Physics 505 Homework No.2 Solution

Transcription

1 Physics 55 Homework No Solution February 3 Problem Calculate the partition function of a system of N noninteracting free particles confined to a box of volume V (i) classically and (ii) quantum mechanically Compare and contrast the results Show the emergence of Gibbs paradox and comment briefly on the resolution of the paradox via the N! factor In particular comment on whether you find the resolution satisfactory or not and say why Consider first only a single particle in a box Classically, the particle has a continuum of states, so instead of sum one might expect the partition function to be an integral over all phase space: Z? = e βe(x,p) d 3 xd 3 p However, this cannot be the correct partition function since it has units of momentum cubed times volume, but the partition function must be dimensionless We can correct for this by setting Z = s 3 e βe(x,p) d 3 xd 3 p where s is a small number which has units of action In a rough sense s 3 represents the volume of a single coarse-grained state in phase space Pure classical theory provides no way to actually calculate the value of s, so for now it will just be left as an undetermined constant For a free particle, E(x, p) = p /m so the partition function is Z = d 3 x e βp /m 4πp dp s 3 = π(mkt ) 3/ s V 4π 3 4 ( ) 3/ πmkt = V s where a Gaussian integral has been used to perform the momentum integration Now consider a single (spinless) quantum mechanical particle in a box with sides L Requiring that the wavefunction be zero at the edges of the box gives the energy levels as: E(n x, n y, n z ) = h 8mL (n x + n y + n z)

2 where n x, n y, and n z are integers The single particle partition function is Z = n x,n y,n z e βe(nx,ny,nz) Since the energy only depends on the the quantity n = n x + n y + n z, the sum can be approximated by an integral over a 3-d volume in n space: Z 8 e βe(n) 4πn dn provided that the integrand varies slowly enough with n, which is true if the size of the box or the temperature is sufficiently large The factor /8 ensures that we count only states with positive n Solving for n in terms of E, we have 8mL n = E h 8mL dn = h E de and substituting these two equations into the integral above converts it into an integral over E: Z e βe D(E)dE where the density of states per unit energy D(E) is D(E) = V π(m)3/ h 3 E Evaluating the integral with the help of a gamma function gives the single particle partition function as Z V (πmkt )3/ h 3 V Λ 3 where the thermal de Broglie wavelength is defined to be Λ = h πmkt For the classical result to be equivalent, the undetermined constant s must be equal to Planck s constant h Now consider what happens when the system is extended to N noninteracting particles The total energy of the system is just the sum of the energies of each particle, E α,β,,γ = E α + E β + + E N γ

3 where the notation Eα n means the energy of the nth particle in a state α The partition function then factors into a product of N single particle partition functions: Z = exp β(eα + Eβ + + Eγ) ] 3 = α,β,,γ ( = Z N e βe α ) ( e βe β ) α β α ( ) e βen γ However, if Z = Z N then the entropy is S = T NkT ln(z ) = Nk ln(v/λ 3 ) + 3 ] Now consider two identical ideal gasses which are initially separated from one another by a partition Suppose both gasses have the same volume V, number N and thermal wavelength Λ Each gas therefore has an entropy S = N k ln(v /Λ 3 ) + 3 ] Now suppose that the partition is removed so that the two gasses are put into contact with one another The temperature of the final gas remains the same and the volume and number both double, so the resulting entropy is S f = N k ln(v /Λ 3 ) + 3 ] However, since the partition can be removed adiabatically and reversibly, the final entropy should be S (ie entropy should be an extensive quantity) This is the Gibbs mixing paradox The paradox can be resolved by insisting that the particles are indistinguishable and setting the many-particle partition function to Z = (Z ) N N! We will try to justify this a bit later, but first let s see if this resolves the mixing paradox With Z = (Z ) N /N!, the entropy becomes S = T kt ln(zn /N!)] = Nk ln(v/λ 3 ) + 3 ] k ln N! ( ) V = Nk ln + 5 ] NΛ 3 3

4 where Stirling s approximation ln N! N ln N N has been used in the last step If we begin with a system with N particles, volume V, and thermal wavelength Λ, and entropy ( ) V S = N k ln + 5 ] N Λ 3 then doubling the system also doubles the entropy: ( ) V S f = N k ln + 5 ] = S N Λ 3 The entropy is thus extensive and the Gibbs paradox is resolved Should you be satisfied with this? Maybe not The problem is that setting Z = (Z ) N /N! is not really correct, and it may be wrong by a significant amount Let s go back and try to re-sum the partition function for indistinguishable particles Indistinguishability implies that permuting the particle labels does not change the physical state, eg the energies and E αβ,γ = E α + E β + + E N γ E β,α,γ = E α + E β + + E N γ (which are equal to one another) correspond to the same physical state and should only be counted in the partition function once Evidently, the sum exp β(eα + Eβ + + Eγ) ] 3 = (Z ) N α,β,,γ counts these states more that once In order to correct for this, terms which correspond to identical physical states should be divided by the number of times they occur in the sum Consider a state with each α, β,, γ different This is only one physical state, but it occurs in the sum exactly N! times, so these terms should be divided by N! The problem arises with terms where more than one of the particles are in the same state For fermions these states are not even allowed because of the Pauli exclusion principle, and they should not be counted at all For bosons, these states are physically allowed, but they occur less than N! times in the sum (In particular, the term with α, β,, γ all equal to a given value α occurs precisely once in the sum) Setting Z = (Z ) N /N! amounts to dividing every term by N! even if it does not occur N! times in the sum Thus (Z ) N /N! is too large for fermions and too small for bosons! Is this a serious problem? Not really What matters is the free energy, which is the logarithm of the partition function For very large numbers (and (Z ) N is a very large number indeed for macroscopic amounts of particles) the logarithm varies very slowly Even if (Z ) N /N! is wrong even by an order of magnitude or so, ln(z ) N /N! may still approximate the logarithm of the partition function to a high degree of accuracy The N! factor can also be obtained as a classical limit of a quantum calculation Recall that the average occupation number n α for a state α for a quantum system is given (approximately) by n α = e β(eα µ) ± 4

5 where the plus sign is for fermions and the minus sign is for bosons Then chemical potential µ is obtained by the requirement that the total number of particles is N = α n α = α e β(eα µ) ± If for every E α, e β(eα µ) then the ± factor in the denominator can be neglected, and the sum over occupation numbers can be approximately calculated as N α e β(eα µ) = e βµ α e βeα = e βµ Z The chemical potential µ is also related to the free energy by µ = F N = N β ln Z N β (ln Z N ln Z N ) where Z N is the partition function for N particles and Z N is the partition function for N particles Exponentiating and combining with the above result This relation holds for all N, so Z N Z N = e βµ = Z N Z = (Z ) / Z 3 = Z Z /3 = (Z ) 3 /6 Z N = (Z ) N /N! This disscussion of the error N! factor is adapted primarily from: R Baierlien, The fraction of all different combinations: Justifying the semi-classical partition function, Am J Phys 65, (997) H Kroemer, How incorrect Is the classical partition function for the Ideal gas, Am J Phys 48, (98) These articles and the references therein provide further enlightening discussion and techniques for estimating how wrong the N! factor is 5