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# PHYSICAL CHEMISTRY CHEM 455

UW

GPA 3.92

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This 30 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 455 at University of Washington taught by William Reinhardt in Fall. Since its upload, it has received 12 views. For similar materials see /class/192569/chem-455-university-of-washington in Chemistry at University of Washington.

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Date Created: 09/09/15

Along come Einstein and de Broglie Einstein explains Planck s ansatz 8V n hv de Broglie suggests that if waves eg light might be particles then fair is fair particles should also be waves By use of analogies with Einstein s theory of relativity de Broglie suggested that x hmv hp Here h is again Planck s constant mv p is the classical momentum of a particle of mass m moving at velocity and 9 is the wavelength of the particle Waveparticle duality X rays and electrons diffracting through Al foil In fact Davission and Germer working at ATampT Bell Labs in New Jersey had already seen diffraction patterns due to the wave nature of electrons but had ignored their data as they assumed that X rays had contaminated their data They were immediately able to calculate h from their previously suspect data confirming the de Broglie hypothesis With the de Broglie formula we are ready to carry out Bohr s quantization of the H atom and give the first correct explanation of the spectrum of an atom or molecule by combining Bohr s stationary orbits with Einstein s photons 7 7 7 Eon Ir 7r 7 ATZL39Z Ze 39e lo the Bohr radius has the numerical value 529 X 10 11 m The velocity 377 corresponding to the orbit with radius 739 is 727 2 7771573 Z92 U77 26071 results for TH and v7 can be substituted into Equation 156 to gi is of energy 7 ZZLAWIE 86872 72 Z2 7 4218 x 10477 1123 Notes on VIT 1 23 for the non degenerate and degenerate cases and a summary of Quantum Measurment tools A The Three Very Imporant Theorems the case of non degeneracy 1 Hermitian operators have real eigenvalues To prove A is Hermetian and has an eigenfucntion a with eigenvalue quotaquot then quotaquot is a real number Proof By assumption TA 3TCET fTA ar arclr and A a r a aT The integrations coordinates and range of integration denoted by r are appropriate to the problem at hand Then Mar a am cur L a ar 53m air or a r a r 6E7 a fT T a r ctr from which a a from which VT1 follows QED 2 Chem455A Lecture 12 13 nb 2 Hermitian linear operators have orthogonal eigenfunctions Let us assume that Hermitian operator A has an eigenfunction a with non degenate eigenvalue quotaquot We need make no assumptions about the possible degeneracy of any of the other eigenfunctions of A To prove f1 r br ch f r a r ch 0 where the pair quotb and brquot are any other eigenvalue eigenfunction of A By assumption of the nondegeneracy of quotaquot we have a 9 b Proof AsA is Hermitian f1 TA ar ch f A b r a rcl r and thus a f r a r d7 b L r a r ch as both eigenvalues are real from WT 1 Subtracting the right side from the left gives a b r ardr 0 As a b by assumption it follows that r a r ch 0 QED 3 Commuting Hermitian linear operaors have simultaneous eigenfunctions Consider the two distinct operators AB such that AB AB BA 0 Now assume that one of the operators say A has an engenfunction b3 with nondegenerate eigenvalue a Note that we need assume nothing about the degeneracy of eigenvalues of B Chem455A Lecture 12 13 nb 3 To prove a is automatically an eigenfunction of operator B Proof as AB is the quotzero operatorquot it give zero when applied to quotany functionquot specifically AB a 0 Thus AB a BA a 0 Using the fact that a is an eigenfunction of A this may be rewritten as A B 155 a B 15a this is equation The nondegeneracy of quotaquot now implies that these is only a single linearly independent function which satisfies Eqn and that is a itself or a constant multiple of itself Note that this is the 2nd time in the proof we have used linearity Thus B a 0C a where quot0C quot denotes proportionality As these funcions B b3 and 755 are proportional there is some constant b such that B a b au Thus 75 is an eigenfunction of operator B QED REMARK note that the eigenvalues a and b are completely unrelated They usually have different units and so can39t even be compared For example one might be an quotenergyquot the other an quotangular momentumquot 4 Chem455A Lecture 12 13 nb B The Three Very lmporant Theorems the case of degeneracy 1 Eigenvalues of Hermitian operators are real This result has nothing to do with degeneracy so the proof is the same PREAMBLE to WT 2 and 3 in the case of degeneracy Before discussing VlT 2 and 3 which are actually quotchangedquot by the presence of degeneracy a few useful remarks about quotdegeneratequot eigenvalues and quotdegeneratequot eigenfuctions may be of use Suppose operator A has two and only two distinct i e linearly independent which simply means one is not simply quota constantquot times the other eigenfuctions say 1a and which have the same eigenvalue quotaquot We then say that quotaquot is doubly degenerate If there and N and only N linearly independent eigenfunctions which share the same eigenvalue then the degeneracy is Nfold We can also say that the eigenfuncions belong to an Nfold degenerate set of eigenfuctions For example the degeneracy of quotporbitalsquot is 3fold and quotdorbitalsquot is 5foldso these are not ideas irrelavent to CHEMISTRY If they were we39d happily skip them as mathematical oddities and simply move on Note also that a set of functions which share an eigenvalue of operator A most often are nondegenerate with respect to the action of other operators say B or even C An important fact which we then prove any linear combination of degenerate functions of operator A is also an eigenfunction of operator A Proof Consider A 2411 cm 2411 aim 2511 cm 2411 cm QED Note that linearity of A was used quotseveral timesquot in this simple demonsteation Note also that if any one of the funcions W in the sum didn39t have eigenvalue quotaquot then the a would not factor outsidet he sum and the result would be false Said specifically quotlinear combinations of nondegenerate eigenfuctionsquot of an operator are NOT eigenfunctions of that opertor Chem455A Lecture 12 13 nb 5 As the above result has nothing to do with quotwhat linear combinationquot has been taken it applies equally to all linear combinations Can we then generate an infintie number of eigenfuctions of operator A Of course not For a double degenerate eigenvalue only two linearly independent eigenfuctions exist and taking linear combinations does not expand the quotspacequot of degenerate functions beyond two it just gives different representaions of the same space An analog in 2D Cartesian space is that any two linearly independent vectors can serve to represent any other vector but this may be done in an infinite number of way simply by quotrotatingquot the xy axies on the pagebut nothing new is actually produced by doing that VIT 2 in the case of degeneracy quotWe can always choose to take all of the eigenfunctions of an Hermetian operator to be orthonormal but we might have to do a bit of work to find the linear combinations of degenerate eigenfunctions that makes this workquot Discussion Where do things not happen quotautomaticallyquot as in the case of a non degenerate eigenvalue of A Supose quotaquot is a multiply degenerate eigenvalue and we examine 75 and note that these superscrips are NOT powers Following the Proof of WT 2 in part A we find that the key result aa f 1r 0537 air 0 does not tell us whether fr r r air is zero or not it might be that 75 and are orthogonal but we haven39t proven it It might also be the case that they are NOT orthogonal BUT and we state this without detailed proof if 75 and are NOT orthogonal it is always possible to find two linearly independent and orthonormal linear combinations of 75 and which ARE orthogonal We will see examples of both cases as Chem 455A develops 6 Chem455A Lecture 12 13 nb VIT3 in the case of degeneracy quotWe can always choose appropraite linear combinations of degenerate eigenfunctions to obtain simultaneous eigenfunctions of two commuting observables but we might have to do a bit of work to find those linear combinationsquot Looking back to part A the key step was B 155 0c 75 which implied that a was an eigenfunction of B In the case of degeneracy we have to replace this by B a oc quotANY linear combination of the degenerate eigenfucntions of A with eigenvalue aquot Thus B 155 is NOT simply a multiple of 755 What do we do We need to then FIND appropriate linear combinations of the degenerate eigenfucntions of A which ARE also eigenfunctions of B This can and we don39t give the proof always be done We will see examples of this also as Chem 455A proceeds C The Measurement Postulate stated in three different ways Statement 1 Given an observable quotAquot the only results of measurments are the eigenvalues of A Commentary this gives us a set often infinite of eigenvalues ai which will39 39 quot IIg we 3 quot g propertA However this doen39t tell us how what we might see in a given experimental situation depends on I r the wavefunction What we can know about a system is contained the wavefunction Postulate 1 so there should be a relationship between 139 and quotwhich of the possible eigenvalues a we might actually expect to findquot if the system is in state 139 So a somewhat more useful statement is Statement 2 The average value of a large number of independent measurements of the property A when the system is always prepared with wave function 139 is ltAgt E Chem455A Lecture 12 13 nb 7 fr 1 A P 139 air where 139 is the normalized wavefunction The uncertainty AA is then defined as lt A2 gt lt A gt2 and gives a measure of the quotspreadquot in expected results Statement 3 However we can go further and actaully predict which eigenvalues will actually be observed and with what probability This is a lot more infomation that just the average standard deviation aka quantum uncertainty This is done by analogy with finding the quotcomponents of a vector in 3D Euclidean Space by projecting the a vector of interest onto the unit vectors chosen to be the quotbasisquot Suppose we have an orthonormal set of unit vectors ijk The components of a vector l7 are given by V1 V2 V3 and the vector itself may be writen as n k K A x x x V V1 I V2 j V3k where V1 iVV2jV and V3 kV Note that we have found the component by quotprojecting the unit vector onto the vector Vquot In QM the analog is The components of a wavefunction I T along the directions defined by the orthonormal eigenfunctions r of an obserable are c T I 15111 and then I r 2x1 c r lf I r is normalized and the r are eigenfunctions of observable quotAquot with eigenvalue a then the probability of observing any one of the eigenvalues say the 1quot 2 Is gIven by P c If an expanSIon coeffIcIent say 014 is zero for 1 we will never observe eigenvalue 14 as there is no quotcomponent of 1 quot in that quotdirectionquot in the function space Bohr realized that spectra were quantum 39qus between energy levels quantized and labeled by n 123 And the frequency or wave length of light associated With these transitions between quantum states was hv hCk AE this is the foundational equation of all of spectroscopy and is called the BohrEinstein frequency rule Applying to a transition m a n in the H atom or ion of nuclear charge Ze AEUn n 2284mg1 1 h rm 4 V Bear m2 n2 A thus 1 1 1 2284ma1 1 n z m2 hc 8gth nZ mz andthus 1 22 e4 mB R R dber Constant Y 9 he Seohz which Bohr thus evaluated in terms of the fundamental constants h c e m9 and ea1 Failings of Bohr Theory ad hoc mixture of old classical and new ad hoc quantum postulates Did not work except for one electron atomsions Failed for He H2 amp periodic table H is flat and has one unit 1 of angular momentum why not zero Heisenberg amp Schrodinger in 192728 started from scratchnew fully quantum postulates All properties of a quantum system for example an electron are contained in knowledge of its wave function 1pxt Classical observables are replaced by quantum operators these often but not always being related to classical counterparts The values of quantum properties are precisely the eigenvalues of the corresponding operators Schrbdinger said that 1pxt has no physical meaning Max Born NOT BOHR said no 1pxt 2 dx is the probability of finding the particle at point Xe xdx at time t So We have a lotto do amp let s get going With the de Broglie formula we are ready to carry out Bohr s quantization of the H atom and give the first correct explanation of the spectrum of an atom or molecule by combining Bohr s stationary orbits with Einstein s photons Fma balance of force and acceleration gives F Ze24rcsor2 mv2r ma Or Z924rcsor mv2 Thus 12mv2 Ze28rcsor A result known as the virial theorem Bohr s quantization Hypothesis which we heuristically derive from de Broglie 2m nk nhp nhmv Or mvr nh2n nh This is Bohr s quantization of angular momentum L mvr reminder L RX P so L mvr Let s actually work things out 1 solve mvr nh for v 2 plug into virial get rn 3 plug virial into ENERGY KE PE 4get Bohr s energy formula 5 The Rydberg formula follows In summary 7 7 7 Eon Ir 7r 739 e 7 7 110 ATZL39Z Ze Z 395 lo the Bohr radius has the numerical value 529 X 10 11 in The velocity 377 corresponding to the orbit with radius 739 is 727 2 7771573 Z92 U77 26071 results for TH and v7 can be substituted into Equation 156 to gi is of energy 7 ZZLAWIE 86872 72 Z2 7 4218 x 10477 1123 Determine h by fitting data to Planck s Formula pAby The visible spectrum of the H atom one proton one electron These are the Balmer lines in the Visible spectrum of excited H The H atom Balmer lines as observed by Angstrom in 186OD Red 6565 A 6565nm Green 4883 A Blue 4342 A nm 1mn 10399m being SI units and giving rise to the name of the whole field of nanotechnology E ESUW Green 4883 A Blue 4342 A What does any of this have to do with BALMER who was Balmer what did he do what happened then Along come Einstein and de Broglie Einstein explains Planck s ansatz 8V n hv de Broglie suggests that if waves eg light might be particles then fair is fair particles should also be waves Waveparticle duality X rays and electrons diffracting through Al foil

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