8.044 Lecture Notes Chapter 9: Quantum Ideal Gases
8.044 Lecture NotesChapter 9: Quantum Ideal GasesLecturer: McGreevy9.1Range of validity of classical ideal gas . . . . . . . . . . . . . . . . . . . . . .9-29.2Quantum systems with many indistinguishable particles . . . . . . . . . . . .9-49.3Statistical mechanics of N non-interacting (indistinguishable) bosons or fermions 9-99.4Classical ideal gas limit of quantum ideal gas . . . . . . . . . . . . . . . . . . 9-139.5Ultra-high temperature (kB T mc2 and kB T µ) limit9.6Fermions at low temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . 9-189.7Bosons at low temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . 9-33Reading: Baierlein, Chapters 8 and 9.9-1. . . . . . . . . . 9-15
9.1Range of validity of classical ideal gasFor a classical ideal gas, we derived the partition function 3/2V2πmkB TZ1N,Z1 3 V,Z N!λthh2where the length scale λth h2πmkB Tis determined by the particle mass and the temperature.When does this break down?1. If ‘idealness’ fails, i.e. if interactions become important. This is interesting and important but out of bounds for 8.044. We’ll still assume non-interacting particles.2. If ‘classicalness’ fails. Even with no interactions, at low enough temperatures, or highenough densities, quantum effects modify Z.An argument that classicalness must fail comes by thinking harder about λth , the “thermal deBroglie wavelength”. Why is it called that? Recall from 8.04 that the de Broglie wavelengthfor a particle with momentum p is λdB hp . For a classical ideal gas, we know the RMSmomentum from equipartitionhppp 23i kB T pRMS h p2 i 3mkB T .2m2We could have definedλ̃th hpRMS h;3mkB Tthere’s no significance to the numerical prefactor of 12πinstead of 1 .3Now recall the significance of λdB in QM:λdB h minimum size of a wavepacket with momentum p .pSo we can infer thatλth minimum size of quantum wavepackets describing atoms in a quantum ideal gas .The classical picture of atoms as billiard balls with well-defined trajectories only makes senseif 1/3Vλth typical spacing between particles N V 1/3And this inequality λth 1T Nis violated at low T or high density. The aboveinequality is the condition for validity of the classical treatment.9-2
System————————–airLiquid 4 HeConduction e in CuT———300K4K300 K V ��——–classicalnot classicalnot classicalA second derivation of the same criterion, this time in momentum space:Consider solutions to the Schrödinger equation in an L L L box. The energy eigenstatesof one particle areφmx) sin (kx x) sin (ky y) sin (kz z) ( πwith k m ,x L xyyzzand m 1, 2, 3.xyzwith energy eigenvalues 2kx2 ky2 kz22mLet’s ask under what circumstances we can ignore the QM.The mean energy per particle in a classical ideal gas is 23 kB T . How many 1-particle stateswith energies like this are there to put our N particles into? The number of states with 23 kB T is 2m 23 kB T 3/241 1/2π83 2volume of octant of sphere in k-space with radius 2mE 2N ( ) (π/L)3volume in k-space per allowed grid point!3r mkB T3·2V V 4 3π 3 hλthπSo: when N λV3 , the number of particles is big compared to the number of possible onethparticle states for the particles to sit in. So we need to worry about things like the PauliExclusion principle, which prevents us from putting more than one particle in each state.For N λV3 , the classical analysis is fine – we needn’t worry about the particles needing tothoccupy the same 1-particle state.9-3
9.2Quantum systems with many indistinguishable particles[This section is about quantum mechanics. You’ve already encountered some of these ideasin 8.04, and will discuss this further in 8.05. We’ll come back in subsection 9.4 and thinkabout when this business reduces to classical mechanics.]Consider two particles. Their state can be described by a wavefunction ψ(x1 , x2 ). (We won’tworry right now about how many coordinates of each particle (e.g. how many dimensions)we have to specify.) If the particles are indistinguishable, then ψ(x1 , x2 ) 2 Prob finding a particle at x1 and a particle at x2Note that we make no specification of which particle is where. Indistinguishability requires: ψ(x1 , x2 ) 2 ψ(x2 , x1 ) 2Swapping the arguments twice should give back the same ψ, not just the same ψ 2 :ψ(x1 , x2 ) ψ(x2 , x1 )9-4
Two choices:Bosons particles for which ψ(x1 , x2 ) ψ(x2 , x1 ). i.e., the wavefunction is symmetric. It is a fact (observed experimentally, understood via quantum field theory) that theyhave integer spin. e.g.: hydrogen atoms, 4 He, photons, phonons, magnons, gluons, Higgs bosons (?). The stat mech of a gas of them was developed by Bose and Einstein, so in the contextof stat mech, these are called “Bose-Einstein statistics”.Fermions particles for which ψ(x1 , x2 ) ψ(x2 , x1 ). i.e., the wavefunction is antisymmetric. It is a fact (observed experimentally, understood via quantum field theory) that theyhave half-integer spin (1/2, 3/2.) e.g.: electron, proton, neutron, 3 He, 7 Li. The stat mech of a gas of them was developed by Fermi and Dirac, hence “Fermi-Diracstatistics”.For non-interacting Bose or Fermi particles (we will always assume this):energy eigenstates are always (symmetric or antisymmetric) linear combinations of productsof single-particle energy eigenstates.[Recall: there is more to QM than energy eigenstates, but they are enough to constructthe partition function.]9-5
Wavefunctions of several bosons or fermionsConsider for example two indistinguishable quantum particles in a box. Label the possiblestates of one quantum particle in the box by a fancy label α which is a shorthand for all itsquantum numbers, e.g., the wavenumbers (mx , my , mz ) of its wavefunction.If they are bosons, the wavefunction must be of the formΨ(x1 , x2 ) φα (x1 )·φβ (x2 ) φβ (x1 )φα (x2 ). {z} {z } {z }an energy eigenstate another energy eigenstate makes it symmetricOK state of bosonsof one particlefor one particle {z}not sym or antisymhence not allowedFor fermions:Ψ(x1 , x2 ) φα (x1 )φβ (x2 ) φβ (x1 )φα (x2 ). {z}makes it antisymmetricOK state of fermionsWorking our way up to 3 indistinguishable particles:Ψ(x1 , x2 , x3 ) φα (x1 )φβ (x2 )φγ (x3 ) φα (x1 )φγ (x2 )φβ (x3 )φβ (x1 )φγ (x2 )φα (x3 ) φβ (x1 )φα (x2 )φγ (x3 )φγ (x1 )φα (x2 )φβ (x3 ) φγ (x1 )φβ (x2 )φγ (x3 )with for bosons and for fermions.This time we count 1, 2, 3, many:9-6
N indistinguishable particles:1. Pick N single particle states α, β, γ.2. There is exactly one symmetric combination. This is an energy eigenstate for N bosons.3. IF α, β, γ. are ALL DIFFERENT then there is exactly one antisymmetric combination. This is an energy eigenstate for N fermions. This fact that “Fermions must all be in different single particle eigenstates” is the PauliExclusion Principle. Many bosons can be in the same single-particle energy eigenstate. If the particles were distinguishable, there would be N ! states for each choice ofα, β, γ. If the particles are indistinguishable, there is (at most) one such state.9-7
Occupation number representation of the many-particle stateFor either bosons or fermions, the state Ψ(x1 , x2 .xN ) is fully specified by indicating1. Which 1-particle states are occupied?2. If bosons, how many particles are in each 1-particle state? (For fermions, this numbercan only be 0 or 1.)Label the 1-particle states α (e.g. mx , my , mz for ideal gas, or n, , m for Hydrogen atoms)So: the state Ψ is specified by a set of integers called OCCUPATION NUMBERS:nα (Ψ) # of particles in 1-particle state αwhen the many-particle state is Ψ.Fermions: nα {0, 1}Bosons: nα {0, 1, 2, 3.}These numbers also specify N, E, ., as follows.total # of particles in state Ψ:N Xnα (Ψ)αThe sum here is over all possible states in which one of the particles could be; many of themwill be unoccupied. This is not a sum over particles.Similarly, if α the energy eigenvalue of the 1-particle energy eigenstate αthen the total energy in the many-particle state Ψ istotal energy in state Ψ:E(Ψ) Xα9-8nα (Ψ) α
9.3Statistical mechanics of N non-interacting (indistinguishable)bosons or fermionsWe’ll use the canonical ensemble: an ensemble of copies of the system, all with the sameN, T (hence, E varies amongst the copies in the ensemble), in contact with a heat bath attemperature T .XhEi i ensemble X X XiEi·p(Ψ ) {z} {z i}total energyprob of findingof N particles the N -particle stateΨi in the ensemble!· p(Ψi )nα (Ψi ) αα!X α ·αnα (Ψi )p(Ψi )i {z} hnα imean value of the occupation numberof the 1-particle state αin the ensemble of N -particle statesSo: if we know hnα i Pinα (Ψi )p(Ψi ), then we knowhEi X α hnα i .αA check on this formalism:!Xαhnα i XXαinα (Ψi )p(Ψi ) X Xnα (Ψi ) p(Ψi ) Nαi i{z N}So: how do we calculate the average occupation numbers hnα i?9-9Xp(Ψi ) N
hnα i for fermionsXhnα i p(Ψ) Z(N ) 1 e βE(Ψ)nα (Ψi )p(Ψi )all N -particle states Ψi 1 Xnα (Ψi )e βEiZ(N ) i 1Z(N )1X1X1Xn1 0n2 0n3 0 nα e β(n1 1 n2 2 .).{z}subject to the constraint n1 n2 . NWe just sum over all occupation numbers, consistent with the fact that there are N particlesaltogether. One of these sums is special, so let’s write the two terms nα 0 and nα 1explicitly:hnα i {z}0 nα 01e β αZ(N )1X1X1Xn 0n 0n 0.2 1{z 3 }doesn’t include nαwhich is 1so the rest aresubject to the constraintn1 n2 . N 1 n2 2 .)e β(n1 1{z}excluding nα αRepackage in a sneaky way which takes advantage of nα 0 or 1:1hnα i e β αZ(N )1X1X1Xn 0n2 0n 0 1.{z 3 }includes nαsubject to the constraintn1 n2 . N 1 n2 2 .)(1 nα ) e β(n1 1{z}includes nα αNotice that the nα 1 term in the summation gives zero; this means it doesn’t matter if weinclude it in the Boltzmann factor. The nα 0 term in the summation gives what we hadbefore.1e β α (Z(N 1) hnα ifor N 1 particles Z(N 1))Z(N )Z(N 1) β α e(1 hnα ifor N 1 particles )Z(N )hnα i 9-10(1)
The Fermi-Dirac distributionWe want to rewrite our expression for hnα ifermions as a function of T, µ.To do this, recall thatµ F (T, V, N ) F (T, V, N 1) kB T (ln Z(N ) ln Z(N 1))Z(N 1) eµ/kB TZ(N ) .Now, in a box with three particles, hnα iN 3 6 hnα iN 1 2 . But for N 1025 ,hnα iN hnα iN 1 for N 1. hnα i eµ/kB T e α /kB T · (1 hnα i)solve this for hnα i: hnα i (1 ex ) ex , with x β (µ α ): hnα i ex1 β( α µ)x1 ee 1for FERMIONSWhat’s µ? It’s determined from the relation between µ and N :N XXhnα i αα1eβ( α µ) 1Notice: each hnα i [0, 1], since e x [0, ] for real x. This is the Pauli Principle in action.The extremes are realized in the T , 0 limits, if we hold µ fixed. More on this below.9-11
hnα i for Bosons (the Bose-Einstein distribution) 1 Xhnα i Z(N ) n 0 1 X Xn2 0n3 0{z· · · nα e β(n1 1 n2 2 .)with n1 n2 . N}one special sum: Xnα e βnα αnα 0No contribution from nα 0. Tricky step: let n0α nα 1 1Z(N ) X X Xn1 0n2 0n3 0 ···{z0(n0α 1) e β α (1 nα ) β(n1 1 .)}with n1 n2 . n0α . N 1N 1 in all states including φα , at least one particle in φα . e β α(Z(N 1) hnα ifor N 1 particles · Z(N 1))Z(N )The only difference from fermions is the red plus sign. Making the same steps followingequation (1)1 hnα i ( α µ)/k Tfor BOSONS.Be 1Notice that the minus sign in the denominator is very significant: with this sign, thedenominator can become small, and so hnα iBOSONS can be larger than unity.Again µ is determined from the relation between µ and N :XX1N hnα i eβ( α µ) 1ααSummary:Xhnα i NXαhnα i α hEiαhnα iFERMIONS hnα iBOSONS 1e( α µ)/kB T 11e( α µ)/kB T 19-12is between 0 and 1.is between 0 and N .
9.4Classical ideal gas limit of quantum ideal gasThe rest of 8.044 will be decoding the physics in the expressions on the previous page (andthere is a lot of physics in them). First let’s check that we can recover our previous classicalresults in the right limit.If e( α µ)/kB T 1 thenhnα ifermions hnα ibosons e ( α µ)/kB T hnα iBoltzmann 1This factor gives the occupation numbers for “Boltzmann statistics”.Claim: this is the classical limit. A weak check on this claim is the fact that the B andF distributions reduce to the same expression in this limit. To show: hnα iBoltzmann are theclassical occupation numbers in thermal equilibrium.[End of Lecture 21.]1. Classical when e µ/kB T 1 e µ/kB T 1. We saw in the previous chapter thata classical ideal gas has V 1µ ln kB TN λ3thSoe µ/kB T 1 V 11N λ3thwhich is our classicality criterion from before (subsection 9.1).2. Now that we know we’re in the right limit, let’s examine the Boltzmann occupationnumber, and show that it’s the classical expectation.hnα iBoltzmann e(µ α )/kB TPreviously we wrote everything in terms of N , so let’s do that now:XXN hnα i eµ/kB Te β αα α {z Z1 eµ/kB T NZ1for Boltzmann statisticsµ kB T ln9-13}Z1N
This is the familiar classical ideal gas result, if we can show that Z1 is indeed theclassical ideal gas 1-particle partition function. We also haveN α /kB Tehnα iBoltzmann Z1which is a familiar fact from the canonical ensemble. That is, it is equivalent to thehappy statemente β αhnα iBoltzmann Z1NAlso, we saw earlier in Chapter 8 that in general, from the definition of chemicalpotential,Z(N ) e µ/kB T Z(N 1)prob (1 atom is in state α) p(α) For the particular case of Boltzmann statistics:Z1Z(N 1)this becomes a recursion relation!Z(N ) NZ1 Z1Z1 Z1Z1 Z(N 2) Z(N 3) .N N 1N N 1N 2(Z1 )NZ(N ) N!We have now derived this result from the classical limit of the quantum result.Finally, we evaluate Z1 :XZ1 e β αwhere α is a label on single-particle energy eigenstatesα kx 2 2mkXXXkyXkxe(kx2 ky2 kz2 )kz2eBT 2mkBk2T x!3πwith kx mx , mx 1, 2, 3.LIn the high temperature or low-density limit, the steps in kx are negligible and we canapproximate the sum as an integral. Z 3 Z 3L 2mk 2 T kx2L 2mk 2 T kx2BBZ1 dkx e dkx eπ2π0 1/2 !3 3/22πmkB TL 2πmkB T V2π 2 2V .λ3thFinally, this is the fulfillment of the promise that the factor of h13 that we apparently arbitrarily used to normalize our classical phase space integrals would come from QM, i.e. it’swhat we get when we start with the correct QM answer and take the classical limit.9-14
9.5Ultra-high temperature (kB T mc2 and kB T µ) limitLet’s evaluate the average energy hEi for an ultrarelativistic gas of fermions or bosons athigh temperature, in particular at temperature bigger than any other energy scale: kTmc2 , kT µ.Z 1X4πk 2 dk8hEi α hnα i 3 (k)hnk i(π/L)0αwhere [8.033] (k) p p2 c2 m2 c4 2 c2 k 2 m2 c4 ck– this last approximation will be valid in the high-temperature regime –and hnk i So:1e( (k) µ)/kB T chEi 2V2πZ0 : fermions : bosons 1 .dkk 3e( ck µ)/kB T 1For kB T µ, we can ignore the contribution from µ. Then we can non-dimensionalize the, in terms of whichintegral by introducing x k ckBThEi(kB T )4 V2π 2 ( c)3 dxx3ex 10{z} 847π , for fermions 154 8π:, for bosons15ZhEiπ 2 (kB T )4 V30 ( c)3(1, for bosons7, for fermions8Observation: the blackbody radiation answer is obtained from this formula with two speciesof massless bosons with µ 0.Reinterpretation: hni for the harmonic oscillator modes of EM field the occupationnumber for k-space modes of these bosons. These bosons are called photons. There are two species because of the two polarizations. They are massless because the dispersion relation for EM waves is ω ck; there is noparameter with dimensions of mass in Maxwell’s equations. This a deep fact. A massterm in the energy density of the EM field would look something like m2 Aµ Aµ ; butthis is not invariant under gauge transformations.9-15
Unlike the number of atoms in a box, the number of photons is not a conserved quantity.This forces the chemical potential to be zero – there’s no need for a constraint fixingN , and hence no need for the Lagrange multiplier µRelatedly, µ 0 says there is no free energy cost to adding one more photon. Thefact that they are massless means we can add photons with k 0 at zero energy cost,since (k 0) 0. (The fact that they are bosons means we can keep doing this overand over, even if the mode already has nonzero occupation number.)PAs T grows, the total number of photons α hnα i also grows. This is distinct fromthe case of e.g. Helium atoms, where the chemical potential µ(T ) must adjust to keepthe number of atoms fixed.Where can we find an example of a ultra-relativistic gas of fermions? Cosmology.Generally, heating something up by an extreme amount is a good way to figure out whatare its constituents. Conveniently for particle physicists, the whole universe somehow gotheated up quite a bit in the past. The early universe is a great source of examples ofultrarelativistic bosons and fermions at high temperature. For example, at kB T 4 108 eV 400M eV (i.e. T 1012 · 5 K), which was the situation at about 1 microsecond afterthe Big Bang, an accounting of the energy density begins by thinking about a gas of photons,electrons, positrons, quarks, antiquarks, gluons. If we treat them as ultra-relativistic andnon-interacting (not necessarily a good idea), we get: π 2 (kB T )4 hEi (# of bosons) 1 {z}V30 ( c)3 2 2 8 {z} {z }photonsgluons2 {z 2}2 spins of q and q̄ (# of fermions) {z}3 {z 3} of 3 colors and 3 flavors2 {z2 5} 7 8 2 spins of e, ē, µ, µ̄ and ν, ν̄ 3People have recently recreated similar conditions by colliding heavy ions at very high speeds(initially at RHIC on Long Island, NY and more recently at the LHC at CERN). They havecreated a new state of matter called Quark-Gluon Plasma, which turns out to be a liquid atthe temperatures and densities that have been studied so far.9-16
Statistical mechanics flowchart9-17
9.6Fermions at low temperaturesSo far we’ve discussed high temperatures. Low temperatures is where quantum statistics –the differences between boson, fermions, and ‘Boltzmons’ – are most dramatic.For fermions in thermal equilibrium, the occupation number of the single-particle state αis1.hnα i ( α µ)/k TBe 1The only property of the single-particle state that enters is its energy, so we may relabel this(as is conventional) asf ( ) 1e( µ)/kB T 1, the Fermi distribution function.This one function contains the physics of metals, semiconductors, white dwarves, neutronstars.Consider first the extreme T 0 limit of this function.It looks like this: A useful fact: f ( µ) 12 at any temperature.How do we understand this answer for the distributionof occupied states at T 0? At T 0 the system is inits groundstate. The groundstate of N free fermions isconstructed by filling the N lowest-energy single-particlestates. The highest energy level that’s filled is called F ,the Fermi energy. (Note that this is a single-particle energy level, not an energy of all N particles.) The answerwe’ve found for f says that, at T 0, all states below thechemical potential are filled (f 1) and all states above the chemical potential are empty(f 0). So: F µ(T 0) .Note that µ will be T -dependent.The value of F is determined by N :ZXN hnα i d D( )f ( )at any T0αZ F µ(T 0) d D( )09-18at T 0
D( )D( ) is the same kind of object that we considered earlierin the context of blackbody radiation. The counting ofmomentum states is the same as in Chapter 7:N (k) # of states with k k 2 {z}spin- 12fermions14πk 3833(π/L).For spin- 21 fermions, there are two spin states (‘up’ and‘down’) for each wavenumber.The only difference is that now we will consider nonrelativistic fermions, which have a dispersion relation 2 2p 2 k .2m2mSoVN ( ) 23π 2m 2 3/2 3/2 3/212md 1/2D( ) N ( ) V22d 2π {z} aZN N ( F ) 0 F223/23/2D( )d aV F aV µ033(2)(here I wrote
9.2 Quantum systems with many indistinguishable particles [This section is about quantum mechanics. You’ve already encountered some of these ideas in 8.04, and will discuss this further in 8.05. We’ll come back in subsection 9.4 and think about when this business reduces to classical mechanics.]
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