Boltzmann, Fermi-Dirac, and Bose–Einstein Statistics

5.62 Lecture #8: Boltzmann, Fermi-Dirac, and Bose–Einstein Statistics THE DIRECT APPROACH TO THE PROBLEM OF INDISTINGUISHABILITY We could have approached the problem of indistinguishability by treating particles as indistinguishable fermions or bosons at the outset. QM tells us 1. All particles are indistinguishable 2. All particles are either fermions or bosons. The odd/even symmetry of a particle's wavefunction with respect to exchange is determined by whether the particle is a fermion or boson. FERMION — a particle which obeys Fermi-Dirac statistics; many-particle wavefunction is antisymmetric (changes sign) with respect to exchange of any pair of identical particles: P12ψ = –ψ 1/2 integer spin e–, proton, 3He Single state occupation number: ni = 0 or ni = 1, no other possibilities! BOSON — a particle which obeys Bose-Einstein statistics; many-particle wavefunction is symmetric (does not change sign) with respect to exchange of any pair of identical particles 4He, H2, photons integer spin ni = any number, without restriction What kind of particle is 6Li, 7Li, D, D+? We are going to figure out how to write ω ( n i ,g i ) N! i=1 where we are considering level i with energy ει and degeneracy gi. Previously we had considered Ω({ni}) for non-degenerate states rather than gi-fold degenerate εi levels. t Ω ({n i }) = ∏ 5.62 Spring 2008 Lecture 8, Page 2 Let us play with some simple examples before generalizing results for each type of system. 3 degenerate states: A, B, C (states could be x, y, z directions for particle in cube) 2 particles: 1, 2 A B C 1 2 1 2 2 1 2 1 1 2 2 1 1,2 1,2 1,2 If particles are distinguishable and there are no restrictions on occupancy, total # of distinguishable arrangements is 32. Note that this is different from the degeneracy of a particular set of occupation numbers for non-degenerate states, N! . ∏ ni ! For each degenerate level occupied by particles, we have a factor: degeneracy of atomic state particles ω(n,g) = gn to correct for particle indistinguishability. We divide by N! t Ω B ({n i }) = ∏ ω(n i ,g i ) i=1 N! t = ∏g ni i i=1 N! 32 gn For our case 2! = 4.5 which is not an integer so can be only an approximation n! to the correct total # of ways. Now go to F–D system occupation # is 0 or 1, indistinguishable particles, therefore g ≥ n. revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 3 ωFD(n,g) = ⎡ g! ⎤ 1 ⎢ (g − n)! ⎥ n! ⎣ ⎦ put excitation first in any of g states, 2nd in any of g–1, then divide by n! for indistinguishability of particles. Finally, divide by (g – n)! for indistinguishability of “holes”. ΩFD ({n i }) = A X X B X X ∏ω i =1 ⎡ t t FD (g i , n i ) N! ∏ ⎢⎣ (g = i =1 gi ! ⎤ ⎥ i − n i )!n! ⎦ N! C g = 3, n = 2 3! ωFD = 2!1! = 3 X X Now for B–E what is the combinatorial factor? n particles g distinguishable states → g–1 indistinguishable partitions arrange n indist. particles and g–1 indist. partitions in all possible orders (n +g–1)! ωBE(N,g) = n!(g–1)! t t ΩBE ({n i }) = for our case ∏ ω BE (n i ,g i ) i=1 N! = ∏ i=1 (n i + g − 1)! n i !(g − 1)! N! (2+3–1)! 4! 2!(3–1)! = 2!2! = 6 A XX B C XX XX X X X X X X for our example ωFD < 3 ωB < 4.5 ωBE 6 revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 4 always true — compare ω(ni,gi) factors in Ω({ni}) term by term. This means that n i for a degenerate level is always largest for BE and smallest for FD. Why? Before starting the general, rigorously derivable result for n i for BE, corrected Boltzmann, and FD, we need to derive a relationship between µ and q for corrected Boltzmann statistics. A = –kT ln Q = –kT ln (qN/N!) = –NkT ln q + kT ln N! for large N, ln N! = N ln N – N. This is Stirling’s approximation. Very, very useful. A = −NkTln q + NkTln N − NkT ⎛ ∂A ⎞ µ≡⎜ = −kTln q + kTln N + NkT (1 N ) − kT ⎝ ∂N ⎟⎠ T,V = −kTln q + kTln N = −kTln ( q / N ) µ = ln(q / N) kT e − µ kT = q / N − q = Ne – µ /kT which is a very convenient form. The probability of finding one particle out of N in level εi is Pi = n i = Ne − εi n i e − εi kT = N q kT which are the standard definition and statisitical mechanical values of Pi q replace q n i = Ne − εi kT ( Ne − µ kT ) = 1 ( εi − µ) kT e This is the corrected Boltzmann result for n i . Notice that when εi < µ, n i > 1 which violates the assumption upon which the validity of corrected Boltzmann depends. Note also, that when T → 0, the only occupied levels are those where εi ≤ µ. When εi > µ and T = 0, n i = 0. Note further, from the derived T dependence of µ µ = –kT ln (q/N) lim µ = 0 T→0 revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 5 Thus, as T → 0, the only occupied level in εi = 0 which has occupancy n ε = 0 = 1. It is clear that we need to replace corrected Boltzmann statistics by BE or FD as T → 0 and whenever niB becomes comparable to 1. Assert (derived for Grand Canonical Ensemble, where µ, V, and T are held constant, pp. 431-439 of Hill) +1 is FD 1 n i = ( εi − µ) kT −1 is BE e ±1 no 1 is B This equation obeys the expectation niFD < niB < niBE In the limit of T → 0 εi = µ εi < µ εi > µ BE < 0 (impossible) 0 ni → ∞ FD 1 1 0 ni → 2 ∞ (illegal) niB → 1 0 First, let's make sure “exact” result for n i is correctly normalized to total number of particles. N = F ∑ ni = F ∑ i i 1 e ( εi − µ )/kT ± 1 normalization correction factor so F= N ∑ i normalization factor 1 e( ε i − µ ) /kT ± 1 ni = Fn i = (e ( ε i −µ ) kT N ± 1)∑ j 1 e (ε j − µ) kT ±1 normalized “exact” result approximate result being checked for normalization ε −µ kT Now make the Boltzmann approximation e( j )  1 for all j revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 6 The factors of ±1 are small and can be neglected, thus: ni = N − ε −µ e(εi −µ) kT ∑ e ( j ) kT = j ni = Ne −εi ∑e kT −ε j kT = Ne −εi q kT j CORRECTLY REDUCES TO BOLTZMANN STATISTICS RESULT! ε −µ — PLOT the FD and BE distribution functions for ni . ni vs. i kT Note that: * n i cannot be larger than 1 for FD * n i goes to ∞ when εi = µ for BE * when n i ≈ 1, we are no longer allowed to use corrected Boltzmann. (εi–µ)/kT — For large εi – µ and consequently large e or ni << 1, revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 7 niFD = niBE and when these occupation numbers are equal — ni << 1 ! Most atoms and molecules at "ordinary" temperatures are in Boltzmann regime where — ni << 1 or q >> N. In this case, there is no difference between FD and BE statistics. Doesn't matter if the molecule is a fermion or boson. So the Boltzmann statistics we have developed is valid over a wide range of molecules and conditions. EXCEPTIONS: 4He and H2 are BOSONS. Must be treated as such at T close to 0K. 3He is FERMION. Must be treated as such at T close to 0K. Notice that the exceptions are lighter atoms and molecules at low T. That's because as you make particle less massive, the spacings in energy between the particle's states get — larger, leading to fewer available states. If fewer states are available, ni goes up, and eventually the difference between FD and BE statistics becomes discernible. ( h2 εi = n 2x + n 2y + n 2z 2 8ma ) ε3 ε7 ε5 ε ε2 ε3 ε1 ε ε1 LIGHT PARTICLE — * ni is larger * may have to use FD or BE statistics HEAVY PARTICLE — * ni is smaller At "normal" temperatures > ~20K, can treat 3He, 4He, H2 with Boltzmann statistics. revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 8 So temperature plays a role here too. We'll talk about this next time. BUT ELECTRONS always have to be treated as FERMIONS for all "normal" — temperatures (<3000K), because their ni ’s  1. The valence electrons of the Au atoms which make up a gold crystal are delocalized throughout the crystal. These electrons can be thought of as an electron gas contained within the crystal. This is called a free electron model where the energy levels of electrons are particle-in-box energy levels. The average number of electrons in each electron state is given by the Fermi-Dirac distribution function ni = 1 e ( ε i − µ ) kT +1 F.D. At T = 0 We fill each state with 1 e– in order of increasing energy. The energy at which we run out of e–s is εi = µ. In Solid State Physics language µ ≡ εF FERMI ENERGY which is the maximum energy that an e– can have at T = 0. AT T > 0K revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 9 Electrons move from occupied to unoccupied states as T is increased. Must move to — unoccupied states because ni > 1 not allowed. This is the origin of conductivity in metals. [More on this in second half of course.] revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 10 Non-Lecture Alternative Derivation of ni for B-E and F-D Particles Canonical p. f. (this is actually a sum over sets of occupation numbers, {ni}, and over states, i) Q = ∑ Ω ({ni }) e−βni εi {ni } =∑ {ni } ∏ ⎡⎣ω ( ni ) e−βni εi ⎤⎦ i ω(ni) is the number of distinguishable ways of arranging ni particles in the εi singleparticle energy level. The form of ω(n) depends on the particle statistics. If it were possible to evaluate this sum, then we could determine ni from ⎛ ∂lnQ ⎞ −kT ∂Q ni ≡ −kT ⎜ = ⎟ Q ∂εi ⎝ ∂εi ⎠V ,T , N =− kT Q ∑ = { i} ∑ {ni } ⎛ −ni ⎞ ⎜ ⎟Ω ({ni }) e−βni εi ⎝ kT ⎠ ni Ω ({ni }) e−βni εi n Q ≡ ni . Now we will evaluate Q approximately by finding the single set of occupation numbers {ni} that gives the maximum term in the sum over occupation numbers that defines Q. The approximation is to set Q equal to the value of this maximum term in the sum. There remains a sum over states, i. This approximation can only be valid if the maximum term in the sum is vastly larger than any term corresponding to a different set of occupation numbers. This is a common and useful approximation in statistical mechanics. Find the maximum term in the sum, call it QM and assume QM ≈ Q A = −kT lnQM −βA = lnQM = ∑′ {ln ⎡⎣ω (n )⎤⎦ − βε n }, i i i i we have kept only the single set of occupation numbers that gives the maximum term in the sum that defines Q. revised 2/25/08 4:35 PM 5.62 Spring 2008 The prime on the Lecture 8, Page 11 ∑′ implies the constraint i N= ∑ ni i Use Lagrange multipliers to impose this constraint so that the constrained sum can be replaced by an unconstrained sum, ⎛ ⎞ −βA = lnQM = ∑ {ln ⎡⎣ω (ni )⎤⎦ − βεi ni } + λ ⎜∑ ni − N ⎟ ⎝ i ⎠ i   =0 unconstrained sum ⎛ ∂A ⎞ −β ⎜ ⎟ = −λ ⎝ ∂N ⎠V ,T value to be chosen ⎛ ∂A ⎞ but ⎜ ⎟ = µ ⎝ ∂N ⎠V ,T Thus λ = βµ = µ . kT Insert the derived specific value for λ, −βA = ∑ {ln ⎡⎣ω (n )⎤⎦ − β (ε i i − µ) ni } − Nβµ , i rearrange { Nβµ − βA = ∑ ln ⎡⎣ω(n i ) − β ( ε i − µ ) n i ⎤⎦ i } and use the thermodynamic identity G = Nµ = A + pV β ( Nµ − A) = βpV = lnQ + Nβµ = ∑ {ln ⎡⎣ω (n )⎤⎦ − β (ε i i − µ) ni } i Now we choose particular forms for ω(ni) and insert them into the above equation for βpV. Note that we have not yet addressed the approximation of Q by QM. A. Corrected Boltzmann gi is the degeneracy of the single-particle εi energy level, and ni is the number of particles in the assembly in the εi energy level. revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 12 gn ωB = n! ln ω B = n ln g − n ln n + n βpV = ∑ {ni ln gi − ni ln ni + ni − β (εi − µ) ni } i to obtain the maximum term in this sum, we take a derivative wrt each ni. For each ni ⎛ ∂ ⎞ = {ln ( gi ) − ln ni −1+1 − β (εi − µ)} = 0, thus ⎜ (βpV ) ⎟ ⎝ ∂ni ⎠V ,T , N niB = gi e−βεi eβµ ∂2 (βpV ) (note = –(1/ni) < 0 which assures that the extremum is a maximum and not a ∂ni2 minimum). Since q = Ne–βµ, we get the standard corrected Boltzmann result when we replace eβµ by N/q, niB gi e−βεi = , N q moreover, when we replace the original sum over sets of {ni} that defines βpV by the specific set {ni} that gives the maximum term in the sum, we get g ⎧ ⎫ βpV = ∑ ⎨ni ln i + ni − β ( ε i − µ) ni ⎬ ni i ⎩ ⎭ gi = eβ ( εi − µ) ni ni ln ( gi ni ) = niβ ( ε i − µ) . Thus βpV = ∑ ni = N i which is the ideal gas law. revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 13 B. Fermi-Dirac ω FD ( g, n ) = βpV = ∑ g! n!( g − n )! {ln ⎡⎣ω (n ) − β (ε i i − µ) ni ⎤⎦} i βpV = ∑ {gi ln gi − gi − ni ln ni + ni − (gi − ni )ln(gi − ni ) + ( gi − ni ) − β (εi − µ) ni } i −gi + ni + ( gi − ni ) = 0 The extremum term in the sum over sets of {ni} is obtained from taking ∂ and setting dn i result = 0. ⎛ ∂ (βpV ) ⎞ = {ln ( gi − ni ) − ln ni − β (εi − µ)} = 0 for all i. ⎜ ⎟ ⎝ ∂ni ⎠V ,T , N ln gi − ni = β (εi − µ) ni gi − ni gi = −1 = eβ (εi − µ) ni ni gi = eβ (εi − µ) +1 ni Thus niFD = gi 1 β (εi − µ) e +1 , replacing the sum in Q by its maximum term, QM βpV = ∑ {gi ln gi − ni ln ni − gi ln ( gi − ni ) + ni ln ( gi − ni ) − β (εi − µ) ni } i ⎧ ⎫ g − ni = ∑ ⎨ gi ln gi − gi ln ( gi − ni ) + ni ln i − β (εi − µ) ni ⎬ ni ⎩ ⎭ i but, for the term in the sum over sets of {ni} that has the maximum value g −n ln i i = β (εi − µ) ni and the last two terms in { } cancel. We obtain revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 14 βpV = −∑ gi ln i ⎛ n⎞ gi − ni = −∑ gi ln ⎜1 − i ⎟ gi ⎝ gi ⎠ i ⎛ ⎞ 1 = −∑ gi ln ⎜1 − β (ε − µ) ⎟ ⎝ e i +1 ⎠ i ⎛ eβ (εi − µ) +1 −1 ⎞ = −∑ gi ln ⎜ β (εi − µ) ⎟ ⎝ e +1 ⎠ i ⎛ eβ (εi − µ) +1 ⎞ = ∑ gi ln ⎜ β (ε − µ) ⎟ = ∑ gi ln [1 + eβ ( µ−εi ) ] ⎝ e i ⎠ i i which is not the ideal gas law. However, at high ε, ln [1+ eβ ( µ−ε) ] → eβ ( µ−ε) because, for x  1 ln (1+ x ) = ln1+ β ( µ−ε) and, for x = e ≈x 1 x +… 1+ 0 when ε  µ eβ( µ−ε)  1 . Thus βpV = ∑ gi eβ ( µ−εi ) = ∑ gi eβµe−βεi i e −βµ i = (q N ) βpV = ∑ i Ngi e−βεi q =N =N q q which is the ideal gas law! C. Bose-Einstein ω BE ( ni ) = ( gi + ni −1)! ni !( gi −1)! Thus ln ω BE ( ni ) = ∑ {( gi + ni −1) ln ( gi + ni −1) − ( gi + ni −1) − ni ln ni + ni − ( gi −1) ln ( gi −1) + gi −1} i =∑ {( gi + ni −1) ln ( gi + ni −1) − ni lni ni − ( gi −1) ln ( gi −1)} i βpV = lnQ + Nβµ = ∑ {ni } {ln ω BE (ni ) − β (εi − µ) ni } . The set of occupation numbers that gives the largest contribution to βpV is obtained from revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 15 ⎛ ∂(βpV ) ⎞ g + n −1 n = 0 = ln ( gi + ni −1) + i i − ln ni − i − β (εi − µ) ⎜ ⎟ gi + ni −1 ni ⎝ ∂ni ⎠V ,T , N thus ln gi + ni −1 = β (εi − µ) ni gi −1 +1 = eβ (εi − µ) ni niBE = ( gi −1) Since gi  1 , we obtain the usual result 1 β (εi − µ) e −1 g niBE = i β (εi − µ) e −1 BE and, when εi = µ, the denominator vanishes and ni can be very large. This is the BoseEinstein condensation. If εi < µ, the nonsense result of a negative occupation number is obtained. D. Accuracy of the Maximum Term Approximation (see Hill, Appendix II, pp. 478-480). ′ Q≡∑ {ni } ∏ ⎡⎣ω ( ni ) e−βεi ni ⎤⎦ ≡ ∑ t ({ni }) {ni } ni where t({ni}) is a typical term in the sum and ∑′ imposes the implicit constraint ∑n i =N. i Expand the value of the typical term t({ni}) in the sum as a power series in deviations from the special set of occupation numbers {n̂i } that give the maximum value of ∏ ⎡⎣ω ( ni ) e−βεi ni ⎤⎦ , ni t M = t ({n̂i }) . We have already used the requirement that 0 = value of tM and the set {nˆi } . ∂ ⎡ −βε n ⎣ω ( ni ) e i i ⎤⎦ for all ni to find the ∂ni Thus 2 ˆi ) 1 2 ∂ ln ω ( n . δn ∑ i 2 2 i ∂ni The first nonzero term in the expansion involves the second derivatives because all of the first derivatives were required to be zero (condition for the maximum term) ln t ({ni }) = ln t ({nˆi }) + revised 2/25/08 4:35 PM 5.62 Spring 2008 Lecture 8, Page 16 ∂ln t ({n̂i }) ∂ni =0 and the e−βεi ni term is “used up” in the extremum condition. Thus the true value of Q is given in terms of the value of the maximum term in the sum over {ni} as ⎛ ⎞ 2 ⎡1 ⎤ ′ 2 ∂ ln ω ( n̂i ) ⎜ ⎟ Q = QM ⎜1+ ∑ exp ⎢ ∑ δni ⎥ +…⎟ 2 ∂ni ⎣2 i ⎦ ⎝ {δni } ⎠ but, since we showed that the set {n̂i } maximizes t({ni}), all of the second derivatives must be negative. This means the factor multiplying QM is (1 + e–x) where x > 0 hence Q ≈ QM. To make this argument stronger we need to compute these second derivatives and also realize that the exp[ ] contains many additive negative terms, thus e–x → 0. The derivation of the second derivatives listed below is left as an exercise for you: Statistics Boltzmann Fermi-Dirac Bose-Einstein ∂2 ln ω ( n̂i ) ∂ni2 -1/ni gi − ni ( gi − ni ) − gi −1 ni ( gi + ni −1) revised 2/25/08 4:35 PM