Σ P[ψ a (1)... ψ z (N)], Permutations

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1 Chemistry 593: Ideal Bose & Fermi Gases David Ronis McGill University 1. Introduction and General Considerations When we consider quantum statistical mechanics of identical particles, we abandon the conventional, classical, notion of states (i.e., as lists of quantum numbers assigned to the particles) and instead focus on "occupation numbers", n i,the number of particles in 1-particle state i (e.g., a single hydrogenic electron or a single particle in a box). Indistinguishably implies that the quantum probability density, ψ (1, 2, 3, 4,...) 2,besymmetric under exchange of any particle labels, and, in turn, this means that ψ (1, 2,...) =±ψ (2, 1,...). Quantum field theory and special relativity show that the + sign is used for integer spin particles, generically called Bosons, while the sign is used for half-odd integer spins, referred to as Fermions. Tw o see how this works, consider the following 2-particle example: each particle can be in one of two states, α, i >or β, i >, i = 1, 2, with energies ε α and ε β,respectively. Ifweignore the consequences of indistinguishability we can construct 4 possible states for this system as shown in the following table: Table 1: Possible States and Energies for a Two Level System with Two Particles State Wavefunction Energy n α n β Particle 1 α,1 > α,2 > 2ε α 2 0 Bosons 2 2 1/2 ( α,1 > β,2 >+ β,1 > α,2 >) ε α + ε β 1 1 Bosons 3 2 1/2 ( α,1 > β,2 > β,1 > α,2 >) ε α + ε β 1 1 Fermions 4 β,1 > β,2 > 2ε β 0 2 Bosons 2 e ε α /k B T + e ε β /k B T, for distinguishable particles (4 states) Q(N = 2, T, V ) = e (ε α +ε β )/k B T, for Fermions (1 state) e 2ε α /k B T + e (ε α +ε β )/k B T + e 2ε β /k B T, for Bosons (3 states). For systems with more than two particles we simply symmetrize or anti-symmetrize the wavefunction (i.e, add all possible permutations of the particle labels together) for Bosons or use the so-called Slater determinant for Fermions. 1 This shows that n i = 0, 1, 2,... for Bosons, while 1 For Bosons this means Ψ(1, 2,..., N) = Σ P[ψ a (1)... ψ z (N)], Permutations where, P permutes the particle labels and where ψ a (1) is an orbital for one particle in 1-particle state a (e.g., for atoms, an atomic orbital). For Fermions we construct the so-called Slater determinant, i.e., ψ a (1) ψ b (1) Ψ(1, 2,..., N) =... ψ z (1) ψ a (2) ψ b (2)... ψ z (2)... ψ a (N)... ψ b (N) ψ z (N) (1)

2 Chemistry Ideal Bose & Fermi Gases n i = 0, 1 for Fermions, as required by the Pauli Exclusion Principle. 1 In either case, the wavefunction is determined, up to an overall phase factor, byspecifying the n i s. Since E = Σi n iε i,itfollows that the canonical partition function is Q(N, V, T ) = 1or Σ 1or Σ... exp Σ ε i n i /k B T i, (2) n α = 0 n β = 0 iσ n i = N where the upper limits of the sums depend on whether Fermions or Bosons are under consideration. The sums aren t easy to do when constrained to have Σi n i = N. Since N fluctuates in the grand canonical ensemble, we expect that the constraint on N to disappear. Moreover, we ve shown that the choice of ensemble does not matter when calculating thermodynamic quantities. Hence, we switch to the grand canonical ensemble, with partition function Ξ= Σ e β µn N = 0 1or Σ 1or n 1 = 0 n 2 = 0 iσ n i = N = Π 1 or i Σ e β (µ ε i)n i n i = 0 Σ... e β iσ ε in i = 1or Σ 1or Σ... e β n 1 = 0 n 2 = 0 iσ (ε i µ)n i = Π ±1 i 1 ± eβ (µ ε i), (3) where we ve gone back to our earlier notation of naming states by a roman index, β and µ have their usual meanings, and where the last equality follows by explicitly carrying out the sums for Fermions (upper signs) or Bosons (lower signs). Once we have the grand canonical partition function, thermodynamic functions are obtained in the usual manner, e.g., and N = β pv = ln(ξ) =± ln(ξ) = β µ T,V where the average occupation numbers are Σ ln i 1 ± eβ (µ ε i), (4) Σ n i = i Σ i 1 e β (ε i µ) ± 1, (5) E = ln(ξ) = β Σ n i ε i, (6) β µ,v i 1 n i = e β (ε i µ) ± 1, (7) and are known as the Fermi-Dirac (+ sign) or Bose-Einstein (- sign) distributions. Note that neither expression is normalized. The Slater determinant is odd under exchange of any two rows and vanishes if two or more rows are the same, thereby having the required symmetry and Pauli exclusion principle.

3 Chemistry Ideal Bose & Fermi Gases Finally consider the heat capacity. From Eqs. (6) and (7) it follows that C V = k B 2 Σ n j 2 e β (ε j µ) n i 2 e β (ε i µ) [β (ε i ε j )] 2 i, j Σ n j 2 e β (ε j µ) j, (8) where the details of calculation are given inappendix A. Note that C V is positive, asexpected. The quantity n i 2 exp[β (ε i µ)] plays a key role, and is shown in Figs. 1 and 2, for Fermions and Bosons, respectively. Anapproximate expression for Fermions for 4 n i 2 e x can be obtained by noting that the Taylor expansion of ln(4 n i 2 e x ) x 2 /4 + O(x 4 ), where x β (ε i µ); hence, 4 n i 2 e x e x2 /4 Fig. 1. Mean occupation numbers, n i,and 4 n i 2 e β (ε µ) for Fermions. The spin degeneracy isnot included. Notice how the orbitals change from filled to empty over a range of β (ε µ) ± 5 or, for T = 300K, over a range (ε µ) ± 2. 5k B T =± ev. The plot of n i 2 β (ε µ) e shows that the main contributions to the heat capacity come from a narrow band of 1-particle states near the Fermi energy. Finally, the dotted brown line is the approximation to 4 n i 2 e β (ε µ) discussed in the text.

4 Chemistry Ideal Bose & Fermi Gases Fig. 2. Mean occupation numbers, n i,and n i 2 e β (ε µ) for Bosons. The spin degeneracy isnot included. Equation (8) leads to an approximate expression for C V. Appendix A shows a more rigorous approach 2 2. Results for Particles in a Cubic Box In order to proceed we need a concrete model for the 1-particle states and energies. The simplest is probably the particle with mass m in a cubical box of side length L. This has energies: ε nx,n y, n z = h2 8mL 2 (n2 x + n 2 y + n 2 z), where n i = 1, 2, 3,.... (9) Note that the ground state energy, ε 1,1,1 = 3h 2 /(8mL 2 ) 0as L. When the sum over single particle states (i.e., over positive n i, i = x, y, z,) is approximated as an integral and the result transformed to polar coordinates, we find that all the examples at the end of the previous section can be written as (2S + 1) π 2 0 dn n2 f (λ, βε n ), (10) where the factor of (2S + 1) accounts for the spin degeneracy ofaparticle with spin S, the factor of π /2 = 4π /8 is the area of the unit sphere in the octant where n i >0, i = x, y, z, and λ e β µ is the proper activity. Next, change variables by letting n = 8m/h 2 Lε 1/2 and rewrite Eq. (10) as where 0 dε g(ε ) f (λ, βε), (11) 2 See J.E. Mayer and M.G Mayer, Statistical Mechanics, (John Wiley &Sons, Inc., 1940), Sec. 16g.

5 Chemistry Ideal Bose & Fermi Gases g(ε ) 2π (2S + 1) 3/2 2m V h 2 ε 1/2 (12) is known as the density of states and can be interpreted as the number of states per unit energy with energies between ε and ε + dε. With this, the general results given atthe end of the preceding section become and β pv =± 0 dε g(ε )ln 1 ± λe βε, N = 0 dε g(ε ) λe βε (13a), (13b) 1 ± λe βε E = dε g(ε )ε λe βε. (13c) 0 1 ± λe βε Note that Eqs. (13b) and (13c) can be obtained by taking the usual derivatives of β pv,i.e., ln(ξ), cf. Eq. (13a). and Next we expand the integrands into Taylor series in λ, specifically, ± ln 1 ± λe βε = + Σ ( + λ) j j j = 1 e jβε λe βε 1 ± λe =± ln(1 ± λe βε ) = + βε β µ Σ ( + λ) j e jβε, β,v,n and change variables yet again by letting ε (k B T / j)x.this gives: and β pv = + 2 π 1/2 N = + 2 π 1/2 E = + 2 π 1/2 V Λ V Λ j = 1 (14a) (14b) (2S + 1) 3 Σ ( + λ) j Γ(3 / 2), (15a) j 5/2 j = 1 (2S + 1) 3 Σ ( + λ) j Γ(3 / 2), (15b) j 3/2 j = 1 V Λ k BT (2S + 1) Σ ( + λ) j Γ(5 / 3), (15c) 3 j 5/2 where Λ h/ 2π mk B T is the thermal de Broglie wavelength, and have expressed the remaining integrals as Γ-functions, 3 specifically, Γ( 3) = 1 π 1/2,for β pv and N, and Γ( 5) = 3 π 1/2 for E When these are used in Eqs. (15a) (15c) and the results rearranged, we find that j = 1 3 One of the ways of defining the Γ function is via its integral representation Γ(z) 0 dx x z 1 e x, which includes all of the remaining integrals. For more properties of these functions, see, e.g., the Handbook of Mathematical Functions, tenth edition, M. Abramowitz and I.S. Stegun, eds., ch. 6.

6 Chemistry Ideal Bose & Fermi Gases and ρλ 3 = (2S + 1)G 3/2 (λ) (16a) where β pλ 3 = 2β E Λ3 3V = (2S + 1)G 5/2 (λ), (16b) G z (λ) 1 Γ(z) dx x z 1 λe x 0 1 ± λe = λφ( x + λ, z,1) = + Σ j z ( + λ) j, j=1 (16c) generalizes Hill s F z (α ), 4 the function Φ(a, b, c) isdiscussed in Gradshetyn and Ryzhik, 5 and ρ = N /V is the number density. The first equality in Eq. (16b) holds for the individual states for the particle in a box, i.e., p n = 2ε n /3V. These sums converge absolutely for λ <1 (µ<0) and can be summed numerically, although the effort to compute the integrals numerically is comparable and must be used for Fermions for λ >1. Some results are shown in Fig. 3. When λ = 1(i.e., µ = 0), the sums in Eq. (16c) reduce to the Riemann zeta functions as shown in the following table: Table 2: Connections to the Riemann- Zeta Functions G z (1): Formulas and Numerical Values z Fermions: (1 2 1 z )ζ (z) Bosons: ζ (z) 3/ / T. L.Hill, An Introduction to Statistical Thermodynamics (Dover Publications), p I.S. Gradshetyn and I.M. Ryzhik, Table of Integrals, Series and Products (Academic Press. New York, 1980), Sec

7 Chemistry Ideal Bose & Fermi Gases Fig. 3. Proper activities, λ, and equations of state for ideal Boltzmann, Fermi-Dirac, and Bose-Einstein gasses. The spin degeneracy factor has not been included. Note that the figure was actually obtained by choosing a range of λ s,calculating ρλ 3,and transposing the plot. The first thing to notice is that the curves are universal, in that all ideal systems properties, once corrections for the spin degeneracy are made, will fall on the same curves. Also notice the positive (Fermions) and negative deviations (Bosons) from ideal gas behavior. For Fermions, these positive deviations arise because the Pauli exclusion principle which disallows two noninteracting particles to occupy the same state or space, and thus has an effect similar to steric repulsion). Bosons don t have to obey the exclusion principle and can have any number of particles in the same state, thereby acting like anattraction The Case For Boltzmann Statistics It turns out that there is a simple fix that allows us to treat the particles as if they were distinguishable. Consider aseparable Hamiltonian which has 1-particle energy levels, ε i = i, for i = 1, 2, 3,.... Let s examine the states for a system comprised of three identical particles, where HΨ(a, b, c) = 9Ψ(a, b, c)

8 Chemistry Ideal Bose & Fermi Gases Table 3: States where HΨ(a, b, c) = 9Ψ(a, b, c) 1-Particle States Nonzero n i s Degeneracy Ω FD Ω Distinguishable Ω BE (7, 1, 1) n 7 = 1and n 1 = (6, 2, 1) n 6 = 1, n 2 = 1, and n 1 = 1 1 3! 1 (5, 3, 1) n 5 = 1, n 3 = 1and n 1 = 1 1 3! 1 (5, 2, 2) n 5 = 1and n 2 = (4, 4, 1) n 4 = 2and n 1 = (4, 3, 2) n 4 = 1, n 3 = 1, and n 2 = 1 1 3! 1 (3, 3, 3) n 3 = 3 0 3! 1 The degeneracy for Fermions, Ω FD,vanishes whenever the Pauli exclusion principle is violated, i.e., whenever n i >1,asexpected. Since the wavefunction for Bosons can always be symmetrized, any choice of the n i s gives asingle wavefunction, and Ω BE = 1. Finally, the degeneracies for the distinguishable cases have 0 < Ω Distinguishable N!, the equality holding when all the 1-particle states are different. This discussion can be summarized by noting that Ω FD Ω Distinguishable Ω N! BE, where equality holds when all the 1-particle states are different. If the number of available states is much larger than the number of particles, then most of the states included in the partition function obey the Pauli exclusion principle, and the distinguishable calculation over-counts these by the same factor of N!. Hence, a corrected partition function becomes Q Boltzmann = Q Distinguishable. N! This approach is known as Boltzmann statistics. Note that each component of a mixture contributes a N! factor. and thus Returning to the particle in a box example, we now have Q Boltzmann = (V Λ 3 ) N β A = log(q) = N log(v Λ 3 ) + N[log(N) 1] = N[log(ρΛ 3 ) 1], where we have used Stirling s formula and ρ N/V is the number density. 6 The Gibbs paradox is resolved! The ideal gas model makes it fairly straightforward to quantify the condition necessary for Boltzmann statistics to be used, namely, that the number of accessible states, N States,greatly exceeds the number of particles, N. As we have just shown, when this holds, the resolution of Gibbs paradox is to simply divide the partition function, calculated as if the particles were distinguishable, by N!. The argument is simple and is essentially that presented in Hill. 7 To begin, note that the particle in a box quantum numbers, n i,i=x,y,z,which when plotted, form a three 6 Hill writes this as a single logarithm as N log(ρλ 3 /e). 7 T.L. Hill, op. cit., Sec N!

9 Chemistry Ideal Bose & Fermi Gases dimensional cubic lattice in the positive octant, each lattice point corresponding to a state. In terms of the lattice cells, each has unit volume and each corresponds to a single state (to be sure, there are eight lattice points per cell, but each is shared with 8 neighbors), Thus, if we ignore edge effects, the number of states with energies less than ε max is just the volume of of an eighth of a sphere with radius, n, less than the n corresponding to the ε max. Byusing Eq. (9), the cutoff radius is n = (8mε max /h 2 ) 1/2 L,and the corresponding number of states is simply N States 1 4π n 3 = 4π mε max h 2 By taking ε max = 3k B T /2 and requiring that N States >> N, wesee that ρλ 3 << 6 π in order to use Boltzmann statistics. Perhaps more physically, the number of particles per volume corresponding a size comparable to the thermal de Broglie wavelength is small, and hence, the number of particles within a thermal de Broglie wavelength, the scale at which quantum effects (e.g., wave-particle duality) become important, is small. Note that raising temperature and/or mass, and/or lowering the density favor the use of Boltzmann statistics. 1/2 3/2 V. (17) Fig. 4. The temperature dependence of the thermal de Broglie wavelength for various gases. Note that with the exception of the electron, Λ de Broglie is comparable to or smaller than the size of an atom at room temperature. Some specific results are given inthe following table: 8 8 From, D.A. McQuarrie, Statistical Mechanics, (Harper and Row Pub., Inc., New York, NY, 1973), Table 4-1, p. 72.

10 Chemistry Ideal Bose & Fermi Gases Table 4: Is Boltzmann Statistics Valid? Element (State) T(K) (π /6) 1/2 ρλ 3 He (l) He (g) He (g) He (g) Ne (l) Ne (g) Ne (g) Ar (l) Ar (g) Kr (l) Kr (g) electrons in metals (Na) Thus, we see that except for ultra-low temperatures and the lightest elements (both probably aren t of much interest to chemists), Boltzmann Statistics should be an excellent approximation. There is one major exception, namely electrons around room temperature, something that has major ramifications for bonding, etc.. 3. Results Specific to Fermions or Bosons 3.1. Fermions As we saw in the preceding section, cf. Fig. 1, the Fermi-Dirac distribution is basically flat for ε <µ F and falls off rapidly for ε >µ F,where µ F is the chemical potential, known as the Fermi energy to honor Enrico Fermi; hence, a reasonable approximation is to cut off the integrations at ε = µ F and let n i 2S + 1for ε <µ F. For example, by using Eqs. (13b) and (13c) it follows that or Similarly, E µ F N µ F 0 0 dε g(ε )ε = dε g(ε ) = µ F h2 8m 4π (2S + 1) 3 6ρ π (2S + 1) 4π (2S + 1) 5 2mµ F h 2 2/3 3/2 V (18). (19) 3/2 2m V µ 5/2 h 2 F = 3 5 N µ F. (20) As the temperature is reduced, the approximations we ve just made become more accurate; hence, the simple results just obtained become more valid and represent the ground state configuration of the Fermi (ideal) gas. Finally, the approximations leading these results don t lend themselves well to quantities like like the heat capacity. Aswesaw above, C V,arises from a small band of energy levels where ε µ F ; these make C V nonzero and its calculation more

11 Chemistry Ideal Bose & Fermi Gases complicated. By using Eq. (8) and the approximate expression for n i e β (ε i µ) it follows that C V k B k BT 8 dx g(k BTx + µ)e x2 /4 dy g(k BTy + µ)e y2 /4 (x y) 2 dx g(k BTx + µ)e x2 /4 k BTg(µ) 4 dx /4 x 2 = k e x2 B Tg(µ)π 1/2 = 2(2S + 1) 3/2 2π mk B T V h 2 where the last expression was obtained by using Eq. (12). By using Eq. (18) to eliminate the volume in favor of N it follows that C V π 1/2 3k B T = 3τ N k B 2µ F 2π, 1/2 where τ π k B T /µ F is a reduced temperature. The more rigorous calculation given inthe Appendix, cf. Eq. (A13), gives πτ/2, i.e., a 46% error. Inany event, both calculations show that that C V T as T 0, which is consistent with the Third Law ofthermodynamics. Also note that k B T /µ F << 1;e.g., if µ F = 1eV, and T = 300K then C V /k B Bosons The Bose gas thermodynamic properties can be obtained from the results given in Eqs. (13a c) or (4 7). In particular, from Eq. (13b), i.e., N = 0 dε g(ε ) e β (ε µ) 1, (21) noting that the particle in a box model has ε 0as L, itfollows that µ <0or λ <1so that the integrand in Eq. (21) be positive and integrable. However, from the data shown in Fig. 3, it follows that λ = 1for ρλ For 4 He, using the experimental density of liquid helium (0. 145g/cm 3 )this occurs at T = 3. 14K. Experiment shows that there is a transition at 2.18K. (What contributes to the discrepancy?) In order to resolve this problem, note that the exact number of particles in the ground state (ε = 0) in the grand canonical ensemble is (2S + 1) µ k B T 1/2, λ (1 λ). (22) On the other hand, the factor of g(ε ) ε 1/2 in Eq. (21) shows that the ground state doesn t contribute at all. This makes sense if the states form a continuum; the probability of finding any exact state (not a narrow band of them) vanishes, and we must introduce probability densities. On the other hand, the ground state contribution to the partition function is the most divergent! The way out of this seeming paradox is to postulate what is known as the two fluid model. We write N 2S + 1 = λ 1 λ + 2π 2m 3/2V h 2 0 dε ε 1/2 e βε 1 (23a)

12 Chemistry Ideal Bose & Fermi Gases λ = 1 λ + V ζ (3/ 2), (23b) Λ3 where the first term is the number of particles in the ground state, the so called Bose condensate, while the second accounts for those in excited states. Finally, ifweassume that µ 0, and expand λ in the denominator of the first term in Eq. (23b) we find that Λ 3 β µ V ρλ3 ζ (3/ 2), (24) where we have taken S = 0with ρλ 3 ζ (3 / 2). Thus, the number of particles in the ground state is λ N 0 1 λ = N ζ (3/ 2) 1 ρλ 3 = N 1 T 3/2 T 0, (25) where T 0 is the transition temperature (i.e., where ρλ 3 = ζ (3 / 2) = ). Notice that when T = T 0 no particles are in the ground state, but this rises as temperature is lowered until 100% of them are in the ground state at absolute zero. Having macroscopic occupation of a single state drastically changes the physical properties of the system. Have a look at the demonstrations by Alfred Leitner for an experimental survey of some of the more striking changes that arise in superfluids or at for superconductors Photons and Black Body Radiation Photons can be wave-like or particle-like. The former means that each photon has a frequency, ω, a wave-vector, k, and amplitudes for the components of the photon s electric, E, and magnetic, H, fields. The amplitudes are arbitrary, except that the radiation must be transverse, i.e., k E = 0, while the wave-vector and frequency are related by the well known dispersion relation ω = kc, (26) where c is the speed of light in vacuum. The condition of transversality means that radiation is made up of two independent polarizations, e.g., left and right circular polarization. From the selection rule for absorption and emission, we know that the photon behaves as a spin-1 particle, modulo transversality, since a molecule s angular momentum changes by l =±h on absorption or emission in all linear spectroscopies. Hence, photons are Bosons. Finally, the energy of the photon is just ε = h ω. Note that the number of photons of a given frequency isn t conserved (unlike the atoms in a chemical reaction) and, in particular, processes like M photons at frequency ω N photons at frequency ω (27) are possible. At equilibrium, the free energy is a minimum, and the usual argument requires that (N M)µ Photon = 0, or µ Photon = 0. With this introduction, consider the radiation in a closed cubical container of side length L. The radiation in the box comes to equilibrium with the walls and forms standing waves in different directions and having different frequencies. If we assume that the electric field vanishes at

13 Chemistry Ideal Bose & Fermi Gases the walls and consider the field as sin-like (just like the particle in a box) it follows that k = π /L(n x, n y, n z ) T,where n i = 1, 2, 3,... Byrepeating the steps used above, itfollows that n(ω )the density of radiation between frequencies ω and ω + dω is n(ω ) = V ω 2 π 2 c 3 e β h ω 1, (28) where remember that the photon energy states are doubly degenerate owing to the two polarization possibilities. The main differences between this and our earlier result are caused by the fact that ε = h ω = h kc. Bymultiplying this result by the photon energy, h ω,gives E(ω ) = V h ω 3 π 2 c 3 e β h ω 1, (29) where E(ω )dω is the energy found in the frequency interval ω to ω + dω and is known as Planck s formula. Limiting behaviors are easy to find; specifically, V k BTω 2 for h π E(ω ) 2 c 3 ω << k B T V h π 2 c 3 ω 3 e β h ω when h ω >> k B T. The low frequency (or high temperature) behavior is known as the Raleigh-Jeans radiation law, and was known before the discovery of quantum mechanics. One of the problems with the classical theory, according to the Raleigh-Jeans formula, is that the total radiation energy in the box diverges; this is known as the ultraviolet catastrophe, and posed a very difficult problem for classical mechanics and electrodynamics. By integrating the full result, we find that E total V = h π 2 c 3 0 dω ω 3 e β h ω 1 = (k BT ) 4 (ch ) 3 π 2 0 dx x 3 e x 1 = 6ζ (4)(k BT ) 4 (ch ) 3 π 2, (31) where the last two results were obtained by letting ω = k B T /h x. The T 4 dependence was deduced classically using thermodynamics by Stephan and Wein. Wein actually proved something more general called Wein s Law; namely, E(ω )/V = ω 3 f (ω /T ), where f is not known. Note that Wein s Law is obeyed by the Planck formula. 4. Appendix A: C V in an Ideal Bose or Fermi Gas The mathematical details leading to the constant volume heat capacity, C V,cf. Eq. (8), are now presented. First, from thermodynamics, recall that C V E = E + E β µ T N,V T β µ,v β µ β,v T N,V = E E T β µ,v β µ β,v (30) N T β µ,v, (A1) N β µ β,v

14 Chemistry Ideal Bose & Fermi Gases where the second equality follows when we consider E to be a function of T or β, β µ, and volume, V,while the last equality is obtained when we use the cyclic rule (implicit function differentiation). Next we use Eqs. (5) (7) to evaluate the derivatives that appear in Eq. (A1); i.e., E = T β µ,v N = T β µ,v 1 k B T 2 1 k B T 2 which when used in Eq. (A1) gives C V = 1 k B T 2 Σ n i 2 e β (ε i µ) ε i 2 ε i i Σ n i 2 e β (ε i µ) ε i 2, i Σ n i 2 e β (ε i µ) ε i, i Σ n j 2 e β (ε j µ) ε j j Σ n j 2 e β (ε j µ) j E = β µ Σ n i 2 e β (ε i µ) ε i, β,v i = i, j N = β µ β,v Σ n i 2 e β (ε i µ), i Σ n i 2 e β (ε i µ) n j 2 e β (ε j µ) ε i (ε i ε j ) k B T 2 Σ n j 2 e β (ε j µ) j Finally, we take half the double sum in the numerator of Eq. (A2), exchange the dummy sum indices, i j and add it to the unmodified half, resulting in Eq. (8). Note that there are alternate ways of writing Eq. (A2). For example, n i 2 e β (ε i µ) = 1 sech( β (ε i µ) / 2) 4 csch( β (ε i µ) / 2) for Fermions for Bosons. An alternate route to C V,istoeliminate µ in favor of N in the energy. see that ρ = D 0 dε ε 1/2 e β (ε µ) + 1,. (A2) (A3) From Eq. (18) we where ρ N /V and D 2π (2S + 1)(2m/h 2 ) 3/2. Byintegrating by parts, we can rewrite Eq. (A4) as ρ = 2D ε 3/2 3 e β (ε µ) + 2D + 1) 3 β dε ε 3/2 β (ε µ) e 0 [(e β (ε µ) + 1], (A5a) 2 = 2Dµ3/2 3 dx β µ 1 + k BTx µ 0 3/2 (A4) e x (e x + 1) 2, (A5b) where the boundary term in Eq. (A5a) vanishes and where we have changed variables to x β (ε µ) to get Eq. (A5b). By using the Taylor expansion we see that (1 + x) α = 1 + Σ α (α 1)...(α n + 1) x n, n! n=1 (A6)

15 Chemistry Ideal Bose & Fermi Gases where ρ 2Dµ3/2 3 Z n 1 n! dx 1 + Σ 3 n=1 2 x n e x (e x + 1) 2 = n k B T µ n Z n, 0 for n odd 2(1 2 n )ζ (n), for n even, where ζ (n) isthe Riemann zeta function. 9 Note that the odd n terms vanish because e x /(e x + 1) 2 is even in x. In obtaining Eq. (A7) we ve extended the lower integration limit to. Onone hand, this only introduces exponentially small errors, since β µ dx xn e x... is exponentially small when β µ>>1;nonetheless, while this makes sense for each term in the Taylor expansion, the sum must (and does) diverge since (1 + k B Tx/µ) 3/2 is imaginary when x < β µ. Our approach generates what is known as an asymptotic expansion. If you plot the individual terms, they will decrease until some minimum value is obtained, and increase thereafter causing the series to diverge. The trick is to sum no further. Another example of this is the Stirling formula for n!. let (A7) (A8) The goal of this calculation is to obtain a low temperature expansion for µ; accordingly, we µ µ 0 (1 + c 1 τ + c 2 τ ), (A9) where µ 0 h 2 (6ρ/π (2S + 1)) 2/3 /8m is the chemical potential at absolute zero (cf. Eq. (A19)) and where τ π k B T /µ 0 is a reduced temperature. When Eq. (A9) is substituted into Eq. (A7),the result expanded into a series in τ,and the coefficients of τ n, n >0,set to zero, it follows that 10 µ = µ τ τ τ τ τ (A10) By starting with the general expression for the energy, i.e., E = DV dε ε 3/2 0 e β (ε µ) + 1, and repeating the steps that led from Eq. (A4) to (A7), it follows that E = 2DV µ5/ Σ 5 n=1 2 Finally, byusing Eq. (A10) to eliminate µ it follows that n k B T µ n Z n. (A11) E = 3 5 N µ τ τ τ τ τ , (A12) 9 See I.S. Gradshetyn and I.M. Ryzhik, op. cit., Eq. ( ), p The algebra becomes horrendous if you want to go much beyond the first correction; nonetheless, it s easy for a symbolic algebra program, e.g., Mathematica or Maxima ( to do so. Maxima was used here.

16 Chemistry Ideal Bose & Fermi Gases which implies that C V = πτ N k B τ τ τ τ (A13) Note that C V T as T 0, which is consistent with the 3rd Law ofthermodynamics. The first correction is O(T 3 ), just like the Debye-T 3 law for vibrations. That being the case, how can we distinguish between vibrations and electronic contributions? Fig. 5. The low temperature, constant volume, heat capacity for a system of ideal Fermions in a box as a function of the reduced temperature, τ, introduced in the text.. The different curves correspond to the number of terms kept in Eq. (A13). As expected, all the curves reduce to the linear one as τ 0. This doesn t happen at higher temperatures, and moreover, it is obvious that something is breaking down since C V >0.