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# Introductory Modern Physics PHYS 2710

Utah State University

GPA 3.6

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This 13 page Class Notes was uploaded by Talon Thompson on Wednesday October 28, 2015. The Class Notes belongs to PHYS 2710 at Utah State University taught by David Peak in Fall. Since its upload, it has received 13 views. For similar materials see /class/230478/phys-2710-utah-state-university in Physics 2 at Utah State University.

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Date Created: 10/28/15

Bare essentials of statistical mechanics Systems of technological importance until very recently have been macroscopic that is involving 10201030 atoms Despite their great size many properties of macroscopic systems depend intimately on the microscopic behavior of their microscopic constituents The proper quantum mechanical description of an Nparticle system is a wavefunction that depends on 3N coordinates 3 ways of moving in general for every particle and 4N quantum numbers 3 motional quantum numbers and 1 spin quantum number for every particle In a macroscopic solid or liquid there are 1023 atoms per cm3 in a gas there are 1020 Thus the amount of information required to specify the wavefunction for such a system is scandalously large A fundamental theory of macroscopic systems cannot be based on microscopic details Instead a new way of thinking is required This is known as statistical mechanics Statistical mechanics is usually based on an independentparticle approximation ie the system s particles are assumed to not interact with each other The wavefunction can then be separated into a product of N single particle wavefunctions each depending on 3 coordinates and 4 quantum numbers Important to the success of statistical mechanics is the assumption that the system interacts with its surroundings These interactions are so frequent we can t keep track of them so we invoke a simpli cation the interactions serve to cause random transitions between the microscopic states Thus if a system is initially prepared with some unique set of single particle quantum numbers in short order 103913 s or less there will be another set and we can t tell which it will be This kind of uncertainty is different from quantum uncertainty it s due to our overwhelming ignorance of all of the possible information Macroscopic statistical equilibrium The measurement of a macroscopic variable such as temperature or pressure for example is assumed to occur slowly compared with the rate of internal transitions The result of such a measurement is therefore a timeaverage thus the meaning of statistical mechanics over many internal states sets of quantum numbers If the average value doesn t change much from macroscopic ie longer than 103913 s momenttomoment the system is said to be in statistical equilibrium Absolute temperature When a system is in statistical equilibrium it can be usefully characterized by a few macroscopic variables Temperature is one of the most important of these The absolute temperature scale measured in kelvins K is de ned as follows 1 T 0 K is the temperature of a macroscopic system found permanently in its ground state Such a system has no excitations it has its lowest possible energy and is completely isolated from the rest of the universe Of course such a condition is impossible to achieve in practice so absolute zero is a conceptual state not a real one 2 T 00 K is the temperature of a macroscopic system found equally likely in any of its allowed states In other words if one could measure the 4N quantum numbers of the system againandagain each possible combination would be observed with equal frequency Again this is really a concept not an achievable reality 3 At nite temperature the probability of nding the system in a 4N quantum number state S is proportional to exp ES kBT where ES is the energy associated with state S and k3 is a fundamental constant of nature known as Boltzmann s constant the value of which is 86l7x10395 eVK In this scale pure water freezes at 273 K and boils at 373 K at a pressure of 1 atm Human body temperature is about 310 K and room temperature is taken to be 300 K though that might be uncomfortably warm At room temperature kBT 00249 eV 140 eV among friends Stat mech 2008 l Average number of particles per state The most profound consequences of quantum mechanics in the macroscopic world arise from the indistinguishability of identical particles A system of N identical noninteracting particles has total energy E5 ENJEIW all 0 where little 0 means a single particle state consisting of a small number of quantum numbers sais the energy of that state and N0 is the average number of particles in that state Provided the system has a xed number of particles then N 2N0 II If the wavefunctions of every particle in the system overlap then a single particle state 0 consists of 4 quantum numbersithree for motion in 3D one for spin direction On the other hand if the wavefunctions don t overlap it is possible to introduce another quantum number 7 namely where in space the center of the wavefunction is This label serves to distinguish one particle from another In this event the system wavefunction describes N distinct particles A criterion for when this quantum number distinguishing centers is relevant compares the average separation of the centers to the spread of the wavefunctions If the spread is larger than the center separation then the particles of the system cannot be distinguished and the system has to be described by one giant wavefunction Consider electrons in atoms In high Z atoms the most tightly bound electrons the ones with the lower n values are on average much closer to the nucleus than the less tightly bound electrons The latter have wavefunctions that spread out over roughly 02 nm or so Thus if atoms are packed about as closely as this the least tightly bound electrons leak out from their own atomic centers and are shared by many possibly all of the atoms in the system This is exactly what happens in solids the atoms sharing electrons with one another hold solids together The tightly bound electrons are distinguishable these are called core electrons but the shared electrons are indistinguishable Often the system consists of weakly interacting particles that move relatively freely about as in a uid Such a system is sometimes called an ideal gas An estimate of the particle wavefunction spread for an ideal gas can be obtained from the Uncertainty Principle AxAp s h 2 or Ax z h2Ap What should we choose for Ap For massive particles we can borrow a result from elementary thermodynamics namely that at temperature T a classical particle has an average kinetic energy for motion in the xdirection equal to KEW kBT If we set Al lz 2 KEW T then we can approx1mate Ax by Ax z ha 2 kBTmC An est1mate of the center m separation is 1 p where p is the number of particles per unit volume Consequently if the ratio 21 kBTmC2 1 pl 3 pl 3 hCZ IkBTm c2 is much greater than 1 then single particle wavefunction spread dominates and the system has to be described by a single giant wavefunction if the ratio is much less than 1 then the single particles are distinct 13 Example Is the system wavefunction of the molecules in a glass of water at room temperature one giant wavefunction Stat mech 2008 2 Solution The mass density of water is 1 gmcc One gm of water corresponds to 118 of a mole so the mass density of water converts into a number density of 33x1022 cm393 or a centertocenter distance of 031 nm On the other hand the wavefunction thermal spread is only 00047 nm so the ratio is 0015 much less than 1 The system wavefunction of molecules in liquid water separates into distinct parts Example Suppose one electron per atom is shared among all of the atoms in a chunk of gold Characterize the system wavefunction of these conduction electrons at room temperature Solution The mass density ofgold is 193 gmcc and the atomic mass is 197 Thus 197 gm of gold 6x1023 atoms so p 193197x6x1023 atomscc x1 elecatom 59 elecnm3 Placing these values into pl 3 ha 21 kBTmc2 yields about 34 Thus the motions of gold s conduction electrons are strongly correlated at room temperature If the system consists of identical overlapping fermions Na can only be 0 or 1 while for identical overlapping bosons Na can be anything from 0 to all of N A bit of mathematics ie maximizing the probability PS subject to the constraints that the system has energy ES and total number of particles N leads to No BT exp5O kHT 1 where B is evaluated by requiring the sum of all Na to equal N and the sign is for fermions and the isign is for bosons Identical fermion ideal gas Let s rst examine the identical noninteracting fermion ideal gas In the limit that T approaches 0 the N lowest energy single particle states will be lled and all higher energy states will be empty that s the Nparticle ground state The highest energy lled state has energy 5F designated as the Fermi energy A convenient way to express BT for fermions is BT exp 5F kBT This leads to 1 exp5I 5FkBT139 Note that if 50 5F lt 0 thenNa gt 1 as T gt 0 whereas if 50 5F gt0 then Na gt 0 as T gt 0 just what we want No The Fermi energy actually decreases as T increases but often it is sufficient to use its value at T 0 To determine 5F at T 0 we need to know what the so are It is often very useful to assume that the N fermions are contained in a cubical in nite well In such a well a single particle state is given by 3 motional quantum numbers 11 ny nz 1 2 3 and one spin quantum number 112 for spin12 fermions If L is the length of a side of the well the single particle energy is 2 2 75 h 2 2 2 WWI 2mL2 nx ny nz each of which is twofold degenerate the two spin states have the same energy We write the Fermi energy as 7rth 2 2 n F 2mL Because N is of order of 1020 or more nF is similarly large Thus to a good approximation N is the total number of states inside an octant because we only want positive n values of a sphere of 5F Stat mech 2008 3 4 3 1 4 3 7m volume frmF1e N 2x7xirmp T the factor oftwo being due to the spin 3 er 5 gr3w But L2 the cubical volume 3 so nally 23p23y degeneracy In other words 5F 1 2m 7139 where p as before is the number density numbervolume of fermions in the well Example Suppose as above one electron per atom is shared among all of the atoms in a chunk of gold Treating these shared electrons as noninteracting fermions in a well what is the associated Fermi energy Solution Using the number density found in the previous example p 59 elecnm3 we nd 5F 55 eV Example A neutron star consists mostly of neutrons packed at nuclear density What is the Fermi energy for noninteracting neutrons in a neutron star cubical in nite well Solution Nuclear density is about 012 nfm3 12x1017 nnm3 Using the neutron mass for m yields 5F 52x107 eV 52 MeV Whether the T 0 value for SF is useful depends on whether the fermion wavefunctions overlap signi cantly We found that the conduction electrons in gold have substantial overlap at room temperature Thus even though T 300 K electrons in gold are cold Similarly suppose T 108 K for a neutron star At that temperature pl 3 hcZ IkBTm c2 comes out to be about 13 so neutrons in a neutron star are cold even at 100 million K The total energy of the N particleinabox system at T 0 is given by 12712 471 12712 n 1 12712 E 250 E 2 n3n2n227 FnAdni n5 L2 y 0 quotwilyquot2 r 2m Z 8 2mL2 0 5 2mL2 To see one of the profound consequences of Fermi energy we introduce the notion of pressure P E V Remembering that L in the denominator above is really V it is straightforward to 2 2 53 2 2 show that PF 21in 21 3 h it 15 2mV5 3 15 1 external pressure needed to appreciably compress the fermion ideal gas 2m ps 3 You can think of this quantity as the Example What is PF for the conduction electrons in gold Solution Again using p 59 elecnm3 we nd PF 140 eVnm3 The latter is not as directly informative as expressing PF in macroscopic unitsiie converting eV to J and nm to m When this is done we nd PF 2x1010 Nmz Since 1 atm 105 Nmz we see that PF 2x105 atm No wonder solids are hard You can t squeeze the electrons very much Example What is PF for a neutron star Solution Substituting neutron mass and neutron star density produces PF 4x1027 atm so it s really difficult to squeeze a neutron star But of course the star s gravity is trying to do just that If the inward gravitational pressure of the star s mass exceeds its Fermi outward pressure then nothing can hold the star up from collapsing further and a black hole results Stat mech 2008 4 Identical boson ideal gas For bosons 1 BTexp50 kBT 139 The minus sign in the denominator makes a huge difference as T approaches 0 At a nite temperature all N bosons go into the same single particle ground state with energy so At low 0 temperatures BT 1exp so kBT so that No gt N We can use the same criteria as above for wavefunction overlap to see at what temperatures such Bose condensation might begin Signi cant overlap occurs when pAx3 gets close to 1 It is known that about 10 of a volume of helium 4 combines into a super uid state no viscosity at T 217 K with density equal to 0146 gcc For those values we nd that pl3 its21 kBTm c2 is about 03 apparently sufficiently large to account for the observed degree of super uidity Photons Photons have no mass so the criterion used previously for overlap of wavefunctions doesn t apply In a container all photon wavefunctions overlap In addition a container of photons at temperature T does not have a xed number N Photons can be absorbed leave and emitted enter by whatever the walls of the container are made of As a consequence the number of photons in a container is constantly changing N does not constrain what the probability is of nding a photon in state a Photons are bosons but with BT 1 1 exp5O kHT 1 For photons as T gt 0 Na gt 0 as well The colder the container the fewer photons are present Photons do not undergo Bose condensationithey just disappear 0 More generally if so kBT ltlt 1 expsI kHT 1sfy kHT and NI CRT50 gtgt 1 That is at nite T low energy photon states are lled with many photons but those states don t have much energy so don t contribute much to the total energy in the container On the other hand if so kBT gtgt 1 NI 0 so high energy photon states that do have energy are empty Most of the container s energy comes from intermediate energy states This situation is described by a blackbody spectrum usually given in terms of how photon energy per unit volume per unit 3 frequency varies with photon frequency uw The maximum in u 7139 c exphw kHT 1 occurs at a photon energy lime 282kHT and is proportional to T8 The maximum of the distribution and the energy per unit volume in the container are determined solely by temperature and are independent of what the container is made of Despite having a pronounced peak at hwmax 282kHT the blackbody distribution still has 5 of the maximum value at an energy ha 93kBT so the tail of the distribution is fairly long and photons albeit few exist with energies well in excess of kBT Example What is the wavelength of a photon with hwmax 282kBT at room temperature Stat mech 2008 5 Solution At room temperature kBT 140 eV so the photon energy is 007 eV This also hcit so the wavelength is about 18 um about the size of a biological cell Example What is the wavelength of a photon with hwmax 282kBT at the temperature of the Cosmic Microwave Background CMB Solution The CMB is a blackbody with T 2725 K about 1100 room temperature Consequently the wavelength is about 100 X larger than in the previous example or about 2 mm Incidentally there are about 3X109 CMB photons for every proton in the universe but because these photons carry so little energy the total me2 for protons is about 2000 times greater than the energy of the CMB High temperatures low densities In a uid of particles at high temperatures andor low densities particle wavefunction overlap decreases We observe this loss of quantumness in the way BT behaves at high temperature or low density it becomes very large so that the number of particles in any given single particle state becomes very small That is 0 gt lexp 5 I kB T regardless of whether the particles in the system BT exp5O kHT 1 B are bosons or fermions When this happens we say we are in the classical regime Stat mech 2008 6 Electromagnetism and photons Classical electromagnetism deals with continuous elds with continuous distributions of energy and momentum Photons on the other hand are particles of electromagnetic energy and momentum Can the two pictures be somehow reconciled The answer is yes but it requires stretching our imagination and our use of mathematics Classical electromagnetism can be summarized succinctly as follows 0 Electric charge makes electric eld g 0 Moving electric charge makes magnetic eld E 0 Timechanging magnetic eld also makes electric eld 0 Timechanging electric eld also makes magnetic eld These four points imply that accelerating charge makes timechanging E that in turn makes timechanging g that in turn makes timechanging E that in turn makes This sequence of events corresponds to electric and magnetic elds called electromagnetic EM radiation propagating away from the accelerating source charges Electromagnetic elds carry energy The energy per unit volume it in electromagnetic elds at a given point in space at a given time is u 526282 1 Maxwell s equations MEs constitute a precise mathematical statement of how EM radiation originates and sustains itself MEs predict that the rate of propagation of EM radiation depends on the magnetic permeability and the electric permittivity of the medium through which it propagates Vacuum permits electric elds so 885x103912 C2 Nmz and can be permeated by magnetic elds M0 470103912 N A2 and the speed of propagation 1 M equals c 30x108 ms the speed of light in vacuum MEs are a set of partial differential equations that describe how g and E vary in space and in time In full generality they are fairly complicated but for one important special case namely the electromagnetic plane wave in a vacuum they simplify greatly The plane wave geometry is depicted to the right Two parallel planar sheets of electric and magnetic elds are shown traveling in the positive xdirection In these sheets g points in the ydirection and E points in the zdirection In a given sheet g is the same at every point and E is the same at every point At one instant the difference in the respective elds is only along the xdirection If we ignore the fact that the elds are vectors MEs for the case in the gure reduce to and ampi2 2313 ax a 3x C a Note that if you differentiate the rst of these equations with respect to x alternatively t and the second with respect to t alt x it is straightforward to show that 325 1 325 923 1 923 alt C2 a2 3x2 2 a2 3x2 3 EMampphotons fall 2008 l Such relations are called the Maxwell wave equations Among the possible solutions to the wave equations are elds that travel or stand Traveling EM waves depend on x and tthrough the combinations x 1 ct where the 7 alt sign corresponds to waves traveling in the alt 7 xdirection Because the wave equation is linear in the eld ie the derivatives of the eld only appears to the 1 power the sum of two solutions is also a solution Thus a wave traveling in the ixdirection plus a wave traveling in the xdirection is also a possible wave under the right conditions such a sum will be a standing waveione whose zero values are always at the same place in space Important special cases of traveling and standing waves are harmonic waves waves represented by sines or cosines or complex exponentials Example 5x t 50 sinkx ct is a sinusoidal electric eld wave traveling in the x direction 50 is the amplitude ofthe wave and kis the wavenumber k 21 where his the wavelength For example the electric eld amplitude of sunlight at the top of Earth s atmosphere is about 860 Vm and an average wavelength is about 550 nmiyielding a wavenumber of about 11x107 radm Note that for EM waves c A wherefis the frequency ofthe wave as a result kc 2nquot a which is the wave s angularfrequency Thus Ext 50 sinkx wt The frequency of a 550 nm wavelength EM wave is 55x1014 Hz and the angular frequency is 34x1015 rads Thus for sunlight Ext 860 Vmsin115107 radmx 341015 radst The rst hint of how to reconcile classical EM with photons is to recall that photon energy is E hj and photon momentum is p hA We can write E as E hwZn39 andp as p hk 27139 Because hZn39 appears so often in quantum mechanics it is given its own symbol h pronounced aitch bar Then a sinusoidal electric eld can be expressed as 5xt 50 sinpx Eth where wavenumber has been replaced by photon momentum and angular frequency by photon energy OK but where is the randomness of photonarrival hiding To see requires understanding that classical ie bright light optics deals with EM energy u The double slit interference pattern for example is the variation of u over the surface of the detector But this pattern is identical to the variation of the probability that a photon will hit the detector when the apparatus is illuminated with very dim light A way of connecting u and the photonarrival probability is to declare that in a onephoton experiment the probability density probability per unit volume of detecting the photon at a given point in space at a given time is px yz t W where the integral is over all space The probability that the photon will M be detected in a small volume sturrounding the point x y z is pd V Note that f pdV 1 which implies that the photon is somewhere at time t To complete the story of how eld and probability are connected we make a seemingly preposterous assumption based on the fact that the sum of squares of two real numbers can be expressed as the product of complex number with its complex conjugate That is a2 b2 111 2 11J95111where 1P aib and i2 71 and 111 a ib This enables us to de ne a photon wavefunction 1P N E icB The normalization factor N is chosen so that EMampphotons fall 2008 2 2 dV 1 This allows 11le to be interpreted as the probability density p above Can you complete the proof of this You might not like the fact that 1P is complex given that physically 2 measurable quantities are real but in a onephoton experiment we never measure 1P only IIJ which is real Finally we note that the photon wavefunction is a solution of the Maxwell wave equation 92111 1 92111 9x2 2 92 39 Since 1P depends on x and I through for example px Eth the wave equation becomes pZIIJ 1 E2111 112 17 112 39 This suggests that a spatial derivative measures momentum while a time derivative measures energy In fact for photons we reexpress the Maxwell wave equation as energymomentum operator equation czpfplv ELI 4 9 9 where pgp 411 and ELp iii the Signs require a fussy d1scuss1on which is unimportant for now EMampphotons fall 2008 3 More about wavefunctions If 1P1 and 1P2 satisfy the Schrodinger Equation then so does IIJ 111111 aZIIJZ 2 IP dx1 itmustbe that alzaz2 1 m In order for f 0 When you make a series of N gtgtl measurements on IF you nd a result corresponding to 1P1 about Nal2 times and one corresponding to 1P2 about Na times The average value of a measurement on state 1P corresponding to the operator Q0p is de ned as Q fjm 1PQ0p11de This implies that for the wavefunction above Q a12Q1a22Q2 Example Suppose Q is energy and 1P1 and 1P2 are the two lowest energy eigenstates of the in nite square well Suppose that a1 1 12 and a2 412 What is in terms of E1 the ground state energy Answer E1 E2 22 E1 25E1 You interpret al2 and a as the probabilities of being in 1P1 and 1P2 respectively After making the measurement if the result corresponded to 1P1 then an immediate subsequent measurement will also correspond to 1P1 the wavefunction 1P is said to be collapsed into 1P1 by the measurement It will stay in 1P1 as long as there are no other interactions from the rest of the universe Thus QM says that before a measurement is made the most that can be said about the state of a system is that it is some superposition ie sum of eigenstates The act of measurement creates the reality of the system being in one of them More generally 1P Ea 1P where the sum can involve any number of terms and Q EaZQ n n Wavefunctions 2710 Fall 2008 Multiparticle Wavefunctions Different particles Introductory examples of quantum mechanical wavefunction calculations involve a single particle moving about in a magic potential energyieg a particle trapped inside a square well But potential energy arises from interaction so these situations must inevitably include more than one particle Even the simplest atomihydrogeniconsists of two particles the electron and the proton So how should the Schrodinger Equation be generalized to account for multiple particles The idea is to introduce a system wave equation and a system wavefunction For concreteness let s consider a system consisting of an electron and a proton in a 1D in nite square well Assume they are electrically neutral and have no spins so they don t interact In this case ih 51 t 2mg 51x32 2m 07x The variables separate The system wavefunction is a product of three individual functionsione in t and one each in the two spatial coordinates Neither the electron nor proton can be found outside the in nite well so 1P has to vanish when either coordinate is 0 or L These boundary conditions produce two spatial quantum numbersing and npiboth of which are positive integers The system wavefunction is labeled by these two quantum numbers Pimp 2 5193de is the probability nmc 1 1P Wnenp XeXVI Ae iBh sin EL 3 sinL Note that E is the system energy The interpretation of 1P is that when of detecting an electron in state 112 in a small length alxe centered on the position x3 ie a small electron detector at x3 and simultaneously at time t detecting a proton in a small volume dxp centered on the position xp a small proton detector Assuming the system actually does consist 2 of an electron and proton then the integral of dxedxp over all values of both coordinates is when 712 n n l The system energy for this example 1s Eng 7 7 The wavefunction given above is incomplete because it lacks reference to the particles spins Like mass and charge spin is an intrinsic property of a particle Unlike mass and charge however spin has direction which in turn can have dynamical consequences Thus a more compete description requires accounting for spin direction Multiplying the system wavefunction by spin states that are 2D vectors accomplishes this In this notation for example an up spin 1 0 state 1s om 0W2 O and down state 1s om oil2 Sometrmes these vectors are more simply represented by T and J respectively As an example a noninteracting electron proton system might have the wavefunction n 71x at 11 t A6713 sink mt sinL 7 L LUMP Filamenan which would correspond to an electron with spatial quantum number 112 and spinup and a proton with spatial quantum number 11p and spindown In general if the system consists of N particles Multiparticle wavefunctions l all moving in 3D then the system wavefunction will depend on 3N spatial quantum numbers and N spin direction quantum numbers Identical particles In the previous example we dealt with two different kinds of particles When the proton detector clicks it s certainly the proton that causes it and the same for the electron Now suppose we replace the proton with a second electron All electrons are identical there is no way to distinguish between them Replace the proton detector with an identical electron detector If the electron wavefunctions ll up the well then when one of the detectors clicks there s no way to tell which electron was detected Now the system wavefunction might be expected to look something like inh Sini quot177951 Sinin h J TILT L J But the labels 1 and 2 don t distinguish the electrons from one another they simply convey the notion that there are two ns two spins and two places to put detectors The quantity wnimsinzm2x1x2tJ A ll dxldxz is the probability that an electron is detected at pos1tion x1 in state m ml and nlmslnzmsz that an electron is detected at position x2 in state m m It shouldn t matter if we switch detector positions or if we switch the states measured in each detector the system probability should be 2 2 2 Wumn2mzxvx2tJJ J Vn2m2nmxvx2tJJ J Vn1mn2mzx2xthJ This string of equalities can be true provided Wnlm nzm x1x2t zwnzm nlm x1x2t and identical In other words 11 lmum x1x2t zwnlm nzm x2xlt That is the wavefunction can either be the same when states are switched or when positions are switched or it can change sign It turns out that Nature employs both options and the consequences are astounding A profound but difficult to prove theoremithe Spin Statistics T hearemistates that identical noninteracting bosons integer spin have the same wavefunction under switch of state or position while the wavefunction for identical noninteracting fermions odd halfinteger spin undergoes a sign change when state or position are switched In other words the proper wavefunction for two or more identical bosons is symmetric if two state labels or two positions are switched but for fermions it is antisymmetric Example Suppose two identical noninteracting spin0 bosons are in a 1D in nite well The spatial quantum numbers are 2 and 3 The proper wavefunction for the system is A Sin2m1sin3m2 Sin37ml SinZmz L L J L L J L L J L not numerical zeroes Suppose now the particles are fermions with spinl2 and both spins are sin 2761 sin3 sinsin l1l2 1P systmi 0102 The zeroes on the end are spin states 14 77 up Then WW L The symmetry or asymmetry of the wavefunction seems like a formal and unimportant detail but it is anything but Suppose in the previous example the two fermions were both in the spatial quantum number 2 state that is they were in the same total spatial and spin state Then 2m1i an 271x12m2 S111 111 7 S11 1 7 S111 L L J L L J L L J L J the asymmetry of their system wavefunction no two identical noninteracting fermions can be in exactly the same state This rule is often referred to as the Pauli Exclusion Principle the atomic the factor would vanish In other words because of Multiparticle wavefunctions 2 periodic table is one direct consequence The two fermions could be in the same n state if their spins were different however For example suppose the fermions both have n 2 Then wsysmn ASin 2161 Sin2m2 lhlz lilzl39 L In general the fermion wavefunction is either an antisymmetric combination of products of spatial functions times symmetric spin combinations or symmetric space combinations times antisymmetric spin combinations In either case if two of the fermion detector positions are the same the wavefunction also vanishes two identical fermions cannot occupy the same position in space at the same time The antisymmetry of the fermion wavefunction implies that fermions don t like one another The asymmetry of the wavefunction for identical fermions produces a kind of repulsive force one that manifests itself for example in the fact that solids are hard to compress When you push on a solid it pushes back because in pushing you are trying to cause the solids electrons to get closer than they like Bosons are different The symmetry of the wavefunction for identical bosons permits more than one boson to be in the same state and to occupy the same position in space In fact the probability that this happens is actually greater than when they are different states This implies that bosons like one another and that there is a kind of attractive force between them Bosonic attraction leads to the strange and practically important phenomena of super uidity and superconductivity Multiparticle wavefunctions 3

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