almost always into two hydrogen atoms 351 HeitlerLondonHL Method The Heitler London approach aims to correct the shortcomings of the LCAO description by ne glecting the ionic terms altogether Therefore the HL wave function of H 2 includes only covalent terms as follows 1 NW5 This wave function gives a better description of the energy as a function of RAB and predicts the proper asymptotic behavior at large internuclear distances mL MDBQY WDQQNHSJDHVHG ampdml 352 Exercise 54 Prove that according to the HL approach J K E1SW with J lt 13A11SB2 H 13A11SB2 gt and K lt 13A11SB2 H 13311SA2 gt 85 36 Homonuclear Diatomic Molecules Other homonuclear diatomic molecules eg Lig 02 H 62 F2 N2 can be described ac cording to the LCAO approach introduced with the study of the H molecule A general feature of the LCAO method is that a combination of two atomic orbitals on different centers gives two molecular orbitals MO One of these molecular orbitals is called bonding and the other one is called antibonding The bonding state is more stable than the system of in nitely separated atomic orbitals On the other hand the antibonding state is less stable than the isolated atomic orbitals The description of the H molecule discussed in previous sections can be summarized by the following diagram 1L m I This diagram introduces the nomenclature of states of homonuclear diatomic molecules which is determined by the following aspects 1 Nature of the atomic orbitals in the linear combination e g IS orbitals in the study of the H molecules 2 Eigenvalue of 3 with z the internuclear aXis eg such eigenvalue is zero for the H molecule and therefore the orbital is called 0 3 Eigenvalue of the inversion operator through the center of the molecule e g g when the eigenvalue is 1 and u when the eigenvalue is 1 4 Stability with respect to the isolated atoms e g an asterisk indicates that the state is unstable relative to the isolated atoms Other homonuclear diatomic molecules involve linear combinations of p orbitals Such linear combinations give rise to 0 type orbitals when there is no component of the angular momentum in the bond aXis eg we choose the bond aXis to be the z aXis An example of such linear combination is represented by the following diagram 86 egg 9o m 2P0A 2P0B 092p In order to classify molecular states according to eigenvalues of 2 we make linear combinations of eigenfunctions of Lz with common eigenvalues There are four possible states 3 2 1111 II 2Pl1B7 739l39u2Pl17 7T2Pl17 m l II 2P1B 739l39u2P17 W32P1 All of these linear combinations are 7r states because A 1 for all of them In order to justify their symmetry properties with respect to inversion we analyze the following particular case 1 z which is represented by the following diagram wu2P1 2P1A 2P1B 52ei AerAsin6A ei 36r3sin63 87 wu2P1 orbital W Z quot TA TB IIIIIEIIQZIII IQIBllliIlIII z nodal line 937fquot39B 5 HR II TA 39 I I I I I l I I I I ll This diagram shows that under inversion through the origin coordinates are transformed as follows TA TB 6A 63 TB TA 63 6A CM 13 Cb Cb w Cb 7T7 1 7T ewe T e7 because 6 Cosw z39 Sin7r V V 1 0 The states constructed with orbitals P1 differ relative to those constructed with orbitals pH only in the sign of phase gb introduced by the following expression 1 z m This function has a nodal xy plane and is described by the following diagram ZTA ZTB 7r 2P1 52e e 2a mstA 6 2a rBstB nodal xy plane Gib 0U v N Since atomic orbitals 2pm and 21931 are linear combinations of atomic orbitals 2p1 and 2p1 molec ular orbitals Wu2p1 and 7m 2p1 can be combined to construct molecular orbitals 7331956 and 73219 as follows 7TIt2py 2pyA 2pyB Note however that molecular orbitals 7331956 and 73219 are not eigenfunctions of 3 The order of increasing energy for homonuclear diatomic orbitals is described by the following diagram 0219 W32 2 2 p 092p 19 219 025 25 0928 25 015 15 0915 15 The electronic structure of homonuclear diatomic molecules can be approximated to zeroth order by lling up the unperturbed states according to the Pauli exclusion principle However we should always keep in mind that we are using the H 5 molecular orbitals ie the unperturbed states and therefore we are neglecting the repulsive interaction between electrons This is the same kind of approximation implemented in the construction of zeroth order wave func tions of atoms according to hydrogenlike atomic orbitals where the repulsion energy between elec trons was disregarded and the electronic con guration was constructed by lling up hydrogenlike atomic orbitals according to the Pauli exclusion principle 361 Exercise 55 A Predict the multiplicity of the ground state of 02 B Show that the ground electronic state of 02 is a singlet 89 37 Conjugated Systems Organic Molecules The Hamiltonian of a molecule containing n electrons and N nuclei can be described according to the Born Oppenheimer approximation as follows n N A E2 2 n n 2 HeIZZ 221 7quot 731 31 7 kzgti 1k This Hamiltonian includes terms that describe both 7T and 0 electrons However the distinctive chemistry of conjugated organic molecules is usually relatively independently of 0 bonds and rather correlated with the electronic structure of 7T616CtI OIIS For example the spectroscopy of conjugated organic molecules as well as ionization potentials dipole moments and reactivity can be described at least qualitatively by the electronic structure of the 7T616Ct1 011 model Therefore we make the approximation that the solution of the eigenvalue problem of a conjugated system can be factorized as follows w Maw where 21 is an antisymmetrization operator upon exchange of 0 and 7T electrons The potential due to the nuclei and the average eld due to 0 electrons can be described by the following Hamiltonian H7 haore i i 7 i1 71 kzgti m where lime includes kinetic energy of 7T electrons interaction of 7T electrons with 0 electrons and shielding of nuclear Charges An approximate solution can be obtained by disregarding the repulsion between 7T electrons in Eq 90 and by approximating the Hamiltonian of the system as follows 2 A n A E2 N z e H79 ZHeffz where Heff 2 V3 Z k 2m 7 731 3 k1 7 17 91 The effective nuclear Charge 2 incorporates the average screening of nuclear charges due to 0 and 7T electrons Since H630 depends only on coordinates of electron j we can implement the separation of vari ables method and solve the eigenvalue problem Hgomgp gt2 Emp gt according to the factorizable solution sz gt 1121 M gt where Hefo j gt2 Ej j gt 92 The energy E is obtained by using the Pauli exclusion principle to ll up the molecular orbitals after nding the eigenvalues ej 9O Eq 92 is solved by implementing the variational method assuming that gbj gt can be written according to a linear combination of atomic orbitals N 1 gt ZCjlek gt7 k1 where m gt represents a 2p orbital localized in atom k and the sum extends over all atoms in the conjugated system Example Consider the ethylene molecule represented by the following diagram The diagram shows 0 bonds in the equatorial plane of the molecule and 7139 orbitals 1 and 2 that are perpendicular to such plane The LCAO for ethylene is l 6 gt 971 1 X1 gt 032 1 X2 gt 93 Therefore the secular equations can be written as follows H11 S1167 Cj1 H12 S1267 Cj2 0 H21 3216le 1 H22 Sgg j ng Z Hiickel Method The Huckel Method is a semi empirical approach for solving the secular equations The method involves making the following assumptions 1 H M oz where oz is an empirical parameter vide infra 2 H jk 6 when j k i 1 and H jk 0 otherwise The constant 6 is also an empirical parameter vide infra 3 Sjk 1 when k j i 1 and Sjk 0 otherwise 91 According to the Huckel model the secular determinant becomes Oz Ej 6 6 Cll Ej Therefore the eigenvalues of the secular determinant are Ej 2 oz i 6 and can be represented by the following diagram Energy E2a 6 l 2gtlx1gt lx2gt Ela6 l 1gt gtlt1gtlx2gtgt E 2E1 2a 26 was gt l 11gtl 12gtoz6 6a The energy difference between ground and excited states is AE 2 E2 E1 26 Parameter 6 is usually chosen to make AE coincide with the peak of the experimental absorption band of the molecule An interactive program to perform electronic structure calculations within the Simple Huckel Molecular Orbital approximation can be found here A tutorial to perform electronic structure calculations and simulations of electronic dynamics within the Extended Huckel Molecular Orbital method approximation can be found here 38 SelfConsistent Field HartreeFock Method The self consistent eld SCF Hartree Fock HF method is a variational approach for nding the Slater determinant of a system of n electrons lltPgt lX1X2Xngt 94 that minimizes the expectation value of the energy E ltltIgtlHlltIgtgt CECE 92 for a xed nuclear con guration The one electron basis functions X7 are typically expressed as linear combinations of spin or bitals gbk as follows x 2 cm 95 k allowing for the variational approach to minimize E with respect to the expansion coef cients cm The energy is computed according to the usual Hamiltonian of a system of N nuclei and n electrons with a xed nuclear con guration Hal 2 En 1160 En En i 96 731 7 kgti mg where the spin orbit coupling interactions are neglected The rst term in Eq 96 is the sum of l electron core Hamiltonians A E2 N 2 W v2 ZZJ 97 27716 n Tji j1 describing a system of n non interacting electrons in the electrostatic potential of the nuclei The second term in Eq 96 is the sum of electron electron interactions As a simple example we consider the H2 molecule with n 2 N 2 and r1 DMD 2 12ltr1lX1gtltI 2lX2gt ltT1lX2gtltT2lX1gta T17F2l ellq gt 2 12 ltT2lX2gtltF1lh1lX1gt F2lX1gtltI 1lfb1lX2gtltF1lX1gtltF2Wle2gt A 98 ltr1lx2gtltr2lh 2lx1gt tltrllxlgtltr2lx2gt ltr1lx2gtltr2lx1gtll and the energy expectation value E mime Z ltX1lh1lX1gtltX2lhle2gtX2lh1lX2gtltX1lh1lX1gt 2 ltgtlt1gtlt2l ltX1X2HHX1X2gt mm lt99 ltX1lh1lX1gt ltX2lhle2gt ltX1X2llX1X2gt 2 2 2 ltX1X2li lx2gtlt1gt X2X1l lX1X2gtltX2X1l lx2gtlt1gtl since ft 2 Ida according to Eq 97 In general n 91 1 n 62 e2 E 200 lXjgt i Z ZltXijlTlXijgt ltXij T Xk2Xjgta 100 j1 j1 1w 12 12 Of n Am 1 n n 62 e2 E Z 200 lXjgt i Z ZltXijlelXijgt ltXjXC E Xk2Xjgt7 101 j1 j1 k1 93 62 62 Since XijlalXin XijlaleX 0 Whenj k To minimize E with respect to Xj subject to the constraint of orthonormal orbitals we apply the Lagrange multiplier method for the following functional LX1 xn E Z Z Ejkulem 31 102 j k where Ejk are the Lagrange multipliers Varying the spin orbitals Xj in an in nitesimal amount 6 with respect to the eXpansion coef cients ij we obtain 5LltX17 my Xn 5E Z Z jkllt5leXkgt le5XIcgtla 103 j1 k1 where A A 5E Z21lt5leh1lgtltjgt lt2leh115gtltjgt 2 Z1 ZZ1lt5XijlelXijgt ltXj5Xklfngijgt 2 2 ltXijlf gl5Xijgt ltXij ng Xj6Xkgt 104 lt6Xijrlr121XkXjgt ltXj5Xkl 1 mg i i ltXijl l m ltgtltjgtltkl le5Xjgt Substituting Eq 104 into Eq 103 and simplifying we obtain 6L SLAMmm 2 ZZ1lt5Xijlfngijgt lt5Xijli 21kaj 105 ij1 Zkzl Ejklt5leXkgt 00 which gives 6L 2 23mm 1lt1gtxjlt1gtgt 2321aml lxkcmm ltxklt2gt11xjlt2gtgtlxklt1gtgt 2319am 0 Since SM is arbitrary it must be that the eXpression in square brackets is equal to zero for all j 106 162 7 12 W 20042 1 P121xklt2gtgtl mu 2 Z 2 am 107 j1 k1 where the operator P12 permutes the states of electrons 1 and 2 To write Eq 107 in the canonical eigenvalue form we change the basis set according to the unitary transformation 1X3 Z 1ka 108 k with I I l 1 Considering that the Lagrange multipliers matrix 6 is Hermitian since the func tional L is real it is always possible to nd a I that diagonalizes 6 according to the similarity transformation 6 I TEI 109 94 Such a transformation de nes the set of canonical Spin orbitals for which f1lgtlt gt egjlxgt 110 for j 1 n where f 1 is the Fock operator fa 31 71 111 where l71HF is the HartreeFoek potential describing the electron electron interactions as follows 171W 2 J1r1 X101 112 where J1r1 is the Coulomb mean eld potential TL 2 Mn 2amij 113 k1 The matrix elements in Eq 113 are integrals over the spatial and spin coordinates of electron 2 Analogously X101 is de ned as the exchange operator TL A 2 A X101 ZltX2l P12lxz2gt 114 6757 Equation 110 de nes a self consistent eld SCF problem since the operator f 1 required to nd the solutions X depends on those functions through J1 and X1 To solve this SCF prob lem we rst obtain approximate solutions x by approximating f 1 by 1311 ie neglecting the electron electron interactions introduced by J1 and X1 or by using a semiempirical method like the Hueckel method described later in these lectures These approximate functions X are then used to compute J1 and X1 giving an approximate f 1 that can be used to obtain improved func tions X The process is repeated until convergence To solve the Hartree Fock Eq 110 by solving a set of matrix equations we substitute x j by a linear combination of atomic orbitals analogous Eq 95 bag 2 2k hawk Making the substitution and multiplying from the left with my l we obtain Zlt k39lfjl kgt0kj EijWk39W ij 115 k k or in matrix form FC 2 sce 116 where we have introduced the overlap matrix Sjk tbj wk the Fock matrix ij tbj l f Wk and the matrix of column eigenvectors ij de ning the canonical molecular orbitals x j with energies Ejj expressed in the basis of atomic orbitals 95 381 Restricted Closed Shell HartreeFock The so called closedShell restricted Hartree Fock method is essentially the Hartree Fock approach implemented for systems with an even number n of electrons with each orbital j populated by 2 electrons with opposite spins one with spin oz described by rbj and the other one with spin 6 described by 1 It is called restricted to indicate that the spin states are restricted to be either oz or 6 and closed Shell to indicate that each shell rbj is full with 2 electrons The system is described by the Slater determinant M X1X2Xngt W1 1m n2 n2gt 117 Where X1 191 X2 191 a Xn 1 lbn2 Xn lbn2 The energy of this closedShell restricted Hartree Fock wave function is computed according to Eq 101 by replacing the sums over n spin orbitals Xj by sums over n 2 spin orbitals with spin oz rbj and n 2 spin orbitals with spin 6 15 as follows E zzgglt jmlt1gtwjgtgzy aiwgwa wa ltWkwmgt 221 gflmgjtagwjtm ltw 3m 118 Z L421lt jhlt1b jgt 221 951 LJ Z kgt W 39 fl jgt 1 EQWMSEMW wag Mm where we can cross out the terms that cancel upon integration over the spin variable to obtain E 2zy lt jw lt1gtwjgtzy Ef ij wa lt jwaiwmgt 119 73912 Analogously to the general case we minimize E with respect to rbj subject to the constraint of orthonormal orbitals by applying the Lagrange multiplier method for the following functional W1 m E Z 2 mij 37 120 j k where ejk are the Lagrange multipliers Varying the spin orbitals rbj in an in nitesimal amount 6 with respect to expansion coef cients ij we obtain 6W1 n 6E 22ij warm 121 j1 k1 Varying the spatial orbitals rbj in an in nitesimal amount Sj with respect to the expansion coef cients Cj we obtain SE 222121lt6 jl 1wjgt 221 2lt5 j kl 62 WWO lt5 j kl 62 k jgt r12 73912 221 2lt j5 kl l j kgt ijk 62 WMM 73912 2 23121 lt jw lt1gtwjgt 23 351 wraiij ijiww r12 712 23121 221 2lt j aiwmgt waiwkw r12 73912 122 96 which gives 6 2 zy nwww 2 2321 221 winwagons winwagon 123 Substituting Eq 123 into Eq 121 we obtain n2 A n2 2 2 A 5L 2 2 Zlt5 jl h1l jgt Z 2lt klfgl kgtl jgt lt kl3P12l kgtl jgt jkl kgt 00 0 3921 k1 J 124 which is satis ed when A n2 62 A h1 ZWHT wl P12ll kgtJ 1 Enchka 125 k21 To write Eq 125 in the canonical eigenvalue form we change the basis set according to the unitary transformation W3 21ka 126 k with I I l 1 Considering that the Lagrange multipliers matrix 6 is Hermitian since the func tional L is real it is always possible to nd a I that diagonalizes 6 according to the similarity transformation 6 2 Pier 127 Such a transformation de nes the set of canonical orbitals for which 13sz 6W9 128 for j 1 n 2 where is the closedshell restricted HartreeFock operator 1353 W W 129 where lAjrhf is the restricted HartreeFock potential describing the interactions between electrons of the same spin as follows A A Vlrhf 2J171 X39lvl7 130 where J1r1 is the Coulombic mean eld potential due to the presence of other electrons of the same spin n2 2 e J1r1 ZWMEWM 131 k1 Analogously X101 is de ned as the exchange operator A n2 62 A X101 Zlt tlt2gt17aanlt2gti 132 kZl 12 97 where the permutation operator P12 interchanges electrons 1 and 2 Equation 128 de nes a self consistent eld SCF problem since depends on rbj through lAlrhf To solve this SCF problem we fSt Obtain aPPYOXimate SOIUtiOHS 1b by apprOXimating f by j ie neglecting the electron electron interactions introduced by J1 and X1 or by using a semiempirical method like the Huckel method described in these lectures The resulting approx imate solutions rbj are then used to compute J1 and X1 giving a better approximation to that can be used to obtain improved functions rbj The process is repeated until convergence In practice the restricted Hartree Fock eigenvalue problem introduced by Eq 128 is solved with a set of matrix equations obtained by substituting rb j by a linear combination of atomic orbitals analogous Eq 95 rb j 21 ij Making the substitution and multiplying from the left with Ck we obtain n2 n2 ZltCk lfvgilzlegtij Ejj ZltCHlegtijj 133 k1 k1 or in matrix form F7quotth sce 134 where we have introduced the overlap matrix Sjk C j K k the restricted Hartree Fock matrix Fth mic 135 and the matrix of column eigenvectors ij de ning the canonical orbitals rb j with energies ejj in the basis of atomic orbitals The electronic density 0r of the system with 2 electrons per orbital populating the lowest n 2 states ie closed shell Hartree Fock density can be computed as follows pltrgt 222421w39ltrgt 39ltrgt 2 Elm ClrCmr 2221 011ka 136 2 Elm ClI39CmI39sz where n2 PM 2 ago 137 k1 de ne the elements of the electronic density matrix P The elements of the density matrix PM are thus computed from the solution of the eigenvalue problem introduced by Eq 134 The resulting elements of the density matrix PM can be used to compute not only the density according to Eq 136 but also the restricted Hartree Fock matrix according to Eq 135 since 2 it Alrhf with lAlrhf 2J1r1 Xr1 where J1ltr1gt 2221 dr2w39ltr2gtw39ltr2gt 21 ctr22 ZZZ immeagokgea 138 7 235 242100m f dr2ltltr2ltlltr2 3 221 Hm I dr2cltr2gtiltzltr2gt r12 98 and fem 2 221 wagik wa A 139 I Pml de2C73kzltT2gtP12ClltF2gt Therefore A1 A n2 n2 e2 62 A Mn 2 hlt1gt Z Z Pm 2ltltmltr2gt1g1ltzltr2gtgt ltltmltr2gt1EP121lt1ltr2gtgt 140 l1 m1 and Fth 2 ltltj1f 1ltkgt 141 Z Hfgre with A 2 ltCjlt11h gt1lt kltlgt 142 and n2 n2 e2 62 A ij Z Z Z Pml 2ltCj1391 mr21E1C1F2Ck1391gt ltCjltF1CmF21EP121C1F2CI F1gt l1 m1 143 To solve Eq 134 we rst diagonalize the overlap matrix by computing the matrix X 8 12 that transforms the oyerlap matrix into the identity matrix as follows XlSX 1 Then we introduce the matrix C as follows C 2 X6 144 that according to Eq 134 satis es the eigenvalue problem Frhfx SXCE 145 or F7quotth 2 Ce 146 with Frhf XTFWX 147 These equations allow for the implementation of the self consistent eld restricted Hartree Fock SCF RHF method for a xed nuclear con guration as follows 1 Calculate the matrix elements Sjk H re and the 2 electron integrals Cj r1Cm r2l 62 1Q r2Ckr1gt and Cjr1CmI39211C1F1Ckr2gt 2 Diagonalize Sjk to obtain X jk 3 Obtain an approximate density matrix ij according to Eq 137 by solVing Eq 134 with f 1 141 or f 1 m where is the semiempirical extended Huckel Hamiltonian of the system 99 10 Compute the matrix elements ij according to Eq 143 using PM and the 2 electron integrals Compute the Fock matriX Fflgf according to Eq 141 by using H re ij and the 2 electron integrals Compute the transformed Fock matriX ngf by using F f and X j 1 according to Eq 147 Obtain C and e by solving Eq 146 Calculate C by using Eq 144 Compute a new density matriX P according to Eq 137 based on C obtained in 8 If P has changed more than a given tolerance relative to the previous iteration go to 4 Otherwise the SCF calculation has converged and the solution is given by the current eigen vectors C and eigenvalues 6 Con guration Interaction Improvement over the one determinant trial wave function can be achieved by using a trial wave function that involves a linear combination of Slater determinants This method is known as con guration interaction The energy correction over the Hartree Fock energy E007 E EHF is known as correlation energy 100 39 Quantum MechanicsMolecular Mechanics Methods A practical approach for describing the electronic structure of a molecular fragment in a complex molecular environment is the so called quantum mechanicsmolecular mechanics QMMM ap proach where the environment represented by sticks in the gure is described in terms of a sum of classical potentials VMM described by a molecular mechanics MM force eld while the molec ular fragment R1 in the Figure below is described by quantum chemistry QM methods as for example by a Slater determinant obtained according to the Hartree Fock method The interaction between the molecular fragment and the environment is usually de ned as the sum of the elec trostatic interaction between the atomic charges of the environment and the nuclear and electronic charges in molecular fragment If the fragment is covalently bound to the environment the bond is broken and the covalency is completed with a link atom usually a H atom The interaction between the fragment and the environment is included as an additional term in the one electron core Hamiltonian 131 as follows A0 E2 2 N 262 N N 262 A0 1 7 j j 7 h 2 W Z 22 hQMMM 148 e j1 37 j1 j1 7 with A Ne N Ne Ne hgvMM Z k 227 meltm 149 161 7 j1 k1 7 k1 where we have considered a molecular fragment with N nuclei embedded in an environment with N6 electrostatic potential atomic charges The third term in Eq 149 is a van der Waals potential that accounts for the interaction of the electrons in the molecular fragment with electrons in the environment that are not explicitly considered Implementing the Hartree Fock method with the one electron core Hamiltonian introduced by Eq 148 we obtain the Hartree Fock energy EQM of the molecular fragment in the electrostatic eld of the surrounding environment yielding the total QMMM energy of the system as follows EQMMM EQM VMM 150 101 In this simple form the molecular fragment is polarized by the surrounding environment Polariza tion of the environment due to the distribution of charges in the molecular fragment can be included by using a polarizable molecular mechanics force eld or a moving domain QMMM approach where the charges in the environment are obtained self consistently Another QMMM approach is the ONIOM methodology as implemented in Gaussian where the total energy is computed as follows EONIOM EQM V s y em 7aigment7 where V131me and V1627 mm are the energies of the complete system and the fragment as described by the molecular mechanics force eld The force elds are usually parametrized to match the experimental or ab initio ground state potential energy surfaces as a function of nuclear coordinates The following section illustrates the parametrization of the potential energy surface for diatomic molecules 40 Empirical Parametrization of Diatomic Molecules The main features of chemical bonding by electron pairs are properly described by the HL model of H 2 According to such model the covalent bond is described by a singlet state linL N104162 61042lXA1XB2 XA2XB1 with energy J K 1E lt1 DHLlHllleL gt2 my where H 11 112 62712 with M1 v i M2 iv 5 J lt XA1XB2lHlXA21XB2 gt Coulomb integral K lt XA1XB2lHlXA2XB1 gt Exchange integral 32 lt XA1XB2lXA2XB1 gt Similarly the triplet state is described as follows a162 61Oz2l 3mm N3XA1XB2 XB1XA2l Oz1042 a MUM and has energy 3 J K E The energies of the singlet and triplet states are parametrized by the internuclear H H distance and can be represented by the following diagram 102 H H bond length The energies 1E and 3E can be approximated by the following analytical functions 7 1EL D 6 2 RR0 26 RR0 E MR 3E2 2 Parameters D and a can be obtained by tting MR to the actual experimental or ab initio ground state potential energy surface Such parametrization allows us to express the Coulombic and Reso nance integrals J and K in terms of available experimental or ab initio data as follows 6 2aR R0 2e aR R0 E J m M M 32M M NIH 1039 K m M M SQMM This parametrization of Hamiltonian matrix elements illustrates another example of semiempirical parametrization that can be implemented by using readily available experimental information re member that in the previous section we described the semiempirical parametrization of the Huckel model according to the absorption spectrum of the molecule The covalent nature of the chemical bond signi cantly changes when one of the two atoms in the molecule is substituted by an atom of different electronegativity Under those circumstances the wave function should include ionic terms e g 1W NXAlt1gtXAlt2gtlalt1gt6lt2gt mac and W330 NXB1XB2O 162 61Oz2l 103 The complete wave function with both covalent and ionic terms can be described as follows 1b 2 01 Cgtbg where the covalent wave function is m oz162 51042lXA1XB2 Mama and the ionic wave function is w oz162 6lt1gtozlt2gtnxllt1gtxllt2gta XBlt1gtXBlt2gtlt1 51 where the parameter 51 is determined by the relative electronegativity of the two atoms For exam ple consider the HF molecule For such molecule 51 1 A represents the F atom and B represents the H atom ie due to the electronegativity difference between the two atoms the predominant ionic con guration is H F Therefore the ground state energy E9 is obtained as the lowest eigenvalue of the secular equation H 11 E H 12 H 12 H 22 E Here we have neglected 312 assuming that such approximation can be partially corrected according to the parametrization of H12 The semiempirical parametrization strategy can be represented by the following diagram 0 152 RH F This diagram represents the following curves H11 M De2 RR05 26 RR05 is a covalent state represented by a Morse poten tial M H22 I EA 146 bp JR 9 is the potential energy surface of the ionic state where the difference between the H ionization energy and the F electron af nity I EA corresponds to the energy of forming the ion pair H F The term is the Coulombic interaction and 146 bp IR 9 is the short range repulsive potential The ground state potential energy surface E9 M De2 RR0 2e RR0 is represented by a Morse potential M Parameters D and R0 can be obtained from the eXperimental bond energy 104 and bond length The parameter a can be adjusted to reproduce the vibrational frequency of the diatomic molecule The parameter DHF xDHHDFFandd 005121 Parameters A and C are adjusted so that the minimum energy of H 22 corresponds to the H F bond length ie the sum of ionic radii of H and F This empirical parametrization allows us to solve Eq 66 for H12 H12 H11 MltH22 M and obtain the Hamiltonian matrix elements in terms of empirical parameters Conclusion Potential energy surfaces parametrized by a few empirical parameters are able to describe bonding properties of molecules associated with atoms of different electronegativity Dipole Moment The dipole moment is one of the most important properties of molecules and can be computed as follows lug lt lbgllllbg gt7 Z Ti ZQZjRj 73 j The rst term of this equation involves electronic coordinates 7 7 and the second term involves nuclear coordinates Rj For example the dipole moment of HF can be computed as follows where a Mg 2 012 lt 1l l 1 gt 022 lt 2W 2 gt 201C2 lt 1W 2 gt7 EH EH EH 0 6R0 0 since 1amp1 represents a covalent state and the overlap between 1 and 2 is assumed to be negligible The dipole moment is usually reported in Debye units where 4803 Debye is the dipole moment of two charges of l au with opposite sign and separated by l A from each other 401 Exercise 56 Evaluate the dipole moment for HF using the following parameters for the semiempirical model of HF potential energy surfaces energies are expressed in kcalmol and distances in A D134 D6l R0O92 a227 A640 b25 C20 1313 EA83 Polarization The electric eld of an external charge 2 located at coordinate R0 along the axis of the molecule does not affect the energy of the covalent state H11 but affects the energy of the ionic state H 22 as 105 follows 26 ze Hi2 2 H22 RHC RF C Therefore the presence of an external charge perturbs the ground state energy of the molecule Such perturbation can be computed by re diagonalizing Eq 66 using H 2 instead of H 22 Solving for the ground state energy we obtain 1 E Z i 17 H11 H 2 H11 4H12212l 402 Exercise 57 1 Plot E9 as a function of the internuclear distance R for the HF molecule in the presence of an external charge located in the axis of the molecule at 10 A to the left of the F atom 2 Compare your results with the analog Gaussian98 calculation by using the scan keyword Hint The Gaussian98 input le necessary to scan the ground state potential energy surface of H 2 is described as follows hf6 3 l G scan potential scan for H2 0 1 H H 1 R R 09 5 01 This input le scans the potential energy of H 2 by performing single point calculations at 5 internuclear distances The output energies are represented by the following diagram 106 Energy W 09 10 11 12 13 RH39H 107 41 Tunneling Current Landauer Formula We consider a 1 dimensional electron tunneling A 712 02 H 153 1 M W evltxgt 1 lt gt problem described by the Hamiltonian A 712 02 H 154 2771 812 Vltgt7 V if 1lt0 V1 Vb if 0lt1lta 155 V if 1gta where Vb de nes the tunneling barrier and AV 2 V V7 de nes the voltage drop across the barrier Outside the tunneling interval 1 lt 1 lt 1 the solutions of the Schrodinger equation are superpositions of plane waves since the potential is constant For energy E gt 6V and E gt 6 there are two independent solutions 1b and 1D for incident electrons from the left and from the right respectively Considering the solution for incidence from the left we obtain qsgf m5 if 1 lt 0 1b1 Aeikbf Be kalC if 0 lt 1 lt a 156 trgb if 1 gt a where gbji kJ12eiikj5 are de ned divided by the square root of kj so they are normalized to carry the unit of current density hm as shown below The labels j 17 indicate the left I and right 7 side of the barrier kl 2mE eVlh2 and k7 2mE gm12 Applying the continuity conditions for 1D and owl311 at 1 0 and 1 a we obtain kflQ 51277 A B Aeikba Be ikba kr l2treikhna7 11320 n kbA B kbA ikba Be ikzba ki2treikzra 157 The transmission amplitude tr re ection amplitude 77 and coef cients A and B can be obtained by solving for them from Eq 157 The probability uX or current density of incoming electrons from the left described by the incident wave 1b1 t kl 2 iklwt with momentum kl and energy E kl 6V 7126 2771 is iat t 7 75 7 158 we 2771201411 6 was a lt gt 108 01 h 0 mm 31 2 am 9 mat 2 3w agave 2 159 Remltxtgtmltmtgti The ux of transmitted electrons described by transmitted wave 4513 75 trkI126iktfwt with momentum k and energy E kr 61 VICE2m is h 8 Z tt CC7 5 7 160 h ltr12 m Therefore the transmission coef cient T jtj de ned as the transmitted uX j over the inci dent uX at energy E is T 17512 The re ection coef cient R 1 T is the re ected uX over the incident uX Analogously we consider incidence from the right of the tunneling barrier as follows gb ngbj if m gt a 1113 AeikblC B ikb if 0 lt 1 lt a 161 m5 if 1 lt 0 Solving for 75 we obtained the transmission coef cient T 2 it 2 due to incidence from the right More generally we can consider incoming waves from both left and right of and gb re spectively with amplitudes cm 2 ca cm that generate outgoing waves to the left and right of and q j respectively with amplitudes cow 2 022 022 The amplitudes of outgoing and incoming waves are related by the linear transformation de ned by the scattering matrix or S matriX S as follows com 2 Sci l l Cout Tl tl Cm Cgut tr T7 162 Due to the conservation of probability the S matriX must be unitary S 1 2 81 Therefore SST 1 T T Tl tl Tl tr tr 7 t 7 1 m at 1 164 which gives 109 In addition 818 1 1 1 Tl t7 7 t1 11 111 13 1 65 716 Tin 1 166 which gives Therefore according to Eqs 164 and 166 we obtain 1 tltll 777 2 Tim For our 1 dimensional case we obtain 17712 17612 R 167 Under stationary state 011075 2 0 with 0 W bl Then according to the continuity equation 011075 2 0j01 we obtain 8303 0 Therefore jl for 1 lt 0 must be equal to jl for 1 gt 61 Also jr for 1 lt 0 must be equal to jr for 1 gt a lt1 11112 1112 168 and 1 17112 175112 169 Dividing Eq 168 by Eq 169 and using 167 we obtain 75 2 tr 170 Therefore T E tr 2 llt gt 1 12 m 1t11 TrEa so the transmission coef cient is the same for both directions of incidence and R T 1 Considering that the number of electrons with energy E incident from the left and right of the barrier are nlE and nrE respectively the net uX of charge from left to right is 00 hk 00 hk I 2 2e dklnlkl lTl 2e dkrnrkr TTT 0 m 0 m 26 00 hkl 8kg hkr 0kquot g 0 dE TltEgt a E nrE m 0E 7 172 26 00 hkl m k7quot m 7T 0 h2 kl m h2kr 7 26 00 z E dE TE ME ME 0 where factor of 2 accounts for the two possible spin states the rst term on the rhs accounts for the forward uX ie from left to right and the second term accounts for the backward uX 110 ie from right to left Note that in the second row of Eq 172 we used the following equality i delEEl 27rfdklkkl At equilibrium the population of energy levels is determined by the Fermi Dirac distribution 1 Z W where E F is the Fermi level and the factor of 2 in the numerator accounts for the 2 possible spin states Considering the potentials for electrons at either side of the barrier we obtain nlE nE 6 and 71E nE 6 Therefore we can eXpand these distributions as follows 0nE 8E ME nE EF 6 nE 173 nlEnEEF Vl 174 E 0E eW and write the Landauer formula giving the current in the form of the Ohm s law as follows I z 000 dE TE ME ME 262 h GEAV where G R l G0 f dE T 87535 is the conductance or inverse of the resistance R with G0 2 2 12906 k91 the quantum unit of conductance Note that Go de nes the maximum conductance minimum resistance per conduction channel with perfect transmission T 1 ie if the transport through the channel is ballistic and therefore the probability for transmitting the electron that enters the channel is unity as observed in eXperiments At low temperature ie 6 e co the Fermi Dirac distributions become step functions 71 E 2HEF E 6 and nrE 2HEF E 6 with the Heaviside function equal to 1 for 1 gt 0 and O for 1 lt 0 Therefore EWE 6EF E and 0nE 8E 175 dE TE AV 8E 262 I dE TE6EF EAV h 262 176 In this low temperature limit the conductance is the transmission times the quantum of conduc tance GE g 411 WKB Transmission The goal of this subsection is to show that the transmission coef cient TE can be estimated under the WKB approximation as follows TE 6 21 dwv 2m1E x E39rlc7127 177 111 A AFM cantilever Dh min Paint Bantam gates Be nteete E E F 3911 m E I I 1 2 1 13 433 V9 Volts Fig 1 A Schematic diagram at the experi mental setup QPC Epnduetanee i3 meeeured 35 a functipn pf AFN tip PDSiti l39l Elk Peirut eeri teet epnduntrtenee t3 1trerglus gete wattage V with me tip FIFESEW at temperature T 1 quot Pie teaue at integer multiples pf 12th are clearly seen The i neet shpws a tppplgrephie image at the ppth tenteet deviEe where 5 1 describes the tunneling barrier according to Eq 155 To derive Eq 177 we consider the WKB approximate solution of Eq 154 with the follow ing functional form W woe WW 178 where 2mE h2 Note that when V1 is constant my corresponds to a particle moving to the right with constant momentum k Substituting mm as de ned in Eq 178 into Eq 154 we obtain Jim 2m 8132 V1rb1 E 1 A 179 with A ik Therefore the WKB solution is a good approximation when k ltlt According to the WKB solution the probability density 2 remains constant on the left of the tunneling barrier when E gt 1 since Maj tb ooeif oo divk for 1 lt 0 Inside the barrier however the probability density decays exponentially W woe KW QW39E Eltr gt39hi 180 112 since E lt 5 In particular at 1 a the probability density is WW2 l 0l26 2f0adquotk3939 181 In the region with 1 gt a the probability density remains constant again since 1b 1b a 6 75 f div39kr and l2 2 WM 2 Therefore estimating the transmission coef cient as the ratio of the proba bility densities to the right and to the left of the barrier we obtain Went2 WOW 182 6 2f0adW TE 113 42 Solutions to Computational Assignments 421 Problem 1 Computational Problem 1 Write a computer program to represent the wave packet introduced by Eq 10 on a grid of equally spaced coordinates xj xmm j 1A with nite resolution A xmm n 1 and Visualize the output Choose x0 0 and p0 0 in the range x 2020 with oz 2 com where m 1 and w 1 To Visualize the output of this program cut the source code attached below save it in a le ImndeKmbn fgmnm e byQmmg f77 Problem1f o Probleml run it by typing Probleml Visualize the output as follows type gnuplot thentype plot archOOOO That will show the representation of the Gaussian state introduced in Eq 6 in terms of an array of numbers associated with a grid in coordinate space To eXit type quit 114 Download from httpXbeamsChemyaleedubatistaP1Problem1f PROGRAM Probleml call Tnitialize CALL SAVEWFO END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE Initialize Wave Packet Initialization Gaussian centered at Xk with momentum pk O IMPLICIT NONE INTEGER nptxnptskk COMPLEX chiEYE REAL omegaxminxmaxdxpimassxkpkxalpha PARAMETERnptslOnptx2npts COMMON wfunc chinptx common Xy xminxmax common packetmassxkpk xmin 20 xmax20 EYE0010 pi acos lO omegal dxXmax Xminrealnptx pk00 Xk00 masslO alphamassomega do kklnptx xxminkkdx chikkalphapi025 l exp alpha2XXk2EYEpkX Xk end do RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTTNE SAVEWFj Save Wave packet in coordinate space 0 IMPLICIT NONE INTEGER nptxnptskkj COMPLEX chiEYE REAL RVomegaxminxmaxdxpimassxkpkxalphaVpotRKE character9 B PARAMETERnptslOnptx2npts COMMON wfunc chinptx common Xy xminxmax common packetmassxkpk writeB Ai44 arch j 115 OPEN 1 FILEB dxxmax Xminrealnptx do kklnptx xxminkkdx WRITE122 Xchikk end do CLOSE1 22 FORMAT6e1362X RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc 116 422 Problem 2 Computational Problem 2 Write a computer program to represent the initial state introduced by Eq 10 in the momentum space by applying the FFT algorithm to the grid based representation generated in Problem 1 and visualize the output Represent the wave packet amplitudes and phases in the range p 44 and compare your output with the corresponding values obtained from the analytic Fourier transform obtained by using dz eXp a212 111 a0 x7ra2 eXpa0 ai4a2 In order to visualize the output of this program cut the source code attached below save it in a le named Problem2f compile it by typing f77 Problem2f o ProblemZ run it by typing Problem2 Visualize the output as follows type gnuplot thentype plot numeOOOO That will show the representation of the amplitude of the Fourier transform of the Gaussian state introduced in Eq 6 in terms of an array of numbers associated with a grid in momentum space In order to visualize the analytic results on top of the numerical values type replot analOOOO In order to visualize the numerically computed phases as a function of p type plot numeOOOO u 13 and to visualize the analytic results on top of the numerical values type replot analOOOO To eXit type quit 117 Download from httpXbeamsChemyaleeduNbatistaPZProblemZf PROGRAM Problem2 call Tnitialize CALL SAVEFT END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE Initialize Wave Packet Initialization Gaussian centered at Xk with momentum pk O IMPLICIT NONE INTEGER nptxnptskk COMPLEX chiEYE REAL omegaxminxmaxdxpirmassxkpkxalpha PARAMETERnptslOnptx2npts COMMON wfunc chinptx common Xy xminxmax common packetrmassxkpk xmin 20 xmax20 EYE00lO pi acos lO omegal dxXmax Xminrealnptx pk00 Xk50 rmasslO alpharmassomega do kklnptx xxminkkdx chikkalphapi025 l exp alpha2XXk2EYEpkX Xk end do RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc subroutine SAVEFT Save wave packet in momentum space 0 IMPLICIT NONE INTEGER nptxkxnxnptsj REAL thetawmpxminxmaxrmassxkpialenxpkrmreri COMPLEX eyechiPsip character9 B1B2 parameternptslOnptx2npts common Xy xminxmax common packet rmassxkpk COMMON wfunc chinptx jO 118 22 writeBl Ai44 OPENlFILEBl writeB2 Ai44 OPEN2FILEB2 CALL fournchinptxl l pi acos lO alenxxmax xmin do kxlnptx ifkxlenth2l nxkx l else nxkx l nptx end if pO ifnxneO p Numerical Solution chikxchikxalenXsqrt2Opinptx rechikx riimagchikx IFreNEO thetaatanrire rmabschikx IFabspLE4 IFnXEQnth2 Analytic Solution CALL FTanalyPsipp rePsip riimagPsip IFreNEO thetaatanrire rmabsPsip IFabspLE4 IFnXEQnth2 end do CALL fournchinptxll FORMAT6e1362X return end anal j nume j then realnx2pialenx WRITE222 WRITE222 prmtheta WRITEl22 WRITEl22 prmtheta CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine FTanalyPsipp Analytic Fourier Transform of the initial Gaussian wave packet IMPLICIT NONE REAL ppialpharmassxkpkomega COMPLEX PsipcOclc2eye common packet rmassxkpk eye00lO omegal alpha rmassomega piacos lO c2alpha2 clalphaxkeyepk p 119 cO alpha2Xk2 eyepkxk Psipsqrtpic2sqrt20pialphapi025 l expcl240c2expc0 return end ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc c Subroutines from Numerical Recipes CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTINE FOURNDATANNNDIMISIGN REAL8 WRWIWPRWPIWTEMPTHETA DIMENSION NNNDIMDATA NTOT1 DO 11 IDIM1NDIM NTOTNTOTNNIDIM II CONTINUE NPREVI DO 18 IDIM1NDIM NNNIDIM NREMNTOTNNPREV IP12NPREV IP2IP1N IP3IP2NREM I2REVI DO 14 IZ1IP2IP1 IFI2LII2REVIHEN DO 13 IlIZIZIPl 22 DO 12 I3IIIP3IP2 I3REVI2REVI3 I2 IEMPRDAIAltI3gt IEMPIDAIAltI3Igt DAIAltI3gtDAIAltI3REVgt DAIAltI3IgtDAIAltI3REVIgt DAIAltI3REVgtIEMPR DAIAltI3REVIgtIEMPI 12 CONTINUE I3 CONTINUE ENDIF IBITIP22 I IF IBITGEIPIANDI2REVGTIBIT THEN I2REVI2REV IBIT IBITIBIT2 GO TO I ENDIF I2REVI2REVIBIT I4 CONTINUE IFPIIPI 2 IFIFPILTIP2THEN IEP22IFPI THETAISIGN628318530717959DOIEPZIPI WPR 2DODSINO5DOTHETA2 WPIDSINTHETA 120 WR1DO WTODO DO 17 1311FP1IP1 DO 16 111313IP1 22 DO 15 T2TlTP3TFP2 K112 K2K1TFP1 TEMPRSNGLWRDATAK2 SNGLWTDATAK21 TEMPTSNGLWRDATAK21SNGLWTDATAK2 DATAK2DATAKl TEMPR DATAK21DATAK11 TEMPT DATAK1DATAK1TEMPR DATAK11DATAK11TEMPT 15 CONTINUE l6 CONTINUE WTEMPWR WRWRWPR WTWPTWR WTWTWPRWTEMPWPTWT l7 CONTINUE TFP1TFP2 GO TO 2 ENDTF NPREVNNPREV l8 CONTINUE RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc 121 423 Problem 3 Computational Problem 3 Write a computer program to compute the expectation values of the position 5130 2 IlolzdeIIO and the potential energy V IlolliMKIIO where Vz is de ned according to Eq 16 for the initial wave packet introduced by Eq 10 with various possible values of 510 and p0 with oz 2 mm where m 1 and w 1 In order to visualize the output of this program cut the source code attached below save it in a le named Problem3f compile it by typing f77 Problem3f o ProblemB run it by typing Problem3 The printout on the screen includes the numerically expectation values and 1115 122 Download from httpXbeamsChemyaleeduNbatistaP3Problem3f PROGRAM Problem3 IMPLICIT NONE REAL XVENERGY CALL Initialize CALL PEVENERGY CALL PXX PRINT quotltPSilVlPSigtquotVENERGY PRINT quotltPsillesigtquotX END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTTNE Initialize 0 Wave Packet Initialization Gaussian centered at Xk with momentum pk TMPLICIT NONE INTEGER nptxnptskk COMPLEX chiEYE REAL omegaxminxmaxdxpimassxkpkxalpha PARAMETERnptsIOnptx2npts COMMON wfunc chinptx common Xy xminxmax common packetmassxkpk xmin 20 xmax20 EYE00IO pi acos IO omegal dxXmax Xminrealnptx pk00 Xk00 massIO alphamassomega do kklnptx xxminkkdx chikkalphapi025 I exp alpha2XXk2EYEpkX Xk end do RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE PERV O Expectation Value of the Potential Enegy IMPLTCIT NONE INTEGER nptxnptsk COMPLEX chi REAL VpotRVxminxmaXdXX PARAMETERnptsIOnptx2npts 123 COMMON wfunc chinptx common Xy xminxmax dxXmax Xminrealnptx RV00 do klnptx xxminkdx CALL VAVpotX RVRVchikVpotconjgchikdx end do RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTTNE PXRV c Expectation Value of the position IMPLICIT NONE INTEGER nptxnptsk COMPLEX chi REAL RVXminXmaXdXX PARAMETERnptslOnptx2npts COMMON wfunc chinptx common Xy xminxmax dxXmax Xminrealnptx RV00 do klnptx xxminkdx RVRVChikXCOngChikdX end do RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE VAVX c Potential Energy Surface Harmonic Oscillator IMPLICIT NONE REAL VXmassxkpkrkomega common packet massxkpk omegalO rkmassomega2 VO5rkxx RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc 124 424 Problem 4 Computational Problem 4 Write a computer program to compute the expectation values of the initial momentum 190 2 Ilol WO and the kinetic energy T 0110 2m110gt by using the Fourier transform procedure where 110 is the initial wave packet introduced by Eq 10 with 510 2 0 p0 0 and oz 2 mm where m 1 and w 1 Compute the expectation value of the energy E IlolliWIJO where H A22m Vi with V1 de ned according to Eq 16 and compare your result with the zero point energy E0 w 2 In order to visualize the output of this program cut the source code attached below save it in a le named Problem4f compile it by typing f77 Problem4f o Problem4 run it by typing Problem4 The printout on the screen includes the numerically expectation values Iltl 13W IIATWJ and IltlH Note that the analytic value of 1115 TWQ is ha 2 05 in agreement with the numer ical solution 125 Download from httpXbeamsChemyaleeduNbatistaP4Problem4f PROGRAM Problem4 CALL Initialize CALL Ppp PRINT quotltPsilplPsigtquotp CALL KERKE PRINT quotltPsiTPSigtquotRKE CALL PERV PRINT quotltPsilHlPsigtquotRKERV END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE Initialize Wave Packet Initialization Gaussian centered at Xk with momentum pk O IMPLICIT NONE INTEGER nptxnptskk COMPLEX chiEYE REAL omegaxminxmaxdxpimassxkpkxalpha PARAMETERnptslOnptx2npts COMMON wfunc chinptx common Xy xminxmax common packetmassxkpk xmin 20 xmax20 EYE00lO pi acos lO omegal dxXmax Xminrealnptx pk00 Xk00 masslO alphamassomega do kklnptx xxminkkdx chikkalphapi025 l exp alpha2XXk2EYEpkX Xk end do RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE PERV Expectation Value of the Potential Enegy O IMPLICIT NONE INTEGER nptxnptsk COMPLEX chi REAL VpotRVxminxmaXdXX PARAMETERnptslOnptx2npts 126 COMMON wfunc chinptx common Xy xminxmax dxXmax Xminrealnptx RV00 do klnptx xxminkdx CALL VAVpotX RVRVchikVpotconjgchikdx end do RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE KERKE O Expectation value of the kinetic energy IMPLICIT NONE INTEGER kknptxkxnxnpts REAL dpRKEpXminXmaXpialenxdxmassxkpk COMPLEX eyechiPsipchic parameternptslOnptx2npts DIMENSION chicnth common Xy xminxmax common packetmassxkpk COMMON wfunc chinptx RKE00 pi acos IO dxXmax Xminnptx dp2piXmax Xmin do kklnptx chickkchikk end do CALL fournchicnptxll do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if pO ifnxneO p realnxdp chickxp22OmasschickXnptx end do CALL fournchicnthI l do kklnptx RKERKEconjgchikkchickkdx end do return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE Pppe 127 O Expectation value of the momentum IMPLICIT NONE INTEGER kknptxkxnxnpts REAL dppepxminxmaxpialenxdxmassxkpk COMPLEX eyechiPsipchic parameternptslOnptx2npts DIMENSION chicnth common Xy xminxmax common packetmassxkpk COMMON wfunc chinptx pe00 pi acos lO dxXmax Xminnptx dp2piXmax Xmin do kklnptx chickkchikk end do CALL fournchicnptxll do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if pO ifnxneO p realnxdp chickxpchickXnptx end do CALL fournchicnptxl l do kklnptx pepeconjgchikkchickkdx end do return end ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE VAVX c c Potential Energy Surface Harmonic Oscillator c implicit none REAL VXmassxkpkrkomega common packet massxkpk omegalO rkmassomega2 VO5rkxx RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc c Subroutines from Numerical Recipes 128 CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTINE FOURNDATANNNDIMISIGN REAL8 WRWIWPRWPIWTEMPTHETA DIMENSION NNNDIMDATA NTOT1 DO 11 IDIM1NDIM NTOTNTOTNNIDIM II CONTINUE NPREVI DO 18 IDIM1NDIM NNNIDIM NREMNTOTNNPREV IP12NPREV IP2IP1N IP3IP2NREM I2REVI DO 14 IZ1IP2IP1 IFI2LII2REVIHEN DO 13 IlIZIZIPl 22 DO 12 I3IIIP3IP2 I3REVI2REVI3 I2 IEMPRDAIAltI3gt IEMPIDAIAltI3Igt DAIAltI3gtDAIAltI3REVgt DAIAltI3IgtDAIAltI3REVIgt DAIAltI3REVgtIEMPR DATA13REV1TEMPI 12 CONTINUE I3 CONTINUE ENDIF IBITIP22 I IF IBITGEIPIANDI2REVGTIBIT THEN I2REVI2REV IBIT IBITIBIT2 GO TO I ENDIF I2REVI2REVIBIT I4 CONTINUE IFPIIPI 2 IFIFPILTIP2THEN IFP22IFP1 THETAISIGN628318530717959DOIFPZIPI WPR 2DODSINO5DOTHETA2 WPIDSINTHETA WR1DO WIODO DO 17 I3IIFPIIPI DO 16 III3I3IPI 22 DO 15 I2IIIP3IFP2 KII2 K2KlIFPl 129 TEMPRSNGLWRDATAK2 SNGLWIDATAK21 TEMPISNGLWRDATAK21SNGLWIDATAK2 DATAK2DATAKl TEMPR DATAK21DATAK11 TEMPI DATAK1DATAK1TEMPR DATAK11DATAK11TEMPI 15 CONTINUE l6 CONTINUE WTEMPWR WRWRWPR WTWPTWR WTWTWPRWTEMPWPTWT l7 CONTINUE TFP1TFP2 GO TO 2 ENDTF NPREVNNPREV l8 CONTINUE RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc 130 425 Problem 5 Computational Problem 5 Expand the exponential operators in both sides of Eq 40 and show that the Trotter expansion is accurate to second order in powers of 739 Expanding the lefthandside lhs of Eq 40 from the lecture notes gives A A 1 A 6 11 2 1 1H7 511272 073 183 where H 132 2m Therefore A 1234 1A 1232A 145 1H7 2 2 2 2 2 3 l H V V V O 184 e Z7 2477127 2 T 22mT 22mTltTgt7 In order to show that the Trotter expansion introduced by Eq 40 is accurate to second order in 7 we must expand the righthandside rhs of Eq 40 and show that such an expansion equals the rhs of Eq 184 Expanding the righthandside rhs of Eq 18 gives 6 iviT2 ip2T2m6 iviT2 lt1 iVTZ V27 24 073 1 i7 72 C73 X 1 iVTZ V27 24 C73 185 6 iviT2 ip2T2m6 iviT2 lt1 iVTZ l27 24 i r V T22 ii 073 2 2m 2m 2 47712 X 1 iVTZ V27 24 C73 186 A M2 39 A A 1 A 152 A 132 1 154 e LV IZT2 1p T2m LVZET2 I 1 ZVT2 V2724 L T V722 72 2 A A A 1 A iVTZ V2T24 p V7 22 V27 24 07 3 2m 2 A A2 A A 1A A 1 p e LV ET2 1p T2m LVZIZT2 I 1 ZVT L T V272 V722 72 A2 19 A 2 3 V 2 O 2m 7 lt7 gt Note that the rhs of Eq 188 is identical to the rhs of E 184 completing the proof that the Trotter expansion introduced by Eq 18 is accurate to second order in 7 131 426 Problem 6 Computational Problem 6 Write a computer program that propagates the initial state IIO2 for a single time increment 739 01 au Use 510 25 p0 0 and oz 2 mm where m 1 and w 1 Implement the SOFT method for the Hamiltonian H 1322771 Vi where V1 is de ned according to Eq 16 Compare the resulting propagated state with the analytic solution obtained by substituting Eq 23 into Eq 22 In order to Visualize the output of this program cut the source code attached below save it in a MemmdeKbbm cmm b bqumg f77 Problem6f o Problem6 run it by typing Problem6 and Visualize the output as follows type gnuplot thentype set dat sty line thentype set yrange06 andthetype plot arch0002 That will show the numerical propagation after one step with 739 01 In order to Visualize the analytic result on top of the numerical propagation type replot arch0002 u 13 To eXit type quit 132 Download from httpXbeamsChemyaleeduNbatistaP6Problem6f PROGRAM Problem6 l D wave packet propagation O IMPLICIT NONE INTEGER NNnptsnthndump INTEGER istepnstep REAL dtXcpc COMPLEX vproptpropxmeanpmean PARAMETERnpts9nptx2nptsNNl DIMENSION vpropnthNNNNtpropnth DIMENSION XmeanNNpmeanNN COMMON class xcpc CALL ReadParamnstepndumpdt call Initialize CALL SetKinPropdttprop CALL SetPotPropdtvprop DO isteplnstepl IFmodistep llOE00 l PRINT quotStepquot istep lquot Final stepquot nstep IFistepGEl CALL PROPAGATEvproptprop IFmodistep lndumpE00 THEN CALL SAVEWFistepndumpdt END IF END DO 22 FORMAT6e1362X END ocococococococccccccccccccccccccococococococococococococoooooooococo subroutine ReadParamnstepndumpdt o c Parameters defining the grid xmin xmax integration time step dt c mass rmass initial position xk initial momentum pk c number of propagation steps nstep and how often to save a pic ndump o IMPLICIT NONE INTEGER ntypenstepnrptireportndumpnlit REAL xminXmaXpkrmassXkdt common packet rmassxkpk common Xy xminxmax o xmin l00 xmax l00 dtOl rmasslO Xk 25 pklO nstepl ndumpl 133 return end CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O O 22 SUBROUTINE Initialize IMPLICIT NONE INTEGER NNnthnptskk COMPLEX chiOchiEYECRV REAL xcpcomegaxk2xminxmaxdxpirmassxkpkxalphaalpha2 PARAMETERnpts9nptx2nptsNNl DIMENSION CRVNNNN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN COMMON iwfunc chiOnthNN COMMON class xcpc EYE00lO pi acos lO omegal dxXmax Xminrealnptx xckk pcpk Wave Packet Initialization alpharmassomega do kklnptx xXminkkdX Chikklalphapi025 expalpha2X Xk2EYEpkXXkgt chiOkklchikkl end do Hamiltonian Matrix CRV do kklnptx xxminkkdx CALL HAMILCRVX WRITEll22 XrealCRVll END DO FORMAT6e1362X RETURN END Gaussian centered at xk with momentum pk CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTINE HAMILCRVX Hamiltonian Matrix IMPLICIT NONE 134 INTEGER NN REAL XVPOTI COMPLEX CRV PARAMETERNNI DIMENSION CRVNNNN CALL VAVPOTIX CRVllVPOTl RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O SUBROUTINE VAVX Potential Energy Surface Harmonic Oscillator implicit none REAL VXrmassxkpkrkomega common packet rmassxkpk omegalO rkrmassomega2 VO5rkxx RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine SetKinPropdttprop Kinetic Energy part of the Trotter Expansion eXp i p 2 dt2 m IMPLICIT NONE INTEGER nptxkxnxnpts REAL xscXminXmaXpropfacxrmassxkpialenxdtpk COMPLEX tpropeye parameternpts9nptx2npts DIMENSION tpropnth common Xy xminxmax common packet rmassxkpk eyeOl pi acos lO alenxxmax xmin propfacx dt2rmass2pi2 do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if xscO ifnXneO xscrealnXalenx tpropkxexpeyepropfacxxsc2 135 end do return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc subroutine SetPotPropdtvprop Potential Energy part of the Trotter Expansion exp i V dt2 O IMPLICIT NONE INTEGER NNiinthnpts REAL xminxmaxdxdtXVPOI COMPLEX vpropeye parameternpts9nptx2nptsNNl DIMENSION vpropnthNNNN common Xy xminxmax eyeOl dxXmax Xminrealnptx do iilnth xxminiidx CALL VAVPOTX vpropiillexp eye05dtVPOIsgrtnptxl0 END DO RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUIINE energiesenergy IMPLICIT NONE INTEGER jNN COMPLEX energyRVRKE PARAMETER NNl DIMENSION RVNNRKENNenergyNN CALL PERV CALL KERKE DO jlNN energYjRVjRKEj END DO RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc FUNCTION Psiaxistepdt c c Analytic wave packet ltxPsiaistepgt obtained by applying the c harmonic propagator to the initial state c ltX lPsiOgt alphapi25exp alpha2X Xk2eyepkX Xk c where the propagator is c ltXeXp beta HX gt A eXp rgammaX2X 2rgammapxx with c A sgrtmomegapiexpbetaomega eXp betaomega beta it c rgamma O5momegacoshbetaomegasinhbetaomega and c rgammap momegasinhbetaomega 136 IMPLICIT NONE INTEGER istep REAL pkrmassxkdtxtomegapialpha COMPLEX eyePsiabetaArgammargammapcOclc2 common packet rmassxkpk eye00lO omegalO alpha omegarmass piacos l0 beta eyedtistep IFabsbetaE00 beta eyelOE 7 A sqrtrmassomegapiexpbetaomega exp betaomega rgammaO5rmassomegaexpbetaomegaexp betaomega l expbetaomega eXp betaomega rgammap2rmassomegaexpbetaomega exp betaomega cO eyepkXk alpha2Xk2 clrgammapxalphaxkeyepk c2rgammaalpha2 c Psia Aalphapi25sqrtpic2 l eXprgammax2expc0c124Oc2 c return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE SAVEWFje2ndumpdt O Dump Time Evolved Wave packet IMPLICIT NONE INTEGER je2nptxnptskkNNncountndumpjj COMPLEX chiCRVenergypsiPsia character9 B REAL VXlclc2claxxminxmaxdXEVALUESdt PARAMETERnpts9nptx2nptsNNl DIMENSION CRVNNNNenergyNNEVALUESNN DIMENSION psiNNNN common Xy xminxmax COMMON wfunc chinthNN CALL energiesenergy jjje2ndump writeB Ai44 arch jj OPEN1FILEB dxxmax Xminrealnptx ncountje2 lndump Save Wave packet components 0 do kklnptx 137 O 33 l l l xxminkkdx clchikklconjgchikkl claPsiaXje2dtconjgPsiaXje2dt writel33 Xsqrtclrealenergyl sqrtclarealenergyl end do writel33 do kklnptx xxminkkdx writel33 X realchikklrealenergyl realPsiaXje2dtrealenergyl end do writel33 Save Adiabatic states do kklnptx xxminkkdx CALL HAMILCRVX writel33 XCRV11 end do CLOSE1 format6e1362x RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O SUBROUTTNE PERV Expectation Value of the Potential Enegy IMPLTCIT NONE INTEGER nptxnptskkNNj COMPLEX chiEYERV REAL Vpotomegaxminxmaxdxpirmassxkpkxalpha PARAMETERnpts9nptx2nptsNNl DIMENSION RVNN COMMON wfunc chinthNN common Xy common packetrmassxkpk xminxmax dxXmax Xminrealnptx DO jlNN RVj00 do kklnptx xxminkkdx IFjEQl CALL VAVpotX RVjRVjchikkjVpotconjgchikkjdx end do END DO RETURN 138 END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine KERKE Expectation value of the kinetic energy IMPLICIT NONE INTEGER NNkknthkXnXnptsj REAL dpthetawmpxminxmaxrmassxkpialenxpkrmreridx COMPLEX eyechiPsipchicRKE parameternpts9nptx2nptsNN1 DIMENSION chicnthRKENN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN pi acos 10 dxXmax Xminnptx dp2piXmax Xmin DO j1NN RKEj00 do kk1nptx chickkchikkj end do CALL fournchicnptx1 1 do kx1nptx ifkxlenptx21 then nxkx 1 else nxkx 1 nptx end if pO ifnxneO p realnxdp chickxp220rmasschickxnptx end do CALL fournchicnptx11 do kk1nptx RKEjRKEjconjgchikkjchickkdx end do END DO return end CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O O O O SUBROUTTNE PROPAGATEvproptprop Split Operator Fourier Transform Propagation Method J Comput Phys 47 412 1982 J Chem Phys 78 301 1983 IMPLICIT NONE INTEGER ijNNiinthnpts 139 COMPLEX chivpropchinlchin2tprop PARAMETERnpts9nptx2nptsNNl DIMENSION chinlnptxchin2nptx DIMENSION tpropnptxvpropnptxNNNN COMMON wfunc chinptxNN c c Apply potential energy part of the Trotter Expansion c DO ilnptx chinli00 DO jlNN chinlichinlivpropiljchiij END DO END DO c c Fourier Transform wave packet to the momentum representation c CALL fournchinlnptxl l c c Apply kinetic energy part of the Trotter Expansion c DO ilnptx chinlitpropichinli END DO c c Inverse Fourier Transform wave packet to the coordinate representation c CALL fournchinlnptxll c c Apply potential energy part of the Trotter Expansion c DO ilnptx DO jlNN chiijvpropijlchinli END DO END DO END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc c Subroutine for FFT from Numerical Recipes CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTINE FOURNDATANNNDIMISIGN REAL8 WRWIWPRWPIWTEMPTHETA DIMENSION NNNDIMDATA NTOTl DO 11 IDIMlNDIM NTOTNTOTNNIDIM II CONTINUE NPREVl DO 18 IDIMlNDIM NNNIDIM NREMNTOTNNPREV 140 IP12NPREV IP2IP1N IP3IP2NREM I2REV1 DO 14 IZ1IP2IP1 IEI2LTI2REVTHEN DO 13 IlIZIZIPl 22 DO 12 I3I1IP3IP2 I3REVI2REVI3 I2 TEMPRDATAI3 TEMPIDATAltI31gt DATAltI3gtDATAltI3REVgt DATAltI31gtDATAltI3REV1gt DATAltI3REVgtTEMPR DATAltI3REV1gtTEMPI CONTINUE CONTINUE ENDIE IBITIP22 IE IBIT GE IPIANDI2REVGTIBIT THEN I2REVI2REv IBIT IBITIBIT2 GO TO I ENDIE I2REVI2REVIBIT CONTINUE IFP1IP1 IFIFP1LTIP2THEN IFP22IFP1 THETAISIGN628318530717959DOIFPZIPl WPR 2DODSINO5DOTHETA2 WPIDSINTHETA WR1DO WIODO DO 17 I31IEPIIPI DO 16 I1I3I3IPI 22 DO 15 I2I1IP3IEP2 K1I2 K2KlIFPl TEMPRSNGLWRDATAK2 SNGLWIDATAK21 TEMPISNGLWRDATAK21SNGLWIDATAK2 DATAK2DATAKl TEMPR DATAK21DATAK11 TEMPI DATAK1DATAK1TEMPR DATAK11DATAK11TEMPI CONTINUE CONTINUE WTEMPWR WRWRWPR WIWPIWR WIWIWPRWTEMPWPIWI CONTINUE 141 IFP1IFP2 GO TO 2 ENDIF NPREVNNPREV l8 CONTINUE RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc 142 427 Problem 7 Computational Problem 7 Loop the computer program developed in Problem 5 with 30 25 and p0 0 for 100 steps with 739 01 au For each step compute the expectation values of coordinates 2t and momenta pt as done in Problems 3 and 4 respectively Compare your calculations with the analytic solutions obtained by substituting Eq 23 into Eq 22 Verify that these correspond to the classical trajectories 1t E 30 Ecoswt and pt 2 p0 30 Exam sinwt which can be computed according to the Velocity Verlet algorithm Pj1 27 FWD Fj1T2 189 17j1 Ziljj In order to visualize the output of this program cut the source code attached below compile it by typing f77 Problem7f o Problem7 run it by typing Problem7 Visualize the output of time dependent expectation values as compared to classical trajectories as follows type gnuplot thentype set dat sty line thentype plot trajOOOO That will show the numerical computation of the expectation value lt 1115 Wilt gt as a function of time In order to visualize the classical result on top of the quantum mechanical expectation value type replot trajOOOO u 14 In order to visualize the output of lt Ilt I 15 1115 gt as a function of time type plot trajOOOO u 13 and to visualize the classical result on top of the quantum mechanical expectation value type replot trajOOOO u 15 The plot of lt M 13th gt vs lt Mimi gt can be obtained by typing 143 plot trajOOOO u 32 and the corresponding Classical results pt vs 1t plot trajOOOO u 54 To eXit type quit The snapshots of the time dependent wave packet can be Visualized as a movie by typing gnuplotltpp7 Where the le named pp7 has the following lines Download from httpXbeamsChemyaleeduNbatistaP7pp7 set yrange06 set xrange 1010 set dat sty 1 plot quotarchOOOlquot u 12 1w 3 pause 1 plot quotarchOOO2quot u 12 1w 3 pause 1 plot quotarchOOO3quot u 12 1w 3 pause 1 plot quotarchOOO4quot u 12 1w 3 pause 1 plot quotarch0005quot u 12 1w 3 pause 1 plot quotarchOOO6quot u 12 1w 3 pause 1 plot quotarchOOO7quot u 12 1w 3 pause 1 plot quotarchOOO8quot u 12 1w 3 pause 1 plot quotarchOOO9quot u 12 1w 3 pause 1 plot quotarchOOlOquot u 12 1w 3 pause 1 plot quotarch0011quot u 12 1w 3 pause 1 plot quotarch0012quot u 12 1w 3 pause 1 plot quotarch0013quot u 12 1w 3 pause 1 plot quotarch0014quot u 12 1w 3 144 pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l 0015quot 0016quot 0017quot 0018quot 0019quot 0020quot 0021quot 0022quot 0023quot 0024quot 0025quot 0026quot 0027quot 0028quot 0029quot 0030quot 0031quot 0032quot 0033quot 0034quot 0035quot 0036quot 0037quot 0038quot 0039quot lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw 145 plot quotarchOO40quot pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch 0041quot 0042quot 0043quot 0044quot 0045quot 0046quot 0047quot 0048quot 0049quot 0050quot 0051quot 0052quot 0053quot 0054quot 0055quot 0056quot 0057quot 0058quot 0059quot 0060quot 0061quot 0062quot 0063quot 0064quot 0065quot u 12 1w 3 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 12 1w 146 pause 1 plot quotarchOO66quot u 12 1w 3 pause 1 plot quotarchOO67quot u 12 1w 3 pause 1 plot quotarchOO68quot u 12 1w 3 pause 1 plot quotarchOO69quot u 12 1w 3 pause 1 plot quotarchOO70quot u 12 1w 3 pause 1 plot quotarchOO71quot u 12 1w 3 pause 1 plot quotarchOO72quot u 12 1w 3 pause 1 plot quotarchOO73quot u 12 1w 3 pause 1 plot quotarchOO74quot u 12 1w 3 pause 1 plot quotarchOO75quot u 12 1w 3 pause 1 plot quotarchOO76quot u 12 1w 3 pause 1 plot quotarchOO77quot u 12 1w 3 pause 1 plot quotarchOO78quot u 12 1w 3 pause 1 plot quotarchOO79quot u 12 1w 3 pause 1 plot quotarchOO80quot u 12 1w 3 pause 1 plot quotarchOO81quot u 12 1w 3 pause 1 plot quotarchOO82quot u 12 1w 3 pause 1 plot quotarchOO83quot u 12 1w 3 pause 1 plot quotarchOO84quot u 12 1w 3 pause 1 plot quotarchOO85quot u 12 1w 3 pause 1 plot quotarchOO86quot u 12 1w 3 pause 1 plot quotarchOO87quot u 12 1w 3 pause 1 plot quotarchOO88quot u 12 1w 3 pause 1 plot quotarchOO89quot u 12 1w 3 pause 1 plot quotarchOO90quot u 12 1w 3 pause 1 147 plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l 0091quot 0092quot 0093quot 0094quot 0095quot 0096quot 0097quot 0098quot 0099quot lw lw lw lw lw lw lw lw lw 148 Download from httpXbeamsChemyaleeduNbatistaP7Problem7f PROGRAM Problem7 l D wave packet propagation and Velocity Verlet propagation on a Harmonic potential energy surface 0000 IMPLICIT NONE INTEGER NNnptsnthndump INTEGER istepnstepjj REAL dtXcpc COMPLEX vproptpropxmeanpmean character9 Bfile PARAMETERnpts9nptx2nptsNNl DIMENSION vpropnthNNNNtpropnth DIMENSION XmeanNNpmeanNN COMMON class xcpc jj0 writeBfile Ai44 traj jj OPENlOFILEBfile CALL ReadParamnstepndumpdt call Initialize CALL SetKinPropdttprop CALL SetPotPropdtvprop DO isteplnstepl IFmodistep llOE00 l PRINT quotStepquot istep lquot Final stepquot nstep IFistepGEl CALL PROPAGATEvproptprop IFmodistep lndumpE00 THEN CALL SAVEWFistepndumpdt CALL XMXmean CALL PMpmean CALL VVdt WRITEIO22 istep ldt l realxmeanlrealpmeanlXcpc END IF END DO CLOSElO 22 FORMAT6e1362X END ccocococccocococccocococccocococccocococccocooccooocoocoooocoocoococ subroutine ReadParamnstepndumpdt Parameters defining the grid xmin xmax integration time step dt rmass rmass initial position Xk initial momentum pk number of propagation steps nstep and how often to save a pic ndump 00000 IMPLICIT NONE INTEGER ntypenstepnrptireportndumpnlit REAL xminXmaXpkrmassXkdt 149 common packet rmassxkpk common Xy xminxmax xmin l00 xmax 100 dtOl rmasslO Xk 25 pk00 nsteplOO ndumpl return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE VVdt c c Velocity Verlet Algorithm J Chem Phys 76 637 1982 c IMPLICIT NONE REAL vdxdtxcpcrmassxkpkaccxtVPOTlVPOT2F COMMON class xcpc common packet rmassxkpk c c Compute Force c dx00l xtxcdx CALL VAVPOT1Xt xtxc dx CALL VAVPOT2Xt F VPOTl VPOT220dX vpcrmass c c Advance momenta half a step c pcpcO5Fdt c c Advance coordinates a step c xcxcvdt05dt2Frmass c c Compute Force c dx00l xtxcdx CALL VAVPOT1Xt xtxc dx CALL VAVPOT2Xt F VPOTl VPOT220dX c 150 c Advance momenta half a step c pcpcO5Fdt c return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE Initialize IMPLICIT NONE INTEGER NNnthnptskk COMPLEX chiOchiEYECRV REAL xcpcomegaxk2xminxmaxdxpirmassxkpkxalphaalpha2 PARAMETERnpts9nptx2nptsNNl DIMENSION CRVNNNN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN COMMON iwfunc chiOnthNN COMMON class xcpc EYE0010 pi acos lO omegal dxXmax XminreaInth xcxk pcpk c c Wave Packet Initialization Gaussian centered at Xk with momentum pk c alpharmassomega do kklnptx xxminkkdx chikklalphapi025 l exp alpha2XXk2EYEpkX Xk chiOkklchikkl end do RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE HAMILCRVX c c Hamiltonian Matrix c IMPLICIT NONE INTEGER NN REAL XVPOTI COMPLEX CRV PARAMETERNNI DIMENSION CRVNNNN c 151 CALL VAVPOTlX CRVllVPOTl RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O SUBROUTINE VAVX Potential Energy Surface Harmonic Oscillator implicit none REAL VXrmassxkpkrkomega common packet rmassxkpk omegalO rkrmassomega2 VO5rkxx RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine SetKinPropdttprop Kinetic Energy part of the Trotter Expansion eXp i p 2 dt2 m IMPLICIT NONE INTEGER nptxkxnxnpts REAL xscXminXmaXpropfacxrmassxkpialenxdtpk COMPLEX tpropeye parameternpts9nptx2npts DIMENSION tpropnth common Xy xminxmax common packet rmassxkpk eyeOl pi acos lO alenxxmax xmin propfacx dt2rmass2pi2 do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if xscO ifnXneO xscrealnXalenx tpropkxexpeyepropfacxxsc2 end do return end CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC subroutine SetPotPropdtvprop 152 0 Potential Energy part of the Trotter Expansion exp i V dt2 IMPLICIT NONE INTEGER NNiinthnpts REAL xminxmaxdxdtxVPOT COMPLEX vpropeye parameternpts9nptx2nptsNNl DIMENSION vpropnthNNNN common Xy xminxmax eyeOl dxxmax Xminrealnptx do iilnth xxminiidx CALL VAVPOTX vpropiillexp eye05dtVPOTsgrtnptxl0 END DO RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE energiesenergy IMPLICIT NONE INTEGER jNN COMPLEX energyRVRKE PARAMETER NNl DIMENSION RVNNRKENNenergyNN CALL PERV CALL KERKE DO jlNN energYjRVjRKEj END DO RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc FUNCTION Psiaxistepdt c c Analytic wave packet ltxPsiaistepgt obtained by applying the c harmonic propagator to the initial state c ltx lPsiOgt alphapi25exp alpha2x Xk2eyepkx xk c where the propagator is c ltXeXp beta HX gt A eXp rgammaX2X 2rgammapxx with c A sgrtmomegapiexpbetaomega eXp betaomega beta it c rgamma O5momegacoshbetaomegasinhbetaomega and c rgammap momegasinhbetaomega c IMPLICIT NONE INTEGER istep REAL pkrmassxkdtxtomegapialpha COMPLEX eyePsiabetaArgammargammapcOclc2 common packet rmassxkpk 153 l l eye00lO omegalO alpha omegarmass piacos l0 beta eyedtistep IFabsbetaEQO beta eyelOE 7 A sqrtrmassomegapiexpbetaomega exp betaomega rgammaO5rmassomegaexpbetaomegaexp betaomega expbetaomega eXp betaomega rgammap2rmassomegaexpbetaomega exp betaomega cO eyepkXk alpha2Xk2 clrgammapxalphaxkeyepk c2rgammaalpha2 Psia Aalphapi25sqrtpic2 eXp rgammax2expc0cl240c2 return end CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O O l SUBROUTINE SAVEWFje2ndumpdt Dump Time Evolved Wave packet IMPLICIT NONE INTEGER je2nptxnptskkNNncountndumpjj COMPLEX chiCRVenergypsiPsia character9 B REAL VXlclc2claxxminxmaXdXEVALUESdt PARAMETERnpts9nptx2nptsNNl DIMENSION CRVNNNNenergyNNEVALUESNN DIMENSION psiNNNN common Xy xminxmax COMMON wfunc chinthNN CALL energiesenergy jjje2ndump writeB Ai44 OPEN1FILEB dxXmax Xminrealnptx ncountje2 lndump arch jj Save Wave packet components do kklnptx xxminkkdx clchikklconjgchikkl claPsiaXje2dtconjgPsiaXje2dt writel33 Xsqrtclrealenergyl sqrtclarealenergyl end do 154 O 33 writel33 do kklnptx xxminkkdx writel33 X realchikklrealenergyl realPsiaXje2dtrealenergyl end do writel33 Save Adiabatic states do kklnptx xxminkkdx CALL HAMILCRVX writel33 XCRV11 end do CLOSE1 format6e1362x RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTTNE XMRV Expectation Value of the Position IMPLTCIT NONE INTEGER nptxnptskkNNj COMPLEX ChiEYERV REAL Vpotomegaxminxmaxdxpirmassxkpkxalpha PARAMETERnpts9nptx2nptsNNl DIMENSION RVNN COMMON wfunc chinthNN common Xy xminxmax common packetrmassxkpk dxXmax Xminrealnptx DO jlNN RVj00 do kklnptx xxminkkdx IFjEQl CALL VAVpotX RVjRVjchikkjxconjgchikkjdx end do END DO RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTTNE PERV Expectation Value of the Potential Enegy 155 IMPLICIT NONE INTEGER nptxnptskkNNj COMPLEX chiEYERV REAL Vpotomegaxminxmaxdxpirmassxkpkxalpha PARAMETERnpts9nptx2nptsNNl DIMENSION RVNN COMMON wfunc chinthNN common Xy xminxmax common packetrmassxkpk dxXmax Xminrealnptx DO jlNN RVj00 do kklnptx xxminkkdx IFjEQl CALL VAVpotX RVjRVjchikkjVpotconjgchikkjdx end do END DO RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine KERKE Expectation value of the kinetic energy IMPLICIT NONE INTEGER NNkknthkXnXnptsj REAL dpthetawmpxminxmaxrmassxkpialenxpkrmreridx COMPLEX eyechiPsipchicRKE parameternpts9nptx2nptsNNl DIMENSION chicnthRKENN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN pi acos lO dxXmax Xminnptx dp2piXmax Xmin DO jlNN RKEj00 do kklnptx chickkchikkj end do CALL fournchicnptxl l do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx 156 end if pO ifnxneO p realnxdp chickxp220rmasschickxnptx end do CALL fournchicnptxll do kklnptx RKEjRKEjconjgchikkjchickkdx end do END DO return end CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine PMRKE Expectation value of the kinetic energy IMPLICIT NONE INTEGER NNkknthkXnXnptsj REAL dpthetawmpxminxmaxrmassxkpialenxpkrmreridx COMPLEX eyechiPsipchicRKE parameternpts9nptx2nptsNNl DIMENSION chicnthRKENN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN pi acos lO dxXmax Xminnptx dp2piXmax Xmin DO jlNN RKEj00 do kklnptx chickkchikkj end do CALL fournchicnptxl l do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if pO ifnxneO p realnxdp chickxpchickXnptx end do CALL fournchicnptxll do kklnptx RKEjRKEjconjgchikkjchickkdx end do 157 END DO return end CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O O O O O O O O O SUBROUTINE PROPAGATEvproptprop Split Operator Fourier Transform Propagation Method J Comput Phys 47 412 1982 J Chem Phys 78 301 1983 IMPLICIT NONE INTEGER ijNNiinptxnpts COMPLEX chivpropchin1chin2tprop PARAMETERnpts9nptx2nptsNN1 DIMENSION chin1nptxchin2nptx DIMENSION tpropnptxvpropnptxNNNN COMMON wfunc chinptxNN Apply potential energy part of the Trotter Expansion DO i1nptx chin1i00 DO j1NN chin1ichin1ivpropi1jchiij END DO END DO Fourier Transform wave packet to the momentum representation CALL fournchin1nptx1 1 Apply kinetic energy part of the Trotter Expansion DO i1nptx chin1itpropichin1i END DO Inverse Fourier Transform wave packet to the coordinate representation CALL fournchin1nptx11 Apply potential energy part of the Trotter Expansion DO i1nptx DO j1NN chiijvpropij1chin1i END DO END DO END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC C Subroutine for FFT from Numerical Recipes CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC 158 II 12 13 I4 SUBROUTINE FOURNDATANNNDIMISIGN REAL8 WRWIWPRWPIWTEMPTHETA DIMENSION NNNDIMDATA NTOTI DO 11 IDIM1NDIM NTOTNTOTNNIDIM CONTINUE NPREV1 DO 18 IDIM1NDIM NNNIDIM NREMNTOTNNPREV IP12NPREV IP2IP1N IP3IP2NREM I2REV1 DO 14 IZ1IP2IP1 IEI2LTI2REVTEEN DO 13 IlIZIZIPl 22 DO 12 I3I1IP3IP2 I3REVI2REVI3 I2 TEMPRDATAI3 TEMPIDATAltI31gt DATAltI3gtDATAltI3REVgt DATAltI31gtDATAltI3REV1gt DATAltI3REVgtTEMPR DATA13REV1TEMPI CONTINUE CONTINUE ENDIE IBITIP22 IE IBIT GE IPIANDI2REVGTIBIT I2REVI2REv IBIT IBITIBIT2 GO TO 1 ENDIE I2REVI2REVIBIT THEN CONTINUE IFPIIPI IFIFPILTIP2THEN IFP22IFP1 THETAISIGN628318530717959DOIFPZIPI WPR 2DODSINO5DOTHETA2 WPIDSINTHETA WR1DO WIODO DO 17 I31IEPIIPI DO 16 I1I3I3IPI 22 DO 15 I2I1IP3IEP2 K1I2 K2KlIFPl TEMPRSNGLWRDATAK2 SNGLWIDATAK21 159 TEMPISNGLWRDATAK21SNGLWIDATAK2 DATAK2DATAKl TEMPR DATAK21DATAK11 TEMPI DATAK1DATAK1TEMPR DATAK11DATAK11TEMPI 15 CONTINUE l6 CONTINUE WTEMPWR WRWRWPR WTWPTWR WTWTWPRWTEMPWPTWT l7 CONTINUE TFP1TFP2 GO TO 2 ENDTF NPREVNNPREV l8 CONTINUE RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc 160 428 Problem 8 Computational Problem 8 Change the potential to that of a Morse oscillator Vf De1 exp af 516 2 with 526 2 0 De 2 8 and a k2De where k mwQ Recompute the wave packet propagation with 510 2 05 and p0 0 for 100 steps with 739 01 au and compare the expectation values 1t and pt with the corresponding classical trajectories obtained by recursively applying the Velocity Verlet algorithm The output of this program is analogous to Problem 6 but for a Morse potential Cut the source a emmdwdmmmaww na bnmm PKmbm3 cmmme bqumg f77 Problem8f o Problem8 run it by typing Problem8 Visualize the output of the time dependent expectation values as compared to classical trajectories as follows type gnuplot thentype set dat sty line thentype plot trajOOOO That will show the numerical computation of the expectation value lt 1115 Wilt gt as a function of time In order to visualize the classical result on top of the quantum mechanical expectation value type replot trajOOOO u 14 In order to visualize the output of lt Ilt 15 1115 gt as a function of time type plot trajOOOO u 13 and to visualize the classical result on top of the quantum mechanical expectation value type replot trajOOOO u 15 The plot of lt M 13th gt vs lt Mimi gt can be obtained by typing plot trajOOOO u 32 and the corresponding classical results pt vs 1t plot trajOOOO u 54 161 To eXit type quit The snapshots of the time dependent wave packet can be Visualized as a movie by typing gnuplotltpp8 Where the le named pp8 has the following lines Download from httpXbeamsChemyaleeduNbatistaPSpp8 set yrange09 set xrange 525 set dat sty 1 plot quotarchOOOlquot u 12 1w 3 pause 1 plot quotarchOOO2quot u 12 1w 3 pause 1 plot quotarchOOO3quot u 12 1w 3 pause 1 plot quotarchOOO4quot u 12 1w 3 pause 1 plot quotarch0005quot u 12 1w 3 pause 1 plot quotarchOOO6quot u 12 1w 3 pause 1 plot quotarchOOO7quot u 12 1w 3 pause 1 plot quotarchOOO8quot u 12 1w 3 pause 1 plot quotarchOOO9quot u 12 1w 3 pause 1 plot quotarchOOlOquot u 12 1w 3 pause 1 plot quotarch0011quot u 12 1w 3 pause 1 plot quotarch0012quot u 12 1w 3 pause 1 plot quotarch0013quot u 12 1w 3 pause 1 plot quotarch0014quot u 12 1w 3 pause 1 plot quotarch0015quot u 12 1w 3 pause 1 plot quotarch0016quot u 12 1w 3 pause 1 plot quotarch0017quot u 12 1w 3 162 pause 1 plot quotarch0018quot pause 1 plot quotarch0019quot pause 1 plot quotarchOO20quot pause 1 plot quotarchOO21quot pause 1 plot quotarchOO22quot pause 1 plot quotarchOO23quot pause 1 plot quotarchOO24quot pause 1 plot quotarchOO25quot pause 1 plot quotarchOO26quot pause 1 plot quotarchOO27quot pause 1 plot quotarchOO28quot pause 1 plot quotarchOO29quot pause 1 plot quotarchOO30quot pause 1 plot quotarchOO31quot pause 1 plot quotarchOO32quot pause 1 plot quotarchOO33quot pause 1 plot quotarchOO34quot pause 1 plot quotarchOO35quot pause 1 plot quotarchOO36quot pause 1 plot quotarchOO37quot pause 1 plot quotarchOO38quot pause 1 plot quotarchOO39quot pause 1 plot quotarchOO40quot pause 1 plot quotarchOO41quot pause 1 plot quotarchOO42quot pause 1 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 163 plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch 0043quot 0044quot 0045quot 0046quot 0047quot 0048quot 0049quot 0050quot 0051quot 0052quot 0053quot 0054quot 0055quot 0056quot 0057quot 0058quot 0059quot 0060quot 0061quot 0062quot 0063quot 0064quot 0065quot 0066quot 0067quot 0068quot lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw 164 pause 1 plot quotarchOO69quot pause 1 plot quotarchOO70quot pause 1 plot quotarchOO71quot pause 1 plot quotarchOO72quot pause 1 plot quotarchOO73quot pause 1 plot quotarchOO74quot pause 1 plot quotarchOO75quot pause 1 plot quotarchOO76quot pause 1 plot quotarchOO77quot pause 1 plot quotarchOO78quot pause 1 plot quotarchOO79quot pause 1 plot quotarchOO80quot pause 1 plot quotarchOO81quot pause 1 plot quotarchOO82quot pause 1 plot quotarchOO83quot pause 1 plot quotarchOO84quot pause 1 plot quotarchOO85quot pause 1 plot quotarchOO86quot pause 1 plot quotarchOO87quot pause 1 plot quotarchOO88quot pause 1 plot quotarchOO89quot pause 1 plot quotarchOO90quot pause 1 plot quotarchOO91quot pause 1 plot quotarchOO92quot pause 1 plot quotarchOO93quot pause 1 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 165 plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l 0094quot 0095quot 0096quot 0097quot 0098quot 0099quot lw lw lw lw lw lw 166 Download from httpXbeamsChemyaleeduNbatistaP8Problem8f PROGRAM Problem8 l D wave packet propagation and Velocity Verlet propagation on a Morse potential energy surface 0000 IMPLICIT NONE INTEGER NNnptsnthndump INTEGER istepnstepjj REAL dtXcpc COMPLEX vproptpropxmeanpmean character9 Bfile PARAMETERnptslOnptx2nptsNNl DIMENSION vpropnthNNNNtpropnth DIMENSION XmeanNNpmeanNN COMMON class xcpc c xo jj0 writeBfile Ai44 traj jj OPENlOFILEBfile CALL ReadParamnstepndumpdt call Initialize CALL SetKinPropdttprop CALL SetPotPropdtvprop DO isteplnstepl IFmodistep llOEQO l PRINT quotStepquot istep lquot Final stepquot nstep IFistepGEl CALL PROPAGATEvproptprop IFmodistep lndumpEQO THEN CALL SAVEWFistepndumpdt CALL XMXmean CALL PMpmean CALL VVdt WRITEIO22 istep ldt l realxmeanlrealpmeanlXcpc END IF END DO CLOSEIO 22 FORMAT6el362X END cococoocococoococoooococococcccoooooooocococooccoococcccoooocooococc subroutine ReadParamnstepndumpdt Parameters defining the grid xmin xmax integration time step dt rmass rmass initial position Xk initial momentum pk number of propagation steps nstep and how often to save a pic ndump 00000 IMPLICIT NONE INTEGER ntypenstepnrptireportndumpnlit REAL xminXmaXpkrmassXkdt 167 common packet rmassxkpk common Xy xminxmax xmin 50 xmax250 dtO2 rmasslO Xk 5 pk00 nsteplOO ndumpl return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE VVdt c c Velocity Verlet Algorithm J Chem Phys 76 637 1982 c IMPLICIT NONE REAL vdxdtxcpcrmassxkpkaccxtVPOTlVPOT2F COMMON class xcpc common packet rmassxkpk c c Compute Force c dx00l xtxcdx CALL VAVPOT1Xt xtxc dx CALL VAVPOT2Xt F VPOTl VPOT220dX vpcrmass c c Advance momenta half a step c pcpcO5Fdt c c Advance coordinates a step c xcxcvdt05dt2Frmass c c Compute Force c dx00l xtxcdx CALL VAVPOT1Xt xtxc dx CALL VAVPOT2Xt F VPOTl VPOT220dX c 168 c Advance momenta half a step c pcpcO5Fdt c return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE Initialize IMPLICIT NONE INTEGER NNnthnptskk COMPLEX chiOchiEYECRV REAL xcpcomegaxk2xminxmaxdxpirmassxkpkxalphaalpha2 PARAMETERnptslOnptx2nptsNNl DIMENSION CRVNNNN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN COMMON iwfunc chiOnthNN COMMON class xcpc EYE0010 pi acos lO omegal dxXmax XminreaInth xcxk pcpk c c Wave Packet Initialization Gaussian centered at Xk with momentum pk c alpharmassomega do kklnptx xxminkkdx chikklalphapi025 l exp alpha2XXk2EYEpkX Xk chiOkklchikkl end do RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE HAMILCRVX c c Hamiltonian Matrix c IMPLICIT NONE INTEGER NN REAL XVPOTI COMPLEX CRV PARAMETERNNI DIMENSION CRVNNNN c 169 CALL VAVPOTlX CRVllVPOTl RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O SUBROUTINE VAVX Potential Energy Surface Morse Potential Phys Rev 1929 3457 implicit none REAL VXrmassxkpkrkomegaDexeqa common packet rmassxkpk xeq00 omegalO De80 rkrmassomega2 asqrtrk20De VDelO exp ax xeq2 RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine SetKinPropdttprop Kinetic Energy part of the Trotter Expansion eXp i p 2 dt2 m IMPLICIT NONE INTEGER nptxkxnxnpts REAL xscXminXmaXpropfacxrmassxkpialenxdtpk COMPLEX tpropeye parameternptslOnptx2npts DIMENSION tpropnth common Xy xminxmax common packet rmassxkpk eyeOl pi acos lO alenxxmax xmin propfacx dt2rmass2pi2 do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if xscO ifnXneO xscrealnXalenx tpropkxexpeyepropfacxxsc2 end do return 170 end coocooooooococococococococococococococococococoooooooooooooooooooooo subroutine SetPotPropdtvprop c Potential Energy part of the Trotter Expansion exp i V dt2 IMPLICIT NONE INTEGER NNiinthnptS REAL xminxmaxdxdtxVPOT COMPLEX vpropeye parameternptslOnptx2nptsNNl DIMENSION vpropnthNNNN common Xy xminxmax eyeOl dxxmax XminreaInth do iilnth xxminiidx CALL VAVPOTX vpropiillexp eye05dtVPOTsqrtnptxl0 END DO RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE energiesenergy IMPLICIT NONE INTEGER jNN COMPLEX energyRVRKE PARAMETER NNI DIMENSION RVNNRKENNenergyNN CALL PERV CALL KERKE DO jlNN energYjRVjRKEj END DO RETURN END ccccccccccccccccccccccccccccccococccccococccccoocccccccccccccccccccccc SUBROUTINE SAVEWFje2ndumpdt c Dump Time Evolved Wave packet IMPLICIT NONE INTEGER je2nptxnptskkNNncountndumpjj COMPLEX chiCRVenergypsiPsia character9 B REAL VXlclc2claxxminxmaxdXEVALUESdt PARAMETERnptslOnptx2nptsNNl DIMENSION CRVNNNNEVALUESNN DIMENSION psiNNNN common Xy xminxmax 171 O O CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC 33 COMMON wfunc chinthNN COMMON ENER energyNN IFje2EQl jjje2ndump writeB Ai44 OPEN1FILEB dxXmax Xminrealnptx ncountje2 lndump CALL energiesenergy arch jj Save Wave packet components do kklnptx xxminkkdx clchikklconjgchikkl writel33 Xsqrtclrealenergyl end do writel33 do kklnptx xxminkkdx writel33 end do writel33 Xrealenergyl Save Adiabatic states do kklnptx xxminkkdx CALL HAMILCRVX writel33 XCRV11 end do CLOSE1 format6e1362x RETURN END SUBROUTINE XMRV Expectation Value of the Position IMPLICIT NONE INTEGER nptxnptskkNNj COMPLEX chiEYERV REAL Vpotomegaxminxmaxdxpirmassxkpkxalpha PARAMETERnptslOnptx2nptsNNl DIMENSION RVNN COMMON wfunc chinthNN common Xy xminxmax common packetrmassxkpk dxXmax Xminrealnptx 172 DO jlNN RVj00 do kklnptx xxminkkdx IFjEQl CALL VAVpotX RVjRVjchikkjxconjgchikkjdx end do END DO RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O SUBROUTINE PERV Expectation Value of the Potential Enegy IMPLICIT NONE INTEGER nptxnptskkNNj COMPLEX chiEYERV REAL Vpotomegaxminxmaxdxpirmassxkpkxalpha PARAMETERnptslOnptx2nptsNNl DIMENSION RVNN COMMON wfunc chinthNN common Xy xminxmax common packetrmassxkpk dxXmax Xminrealnptx DO jlNN RVj00 do kklnptx xxminkkdx IFjEQl CALL VAVpotX RVjRVjchikkjVpotconjgchikkjdx end do END DO RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O subroutine KERKE Expectation value of the kinetic energy IMPLICIT NONE INTEGER NNkknthkXnXnptsj REAL dpthetawmpxminxmaxrmassxkpialenxpkrmreridx COMPLEX eyechiPsipchicRKE parameternptslOnptx2nptsNNl DIMENSION chicnptxRKENN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN 173 pi acos lO dxXmax Xminnptx dp2piXmax Xmin DO jlNN RKEj00 do kklnptx chickkchikkj end do CALL fournchicnptxl l do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if pO ifnxneO p realnxdp chickxp220rmasschickxnptx end do CALL fournchicnptxll do kklnptx RKEjRKEjconjgchikkjchickkdx end do END DO return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc subroutine PMRKE O Expectation value of the kinetic energy IMPLICIT NONE INTEGER NNkknthkXnXnptsj REAL dpthetawmpxminxmaxrmassxkpialenxpkrmreridx COMPLEX eyechiPsipchicRKE parameternptslOnptx2nptsNNl DIMENSION chicnthRKENN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN pi acos lO dxXmax Xminnptx dp2piXmax Xmin DO jlNN RKEj00 do kklnptx chickkchikkj end do 174 CALL fournchicnptx1 1 do kx1nptx ifkxlenptx21 then nxkx 1 else nxkx 1 nptx end if pO ifnxneO p realnxdp chickxpchickXnptx end do CALL fournchicnptx11 do kk1nptx RKEjRKEjconjgchikkjchickkdx end do END DO return end CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC 0000 O O O SUBROUTINE PROPAGATEvproptprop Split Operator Fourier Transform Propagation Method J Comput Phys 47 412 1982 J Chem Phys 78 301 1983 IMPLICIT NONE INTEGER ijNNiinthnpts COMPLEX chivpropchin1chin2tprop PARAMETERnpts10nptx2nptsNN1 DIMENSION chin1nptxchin2nptx DIMENSION tpropnptxvpropnthNNNN COMMON wfunc chinthNN Apply potential energy part of the Trotter Expansion DO i1nptx chin1i00 DO j1NN chin1ichin1ivpropi1jchiij END DO END DO Fourier Transform wave packet to the momentum representation CALL fournchin1nptx1 1 Apply kinetic energy part of the Trotter Expansion DO i1nptx chin1itpropichin1i END DO 175 c Inverse Fourier Transform wave packet to the coordinate representation CALL fournchinlnptxll Apply potential energy part of the Trotter Expansion 0 DO ilnptx DO jlNN chiijvpropijlchinli END DO END DO END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc c Subroutine for FFT from Numerical Recipes CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTTNE FOURNDATANNNDTMTSTGN REAL8 WRWTWPRWPTWTEMPTHETA DIMENSION NNNDIMDATA NTOTl DO 11 TDTMlNDTM NTOTNTOTNNTDTM II CONTINUE NPREVl DO 18 TDTMlNDTM NNNTDTM NREMNTOTNNPREV IPl2NPREV TP2TP1N TP3TP2NREM I2REVI DO 14 TZlTP2TPl IFI2LII2REVIHEN DO 13 III2I2IPI 22 DO 12 I3IIIP3IP2 I3REVI2REVI3 I2 TEMPRDATATB IEMPIDAIAltI3Igt DAIAltI3gtDAIAltI3REVgt DAIAltI3IgtDAIAltI3REVIgt DATATBREVTEMPR DAIAltI3REVIgtIEMPI 12 CONTINUE l3 CONTINUE ENDIF IBITIP22 1 IF IBITGEIPlANDI2REVGTIBIT THEN I2REVI2REV IBIT IBITIBIT2 GO TO I ENDIF I2REVI2REVIBIT 176 14 CONTINUE IFP1IP1 2 IFIFP1LTIP2THEN IFP22IFP1 THETAISIGN628318530717959DOIFPZIPI WPR 2DODSINO5DOTHETA2 WPIDSINTHETA WR1DO WIODO DO 17 1311FP1IP1 DO 16 111313IP1 22 DO 15 1211IP31FP2 K112 K2KlIFPl TEMPRSNGLWRDATAK2 SNGLWIDATAK2I TEMPISNGLWRDATAK21SNGLWIDATAK2 DATAK2DATAKl TEMPR DATAK21DATAK1I TEMPI DATAK1DATAK1TEMPR DATAK11DATAK11TEMPI 15 CONTINUE I6 CONTINUE WTEMPWR WRWRWPR WIWPIWR WIWIWPRWTEMPWPIWI I7 CONTINUE IEPIIEP2 GO TO 2 ENDIE NPREVNNPREV l8 CONTINUE RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC 177 429 Problem 9 Computational Problem 9 Simulate the propagation of a wave packet with 510 2 55 and initial momentum p0 2 colliding with a barrier potential V1 3 if abs1 lt 05 and V1 0 otherwise Hint In order to avoid arti cial recurrences you might need to add an absorbing imaginary potential Va1 iabs2 104 if abs1 gt 10 and Va1 0 otherwise The output of this program can be generated and Visualized as follows Cut the source code attached below save it in a le named Problem9f compile it by typing f77 Problem9f o Problem9 run it by typing Prob1em9 The snapshots of the time dependent wave packet can be Visualized as a movie by typing gnuplotltpp9 where the le named pp9 has the following lines Download from httpXbeamschemyaleeduNbatistaP9pp9 set yrange04 set xrange 1010 set dat sty 1 plot quotarchOOOlquot u 12 1w 3 pause 1 plot quotarchOOO2quot u 12 1w 3 pause 1 plot quotarchOOO3quot u 12 1w 3 pause 1 plot quotarchOOO4quot u 12 1w 3 pause 1 plot quotarch0005quot u 12 1w 3 pause 1 plot quotarchOOO6quot u 12 1w 3 pause 1 plot quotarchOOO7quot u 12 1w 3 pause 1 plot quotarchOOO8quot u 12 1w 3 pause 1 plot quotarchOOO9quot u 12 1w 3 pause 1 plot quotarchOOlOquot u 12 1w 3 pause 1 plot quotarch0011quot u 12 1w 3 178 pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l plot quotarch pause l 0012quot 0013quot 0014quot 0015quot 0016quot 0017quot 0018quot 0019quot 0020quot 0021quot 0022quot 0023quot 0024quot 0025quot 0026quot 0027quot 0028quot 0029quot 0030quot 0031quot 0032quot 0033quot 0034quot 0035quot 0036quot lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw lw 179 plot quotarchOO37quot pause 1 plot quotarchOO38quot pause 1 plot quotarchOO39quot pause 1 plot quotarchOO40quot pause 1 plot quotarchOO41quot pause 1 plot quotarchOO42quot pause 1 plot quotarchOO43quot pause 1 plot quotarchOO44quot pause 1 plot quotarchOO45quot pause 1 plot quotarchOO46quot pause 1 plot quotarchOO47quot pause 1 plot quotarchOO48quot pause 1 plot quotarchOO49quot pause 1 plot quotarch0050quot pause 1 plot quotarch0051quot pause 1 plot quotarch0052quot pause 1 plot quotarch0053quot pause 1 plot quotarch0054quot pause 1 plot quotarch0055quot pause 1 plot quotarch0056quot pause 1 plot quotarch0057quot pause 1 plot quotarch0058quot pause 1 plot quotarch0059quot pause 1 plot quotarchOO60quot pause 1 plot quotarchOO61quot pause 1 plot quotarchOO62quot u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 u 12 1w 3 180 pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarchOO80quot pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 plot quotarch pause 1 0063quot 0064quot 0065quot 0066quot 0067quot 0068quot 0069quot 0070quot 0071quot 0072quot 0073quot 0074quot 0075quot 0076quot 0077quot 0078quot 0079quot 0081quot 0082quot 0083quot 0084quot 0085quot 0086quot 0087quot 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 u 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 181 plot quotarchOO88quot pause 1 plot quotarchOO89quot pause 1 plot quotarchOO90quot pause 1 plot quotarchOO91quot pause 1 plot quotarchOO92quot pause 1 plot quotarchOO93quot pause 1 plot quotarchOO94quot pause 1 plot quotarchOO95quot pause 1 plot quotarchOO96quot pause 1 plot quotarchOO97quot pause 1 plot quotarchOO98quot pause 1 plot quotarchOO99quot pause 1 u 12 1w 3 u 12 1w 3 u 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 12 1w 3 182 0 Download from httpXbeamsChemyaleeduNbatistaP9Problem9f PROGRAM Problem9 l D wave packet propagation of tunneling through a barrier IMPLICIT NONE INTEGER NNnptsnthndump INTEGER istepnstepjj REAL dtXcpc COMPLEX vproptpropxmeanpmean PARAMETERnptslOnptx2nptsNNl DIMENSION vpropnthNNNNtpropnth DIMENSION XmeanNNpmeanNN COMMON class xcpc CALL ReadParamnstepndumpdt call Initialize CALL SetKinPropdttprop CALL SetPotPropdtvprop DO isteplnstepl IFmodistep llOEQO PRINT quotStepquot istep lquot Final stepquot nstep IFistepGEl CALL PROPAGATEvproptprop IFmodistep lndumpEQO THEN CALL SAVEWFistepndumpdt END IF END DO END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O O O O O subroutine ReadParamnstepndumpdt Parameters defining the grid xmin xmax integration time step dt rmass rmass initial position Xk initial momentum pk number of propagation steps nstep and how often to save a pic ndump IMPLICIT NONE INTEGER ntypenstepnrptireportndumpnlit REAL xminXmaXpkrmassXkdt common packet rmassxkpk common Xy xminxmax xmin 130 xmax130 dtOl rmasslO Xk 45 pkl nsteplOO ndumpl 183 return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE Initialize IMPLICIT NONE INTEGER NNnthnptskk COMPLEX chiOchiEYECRV REAL xcpcomegaxk2xminxmaxdxpirmassxkpkxalphaalpha2 PARAMETERnptslOnptx2nptsNNl DIMENSION CRVNNNN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN COMMON iwfunc chiOnthNN COMMON class xcpc EYE0010 pi acos lO omegal dxXmax Xminrealnptx xcxk pcpk c c Wave Packet Initialization Gaussian centered at Xk with momentum pk c alpharmassomega do kklnptx xxminkkdx chikklalphapi025 l exp alpha2XXk2EYEpkX Xk chiOkklchikkl end do RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE HAMILCRVX c c Hamiltonian Matrix c IMPLICIT NONE INTEGER NN REAL XVPOTI COMPLEX CRV PARAMETERNNI DIMENSION CRVNNNN c CALL VAVPOTIX CRVllVPOTl c RETURN 184 END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE VAVX 0 Potential Energy Surface Barrier implicit none REAL VXrmassxkpkrkomega common packet rmassxkpk V00 IFabsXLE5 V3 RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc subroutine SetKinPropdttprop Kinetic Energy part of the Trotter Expansion eXp i p 2 dt2 m 0 IMPLICTT NONE INTEGER nptxkxnxnpts REAL xscXminXmaXpropfacxrmassxkpialenxdtpk COMPLEX tpropeye parameternptslOnptx2npts DIMENSION tpropnth common Xy xminxmax common packet rmassxkpk eyeOl pi acos lO alenxxmax xmin propfacx dt2rmass2pi2 do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if xscO ifnXneO xscrealnXalenx tpropkxexpeyepropfacxxsc2 end do return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc subroutine SetPotPropdtvprop c Potential Energy part of the Trotter Expansion exp i V dt2 TMPLTCIT NONE INTEGER NNiinthnptS 185 REAL xminxmaxdxdtxVPOTxa COMPLEX vpropeye parameternptslOnptx2nptsNNlxalO DIMENSION vpropnthNNNN common Xy xminxmax eyeOl dxxmax Xminrealnptx do iilnptx xxminiidx CALL VAVPOTX vpropiiIlexp eye05dtVPOTsqrtnptxl0 IFabsxGTxa vpropiillvpropiillexp absx xa4 END DO RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTINE energiesenergy IMPLICIT NONE INTEGER jNN COMPLEX energyRVRKE PARAMETER NNI DIMENSION RVNNRKENNenergyNN CALL PERV CALL KERKE DO jlNN energYjRVjRKEj END DO RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC O SUBROUTINE SAVEWFje2ndumpdt Dump Time Evolved Wave packet IMPLICIT NONE INTEGER je2nptxnptskkNNncountndumpjj COMPLEX chiCRVenergypsiPsia character9 B REAL VXlclc2claxxminxmaxdXEVALUESdt PARAMETERnptsIOnptx2nptsNNl DIMENSION CRVNNNNEVALUESNN DIMENSION psiNNNN common Xy xminxmax COMMON wfunc chinthNN COMMON ENER energyNN IFje2EOI CALL energiesenergy jjje2ndump writeB Ai44 arch jj 186 O O 33 OPEN1FILEB dxXmax Xminrealnptx ncountje2 lndump Save Wave packet components do kklnptx xxminkkdx clchikklconjgchikkl writel33 Xsqrtclrealenergyl end do writel33 do kklnptx xxminkkdx writel33 X realchikklrealenergyl end do writel33 Save Adiabatic states do kklnptx xxminkkdx CALL HAMILCRVX writel33 XCRV11 end do CLOSE1 format6e1362x RETURN END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTINE PERV Expectation Value of the Potential Enegy IMPLICIT NONE INTEGER nptxnptskkNNj COMPLEX chiEYERV REAL Vpotomegaxminxmaxdxpirmassxkpkxalpha PARAMETERnptslOnptx2nptsNNl DIMENSION RVNN COMMON wfunc chinthNN common Xy common packetrmassxkpk xminxmax dxXmax Xminrealnptx DO jlNN RVj00 do kklnptx xxminkkdx IFjEQl CALL VAVp0tX 187 RVjRVjchikkjVpotconjgchikkjdx end do END DO RETURN END cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc subroutine KERKE O Expectation value of the kinetic energy IMPLICIT NONE INTEGER NNkknthkXnXnptsj REAL dpthetawmpxminxmaxrmassxkpialenxpkrmreridx COMPLEX eyechiPsipchicRKE parameternptslOnptx2nptsNNl DIMENSION chicnthRKENN common Xy xminxmax common packet rmassxkpk COMMON wfunc chinthNN pi acos lO dxXmax Xminnptx dp2piXmax Xmin DO jlNN RKEj00 do kklnptx chickkchikkj end do CALL fournchicnptxl l do kxlnptx ifkxlenptx2l then nxkx l else nxkx l nptx end if pO ifnxneO p realnxdp chickxp220rmasschickxnptx end do CALL fournchicnptxll do kklnptx RKEjRKEjconjgchikkjchickkdx end do END DO return end cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc SUBROUTINE PROPAGATEvproptprop c Split Operator Fourier Transform Propagation Method 188 O O O O O J Comput Phys 47 412 1982 J Chem Phys 78 301 1983 IMPLICIT NONE INTEGER ijNNiinptxnpts COMPLEX chivpropchin1chin2tprop PARAMETERnpts10nptx2nptsNN1 DIMENSION chin1nptxchin2nptx DIMENSION tpropnptxvpropnptxNNNN COMMON wfunc chinptxNN Apply potential energy part of the Trotter Expansion DO i1nptx chin1i00 DO j1NN chin1ichin1ivpropi1jchiij END DO END DO Fourier Transform wave packet to the momentum representation CALL fournchin1nptx1 1 Apply kinetic energy part of the Trotter Expansion DO i1nptx chin1itpropichin1i END DO Inverse Fourier Transform wave packet to the coordinate representation CALL fournchin1nptx11 Apply potential energy part of the Trotter Expansion DO i1nptx DO j1NN chiijvpropij1chin1i END DO END DO END CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC C Subroutine for FFT from Numerical Recipes CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC SUBROUTINE FOURNDATANNNDIMISIGN REAL8 WRWIWPRWPIWTEMPTHETA DIMENSION NNNDIMDATA NTOT1 DO 11 IDIM1NDIM NTOTNTOTNNIDIM CONTINUE 189 NPREV1 DO 18 IDIM1NDIM NNNIDIM NREMNTOTNNPREV IP12NPREV IP2IP1N IP3IP2NREM I2REV1 DO 14 IZ1IP2IP1 IEI2LTI2REVTHEN DO 13 IlIZIZIPl 22 DO 12 I3I1IP3IP2 I3REVI2REVI3 I2 TEMPRDATAI3 TEMPIDATAltI31gt DATAltI3gtDATAltI3REVgt DATAltI31gtDATAltI3REV1gt DATAltI3REVgtTEMPR DATA13REV1TEMPI CONTINUE CONTINUE ENDIE IBITIP22 IE IBIT GE IPIANDI2REVGTIBIT THEN I2REVI2REv IBIT IBITIBIT2 GO TO 1 ENDIE I2REVI2REVIBIT CONTINUE IFP1IP1 IFIFP1LTIP2THEN IFP22IFP1 THETAISIGN628318530717959DOIFPZIPl WPR 2DODSINO5DOTHETA2 WPIDSINTHETA WR1DO WIODO DO 17 I31IEPIIPI DO 16 I1I3I3IPI 22 DO 15 I2I1IP3IEP2 K1I2 K2KlIFPl TEMPRSNGLWRDATAK2 SNGLWIDATAK21 TEMPISNGLWRDATAK21SNGLWIDATAK2 DATAK2DATAKl TEMPR DATAK21DATAK11 TEMPI DATAK1DATAK1TEMPR DATAK11DATAK11TEMPI CONTINUE CONTINUE 190 WTEMPWR WRWRWPR WTWPTWR WTWTWPRWTEMPWPTWT l7 CONTINUE TFP1TFP2 GO TO 2 ENDTF NPREVNNPREV l8 CONTINUE RETURN END ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc 191 CHAPTER 1 AN INTRODUCTION To CHEMISTRY I woufofwatcfi trie Ends sweffin spring trie mica int in tile granite my own lianofs and I woufcfesearyewemysegf I wif unc arstancf triis too I wiffunc arstandeveryt ing Primo Levi 1919 87 Italian chemist and author The Periodic Table t is human nature to wonder about the origins of the universe and of life about their workings and even about their meanings We look for answers in physics biology and the other sciences as well as in philosophy poetry and religion Primo Levi could have searched for understanding in many ways but his wondering led him to study chemistry and so has yours though you may not yet know why T 11 What is Chemistry and What Can Chemistry Do for You One thing is certain Once you start studying chemistry all kinds of new questions will begin to occur to you Let s consider a typical day Your alarm rings early and you are groggy from sleep but eager to begin working on a chemistry assignment that s coming due Chemistry has taught you that there are interesting answers to questions you might once have considered silly and childish Preparing tea for example now makes you wonder why the boiling water bubbles and produces steam while the teakettle retains its original shape How do the tea leaves change the color of the water while the teabag remains as full and plump as ever Why does sugar make your tea sweet and why is the tea itself bitter You settle down with tea and newspaper and the wondering continues An article about methyl bromide a widely used pesticide says some scientists think it damages the ozone layer What are methyl bromide and ozone How does one destroy the other and why should we care How can we know if the ozone really is being depleted Later as you drive to the library to get some books you need to complete that chemistry assignment you wonder why gasoline burns 7 and propels your car down the road How does it pollute the air we breathe and what does the catalytic converter do to minimize the pollution a window to the world around you An understanding of chemistry opens up 11 12 13 14 15 0 What Is Chemistry and What Can Chemistry Do for You Suggestions for Studying Chemistry The Scientific Method Measurement and Units Reporting Values from Measurements 4 Chapter 1 J An Introduction to Chemistry At the library you wonder why some books that are hundreds of years old are still in good shape while other books that are only 50 years old have pages that are brown brittle and crumbling Can the books with damaged pages be saved Chemists can answer all these questions and others like them They are scientists who study the structure of material substances collectively called matter and the changes that they undergo Matter can be solid like sugar liquid like water or gaseous like the exhaust from your car s tailpipe Chemistry is often de ned as the study of the structure and behavior of matter Chemists do a lot more than just answer questions Industrial chemists are producing new materials to be used to build lighter and stronger airplanes more environmentally friendly disposable cups and more ef cient anti pollution devices for your car Pharmaceutical chemists are developing new drugs to ght cancer control allergies and even grow hair on bald heads In the past the chemists creations have received mixed reviews The chlorofluorocarbons CFCs used as propellants in aerosol cans are now known to threaten the Earth s protective ozone layer The durable plastics that chemists created have proved too durable so when they are discarded they remain in the environment for a long long time One of the messages you will nd in this book is that despite the occasional mistakes and failures most chemists have a strong social conscience Not only are they actively developing new chemicals to make our lives easier safer and more productive but they are also working to clean up our environment and minimize the release of chemicals that might be harmful to our surroundings see Special Topic 11 Gram Chemistry As you read on in this book you will nd perhaps to your disappointment that only limited portions of each chapter provide direct answers to real life questions An introductory chemistry teXt such as this one must focus instead on teaching basic principles and skills Some of the things you need to learn in order to understand chemistry may seem less than fascinating and it will not always be easy to see why they are useful Try to remember that the fundamental concepts and skills will soon lead you to a deeper understanding of the physical world Before you could run you needed to learn how to walk Before you could read a book or write a paper you needed to learn the alphabet build your vocabulary and understand the basic rules of grammar Chemistry has its own alphabet and vocabulary as well as many standards and conventions that allow chemists to communicate and to do ef cient safe and meaningful work Learning the symbols for the common chemical elements the rules for describing measurements or the conventions for describing chemical changes might not be as interesting as nding out how certain chemicals in our brains affect our moods but they are necessary steps in learning chemistry This chapter presents some suggestions for making your learning process easier and introduces you to some of the methods of scienti c measurement and reporting You will then be ready for Chapter 2 which gives you a rst look at some of chemistry s underlying concepts 12 Suggestions for Studying Chemistry 5 9 SPECIAL TOPIC 11 Green Chemistry With Emmaedge comes t e Eurckn of into the water various plants and animals begin to grow on responsi il39ity its bottom This growth slows the ship s progress and causes PaulTAnaStaS and Tracy C WilliamSOIl other problems as well ultimately costing the shipping GREEN CHEMISTRY industry an estimated 3 billion per year Substances called The Chemical industry has made many positive contributions antifoulants are painted on the bottom ofships to prevent the unwanted growth but most of them can be harmful to the to modern life but these improvements have come at a price marine environment Sea Nine is a more environmentally The chemical products themselves the chemicals used to produce them and the byproducts of their production have friendly chemical than the alternatives because it breaks down sometimes been harmful to our health or the environment In the ocean more quICkly and does not accumulate m marme Concern about these dangers has lead to a movement in the orgamsms39 chemical industry that is sometimes called Green Chemistry Thls 18 only one example Of the pmgress that Green or Environmentally Benign Chemistry Its goal is to produce Chemistry is making Others Will be presented in later process and use chemicals in new ways that pose fewer risks Chapters39 to humans and their environment Source P T Anastas and T C Williams Eds Green Chemistry In support of this goal Pres1dent Clinton inaugurated the Washington DC American chemi C a1 Sociew1996 Green Chemistry Program in March 1995 to be coordinated by the Environmental Protection Agency As part of this program the Green Chemistry Challenge Awards were set up as part of the Green Chemistry Program to recognize and promote fundamental and innovative chemical methodologies that accomplish pollution prevention and that have broad application in industry Among the rst award recipients was the Rohm and Haas Company which received the Green Chemistry Challenge s 1996 Designing Safer Chemicals Award for the development 0f 4a5 diCthIO Z H OCW1 4 isothiamlin 3 Onea otherWise New environmentally friendly marine antifoulants known as Sea Nine antifoulant As soon as a new ship goes protect ships without polluting the water 12 Suggestions for Studying Chemistry T Tfie wiff t0 succeedis importantn uLwiat s more important is tfre wiff to prepare T Basketball Coach Bobby Knight Let s face it Chemistry has a reputation for being a difficult subject One reason is that it includes so many different topics Individually they are not too difficult to understand but collectively they are a lot to master Another reason is that these topics must be learned in a cumulative fashion Topic A leads to topic B which is important for understanding topic C and so on If you have a bad week and do not study topic B very carefully topic C will not make much sense to you Because chemistry is time consuming and cumulative you need to be very organized and diligent in studying it There is no correct way to study chemistry Your study technique will be decided by your current level of chemical knowledge the time you have available your strengths as a student and your attitude toward the subject The following is a list of suggestions 6 Chapter 1 J An Introduction to Chemistry to help you take chemistry s special challenges in stride Use the Review Skills sections in this texthook Starting with Chapter 2 each chapter begins with a section called Review Skills complemented at the end of the chapter by a set of review questions that test those skills The Review Skills section and the review questions identify the speci c skills from earlier chapters that are necessary for success in the new chapter If you have trouble with the tasks on the Review Skills list you will have trouble with the chapter so promise yourself that you will always review the topics listed in the Review Skills section before beginning a new chapter Reaa each chapter in the texthook hefore it is covered in lecture There are several good reasons why a relaxed pre lecture reading of each chapter is important It provides you with a skeleton of knowledge that you can esh out mentally during the lecture at the same time guaranteeing that there will be fewer new ideas to absorb as you listen to your teacher talk If you already know a few things you will be better prepared to participate in class Attena39 class meetings take notes ana participate in class discussions You will get much more out of a lecture discussion section or laboratory and will enjoy it more if you are actively involved Don t hesitate to ask questions or make comments when the lecture confuses you If you have read the chapter before coming to class and paid attention to the lecture and you still have a question you can be fairly certain that other students have that question too and will appreciate your asking for clari cation Rereaa39 the chapter marking important sections ana working the practice exercises In the second reading mark key segments of the chapter that you think you should reread before the exam Do not try to do this in the rst reading because you will not know what parts are most important until you have an overview of the material the chapter contains It is a good idea to stop after reading each section of text and ask yourself What have I just read This will keep you focused and help you to remember longer You will nd Examples in each chapter that show how to do many of the tasks that you will be asked to do on exams The examples are followed by Exercises that you should work to test yourself on these tasks The answers to these exercises are found at the end of the book If you have trouble with a practice exercise look more closely at the example that precedes it but don t get bogged down on any one topic When a new concept or an exercise gives you trouble apply the 15 minute rule If you have spent more than 15 minutes on an idea or problem and you still do not understand it write down what you do not understand and ask your instructor or another student to explain it Use the chapter ohjectioes as a focus of study At the end of each chapter is a list of Chapter Objectives that provides a speci c description of what you should be able to do after studying the chapter Notations in the text margins show you where to nd the information needed to meet each objective Your instructor may wish to add to the list or remove objectives from it Many of the objectives begin with Explain or Describe You might be tested on these objectives in short essay form so it is a good idea to actually write your responses down This will force you to organize your thoughts and develop a concise explanation that will not take too much time to write on the exam If you do it conscientiously perhaps using color to highlight key phrases you will be able to Visualize your study sheet while taking the exam 13 The Scientific Method 7 Many of the objectives refer to stepwise procedures This book will suggest one procedure and sometimes your instructor will suggest another Write out the steps that you think will work best for you The act of writing will help you remember longer and you will nd it much easier to picture a study sheet written out in your own handwriting than one printed in this book Use the compater hasea tools that accompany the coarse This textbook is accompanied by an Internet site that supplements the text If you have access to it you will nd it useful at this stage in your studying VVorh some of the prohlems at the ena of the chapter Do not try the end of chapter problems too soon They are best used as a test of what you have and have not mastered in the chapter after all the previous steps have been completed Note that many of the problems are accompanied by the number of the learning objective to which they correspond In fact the problems serve as examples of ways in which you might be tested on various chapter objectives Use the same 15 minute rule for the problems as for the chapter concepts and exercises If you have spent more than 15 minutes on any one problem it is time to seek help Ase for he o when you need it Don t be shy Sometimes ve minutes in the instructor s office can save you an hour or more of searching for answers by other means You might also consider starting a study group with fellow students It can bene t those of you who are able to give help as well as those who need it There is no better way to organize one s thoughts than to try to communicate them to someone else There may be other ways to get help Ask your instructor what is available Review for the exam Read the list of objectives asking yourself whether you can meet each one Every time the answer is No spend some time making it Yes This might mean meeting with your instructor or study group rereading this text or reviewing your notes Ask your instructor which objectives are being emphasized on the exam and how you are going to be tested on them Your lecture notes can also provide clues to this Lastly work some more of the end of chapter problems This will sharpen your skills and improve your speed on the exam I hear and I forge t I see and I rememher I 0 and I understanc Chinese proverb He who neglects to tirinh of the spring of ex erience is ihefy to die 0 thirst in the desert of ignorance Ling Po 13 The Scientific Method T I is as simpfe as seeing a hug that intrigues you you want to hnow where it goes at night what it eats David Cronenberg Canadian lmmaker Before beginning our quest for an understanding of chemistry let s look at how science in general is done There is no one correct way to do science Different scienti c disciplines have developed different procedures and different scientists approach their pursuit of knowledge in different ways Nevertheless most scienti c work has certain characteristics in common We can see them in the story of how scientists discovered the rst treatment for Parkinson s disease a neurological condition that progressively Chapter 1 J An Introduction to Chemistry OBJECTIVE 2 OBJECTIVE 2 OBJECTIVE 2 OBJECTIVE 2 affects muscle control The principal steps in the process are summarized in Figure 11 Like most scienti c work the development of a treatment for Parkinson s disease began with ohseroation ana the collection of data In the 1960 s scientists observed that South American manganese miners were developing symptoms similar to the muscle tremors and rigidity seen in Parkinson s disease Next the scientists made an initial hypothesis on the hasis of their ohseroations Perhaps the symptoms of the manganese miners and of Parkinson s sufferers had a common cause The initial hypothesis led to a more purposeful collection of information in the form of systematic research or experimentation Systematic study of the manganese miners brain chemistry showed that manganese interferes with the work of a brain chemical called dopamine Because dopamine is important in the brain s control of muscle function anyone absorbing abnormally high levels of manganese would be expected to have troubles with movement 7he hypothesis was re ned based on the new information and research was designed to test the hypothesis Speci cally the researchers hypothesized that the brains of Parkinson s sufferers had low levels of dopamine Brain studies showed this to be the case 7he results were pahlishea39 so that other scientists might repeat the research ana con rm or refute the conclusions Because other scientists con rmed the results of the dopamine research the hypothesis hecame accepted in the scienti c community The discovery of the dopamine connection started a search for a drug that would elevate the levels of dopamine in the brain This provides an example of what is very often the next step of the scienti c method a search for useful applications of the new ideas Dopamine itself could not be used as a drug because it is unable to pass from the blood stream into the brain tissue Instead the researchers looked for a compound that could penetrate into the brain and then be converted into dopamine Levodopa or L dopa met these requirements It should be noted that signi cant scienti c research does not always lead directly to applications or even to a published paper One of the main driving forces of science is just the desire to understand more about ourselves and the world around us Any research that increases this understanding is important 7he development of applications often leads to another round of hypothesizing ana testing in order to re ne the applications There was some initial success with L dopa It caused remission of Parkinson s disease in about one third of the patients treated and improvements in one third of the others but there were also problematic side effects including nausea gastrointestinal distress reduced blood pressure delusions and mental disturbance The drug s effects on blood pressure seem to be caused by the conversion of L dopa to dopamine outside the brain For this reason L dopa is now given with levocarbidopa which inhibits that process And the cycle of hypothesis experimentation and nding new applications which leads to further re nement of the hypotheses continues 14 Measurement and Units 9 Figure 11 The Scientific Method OBJECTIVE 2 Collection of information leads to The hypothesis a testable hypothesrs is subjected to Disproof The ex fails Further researc is done to re ne the applications Experimentation Application To rmation 39ji ent is repeted successfully 14 Measurement and Units T asura fe w at is not so Measure wliat is m39 r Galileo Galilei The practice of chemistry demands both accuracy and clarity The properties of matter must be measured correctly and reported without ambiguity For example a chemist has a lot of measuring to do while testing to see whether the pesticide methyl bromide might be destroying the protective layer of ozone gas in the earth s upper atmosphere The ozone in the upper atmosphere lters out harmful radiation from the sun She or he might add a carefully measured amount of methyl bromide to a reaction vessel that already contains ozone and perhaps other chemicals found in our atmosphere In such an experiment the temperature too would be carefully measured and adjusted to duplicate the average temperature in the ozone layer and then the substances in the vessel might be subjected to the same measured amounts and kinds of radiation to which ozone is exposed in the atmosphere Chemical changes would take place after which the amounts of various substances found in the resultant mixture would be measured and reported at various times All of these measurements of amounts temperatures and times must be carefully reported in a way that enables other scientists to judge the 1 0 Chapter 1 a An Introduction to Chemistry A longjump can be won by a centimeter This is US Paralympian Marlon Shirley competing in the men s F44 long jump final at the IPC 2002 Athletics World Championships Photo Courtesy of International Paralympic Committee IPC value of the experiment A measurement is always reported as a value a quantitative description that includes both a number and a unit For example before a IOO meter race is run the distance must be measured as precisely as possible Its value is 100 meters In this value the unit is meters de ned below and the number of units is 100 Units are quantities de ned by standards that people have agreed to use to compare one event or object to another For example at one time the units for length were based on parts of the body An inch the unit was the width of the thumb the standard and a foot was the length of a typical adult foot Centuries ago you might have described the length of a cart you wanted made as eight feet meaning eight times the length of your foot Although the length of a foot varies for different people this description would be accurate enough to allow the carpenter to make a cart that would t your purpose As measuring techniques became more precise and the demand for accuracy increased the standards on which people based their units were improved In the 18th century the French invented the metric system based on a more consistent systematic and carefully de ned set of standards than had ever been used before For example the meter or metre from the Greek metron a measure became the standard for length The rst de nition for the standard meter was one ten millionth of the distance from the North Pole to the Equator This became outdated as the precision of scientist s measuring instruments improved Today a meter is de ned as the distance light travels in a vacuum in 1299792458 second Technical instruments for measuring length are calibrated in accordance with this very accurate de nition The International System of Measurement The International System of Measurement SI for Systeme International el Unz39tes a modern elaboration of the original metric system was set up in 1960 It was developed to provide a very organized precise and practical system of measurement that everyone in the world could use The SI system is constructed using seven base units from which all other units are derived Table 11 The chemist is not usually interested in electric currents or luminous intensity so only the rst ve of the base units on Table 11 will appear in this text The meaning of mole the base unit for amount of substance is explained in Chapter 9 Until then we will use the rst four base units meter m kilogram kg second 3 and kelvin Table 11 14 Measurement and Units Base Units for the International System of Measurement OBJECTIVE 3 Type of Measurement Base Unit Abbreviation length meter m mass kilogram kg time second s temperature kelVin K amount of substance mole mol electric current ampere A luminous intensity candela cd 0 SPECIAL TOPIC 12 One by one all of the SI base units except the kilogram have been de ned in terms of constants of nature For example the speed of light in a vacuum is used to de ne the meter The kilogram is the last of the standards to be based on a manufactured object A kilogram is de ned as the mass of a cylinder made of platinum iridium alloy 39 mm 15 in tall and 39 mm in diameter This cylinder has been kept in an airtight vault in S vres France since 1883 In order to avoid using and possibly damaging the standard itself forty replicas were made in 1884 and distributed around the world to serve as a basis of comparison for mass The United States has kilogram number 20 The scientists at the Bureau International des Poids et Mesures BIPM are so careful with the standard kilogram that it was only used three times in the 20th century There are problems associated with de ning a standard in terms of an object such as the standard kilogram If the object loses or gains mass as a result of contact or contamination measurements that relate to the object change For years the standard kilogram was maintained by a French technician named Georges Girard who kept the cylinder clean using a Chamois cloth and a mixture of ethanol and ether Girard seemed to have just the right touch for removing contaminants without removing any metal Now that he has retired no one seems to be able to duplicate his procedure with the correct amount of pressure As the ability to measure mass becomes more exact the need for an accurate and easily reproduced standard becomes more urgent Already modern instruments have Wanted A New Kilogram shown that the copies of the standard kilogram vary slightly in mass from the original cylinder in France Many scientists feel that the time has come to nd a natural replacement for the kilogram The International Bureau of Weights and Measures may be choosing one soon The copy of the standard kilogram in the United States is kilogram 20 produced in 1884 Photo Courtesy of National Institute of Standards and Technology 11 12 Chapter 1 J An Introduction to Chemistry SI Units Derived from Base Units Many properties cannot be described directly with one of the seven SI base units For example chemists often need to measure volume the amount of space that something occupies and volume is not on the list of SI base units Table 11 Rather than create a new de nition for volume we derive its units from the base unit for length the 3 can be used as a OBJECTIVE 3 meter Volume can be de ned as length cubed so cubic meters m volume unit Various other units are derived in similar ways If you have long arms a meter is approximately the distance from the tip of your nose to the end of your ngers when you are looking forward and extending your arm fully A cubic meter is therefore a fairly large volume In fact it s inconveniently large for many uses You d nd it awkward to buy your milk by a small fraction of a cubic meter such as 1 1000 m3 Scientists too need smaller volume units Chemists prefer to use the liter as the base unit for volume A liter L is 11000 or 10 3 of a cubic HOW many terS meter so there are 1000 or 103 liters per cubic meterl See Figure 12 OBJECTIVE 4 1 L 10 3 m3 or 103 L 1 m3 Figure 12 Liters and Cubic Meters 1 cubic meter 1000 liters cubic meter liter 4 one meter SI Units Derived from Metric Pre xes Because the SI base units and derived units such as the liter are not always a convenient size for making measurements of interest to scientists a way of deriving new units that are larger and smaller has been developed For example the meter is too small to be convenient for describing the distance to the moon and even the liter is too large for measuring the volume of a teardrop Scientists therefore attach pre xes to the base units which have the effect of multiplying or dividing the base unit by a power of 10 Table 12 lists some of the most common metric pre xes and their abbreviations 1 If you are unfamiliar with numbers such as 103 and 103 that are expressed using exponents and scienti c notation you should read Appendix B at the end of this text 14 Measurement and Units The unit kilometer for example is composed of the pre x leilo and meter the base unit for length Kilo means 103 or 1000 so a kilometer is 103 or 1000 meters A kilometer is therefore more appropriate for describing the average distance to the moon 384403 kilometers rather than 384403000 meters Units derived from metric pre xes are abbreviated by combining the abbreviation for the pre x Table 12 with the abbreviation for the unit to which the pre x is attached The abbreviation for leilo is k and for meter is m so the abbreviation for kilometer is km 1 kilometer 103 meter 1 km 103 m Likewise mz39ero means 10 6 or 0000001 or 11000000 so a micrometer is 10 6 meters The abbreviation for micrometer is pm The symbol u is the Greek letter 01 mu The micrometer can be used to describe the size of very small objects such as the diameter of a typical human hair which is 3 pm This value is easier to report than 0000003 m 1 micrometer 10 6 meter or 1 pm 10 6 m Table 12 Some Common Metric Prefixes 13 OBJECTIVE 6 OBJECTIVE 7 OBJECTIVE 6 OBJECTIVE 7 OBJECTIVE 5 Pre xes for units larger than the base unit than the base unit Pre xes for units smaller Abbreviation De nition Pre x Abbreviation De nition Pre x G 1000000000 or 109 centi c 001 or 102 giga mega M 1000000 or 106 milli m 0001 or 103 kilo k 1000 or 103 micro 11 0000001 or 106 nano I1 0000000001 or 109 pico P 0000000000001 or 1012 0 EXAMPLE 11 Units Derived from Metric Prefixes Complete the following relationships Rewrite the relationships using abbreviations for the units a 1 gigagram gram b 1 centimeter meter c 1 nanometer meter Solution Refer to Table 12 until you have the pre xes memorized a The pre x giga G means 109 so 1 gigagram 109 gram or 1 Gg 109 g b The pre x centi c means 1072 so 1 centimeter 10 2 meter or 1 cm 10 2 m c The pre x nano n means 10 9 so 1 nm 10 9 m 1 nanometer 10 9 meter or OBJECTIVE 6 OBJECTIVE 7 14 Chapter 1 J An Introduction to Chemistry 0 EXERCISE 11 Units Derived from Metric Prefixes OBJECTIVE 5 Complete the following relationships Rewrite the relationships using abbreviations OBJECTIVE 6 for the units OBJECTIVE 7 a 1 megagram gram b 1 milliliter liter More about Length Units Although scientists rarely use the centuries old English system of measurement it is still commonly used in the United States to describe quantities in everyday life Figure 13 might help you to learn the relationships between metric and English length unitsz OBJECTIVE 8 A kilometer is a little more than 12 mile The distance between the floor and a typical doorknob is about 1 meter the width of the ngernail on your little nger is probably about 1 centimeter and the diameter of the wire used to make a typical paper clip is about 1 millimeter Figure 14 shows the range of lengths Figure 13 English and Metric Length Units 1 mi 2 1 609 km OBJECTIVE 8 1 km 06214 mi A mile is four kllometersIIIIIIIIIIIIIIIII I times around a 1 lim typical high miles I I I I I i I I I I school track 1 mi 1 m 3281 ft 1 ft 03048 In meters I I 39 1 meter 1 m feet I I I I 1 ft 1 in 2 54 cm 254 mm 1 cm 03937 in 1 centimeter 1 mm 003937 in centimeters IIIIIIIIIIIIIIIIIIIIIII 1 mm 1 cm I I I inchesIlIl 1 millimeter 1 1n 2 These relationships between these units and many others are found in Appendix A 14 Measurement and Units 15 Figure 14 The Range of Lengths The relative sizes of these measurements cannot be shown on such a small page The wedge and the numbers of increasing size are to remind you that each numbered measurement on the scale represents 100000 times the magnitude of the preceding numbered measurement More About Volume Units The relationships in Figure 15 might help you to develop a sense of the sizes of typical metric volume units A liter is slightly larger than a quart There are 493 milliliters in a teaspoon so when the label on the bottle of a typical liquid children s pain reliever suggests a dosage of one teaspoon the volume given will be about 5 milliliters There are 2957 milliliters per uid ounce oz A typical bottle of nail polish contains 05 02 Figure 16 on the next page shows the range of volumes Another common volume unit is the cubic centimeter cm3 which is equivalent to a milliliter 1 cm3 1 mL OBJECTIVE 10 Figure 15 OBJECTIVE 9 English and Metric Volume Units 1 L 1057 qt 1 gal 3785 L 1 qt 09464 L 02642 gal 1 uid ouncem OZ 1 mL 003381 oz 57 iii V ninja J 39 1 milliliter mL g 07 quot r 1 oz 2957 mL about 20 drops 1 gallon gal or 4 quarts qt 1 qt or 32 oz 1 liter L or 1000 mL 16 Chapter 1 J An Introduction to Chemistry Figure 16 The Range of Volumes 10 The relative sizes of these measurements cannot be shown on such a small page The wedge and the numbers of increasing size are to remind you that each numbered measurement on the scale represents 10000000000 times the magnitude of the preceding numbered measurement OBJECTIVE 11 OBJECTIVE 12 Mass and Weight The terms mass and weight are often used as if they meant the same thing The two properties are indeed related but they are not identical Mass is usually de ned as a measure of the amount of matter in an object matter is de ned as anything that has mass and takes up space In contrast the weight of an object is a measure of the force of gravitational attraction between it and a signi cantly large body such as the earth or the moon An object s weight on the surface of the earth depends on its mass and on the distance between it and the center of the earth In fact mass can be de ned as the property of matter that leads to gravitational attractions between objects and therefore gives rise to weight As an object s mass increases its weight increases correspondingly However as the distance between an object and the earth increases the object s weight decreases while the amount of matter it contains and therefore its mass stays the same In the SI system units such as gram kilogram and milligram are used to describe mass People tend to use the terms mass and weight interchangeably and to describe weight with mass units too However because weight is actually a measure of the force of gravitational attraction for a body it can be described with force units The accepted SI force unit is the newton N If your mass is 65 kg your weight on the surface of the earth the force with which you and the earth attract each other is 637 N The chemist is not generally concerned with the weight of objects so neither weight nor its unit will be mentioned in the remaining chapters of this book Figure 17 might help you to develop a sense of the sizes of typical metric mass units and Figure 18 shows the range of masses 14 Measurement and Units 17 1 lb 4536 g lkg22051b 1Mg1000kg1t 39r About 25 grams g or Ei39 about 0088 ounce oz 7 About 1 kilogram kg or About 1 megagram Mg or 1 metric ton t about 22 pounds lb OBJECTIVE 13 OBJECTIVE 14 Figure 17 English and Metric Mass Units University ofAIaska Fairbanks Basketball photo by Duncan Tipton e Qo x6 qb x6 339 9 Q Q I 939 7367 s 6 9 x k 39l Q00 63 10 30 103920 10 10 Figure 18 The Range of Masses The relative sizes of these measurements cannot be shown on such a small page The wedge and the numbers of increasing size are to remind you that each numbered measurement on the scale represents 10000000000 times the magnitude of the preceding numbered measurement Temperature It may surprise you to learn that temperature is actually a measure of the average motion of the particles in a system Instead of saying that the temperature is higher today than yesterday you could say that the particles in the air are moving faster By the time we return to a description of temperature measurements and temperature calculations in Chapters 7 and 8 you will be much better prepared to understand the physical changes that take place when the temperature of an object changes as well as how we measure these changes For now it is enough to have some understanding of the relationships between the three common temperature scales Celsius Fahrenheit and Kelvin 18 Chapter 1 Q An Introduction to Chemistry OBJECTIVE 15 OBJECTIVE 16 OBJECTIVE 15 OBJECTIVE 16 OBJECTIVE 15 For the Celsius scale the temperature at which water freezes is de ned as 0 0C and the temperature at which water boils is de ned as 100 0C Thus a degree Celsius 0C is 1 100 of the temperature difference between freezing and boiling water Figure 19 For the Fahrenheit scale which is still commonly used in the US the temperature at which water freezes is de ned as 32 0F and the temperature at which water boils is de ned as 212 F There are 180 F between freezing and boiling water 212 32 180 so a degree Fahrenheit 0F is 1180 of the temperature difference between freezing and boiling water Figure 19 Note that there are 100 0C between freezing and boiling water but there are 180 F for the same temperature difference Thus a degree Fahrenheit is smaller than a degree Celsius There are 180 F per 100 0C or 18 F per 1 C Figure 19 Celsius and Fahrenheit Thermometers Boiling water Ice water The thermometers that scientists use to measure temperature usually provide readings in degrees Celsius but scientists usually convert Celsius values into Kelvin values to do calculations The unit of measurement in the Kelvin scale is called the kelvin K not degree kelvin just kelvin The value 0 K is de ned as absolute zero the lowest possible temperature Because the temperature of an object is a measure of the degree of motion of its particles as the motion of the particles decreases the temperature of the object decreases Absolute zero is the point beyond which the motion of the 14 Measurement and Units particles and therefore the temperature cannot be decreased Absolute zero is 0 K 273 1 5 C and 45967 F Because the zero point for the Kelvin scale is absolute zero all Kelvin temperatures are positive The kelvin is de ned so that its size is equal to the size of a degree Celsius Figure 110 summarizes the relationships between the three common temperature scales In Chapter 8 we will use these relationships to convert measurements from one temperature scale to another Celsius Kelvin Fahrenheit Boiling water 100 C r 37315 K 212 F 100 units 4 100 units i 180 units Freezing water L 0 C L 27315 K 32 F Absolute zero 27315 C 0 K 45967 F Figure 110 Comparing Temperature Scales OBJECTIVE 15 OBJECTIVE 16 The highest temperatures in the universe are thought to be inside some stars where theory predicts temperatures of about 109 K a billion kelvins Probably the hottest thing in your home is the tungsten lament in an incandescent light bulb with a temperature of about 2800 K Molten lava is about 2000 K Normal body temperature is 3102 K 986 F or 370 0C Normal room temperature is about 20 C 68 0F or 293 The lowest temperature achieved in the laboratory is about 2 X 10 8 K 000000002 K Figure 111 shows the range of temperatures 19 OBJECTIVE 16 Kelvin 6000 Surface of the sun 5000 4000 1 3000 quot quot Light bulb ament 2000 1000 Salt NaCl melts 0 Absolute zero Figure 111 The Range of Temperatures 20 Chapter 1 J An Introduction to Chemistry T 15 Reporting Values from Measurements OBJECTIVE 17 Figure 112 Precision and Accuracy All measurements are uncertain to some degree Scientists are very careful to report the values of measurements in a way that not only shows the measurements magnitude but also re ects its degree of uncertainty Because the uncertainty of a measurement is related to both the precision and the accuracy of the measurement we need to begin by explaining these terms Accuracy and Precision Precision describes how closely a series of measurements of the same object resemble each other The closer the measurements are to each other the more precise they are The precision of a measurement is not necessarily equal to its accuracy Accuracy describes how closely a measured value approaches the true value of the property To get a better understanding of these terms let s imagine a penny that has a true mass of 2525 g If the penny is weighed by ve different students each using the same balance the masses they report are likely to be slightly different For example the reported masses might be 2680 g 2681 g 2680 g 2679 g and 2680 g Because the range of values is i0001 g of 2680 g the precision of the balance used to weigh them is said to be i0001 g but notice that none of the measured values is very accurate Although the precision of our measurement is i0001 g our measurement is inaccurate by 0155 g 2680 2525 0155 Scientists recognize that even very precise measurements are not necessarily accurate Figure 112 provides another example of the difFerence between accuracy and precision This archer is precise but not accurate This archer is imprecise and inaccurate This archer is precise and accurate 14 Measurement and Units Describing Measurements Certain standard practices and conventions make the taking and reporting of measurements more consistent One of the conventions that scientists use for reporting measurements is to report all of the certain digits and one estimated and thus uncertain digit Consider for example how you might measure the volume of water shown in the graduated cylinder in Figure 113 Liquids often climb a short distance up the walls of a glass container so the surface of a liquid in a graduated cylinder is usually slightly curved If you look from the side of the cylinder this concave surface looks like a bubble The surface is called a meniscus Scientists follow the convention of using the bottom of the meniscus for their reading The graduated cylinder in Figure 113 has rings corresponding to milliliter values and smaller divisions corresponding to increments of 01 mL When using these marks to read the volume of the liquid shown in Figure 113 we are certain that the volume is between 87 mL and 88 mL By imagining that the smallest divisions are divided into 10 equal parts we can estimate the hundredth position Because the bottom of the meniscus seems to be about four tenths of the distance between 87 mL and 88 mL we report the value as 874 mL Because we are somewhat uncertain about our estimation of the hundredth position our value of 874 mL represents all of the certain digits and one uncertain digit 9 Comparing the position of the bottom of the meniscus and 8 the milliliter scale yields a measurement of 874 mL Scientists agree to assume that the number in the last reported decimal place has an uncertainty of i1 unless stated otherwise Example 12 shows how this assumption is applied 2quot Figure 113 1quot 39 Measuring Volume with a quot Graduated Cylinder 0 EXAMPLE 12 Uncertainty If you are given the following values that are derived from measurements what will you assume is the range of possible values that they represent a 54 mL c 234 x 103 kg b 64 cm Solution We assume an uncertainty of i1 in the last decimal place reported a 54 mL means 54 i 01 mL or 53 to 55 mL b 64 cm means 64 i 1 cm or 63 to 65 cm c 234 x 103 kg means 234 i 001 x 103 kg or 233 X 103 kg to 235 X 103 kg OBJECTIVE 18 OBJECTIVE 19 OBJECTIVE 19 21 22 Chapter 1 J An Introduction to Chemistry OBJECTIVE 19 4 EXERCISE 12 Uncertainty If you are given the following values that are derived from measurements what will you assume is the range of possible values that they represent a 72 mL b 823 m c 455 gtlt105 g Sometimes it is necessary to use trailing zeros to show the uncertainty of a measurement If the top of the liquid in the graduated cylinder shown in Figure 114 is right at the 8 mL mark you would report the measurement as 800 mL to indicate that the uncertainty is in the second decimal place Someone reading 800 mL would recognize that the measured amount was between 799 mL and 801 mL Reporting 8 mL would suggest an uncertainty of i1 mL in which case the amount would be assumed to lie anywhere between 7 mL and 9 mL Figure 114 Reporting Values with Trailing Zeros OBJECTIVE 18 OBJECTIVE 18 We report 800 mL to show an uncertainty of i001 mL There are other ways to decide how to report values from measurements For example the manufacturer of the graduated cylinder in Figure 114 might inform you that the lines have been drawn with an accuracy of i01 mL Therefore your uncertainty when measuring with this graduated cylinder will always be at least i01 mL and your values reported should be to the tenth position In this case the volumes in Figure 113 and 114 would be reported as 87 mL and 80 mL In general though unless you are told to do otherwise the conventional practice in using instruments such as a graduated cylinder is to report all of your certain digits and one estimated digit For example if you were asked to measure the volumes shown in Figures 113 and 114 you would report 874 mL and 800 mL unless you were told to report your answer to the tenth place 15 Reporting Values from Measurements EXAMPLE 13 Uncertainty in Measurement Consider a laboratory situation where ve students are asked to measure the length of a piece of tape on a lab bench with a meter stick The values reported are 6194 cm 6201 cm 6212 cm 6198 cm and 6210 cm The average of these values is 6203 cm How would you report the average measurement so as to communicate the uncertainty of the value Solution If we compare the original measured values we see that they vary in both the tenth and hundredth decimal places Because we only report one uncertain decimal place we report our answer as 620 cm The nal zero must be reported to show that we are uncertain by i01 cm When we also consider the uncertainties that arise from the dif culty in aligning the meter stick with the end of the tape and the dif culty estimating between the lines for the very tiny 01 cm divisions it is reasonable to assume that our uncertainty is no better than i01 cm 9 EXERCISE 13 Uncertainty in Measurement Let s assume that four members of your class are asked to measure the mass of a dime The reported values are 2302 g 2294 g 2312 g and 2296 g The average of these values is 2301 g Considering the values reported and the level of care you expect beginning chemistry students to take with their measurements how would you report the mass so as to communicate the uncertainty of the measurement Digital Readouts The electronic balances found in most scienti c laboratories have a digital readout that reports the mass of objects to many decimal positions For example Figure 115 shows the display of a typical electronic balance that reports values to the ten thousandth of a gram or to a tenth of a milligram As you become more experienced you will realize that you are not always justi ed in reporting a measurement to the number of positions shown on a digital readout For the purposes of this course however if you are asked to make a measurement using an instrument that has a digital readout you should report all of the digits on the display unless told to do otherwise Figure 115 The Display of a Typical Electronic Balance OBJECTIVE 18 OBJECTIVE 18 OBJECTIVE 18 23 24 Chapter 1 J An Introduction to Chemistry Chapter Chemistry The study of the structure and behavior of matter Glossary Value A number and unit that together represent the result of a measurement or calculation The distance of a particular race for example may be reported as a value of 100 meters Unit A de ned quantity based on a standard For example in the value 100 meters meter is the unit Base units The seven units from which all other units in the SI system of measurement are derived Mass The amount of matter in an object Mass can also be de ned as the property of matter that leads to gravitational attractions between objects and therefore gives rise to weight Weight A measure of the force of gravitational attraction between an object and a signi cantly large object such as the earth or the moon Matter Anything that has mass and occupies space Absolute zero Zero kelvins 0 K the lowest possible temperature equivalent to 27315 C It is the point beyond which motion can no longer be decreased Precision The closeness in value of a series of measurements of the same entity The closer the values of the measurements the more precise they are Accuracy How closely a measured value approaches the true value of a property You can test yourself on the glossary terms at the textbook s Web site Chapter Notations in the text margins show where each objective is addressed in the Objectives chapter If you do not understand an objective or if you do not know how to meet it nd the objective number in the chapter body and reread the corresponding segment of text The goal of this chapter is to teach you to do the following 1 De ne all of the terms in the Chapter Glossary Section 13 The Scienti c Method 2 Describe how science in general is done Section 14 Measurement and Units 3 Use the International System of Measurements SI base units and their abbreviations to describe length mass time temperature and volume 4 Describe the relationship between liters and cubic meters 5 State the numbers or fractions represented by the following metric pre xes and write their abbreviations giga mega kilo centi milli micro nano and pico For example kilo is 103 and is represented by k 6 Describe the relationships between the metric units that do not have pre xes such as meter gram and liter and units derived from them by the addition of pre xes For example 1 km 103 m 10 11 12 13 14 15 16 Chapter Objectives 25 Given a metric unit write its abbreviation given an abbreviation write the full name of the unit For example the abbreviation for milligram is mg Use everyday examples to describe the approximate size of a millimeter a centimeter a meter and a kilometer Use everyday examples to describe the approximate size of a milliliter a liter and a cubic meter Describe the relationship between cubic centimeters and milliliters Describe the relationship between mass and weight Name the two factors that cause the weight of an object to change Use everyday examples to describe the approximate size of a gram a kilogram and a megagram Describe the relationships between metric tons kilograms and megagrams Describe the Celsius Fahrenheit and Kelvin temperature scales Describe the relationship between a degree Celsius a degree Fahrenheit and a kelvin Section 15 Reporting Values from Measurements 17 18 19 Given values for a series of measurements state the precision of the measurements Report measured values so as to show their degree of uncertainty Given a value derived from a measurement identify the range of possible values it represents based on the assumption that its uncertainty is il in the last position reported For example 80 mL says the value could be from 79 mL to 81 mL 26 Chapter 1 J An Introduction to Chemistry T Key Ideas Complete the following statements by writing one of these words or phrases in each blank 0 OC kg 100 0C magnitude 32 F m 212 F mass accuracy meter applications observation base units precision Celsius pre xes certain published data research derive research or experimentation distance second estimated standard hypothesis temperature hypothesizing and testing time K uncertain kelvin uncertainty kilogram 1 Complete this brief description of common steps in the development of scientific ideas The process begins with and the collection of Next scientists make an initial This leads to a more purposeful collection of information in the form of systematic The hypothesis is refined on the basis of the new information and is designed to test the hypothesis The results are so that other scientists might repeat the research and confirm or refute the conclusions If other scientists confirm the results the hypothesis becomes accepted in the scientific community The next step of this scientific of the new ideas This often leads in order to refine the applications method is a search for useful to another round of In the past as measuring techniques became more precise and the demand for accuracy increased the on which people based their units were improved The which has an abbreviation of is the accepted SI base unit for length The which has an abbreviation of is the accepted 31 base unit for mass The which has an abbreviation of s is the accepted 31 base unit for The kelvin which has an abbreviation of is the accepted 31 base unit for Many properties cannot be described directly with one of the seven SI Rather than create new definitions for new units we units from the units of meter kilogram second kelvin mole ampere and candela Chapter Problems 27 8 Because the SI base units such as the meter and derived units such as the liter are not always a convenient size for making measurements of interest to scientists a way of deriving new units that are larger and smaller has been developed Scientists attach to the base units which have the effect of multiplying or dividing the base unit by a power of 10 9 An object s weight on the surface of the earth depends on its and on the between it and the center of the earth 10 For the Celsius scale the temperature at which water freezes is defined as and the temperature at which water boils is defined as For the Fahrenheit scale which is still commonly used in the United States the temperature at which water freezes is defined as and the temperature at which water boils is defined as 11 The thermometers that scientists use to measure temperature generally provide readings in degrees but scientists usually convert these values into values to do calculations 12 All measurements are to some degree Scientists are very careful to report the values of measurements in a way that not only shows the measurements but also reflects its degree of The uncertainty of a measurement is related to both the and the of the measuring instrument 13 One of the conventions that scientists use for reporting measurements is to report all of the digits and one and thus uncertain digit Chapter Section 13 The Scienti c Method Problems 14 Describe how science in general is done OBJECTIVE 2 Section 14 Measurement and Units 15 Complete the following table by writing the property being measured mass OBJECTIVE 3 length volume or temperature and either the name of the unit or its OBJECTIVE 7 abbreviation Type of Type of Unit measurement Abbreviation Unit measurement Abbreviation megagram nanometer mL K 16 Complete the following table by writing the property being measured mass OBJECTVE 3 length volume or temperature and either the name of the unit or its OBJECTIVE 7 abbreviation Type of Type of Unit measurement Abbreviation Unit measurement Abbreviation GL kilogram micrometer 0C 28 Chapter 1 J An Introduction to Chemistry OBJECTIVE 6 OBJECTIVE 1O OBJECTIVE 14 OBJECTIVE 4 OBJECTIVE 6 OBJECTIVE 14 OBJECTIVE 8 OBJECTIVE 9 17 18 19 20 21 22 23 24 25 26 27 28 Convert the following ordinary numbers to scienti c notation See Appendix B at the end of this text if you need help with this C 0001 d 0000000001 a 1000 b 1000000000 Convert the following ordinary numbers to scienti c notation See Appendix B at the end of this text if you need help with this c 00001 d 001 a 10000 b 100 Convert the following numbers expressed in scienti c notation to ordinary numbers See Appendix B at the end of this text if you need help with this a 107 c 10 7 b 1012 d 1012 Convert the following numbers expressed in scienti c notation to ordinary numbers See Appendix B at the end of this text if you need help with this a 105 c 10 5 b 106 d 106 Complete the following relationships between units a m 1km d cm31mL b L 1 mL e kg 1 t t metric ton c g 1 Mg Complete the following relationships between units a g1Gg dL1m3 b L 1 ML e Mg 1 t t metric ton c m 1 nm Would each of the following distances be closest to a millimeter a centimeter a meter or a kilometer a the width of a bookcase b the length of an ant c the width of the letter t in this phrase d the length of the Golden Gate Bridge in San Francisco Which is larger a kilometer or a mile Which is larger a centimeter or an inch Which is larger a meter or a yard Would the volume of each of the following be closest to a milliliter a liter or a cubic meter a a vitamin tablet b a kitchen stove and oven c this book Which is larger a liter or a quart Chapter Problems 29 29 Which is larger a milliliter or a uid ounce 30 Would the mass of each of the following be closest to a gram a kilogram or a OBJECTIVE 13 metric ton a a Volkswagen Beetle b a Texas style steak dinner c a pinto bean 31 Explain the difference between mass and weight OBJECTIVE 11 32 Which is larger a gram or an ounce 33 Which is larger a kilogram or a pound 34 Which is larger a metric ton or an English short ton There are 2000 pounds per English short ton 35 On July 4 1997 after a 7 month trip the Path nder spacecraft landed on Mars OBJECTIVE 12 and released a small robot rover called Sojourner The weight of an object on Mars is about 38 of the weight of the same object on Earth a Explain why the weight of an object is less on Mars than on Earth b Describe the changes if any in the mass and in the weight of the rover Sojourner as the Path nder spacecraft moved from Earth to the surface of Mars Explain your answer 36 Which is larger a degree Celsius or a degree Fahrenheit OBJECT39VE 16 37 How does a degree Celsius compare to a kelvin OBJECTIVE 16 38 Which is the smallest increase in temperature 10 C such as from 100 C to OBJECTIVE 16 110 C 10 K such as from 100 K to 110 K or 10 F such as from 100 F to 110 F 39 Is the temperature around you now closer to 100 K 200 K or 300 K Section 15 Reporting Values from Measurements 40 You nd an old bathroom scale at a garage sale on your way home from getting a physical exam from your doctor You step on the scale and it reads 135 lb You step off and step back on and it reads 134 lb You do this three more times and get readings of 135 lb 136 lb and 135 lb a What is the precision of this old bathroom scale Would you consider this OBJECTIVE 17 adequate precision for the type of measurement you are making b The much more carefully constructed and better maintained scale at the doctor s of ce reads 126 lb Assuming that you are wearing the same clothes that you wore when the doctor weighed you do you think the accuracy of the old bathroom scale is high or low 41 Given the following values that are derived from measurements what do you OBJECTIVE 19 assume is the range of possible values that each represents a 305 m the length of a whale b 612 g the mass of a basketball c 198 m Michael Jordan s height d 91096 x 10 28 g the mass of an electron e 15 X 1018 m3 the volume of the ocean 30 Chapter 1 J An Introduction to Chemistry OBJECTVE 19 42 If you are given the following values that come directly or indirectly from measurements what will you assume is the range of possible values that they represent a 22 L the volume of a basketball b 3 pm the diameter of a hair c 20 X 10 6 nm the diameter of a proton d 198 X 105 kg the mass ofa whale 43 The accompanying drawings show portions of metric rulers on which the numbers correspond to centimeters The dark bars represent the ends of objects being measured OBJECTIVE 18 39 39 7 8 9 7 8 9 a If you were not given any speci c instructions for reporting your values what length would you record for each of these measurements b If you were told that the lines on the ruler are drawn accurately to i0l cm how would you report these two lengths OBJECTIVE 18 44 The images below show a typical thermometer at two different temperatures The units are degrees Celsius 30 20 a If you were not given any speci c instructions for reporting your values what temperature would you record for each of these readings If you were told that the lines on the thermometer are drawn accurately to i1 C how would you report these two temperatures 45 At a track meet three difFerent timers report the times for the winner of a 100 m sprint as 1051 s 1032 s and 1043 s The average is 1042 s How would you report the time of the sprinter in a way that re ects the uncertainty of the measurements 46 Suppose that ve students read a thermometer and reported temperatures of 866 C 868 C 866 C 868 C and 870 C The average of these values is 868 C How would you report this average to re ect its uncertainty 47 The image below represents the digital display on a typical electronic balance 111 11 W lIlI l I I a If the reading represents the mass of a solid object that you carefully cleaned and dried and then handled without contaminating it how would you report this mass b Now assume that the reading is for a more casually handled sample of a liquid and its container Let s assume not only that you were less careful with your procedure this time but also that the liquid is evaporating rapidly enough for the reading to be continually decreasing In the amount of time that the container of liquid has been sitting on the pan of the balance the mass reading has decreased by about 0001 g How would you report the mass Discussion Topic 48 Develop your own system of measurement for length mass and volume based on the objects in your immediate surroundings Your system should have clearly de ned units and abbreviations a What unit could you use to measure the length of a desk the distance from here to the moon the diameter of an atom What abbreviations could you use for each of these units b What unit could you use to measure the mass of this bookthe mass of Godzilla the mass of a cat s whisker What abbreviations could you use for each of these units c What unit could you use to measure the volume of the water that a bathtub holdsthe volume of the planet Jupiterthe volume of a pencil What abbreviations could you use for each of these units Chapter Problems OBJECTIVE 18 OBJECTIVE 18 OBJECTIVE 19 31 CHAPTER 2 MEASUREMENT AND PROBLEM SOLVING Problems 164 6988 91 120 123 124 21 Measuring Global Temperatures measurement a number with attached units When scientists collect data it is important that they record the measurements as accurately as possible and the measurements must reflect the accuracy and precision of the instruments used to collect that data Consider the following plot of global landocean temperatures based on measurements taken from meteorological stations and ship and satellite temperature SST measurements Global LandsOcean TEmpeI allure Anomaly 9C 6 I I i I I L IIdu1a II400IruIImuImudulad4IIrLIEIIIIII gaIII1ILII0Iduudn IIIAHIIE1IIII1LIIIQJIIR15i 10 i I I I II I I Hi I I I I 1 7l I i E I E E I I 39I E I I i I i i l I I I II F I F39 1 l i E n 39 s I El i i Ii I E 39u quotquotquotquot39T3939quot39quotquotquotiquotquotquot39quot39T39 quotquot39quotquotquotquotquotquotquot39quotquot39 39quotquot quot39 I 11 IgI II no IIIII IUIIIIIIIIIEIui111 IIIIE I I i1 l i 1 h r I Ig I F l i I III I I I hi I I El i I 7 a Lquotquot539Y E39l 1EEETEFEETIEEEEE I E39ILquotEE quot awe I II I l I II I n I l I ll I i i I I g I A 39 a I I I I I l I I I l I II I IiKI m g 1 m I I4 I 11 I 39 39 a ll 7 l l I I II I i I II I II I I I I I I i H 1 l I I a 39 39l l III 1 I V II I I Al I n I quot quot l ll i u I lrll l 39I I I E z I IhT l l 1330 19an 15920 1940 39 1960 1931 20 Source Hansen J Mki Sato R Ruedy K Lo DW Lea and M MedinaElizade 2006 Global temperature change Proc Natl Acad Sci 103 1428814293 doi101073pnas0606291103 httppubsgissnasagovabstracts2006HansenetaL 1html 39 ll VJ The plot above shows annual mean average temperatures in black 5year mean temperatures in red and the uncertainty as green bars Ex 1 Based on this plot how have global landocean temperatures changed since the 19503 CHEM 121 Chapter 2 v0912 page 1 of 17 23 SIGNIFICANT FIGURES or SIG FIGS Writing Numbers to Reflect Precision To measure one uses instruments tools such as a ruler balance etc All instruments have one thing in common UNCERTAINTY 9 INSTRUMENTS CAN NEVER GIVE EXACT MEASUREMENTS When a measurement is recorded all the given numbers are known with certainty given the markings on the instrument except the last number is estimated gt The digits are significant because removing them changes the measurement39s uncertainty Thus when measurements are recorded they are recorded to one more decimal place than the markings for analog instruments they are recorded exactly as displayed on electronic digital instruments LENGTH generally reported in meters centimeters millimeters kilometers inches feet miles Know the following EnglishEnglish conversions 1 foot 5 12 inches 1 yard 5 3 feet Example Using Rulers A B and C below indicate the measurement to the line indicated for each ruler Assume these are centimeter rulers so show the each measurement has units of cm Circle the estimated digit for each measurement Ru39erA I I I I I I 0 1 2 3 4 5 IIIIIIIIIIIIIIIIIIIIIIIlllllllllllll 39 4 0 1 2 4 5 RulerC 41 42 43 44 Measurement Increment of the smallest markings on ruler of sig figs Thus a measurement is always recorded with one more digit than the smallest markings on the instrument used and measurements with more sig figs are usually more accurate CHEM 121 Chapter 2 v0912 page 2 of 17 Guidelines for Sig Figs if measurement is given Count the number of digits in a measurement from left to right 1 When there is a decimal point For measurements greater than 1 count all the digits even zeros 624 cm has 3 sig figs 50 m has 2 sig figs 186100 g has 6 sf For measurements less than 1 start with the first nonzero digit and count all digits even zeros after it 0011 mL and 000022 kg each have 2 sig figs 2 When there is no decimal point Count all nonzero digits and zeros between nonzero digits eg 125 g has 3 sig figs 1007 mL has 4 sig figs Placeholder zeros may or may not be significant eg 1000 may have 1 2 3 or4 sig figs Example Indicate the number of significant digits for the following a 1653g c 9040m e 019600g b 105 cm d 10000L f 00050 cm 25 THE BASIC UNITS OF MEASUREMENT VOLUME Amount of space occupied by a solid gas or liquid generally in units of liters L milliliters mL or cubic centimeters cm3 Know the following 1 L a 1 dm3 1 mL 5 1 cm3 These are both exact Note When the relationship between two units or items is exact we use the to mean equals exactly rather than the traditional sign also know the following equivalents in the English system 1 gallon a 4 quarts 1 quart a 2 pints 1 pint a 2 cups MASS a measure of the amount of matter an object possesses measured with a balance and NOT AFFECTED by gravity usually reported in grams or kilograms WEIGHT a measure of the force of gravity usually reported in pounds abbreviated lbs mass 16 weight mass gtlt acceleration due to gravity CHEM 121 Chapter 2 v0912 page 3 of 17 Mass is not affected by gravity Mass 2 68 7 Weight 1501b I i Mass 2 68 kt WE I Th i E 25 lb I l L Mass 3 63 kg Earth Moon Weight 1 lb 22 SCIENTIFIC NOTATION Some numbers are very large or very small gt difficult to express Avogadro s number 602000000000000000000000 an electron s mass 0000 000 000 000 000 000 000 000 000 91 kg To handle such numbers we use a system called scientific notation Regardless of their magnitude all numbers can be expressed in the form Nx10quot where Ndigit term a number between 1 and 10 so there can only be one number to the left of the decimal point n an exponent a positive or a negative integer whole To express a number in scientific notation Count the number of places you must move the decimal point to get N between 1 and 10 Moving decimal point to the right if lt 1 gt negative exponent Moving decimal point to the left if gt 1 gt positive exponent Example Express the following numbers in scientific notation to 3 sig figs 555000 gt 0000888 gt 602000000000000000000000 gt CHEM 121 Chapter 2 v0912 page 4 of 17 Some measurements may be rounded to a number of sig figs requiring scientific notation For example Express 1000 g to 3 sig figs gt Express 1000 g to 2 sig figs gt Express 1000 g to 1 sig fig gt UNBIASED ROUNDING or ROUNDTOEVEN METHOD How do we eliminate nonsignificant digits It first nonsignificant digit lt 5 just drop the nonsignificant digits It first nonsignificant digit 2 5 raise the last sig digit by 1 and drop nonsignificant digits eg 314501 L 315 since the nonsig figs are 501 in 314501 Express each of the following with the number of sig figs indicated a 64675 M b 236500 3si9fi93 gt c 6455 3391193 gt d 000123456 3591195 gt e 1234567 M f 1975 2391195 gt When necessary express measurements in scientific notation to clarify the number of sig figs 24 SIGNIFICANT FIGURES IN CALCULATIONS ADDINGISUBTRACTING MEASUREMENTS When adding and subtracting measurements your final value is limited by the measurement with the largest uncertainty Le the measurement with the fewest decimal places MULTIPLYINGIDIVIDING MEASUREMENTS When multiplying or dividing measurements the final value is limited by the measurement with the least number of significant figures CHEM 121 Chapter 2 v0912 page 5 of 17 EX 1 74333 g 825 g 10781 g Ex 2 1350 cm x 795 cm x 400 cm EX 3 975 mL 735 mL 101755 g 2575 cm x 1025 cm x 850 cm Ex 4 MULTIPLYINGIDIVIDING WITH EXPONENTIAL NUMBERS When multiplying or dividing measurements with exponents use the digit term N in Ngtlt10quot to determine number of sig figs Ex 1 602x10234155gtlt109 250131x1033 How do you calculate this using your scientific calculator Step 1 Enter 602x 1023 by pressing 602 then EE or EXP which corresponds to gtlt10 then 23 gt Your calculator should look similar to 602 X1023 Step 2 Multiply by pressing x Step 3 Enter 4155x 109 by pressing 4155 than EE or EXP which corresponds to gtlt10 then 9 gt Your calculator should look similar to 4155 x109 Step 4 Get the answer by pressing 250131 X1033 gt Your calculator should now read The answer with the correct of sig figs Be sure you can do exponential calculations with your calculator Many calculations we do in chemistry involve numbers in scientific notation Ex 2 325x101286x1042795x1017 t rre tt fsi9 gs gt 15 Ex 3 M4357931435x1010 quote fsig gs gt 8605 x104 CHEM 121 Chapter 2 v0912 page 6 of 17 SIGNIFICANT DIGITS AND EXACT NUMBERS Although measurements can never be exact we can count an exact number of items For example we can count exactly how many students are present in a classroom how many MampMs are in a bowl how many apples in a barrel We say that exact numbers of objects have an infinite number of significant figures 26 CONVERTING FROM ONE UNIT TO ANOTHER or DIMENSIONAL ANALYSIS UNIT EQUATIONS AND UNIT FACTORS Unit equation Simple statement of two equivalent values Conversion factor unit factor eguivalents Ratio of two equivalent quantities Unit eguation Unit factor 1 dollar 10 dimes 10 dimes 1 dollar 1 dollar E10 dimes Unit factors are exact if we can count the number of units equal to another For example the following unit factors and unit equations are exact 7 days 24 hours 1 gallon 100 cm 1 week 1 day 4 quarts 1 m and 1 yard 2 3 feet Exact equivalents have an infinite number of sig figs gt never limit the number of sig figs in calculations Other equivalents are inexact or approximate because they are measurements or approximate relationships such as 161 km 55 miles 454 g 1 mile 1 hour lb Approximate equivalents do limit the sig figs for the final answer 27 SOLVING MULTSTEP CONVERSION PROBLEMS or DIMENSIONAL ANALYSIS PROBLEM SOLVING 1 Write the units for the answer 2 Determine what information to start with 3 Arrange all unit factors showing them as fractions with units so all of the units cancel except those needed for the final answer 4 Check for the correct units and the correct number of sig figs in the final answer CHEM 121 Chapter 2 v0912 page 7 of 17 Example 1 If a marathon is 262 miles then a marathon is how many yards 1 mile25280 feet 1 yardE3 feet Example 2 You and a friend decide to drive to Portland which is about 175 miles from Seattle If you average 99 kilometers per hour with no stops how many hours does it take to get there 1 mile 1609 km Example 3 The speed of light is about 2998gtlt108 meters per second Express this speed in miles per hour 1 mile1609 km 1000 mE1 km 25 Basic Units of Measurement International System or SI Units from French quotIe ysteme1nternational d Unit s standard units for scientific measurement Metric system A decimal system of measurement with a basic unit for each type of measurement quantity basic unit symbol quantity SI unit symbol length meter m length meter m mass gram g mass kilogram kg volume liter L time second s time second s temperature Kelvin K CHEM 121 Chapter 2 v0912 page 8 of 17 Metric Prefixes Multiples or fractions of a basic unit are expressed as a prefix gt Each prefix power of 10 gt The prefix increases or decreases the base unit by a power of 10 Prefix Symbol MultipleFraction kilo k 1000 deci d 01 E i 10 centi c 001 a L 100 milli m 0001 E L 1000 1 39 0000 001 E mmquot M Greek m 1000000 KNOW the metric units above Metric Conversion Factors Ex 1 Complete the following unit equations a 1dolarE cents gt 1 ms cm b 1 dollar dimes gt 1 m 2 dm Note To help remember the number of centimeters or decimeters in a meter just think of the number of cents or dimes in a dollar Ex 2 Complete the following unit equations a 1kg g c1LE mL e 1m mm b 1g cg d 132 ds f 1g ug Note Although scientists use pg to abbreviate microgram hospitals avoid using the Greek letter 11 in handwritten orders since it might be mistaken for an m for mii ie an order for 200 ug might be mistaken to be 200 mg which would lead to an overdose that s 1000 times more concentrated Instead hospitals use the abbreviation mcg to indicate micrograms CHEM 121 Chapter 2 v0912 page 9 of 17 Writing Unit Factors Example Complete the following unit equations then write two unit factors for each equation a 1km m b 1g mg MetricMetric Conversions Solve the following using dimensional analysis Ex 1 Convert 175 ms into units of seconds EX 2 Convert 0120 kilograms into milligrams Ex 3 Convert 3OOgtlt1O8 ms into kilometers per hour Ex 4 Convert 350gtlt1O7 cm to units of kilometers CHEM 121 Chapter 2 v0912 page 10 of 17 MetricEnglish Conversions Enqlish system Our general system of measurement Scientific measurements are exclusively metric However most Americans are more familiar with inches pounds quarts and other English units gt Conversions between the two systems are often necessary These conversions will be given to you on quizzes and exams Quantity English unit Metric unit English Metric conversion length 1 inch in 1 cm 1 in a 254 cm exact mass 1 pound lb 1 g 1 lb 4536 g approximate volume 1 quart qt 1 mL 1 qt 946 mL approximate Ex 1 What is the mass in kilograms ofa person weighing 155 lbs Ex 2 A 20L bottle can hold how many cups of liquid 1 qt 2 2 pints 1 pint E 2 cups Ex 3 A lightyear about 588x1012 miles is the distance light travels in one year Calculate the speed of light in meters per second 1 mile1609 km CHEM 121 Chapter 2 v0912 page 11 of 17 Temperature A measure of the average energy of a single particle in a system The instrument for measuring temperature is a thermometer Temperature is generally measured with these units Fahrenheit scale F Celsius scale C References English system Metric system freezing point for water 32 F 0 C boiling point for water 212 F 100 C Nice summer day in Seattle 77 F 25 C Conversion between Fahrenheit and Celsius scales C F 32 39 18 Kelvin Temperature Scale There is a third scale for measuring temperature the Kelvin scale The unit for temperature in the Kelvin scale is Kelvin K NOT K The Kelvin scale assigns a value of zero kelvins 0 K to the lowest possible temperature which we call absolute zero and corresponds to 27315 C The term absolute zero is used to indicate the theoretical lowest temperature Conversion between C and K Ex 1 Liquid nitrogen is so cold it can be used to make a banana hammer If liquid nitrogen s temperature is 77 K calculate the equivalent temperature in C and in F K C273 F Cx18 32 CK273 CHEM 121 Chapter 2 v0912 page 12 of 17 Determining Volume Volume is determined in three principal ways 1 The volume of any liquid can be measured directly using calibrated glassware in the laboratory eg graduated cylinder pipets burets etc 2 The volume of a solid with a regular shape rectangular cylindrical uniformly spherical or cubic etc can be determined by calculation 3 Volume of solid with an irregular shape can be found indirectly by the amount of liquid it displaces This technique is called volume by displacement VOLUME BY CALCULATION The volume of a rectangular solid can be calculated as follows volume length x width x thickness Ex 1 What is the volume of a gold bar that is 525 cm long 350 cm wide and 275 cm thick Ex 2 A rectangular bar of gold with a volume of 355 cm3 is 750 cm long and 350 cm wide How thick is the bar VOLUME BY DISPLACEMENT a Fill a graduated cylinder halfway with water and record the initial volume b Carefully place the object into the graduated cylinder so as not to splash or lose any water c Record the final volume d Volume of object final volume initial volume CHEM 121 Chapter 2 v0912 page 13 of 17 Example What is the volume of the piece of green jade in the figure below W llllllllllllllllll 29 DENSITY The amount of mass in a unit volume of matter density w or d generally in units of glcm3 or gImL volume 3 m g 3 For water 100 g of water occupies a volume of 100 cm d 2100 gcm v 100 cm3 Density also expresses the concentration of mass ie the more concentrated the mass in an object gt the heavier the object gt the higher its density Sink or Float Some objects float on water eg a cork but others sink eg a penny Thus objects with a higher density than a liquid will sink in the liquid but those with a lower density than the liquid will float L1 Since water39s density is about 100 gcm3 a cork39s density must be less than 100 gcm and a penny39s density must be greater EX Consider the figure at the right and the following solids and liquids and their densities ice d0917 gcm3 honey d150 gcm3 iron cube 787 gcm3 hexane d065 gcm3 L2 rubber cube d119 gcm3 S ldentify L1 L2 81 and 82 by filling in the blanks below 3 L1 and L2 S1 2 and S3 CHEM 121 Chapter 2 v0912 page 14 of 17 Applying Density as a Unit Factor Given the density for any matter you can always write two unit factors For example the density of ice is 0917 gcm3 0917g cm3 or Two unit factors would be 3 cm 0917g EX 1 Give 2 unit factors for each of the following a density of lead 113 gcm3 b density of chloroform 148 gmL Ex 2 Aluminum has a density of 270 gcm3 What is the volume in mL of a piece of aluminum with a mass of 0125 kg Ex 3 Ethanol is used in alcoholic beverages and has a density of 0789 gmL What is the mass of ethanol that has a volume of 150 L EX 4 A chunk of silver metal weighing 168 g is placed in a graduated cylinder with 210 mL of water The volume of water now reads 370 mL Calculate the density of silver CHEM 121 Chapter 2 v0912 page 15 of 17 CALCULATING PERCENTAGES Percent Ratio of parts per 100 parts gt 10 is 25 is 2 5 etc 100 100 To calculate percent divide one quantity by the total of all quantities in sample onepa x100 total sample Percentage Ex 1 In a chemistry class with 25 women and 20 men what percentage of the class is female What percentage is male Express your answers to 3 sig figs Writing out Percentage as Unit Factors Ex 1 Water is 888 oxygen by mass Write two unit factors using this info Ex 2 Pennies cast between 1963 and 1982 are a mixture of 950 copper and 50 zinc by mass Write four unit factors using this information CHEM 121 Chapter 2 v0912 page 16 of 17 Percentage Practice Problems Ex 1 An antacid sample was analyzed and found to be 100 aspirin by mass What mass of aspirin is present in a 350 g tablet of antacid Ex 2 Water is 888 oxygen and 112 hydrogen by mass How many grams of hydrogen are present in 2500 g about a cup of water Ex 3 Pennies cast between 1963 and 1982 are a mixture of 950 copper and 50 zinc Calculate the mass of copper present in a 2495 g penny cast in 1968 Ex 4 Calculate the mass of pennies cast in the 1970s that contains 100 lbs of copper 1 lb 4536 g CHEM 121 Chapter 2 v0912 page 17 of 17 CHAPTER 4 ATOMS AND ELEMENTS Problems 170 then after Chapter 9 complete 7194 103104 107108 113114 41 Experiencing Atoms at Tiburon atom smallest identifiable unit of an element A matter is made up of atoms gt The properties of specific atoms determine the properties of matter with those atoms There are currently 91 naturally occurring elements and 20 manmade elements 42 lndivisible The Atomic Theory Greek philosophers were the first to propose explanations for what was observed in nature Surprisingly some of these Greek ideas led to similar modern ideas Democritus 462370 BC proposed that all matter was made up of tiny indivisible particles called atomos meaning indivisible or atoms Empedocles 490430 BC suggested all matter was composed of four basic elements air water fire and earth Aristotle 384321 BC accepted Empedocles idea and added a fifth element heavenly ether which is perfect eternal and incorruptible Even Democritus ideas were more correct Aristotle s idea of five basic elements was accepted for over 2000 years until John Dalton proposed the modern theory of atoms in 1808 7 7quot 7 John Dalton s Modern Atomic Theory El I I S 1 An element is composed of tiny indivisibe particles 1 W 39 called atoms 2 All atoms of an element are identical and have the 1391 I t l 1 u same mass and properties 3 Atoms of one element will differ in mass and properties j 7 m from atoms of another element H l 4 Atoms combine in small whole number ratios to form compounds eg a H20 molecule has one 0 atom and 2 H atoms 5 Atoms can combine to form different compounds eg carbon and oxygen combine to form C02 or CO Later proven wrong g 151U1Hy m CHEM 121 TM Chapter 4 v0912 page 1 of 11 43 THE NUCLEAR ATOM Subatomic Particles Michael Faraday William Crookes and many other scientists carried out experiments gt discovery of electrons e39 tiny negatively charged subatomic particles JJ Thomson was given credit for discovering electron although evidence had accumulated for 20 years before his research group s determination of the electron s charge and mass Eugen Goldstein late 18803 carried out experiments on canal rays and found they consisted of positively charged subatomic particles gt discovery of protons p Sphere uf peeitive charge Eectren PLUMPUDDING MODEL OF THE ATOM a f Thomson proposed that the atom was a uniform sphere of I g positively charged matter in which electrons were embedded V 39 gt Electrons are like raisins in a pudding of protons THE NUCLEAR ATOM PROTONS AND THE NUCLEUS Ernest Rutherford was a scientist who did many pioneering experiments in radioactivity He had members of his research group test Thomson s PlumPudding Model using radioactive alpha or particles The 0L particles are positively Seine c perticlee are scattered Ch 3 rg ed h el I u m ato ms39 fat Meet ct particles gen etreiglht tlhrcugh 39 r Rutherford39s AlphaScattering Experiment Alpha 0L particles were shot at a thin gold foil only a few atoms thick A circular detector was set up Beam of e around the foil to see what WW happens to the or particles I lf Plumpudding Model was correct the or particles which are much bigger than electrons should go through the foil like bullets through tissue paper quot 3 Geld feiil cnly a few eteme thick Screen tc detect e perticlee quotSource of c perticlee Experimental results Most of the or particles went straight through but some were deflected and a few even bounced back CHEM 121 TM Chapter 4 v0912 page 2 of 11 be particles Nucleus quotif W gquot lf39 ll llquot if atom based on Plumpudding Model atom based on the Nuclear Medial Rutherford s interpretation of the results Most alpha 0c particles pass through foil gt An atom is mostly empty space with electrons moving around that space A few 0L particles are deflected or even bounce back gt Atom must also contain a very small dense region and particles hitting this region are deflected or bounce back towards source gt small dense region atomic nucleus contains atom s protons gt Why this is called the Nuclear Model of the Atom Rutherford also estimated the size of the atom and its nucleus nucleus d1015 m atom diameter 101O m But these dimensions are difficult for us to imagine Ex 1 An atom is 100000 times 105 or 5 orders of magnitude bigger than its nucleus If a nucleus size of a small marble 1 cm in diameter indicate the length in meters then identify a common item that corresponds to that size for the following a 10 times bigger dm m b 100 times bigger m c 1000 times bigger m d 10000 times bigger m e 100000 times bigger m km CHEM 121 TM Chapter 4 v0912 page 3 of 11 44 The Properties of Protons Neutrons and Electrons Decades later James Chadwick won the Nobel Prize winner for his discovery 1935 gt neutron n neutral subatomic particle awn I al39 5 I E 39E I warren if 39 e Atoms are made up of subatomic particles 5251ng L L I 39 is rahLrnJ 5 J J 1 g r a v we 1 electron e39 negatively charged subatomic particle charge 1 HEUTRQQZ I proton p positively charged subatomic particle charge 1 I I I5 1 Fa a neutron n neutral subatomic particle charge0 it w l g i 5 Particle Symbol Location Charge Relative Mass amu electron e39 outside nucleus 1 11836 z 0 proton p inside nucleus 1 1 neutron n inside nucleus 0 1 amuatomic mass units Thus most of the mass of an atom comes from the protons and neutrons in the nucleus What is electrical charge There are 4 fundamental forces gravity electromagnetic force strong force weak force Let s focus on the electromagnetic force which consists of electricity and magnetism electrostatic force the force resulting from a charge on an object Two objects with the same charge both negative or both positive repel one another Two objects with unlike charges one negative and one positive attract one another Electrical charge is a fundamental property of protons and electrons Positive and negative charges cancel one another gt When paired a proton and an electron cancel one another s charges gt neutral Note that matter is usually chargeneutral or neutral If charge imbalances occur they are usually equalized often in dramatic ways Charge imbalance may occur if you walk across carpet and it s equalized when you get a shock touching a doorknob or other piece of metal During electrical storms charge imbalances are usually equalized with stunning displays of lightning CHEM 121 Tro Chapter 4 v0912 page 4 of 11 45 Elements Defined by Their Number of Protons NAMES amp SYMBOLS OF THE ELEMENTS Every element or atom has an individual name symbol and number Some names come Greek hydrogen comes from hydro water former argon comes from argon inactive Convention for writing chemical symbols Use first letter capitalized of element name hydrogen gt H carbon gt C If symbol already used include second letter in lower case of name helium gt He calcium gt Ca cobalt gt Co Some symbols come from Latin names eadplumbum gt Pb god shining dawn aurum gt Au Know the names and symbols for the first 20 elements on the Periodic Table for Exam 1 Element names and symbols for the all of the elements are given in the front cover You will be given a Periodic Table with only the symbols Given the name know the symbol or given the symbol be able to write the name Spelling of elements names counts COMPOUNDS 8 CHEMICAL FORMULAS chemical formulas Symbolically express the number of atoms of each element in a compound Number of atoms is indicated by a subscript following the element s symbol If there is no subscript only one atom of that element is in the compound Example water H20 gt 2 H atoms 1 O atom sodium carbonate NazCO3 gt Na C O atoms Some chemical formulas use parentheses gt more than one subunit present in compound Example Ba3PO42 gt 3 Ba and 2 P04 2 P and 8 O gt TOTAL 2 Ba 2 P and 8 0 How many atoms of each element are present in TNT C7H5N023 C H N and O CHEM 121 TM Chapter 4 v0912 page 5 of 11 48 ISOTOPES WHEN THE NUMBER OF NEUTRONS VARIES An element can be identified using its element name element symbol or its atomic number which indicates the number of protons gt An element will always have the same number of protons eg carbon always has 6 protons oxygen always has 8 protons etc However the number of neutrons may vary gt Atoms with differing numbers of neutrons are called isotopes The convention for distinguishing elements with various isotopes is to give the element name followed by the mass number eg carbon12 C12 carbon13 C13 and carbon14 014 are isotopes of carbon Nuclear Symbol also called Atomic Notation shorthand for keeping track of number of protons and neutrons in an atom s nucleus atomic number whole number of protons number of electrons in a neutral atom mass number of protons of neutrons in an atom s nucleus of protons of neutrons mass number A of protons of electrons atomic number 2 E 39 element symbOI Ex 1 a Write the atomic notation for sodium23 at the right b How many neutrons are in each neutral sodium23 atom Ex 2 a Write the atomic notation for magnesium26 at the right b How many neutrons are in each neutral magnesium26 atom EX 3 Fill in the table below Isotope of carbon mass of protons of neutrons of electrons carbon12 carbon13 carbon14 argon39 Fe59 CHEM 121 TM Chapter 4 v0912 page 6 of 11 49 ATOMIC MASS Atoms are too small to weigh directly eg one carbon atom has a mass of 199gtlt103923 g too inconvenient an amount to use gt need more convenient unit for an atom s mass gt atomic mass unit amu Carbon12 was chosen as the reference and given a mass value of 12 amu gt 1 amu 112th the mass of a carbon12 atom gt Masses for all other elements are measured relative to mass of a carbon12 atom Weighted Average Atomic Mass of an Element Why is carbon s mass on the Periodic Table 1201 amu NOT 1200 amu Atomic masses reported on the Periodic Table are weighted averages of the masses of all the naturally occurring isotopes for each element based on their percent natural abundance ie the percentage of existing atoms that are a specific naturally occurring isotope Ex 1 Use the atomic mass reported on the Periodic Table to determine which one of the naturally occurring isotopes is most abundant for each element below a The two naturally occurring isotopes for lithium are Circle one lithium6 lithium7 b The three naturally occurring isotopes for argon are Circle one argon36 argon 38 argon 40 c The two naturally occurring isotopes for silver Ag are Circle one Ag107 Ag109 Note You cannot simply round the atomic mass of an element to a whole number and assume that s the most abundant isotope because that isotope may not exist For example in Ex 1c above rounding the atomic mass for silver to the nearest whole number would indicate that Ag108 is the most abundance isotope but the only two naturally occurring isotopes for silver are Ag107 and Ag109 Some elements have naturally occurring isotopes that are radioactive and unstable gt distinguished on the Periodic Table with parentheses around a mass number for the most abundant radioactive isotope instead of a weighted average of the atomic masses for all naturally occurring isotopes eg the mass number is 222 for the most abundant isotope of radon Rn and the mass number is 209 for the most abundant isotope of polonium Po CHEM 121 TM Chapter 4 v0912 page 7 of 11 46 Looking for Patterns The Periodic Law and the Periodic Table PERIODIC TABLE A vertical column is called a group or family Elements belonging to the same group exhibit similar chemical properties A horizontal row is called a period or series MainGroup Representative or A Group Elements Those elements in groups 1 2 13 14 15 16 17 18 or IA to VlllA Group 1 or IA alkali metals Group 2 or IIA alkaline earth metals Group 17 or VIIA halogens Group 18 or VlllA noble gases because they are all gases that do not react Transition Metals or B Group Elements Elements in Groups 3 to 12 middle of the Periodic Table Inner Transition Elements beneath the main body of Periodic Table Lanthanide series CeLu also called rare earth metals make up lt0005 of Earth39s crust Actinide series ThLr also called transuranium elements generally all manmade and exist for only very short periods of time before decaying to other elements periodic law Elements can be arranged to display recurring properties gt We can use the Periodic Table to predict the properties of elements Dimitri Mendeleev proposed that elements display recurring properties according to increasing atomic mass gt The first Periodic Table arranged elements according to increasing atomic mass Henry G J Moseley s highenergy Xray radiation experiments of atomic nuclei gt Repeating properties of elements are more clearly reflected if elements are arranged according to increasing atomic number not increasing atomic mass gt Periodic Table s arrangement today Trends for increasing atomic mass are identical with those for increasing atomic number except for Ni amp Co Ar amp K Te amp Example Which of the following elements will behave similarly to calcium Ca Na Cl Mg 8 Sr Al Ar P CHEM 121 TM Chapter 4 v0912 page 8 of 11 METALS NONMETALS and SEMIMETALS or METALLOIDS Properties of Metals Properties of Nonmetals 0 shiny appearance 0 dull appearance 0 malleable can be pounded into flat sheets brittle Wm Shatter under 0 ductile can be drawn into a fine wire pressure 0 poor or nonconductor of 0 can conduct heat and electrICIty heat and electricity 0 Examples are gold copper aluminum etc Examples are sum graphite carbon oxygen nitrogen Properties of Metalloids or Semimetals Have properties of metals and nonmetals eg silicon Si has a shiny appearance and conducts electricity like a metal Metals are often used for pots and pans because they conduct heat they are hammered into tools and armor and they are used for wiring because they conduct electricity METALS NONMETALS and SEMIMETALS on the Periodic Table Know which elements are metals semimetals nonmetals using the Periodic Table 5 Vlliii 13 M m l 2 He A n W A 7 H i ll ll if Si m Wequot VI l 3391 39 5 E 39f E ii B C l N D F El i E T a u 13 n 3912 Iii Hquot 15 quot715 liiB we VB FIB Witi um um um m nu M 5quot P 539 31 22 El iii 26 2quot Eli Iiii FiLi 3 3 2 E3 134 3 35 SE Ti if Cr Mn Fe Cu Ni E11 2131 Ga Ge A5 Se Br tin r I 5 Nb TE Elli Rh Pd Ag Cd In Silt Slit TIE I r3 r4 5 at r FE quotm 7 m a m 35 r V 7 l 39u 39 2 Dis CHEM 121 Tro Chapter 4 v0912 page 9 of 11 SOLIDS LIQUIDS AND GASES KNOW the physical state of each element at 25 C At standard state conditions 25 C and 1 atm Only mercury Hg and bromine Br are liquid H N O F Cl and all Noble gases group VlllA are gases All other elements are solids ii 2 p x i3 i3 13 13 ii m A mm iiLJ WA 333 333 WM 3 4 39 3 3 g 3 a 3 7 3 i137 Li Be 3 2 N on 3 313 39l39l 3 Q 5 E 3 9 1D 11 12 13 V 114 15 IE 1 113 M5 1113 1333 3393 33913 WE VIII W11 Will 13 33 M 5 P S C M 133 33 23 33 r 337 23 3339 33 33 7 23 33 31 337 33 33 33 33 1K Ea Si Ti Erquot Min Fe Co Ni Cu 213 Ga Ga As 52 Eur Kr 33 33 33 33 3 1 33 33 33 33 33 V 33 33 43 3 33 32 33 31 Rh 5quot Z M It Ru I39d Ag C Sift TE I ELEMENTAL SYMBOLS AND THE PERIODIC TABLE Know the names and symbols for all the elements included below as well as Sr strontium titanium Ti and U uranium Spelling counts WE HQ M 7 I33 CHEM 121 TM Chapter 4 v0912 page 10 of 11 3 Be sure the learn the American spelling shown below NOT the British spelling included on Commits Elements some tables found on the Web Altnil iu Atomic Mu mini Number Symbol Element Number Symbol Element Number Symbol Element l H H ydrugcn l A I t l M Hill in 2 3 Ni N ickcl 12 Hi Hi l tum l 4 5 ll ltl 2 9 Cu Cu piper Li LL i l h l u m l 5 IF Ph mi thorn 339 ll 1 El nit Bi Beryllium lt r Stillli ur 35v Br Bromine B E ornn l 739 l C I1 loriiiii If Kiquot K rjr pt o n in C Carbon l 8 At A rg on 394 g 3 39i I in r T N N M rtth 1 l 2 K Future um 5 H Sin Ti 1 8 U U xylgun Ell C ti Cu l El u 5 3 l ntll inn inquot F F tiler rIlL39 391 C It to mitt m 5 3 Eli Bari u ll i I ll Ni New 25 Mn Mungtine an H Mercury il l N it E l lllll39l39i El n Fl l run 37 39F39h Lead 39 2 M g ugliest u 39ITl E C in C l l lit Note We skipped the sections covering ions charged particles but we will return to those sections after we finish Chapter 9 CHEM 121 TM Chapter 4 v0912 page 1101 11 CHAPTER 1 The Chemical World Problems 118 2122 science the study of nature to explain what one observes Consider the following On the first day of school you get in your car and turn the key on the ignition but nothing happens What could be the problem 14 THE SCIENTIFIC METHOD How Chemists Think Applying the Scientific Method 1 Propose a hypothesis to explain what is observed 2 Test the validity of the hypothesis by carrying out experiments controlled observations designed specifically to verify or disprove a hypothesis Record observations and analyze the data on the system being studied gt It s important to keep good records so others can reproduce the work 3 Conduct additional experiments to test the hypothesis under various conditions If all or part of the hypothesis does not hold up to testing then it is adjusted or a new hypothesis is proposed to explain the observations lf hypothesis holds up to extensive testing it can lead to the development of a scientific or natural law andor a scientific theory scientific or natural law a simple statement or equation that summarizes past observations and predicts future ones scientific theory a tested broader and deeper explanation of observed natural phenomena Thus a scientific law summarizes what happens a scientific theory explains why it happens Example Many news reports and articles claim that global warming is just a theory How does this demonstrate the lack of understanding by the media and general public regarding scientific theories CHEM 121 TM Chapter 1 page 1 of 3 12 Chemicals Compose Ordinary Things Ex 1 What is your major and why are you taking this class Why do you think knowledge in chemistry is needed in so many different fields Ex 2 What comes to mind first when you hear the word chemical Is it positive or negative Ex 3 Are there chemicals in a cup of coffee Give some examples Thus chemicals are not necessarily hazardous In fact almost everything consists of chemicals since any substance consisting of more than one type of atom is a chemical J 39 r 139 Copyright EDGE Pearson Prentice Halli inc For example the glass of soda above contains water H20 molecules and carbon dioxide C02 molecules the lead in a pencil is actually graphite which consists of carbon atoms and we are made of DNA and various other biological molecules or chemicals CHEM 121 TM Chapter 1 page 2 of 3 13 All Things Are Made of Atoms and Molecules Consider molecules of water and carbon dioxide Water molecule Carbon dioxide molecule f 39 r 39 Oxygen Carbon Oxygen y r gen Kygen r gE atom t m atom atom atom atom lmFlaw DEW 39mlw Hid In These molecules consist of atoms and their structure at the atomic or molecular level determines their properties and behavior at the macroscopic level ie what we can observe with the naked eye The Structure and Properties of Ice Ice floats on water because ice is less dense than liquid water The strong attraction and repulsion between two different water molecules in ice results in quotholesquot or empty space in the ice crystal When ice melts the water molecules fill in the holes so liquid water is more dense than ice Note in ice s molecularlevel structure the holes have a hexagonal sixsided shape gt Snowflakes have hexagonal symmetry because of the hexagonal holes formed by the arrangement of H20 molecules in ice chemistry the science that studies how matter behaves by understanding the properties and behavior of atoms and molecules that make up the matter CHEM 121 TM Chapter 1 page 3 of 3

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I made $350 in just two days after posting my first study guide."

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.