Quantum statistics: properties of the Fermi-Dirac distribution.

Statistical Mechanics Phys54 Fall 26 Lecture #11 Anthony J. Leggett Department of Physics, UIUC Quantum statistics: properties of the Fermi-Dirac distribution. In the last lecture we discussed the properties of the wave function Ψ(ξ 1, ξ 2,... ξ N ) where ξ i is a shorthand for all the variables of the i-th particle, and showed that it must be symmetric (antisymmetric) for bosons (fermions) under the exchange ξ i ξ j ; this result is of course completely independent of the choice of basis for Ψ. It is now convenient to choose a particular basis, namely that in which the basis vectors (wave functions) for the N- body system are each just a product of a set of functions ϕ i (ξ 1 )ϕ j (ξ 2 )... ϕ s (ξ N ), where the single-particle functions ϕ i (ξ), i = 1, 2... form a complete orthonormal set 1 ; note that for the moment we do not assume that they are necessarily eigenfunctions of any particular operator (e.g. the KE). Of course, the product functions are not eigenstates of the N-body Hamiltonian in the general case, but they nevertheless form a convenient basis for further work (in particular, they can be used to set up the second-quantization formalism, which however we will not need explicitly in this course). In this basis it is clear that if the many-body wave function has a component containing ϕ i (ξ 1 )ϕ j (ξ 2 ), then it must also contains a term ϕ j (ξ 1 )ϕ i (ξ 2 ), with a + or sign depending on whether the particles in question are bosons or fermions; in the Fermi case this observation implies the Pauli principle (cf. last lecture). What about the normalization of the manybody wave function? Consider for definiteness the case N = 2, and suppose first that i j. Then the two-body basis wave function is Ψ(ξ 1, ξ 2 ) = N {ϕ i (ξ 1 )ϕ j (ξ 2 ) ± ϕ j (ξ 1 )ϕ i (ξ 2 )} (1) where N is the normalization constant, which we wish to determine. Now by construction the ϕ i (ξ) form an orthonormal set, so the diagonal terms ( ϕ i (ξ 1 ) 2 ϕ j (ξ 2 ) 2 etc.) yield 1 when integrated, while the cross terms (ϕ i (ξ 1 )ϕ j(ξ 2 )ϕ j (ξ 1 )ϕ i (ξ 2 ) etc.) yield zero. Thus for i j the squared norm of quantity in { } is 2, hence the normalization constant N = 2 1/2. On the other hand for i = j (only possible for bosons) there is only a single term ϕ i (ξ 1 )ϕ i (ξ 2 ), 1 In technical mathematical language, the N-particle Hilbert space is regarded as the tensor product of N single-particle Hilbert spaces. 1

and the normalization constant is simply unity. Generalizing the argument to arbitrary N, we can easily convince ourselves that for fermions (where the i, j... must all be different), the normalization factor is always simply (N!) 1/2, while for a state of N bosons with n i particles in state i it is ( i N = n ) 1/2 i! (2) N! Now we shall verify below that at sufficiently high temperature the probability of finding any n i larger than 1 is negligible, so we have N = (N!) 1/2 independently of the statistics. This factor would not occur for distinguishable particles (in that case we would have to count the normalized states ϕ i (ξ 1 )ϕ j (ξ 2 ) and ϕ j (ξ 1 )ϕ i (ξ 2 ) as physically distinct); it is exactly the factor that Gibbs needed to resolve his paradox (though of course he could not have guessed its origin). More generally, the bottom line of the above discussion is this: independently of whether we are talking about bosons or fermions, a given (allowed) 2 distribution {n i } of the N particles among possible other normal states i corresponds to one and only one many-body state. It is this observation which is crucial for the argument which now follows. In the rest of this lecture I specialize to the case of identical noninteracting particles. In this case it is natural to guess (and we shall verify) that the many-body density matrix will be diagonal in the product basis provided we choose our single-particle functions ϕ i (ξ) to be eigenfunctions of the single-particle Hamiltonian, and the physically interesting question is: what is the distribution of particles over the states i? It is the answer to this question which is conventionally described as the (FD or BE) statistics of the particles in question. Since we eventually wish to describe not only systems which are approximately translation-invariant (such as gaseous 4 He) but also for example the magnetically trapped atomic alkali gases and the electrons in amorphous (noncrystalline) metals, it is convenient to allow our particles to move in an arbitrary bounded external one-particle potential which in general is a function of both the coordinate r and the spin projection 3 σ. However complicated this potential, we can always find a complete set of orthonormal eigenfunctions 2 Of course for fermions n i > 1 is forbidden. 3 In principle it could be a matrix with respect to σ and even depend on velocity; this in no way affects the argument. 2

ϕ i (r, σ) of the single-particle Hamiltonian and the corresponding eigenvalues ɛ i ; it will be convenient to choose the zero of energy to coincide with the minimum value of ɛ i. Note that the eigenfunctions ϕ i (r, σ) need not be simple products of a coordinate-space and a spin-space function (this is true only if the potential is a sum of terms which depend only on coordinates and only on spin respectively). For the moment I will assume that the only conservation law which is relevant to the thermodynamics is that of the total particle number N: see below. It follows from the basic principles of quantum statistical mechanics that for a set of N particles under the above conditions the density matrix of the many-body system is diagonal in the energy representation and given by the standard macrocanonical Gibbs distribution; that is, the probability of occurrence of a many-body state E α is p α = Z 1 exp βe α, Z α exp βe α (3) where β 1/k B T as usual. Since the total energy E α is given for a noninteracting system by E α = i n i ɛ i (4) and the total particle number N is constrained to satisfy the condition N = i n i (5) it follows that we can rewrite (3) in the form p{n i } = Z 1 exp β i n i ɛ i, Z = {n i } exp β i n i ɛ i (6) provided that the values of n i are constrained by the condition (5). This constraint prevents the n i from being regarded as completely independent, and is very inconvenient. As is well known, the standard solution is to relax the constraint that the total particle number N is fixed, replacing it by the condition that the system is held at a constant chemical potential µ which is then obtained from the condition that the calculated quantity i n i is equal to the actual number N of particles in the system. The effect is to replace (3) by the grand canonical distribution p α = Z 1 G exp (E α µn α ), Z G α exp β(e α µn α ) (7) 3

where the quantity N α is the value of i n i in the state α. One can now write p α in the form p α {n i } = Z 1 G exp β i n i (ɛ i µ), Z G {n i } exp β i n i (ɛ i µ) (8) where the n i may now be taken as independent variables. This means that Z G factorizes, Z G = i Z i, Z i n i exp βn i (ɛ i µ) (9) and we can rewrite p{n i }, which no longer depend explicitly on α, as a product: p{n i } = i p i (n i ), p i (n i ) = exp β(ɛ i µ)n i / i exp β(ɛ i µ) (1) where the allowed values of n i are and 1 for fermions, and,1,2...for bosons. It is worth noting explicitly that the fact that the grand canonical distribution factorizes in this simple way is a consequence of the fact, noted above, that in each case there is one and only one many-particle state which corresponds to any given set of values {n i }, and that in the Bose case this is itself a consequence of the requirement that the wave particle be totally symmetric under the exchange of any pair of particles. From (1) we immediately obtain the standard Fermi-Dirac (or Fermi) and Bose-Einstein (Bose) formula for the average number n i (T ) of fermions and bosons respectively in the single particle state i at temperature T : n i (T ) = (exp β(ɛ i µ) + 1) 1 fermions (11) n i (T ) = (exp β(ɛ i µ) 1) 1 bosons (12) where in each case the chemical potential µ(t : N) is implicitly defined by the condition n i (µ, T ) = N (13) i Note that in the Bose case the formula (12) makes sense only if the chemical potential µ is negative: we return in lecture 13 to the question of what happens if and when this condition becomes inconsistent with the constraint (13) 4. In the remainder of this lecture and the next two I shall explore the properties, and some applications, of the distributions (11) and (12). 4 In the case of photons the total number N is not conserved and we must therefore set µ = from the start: this gives of course the standard Planck distribution, which is then trivially consistent with (12). 4

Let s start by examining the conditions under which these distributions reduce to the classical (generalized Maxwell) distribution n i (T ) = const exp βɛ i (14) It is clear that this will be a good approximation whenever, for all of the ɛ i, the quantity exp β(ɛ i µ) is large compared to unity, and this will certainly be true 5, independently of the statistics, if (and only if) e βµ (i.e. µ ). Under these conditions both (11) and (12) reduce to (14) with the constant equal to e βµ, and eqn. (13) becomes N = e βµ i e βɛ i e βµ Z 1 (15) where Z 1 is just the single-particle classical PF for the system in question, and can be evaluated for any specific case of interest. moving freely in volume V in l.7 6 : Inserting this expression into (15), we get In particular, we calculated it for a particle Z 1 = (V/λ 3 T )g (16) e βµ = Nλ 3 T /(V g) nλ 3 T /g (17) Thus the condition for the classical statistics to be a good approximation, e βµ 1, reduces simply to nλ 3 T /g 1. More generally, since Z 1 is of the order of the number of states with energies k B T, an order-of-magnitude criterion for classical statistics for classical statistics to be applicable is that this number should be large compared to the total number of particles N. This is rather natural, since under these conditions the probability (using classical statistics) of more than one particle occupying a given state i is negligible, and the only effect of their indistinguishability is the Gibbs factor 1/N!, which as we have seen does not affect most thermodynamic quantities. Let s briefly consider the first corrections to the classical ideal-gas formulae coming from quantum statistics. If is probably easier to work from the expression for the grand potential Ω β 1 ln Z G, which according to eqn. (9) is given by Ω = β 1 i ln Z i (18) 5 Since we choose the zero of energy to coincide with the lowest ɛ i. 6 I include a factor g for the number of available internal states. 5

Since the allowed values of n i are all positive integers for bosons and and 1 only for fermions, eqn. (9) gives quite generally [ Ω = β 1 ± ln ( 1 ± e )] β(ɛ i µ) (19) i where the + sign refers to fermions and the sign to bosons. Expanding the logarithm up to second order in the quantity e βµ, we have ( ) ] [ Ω = β 1 e β(ɛi µ) + 1 ± 2 i i e 2β(ɛ i µ) +... (2) This formula is general. For a gas of particles moving in volume V, we recall that Ω = P V and i e βɛ i = gv/λ 3 T, where g is the number of internal states available (e.g. 1 for 4 He, 2 for spin-1/2 fermions), so (since β 2β in the second term) P V k B T = eβµ (gv/λ 3 T ) + 1 ( ±e 2βµ 2 3/2 (gv/λ 3 T ) +... ) (21) 2 But the first term on the RHS is just the classical free-gas result N and in evaluating the second term we can consistently make the approximation (cf. eqn. (17)) e βµ nλ 3 T /g, so finally P V = Nk B T ( 1 + (±) 1 ) 2 2 3/2 (nλ 3 T /g) +... The contribution of quantum statistics to the second virial coefficient B(T ) is therefore ± 1 2g 2 3/2 λ 3 T where the + sign refers to fermions and the sign to bosons7. Thus the pressure of a free Fermi gas at a given temperature is somewhat greater then of its classical counterpart, that of a free Bose gas somewhat less. For the rest of this lecture I specialize to the Fermi case, where the equilibrium distribution of the particles over states i is given by formula (11), which I write out again for convenience: (22) n i (T ) = (exp β(ɛ i µ) + 1) 1 (23) Let s first observe that the distribution (23) satisfies the principle of detailed balance, which we can define as follows: suppose that the particles in question are not totally noninteracting, but very occasionally undergo collisions, as a result of which two particles which start in 7 This agrees with LL eqn. (55.15). 6

states 1 and 2 end in states 3 and 4. Such (real) collision processes must evidently satisfy the condition of energy conservation ε 1 + ε 2 = ε 3 + ε 4 (24) The principle of detailed balance then states that in thermal equilibrium, for each possible choice of 1, 2, 3 and 4, the average number of collisions in which 1,2 3,4 is equal to that of the reverse process 3,4 1,2. For a system with a Hamiltonian (including the interaction terms) which satisfies time reversal invariance, the principle is a special case of a more general QM theorem; for present purposes let us take it as given. Now the rate of processes in which 1,2 3,4 is clearly proportional to the probability of both initial states being occupied, i.e. to n 1 n 2. However, because of the Pauli principle it is also proportional to the probability of both final states being unoccupied, i.e. to (1 n 3 )(1 n 4 ). Apart from these factors, there are of course matrix elements etc. which enter the rate, but (for a time-reversal invariant Hamiltonian) these cancel between the forward and reverse processes. Hence satisfaction of the principle of detailed balance requires that whenever eqn. (24) is satisfied then n 1 n 2 (1 n 3 )(1 n 4 ) = n 3 n 4 (1 n 1 )(1 n 2 ) (25) Rewriting this relation in the form n 1 1 n 1 n 2 1 n 2 = n 3 1 n 3 n 4 1 n 4 (26) we see that the distribution (23) indeed satisfies it whenever (24) holds. In fact, one can actually run the argument backward and derive (23) from the principle of detailed balance. At zero temperature the RHS of Eqn. (11) reduces to the Heaviside step function θ(µ ɛ i ), so the chemical potential at zero temperature (the Fermi energy ɛ F ) is simply given by the condition i(ɛ i <µ) = N (27) or equivalently, introducing the single-particle density of states 8 ρ(ɛ) by ρ(ɛ) i δ(ɛ ɛ i ) (28) 8 I have assumed as usual that N is very large compared to unity, so that the relevant levels may be taken as continuously distributed. 7

by the formula µ ρ(ɛ)dɛ = N (29) The numerical value of µ(t = ) depends on the density of states ρ(ɛ), which in turn depends on the nature of the confining potential (if any) and the number of available values of the spin projection, g (cf. below). I quote two useful special cases: For a set of particles of mass m moving freely in a volume Ω we have ρ(ɛ) = Ω g 4π 2 (2m) 3 3 2 ɛ 1 2. and hence µ(t = )( ɛ F ) k B T F = ( 2 /2m)(6π 2 n/g) 2/3 7.6g 2 2 2 3 n 3 m (n N/Ω) while in the case of fermions confined in a three-dimensional isotropic harmonic trap with frequency ω we have ρ(ɛ) = gɛ 2 /2( ω ) 3 and hence (3) µ(t = ) = (6N/g) 1/3 ω (31) It should be emphasized that all the above formulae are based on the assumption that the only conservation law relevant to the thermodynamics is that of total particle number; in particular, in the case of more than one spin (hyperfine) species it is assumed that interspecies conversion can take place on a time scale fast compared to that of the relevant experiments. This is overwhelmingly so for (e.g.) electrons in metals, but need not be so for the alkali gases, where inter-species conversion is a very slow process. In such a case it may be necessary to treat the different spin species as effectively separate physical systems (just as one would different chemical species) and to assign to each species α its own chemical potential µ α, which is then fixed by eqn. (31) with N N α and g 1. The mean energy per particle of a noninteracting Fermi gas at T = is evidently given by the expression ɛ = µ ɛρ(ɛ) dɛ / µ ρ(ɛ) dɛ (32) and hence from the relevant form of the DOS ρ(ɛ) is 3/5 ɛ F for a free Fermi gas in 3D and 3/4 ɛ F for a Fermi gas in a 3D harmonic trap. Moreover, it is easy to obtain an expression for the compressibility K dn/dµ of the free gas at T = : since the chemical potential will rise just enough to accommodate the dn extra particles in the extra states made available between µ and µ + dµ and the number of these is ρ(ɛ F ) dµ, we have simply K = ρ(ɛ F ) (33) 8

A similar calculation can be done for the spin susceptibility, in the case that the single-particle energy eigenstates can be chosen to be eigenstates of spin. For example, in the standard case of spin 1/2 the effect of an external magnetic field H giving rise to Zeeman energy µ σb is to shift the energies of the up-spin (σ z = +1) spin states down by µb and these of the down-spin (σ z = 1) states up by the same amount; however, in equilibrium the Fermi energies of and must be equal, and to account this a number 1ρ(ɛ 2 F ) µh of spins must migrate to (remember ρ(ɛ F ) is the total DOS, of both spins!), so the induced magnetization is µ ρ(ɛ F )µh. Hence we have for the susceptibility χ χ = µ 2 ρ(ɛ F ) (34) Note that the result (33) and (34) are independent of the overall behavior of the DOS ρ(ɛ) (and thus of the dimensionality, etc., of the system); they depend only on the DOS at the Fermi energy, ρ(ɛ F ). We will see later that this is a special case of a more general state of affairs. To discuss the low-temperature specific heat we need to generalize our consideration to nonzero T ; we now do this. At nonzero temperature the equation which implicitly determines the chemical potential µ(t ) is ρ(ɛ) (exp β(ɛ µ) + 1) 1 dɛ = N (35) and is not generally analytically solvable. An exception is the special case of a free gas in 2D, for which ρ(ɛ) is a constant ρ (= gv (m/π 2 )). In this case the LHS is just ρ β 1 ln ( 1 + e βµ), and since µ() = Nρ 1, this gives µ(t ) = k B T ln ( e µ()/k BT 1 ) (36) We see that the deviation of µ(t ) for k B T µ() from its zero-temperature value µ() is of order exp µ()/k B T, that is, exponentially small. As we will see, this behavior is peculiar to a flat DOS; for most cases of physical interest, where ρ(ɛ) is not a constant, the deviation is of power-law type. Let us study the general case in the high- and low-temperature limits. As regards the high-temperature case, we have seen above that classical (Maxwell-Boltzmann) statistics is recovered when the number of single-particle states of energy the thermal energy k B T is 9

large compared to N. But since ɛ F is determined by the statement that the number of states with energy less then ɛ F is N, it follows that in the absence of pathological behavior of ρ(ɛ) the condition for classical statistics to apply is, to an order of magnitude, T ɛ F /k B T F (the Fermi temperature). The details of the crossover to classical behavior will depend on the specific form of ρ(ɛ). At low temperatures (k B T ɛ F ) a little more care is needed. We first need to determine the temperature-dependence of the chemical potential µ(t ). First, a qualitative argument: if we could assume µ(t ) stayed at its zero-temperature value µ() ɛ F, then the effect of noninifinite β in eqn. (11) is to smear the T = step discontinuity at ɛ = ɛ F over an energy range k B T. But if this is all we do, and if (for example) single-particle DOS ρ(ɛ) is an increasing function of energy ɛ, then we will add more particles above ɛ F then we subtract below; so to compensate for this effect µ must decrease somewhat below its T = value. (Correspondingly, if ρ(ɛ) is a decreasing function of ɛ as in the case of a 1D free gas, µ must increase). We may estimate the order of magnitude of the change in µ as follows: at constant µ the imbalance created is of order (k B T ) 2 ρ (ɛ F ) where ρ (ɛ F ) dρ(ɛ)/dɛ ɛ=ɛf. On the other hand, a shift of µ by µ changes the T = value by ρ(ɛ F ) µ. Since these two effects must add to zero µ (k B T ) 2 ρ (ɛ F )/ρ(ɛ F ) (37) A more quantitative argument goes as follows: since N is unchanged dµ/dβ 1 = ( I(β, µ) / β 1 )/( I(β, µ)/ µ) (38) where I(β, µ) is the integral on LHS of (35). After performing the differentiation we may take the limit T. Now we have (cf. the argument for the compressibility K) that at T = I/ µ is simply ρ(ɛ F ). As for I/ β 1, this is explicitly the integral I = β2 β 1 dɛ ρ(ɛ)(ɛ µ) 1 4 sech2 β(ɛ µ)/2 (39) Because of the sech 2 factor, the integral is appreciable only for ɛ µ k B T, and in this region we can expand ρ(ɛ) in a Taylor series around ɛ F and take the integral from to. The zero-order term then vanishes from symmetry, and the first-order one is ( β 2 /4 ) ρ (ɛ F ) x 2 sech 2 βx/2 dx = (π 2 /3)ρ (ɛ F )k B T 1 (4)

Inserting this and the expression for I/ µ in eqn.(38) and integrating with respect to T, we finally get µ(t ) = µ() (π 2 /6)(k B T ) 2 ρ (ɛ F )/ρ(ɛ F ) + O(T 4 ) (41) Note that the correction depends only on quantities defined at ɛ F and does not depend on the behavior of DOS ρ(ɛ) at points distant from ɛ F. Also note that the relative shift of µ(t ) is of order (k B T/ɛ F ) 2. This result is actually a special case of a more general state of affairs: generally speaking, to calculate a physical property of a Fermi gas at low temperatures (T T F ) it is sufficient, when evaluating the relevant integrals, to set µ(t ) = µ() ɛ F and ρ(ɛ) = ρ(ɛ F ) = const. Any calculations will generally be of relative order (k B T/ɛ F ) 2 (T/T F ) 2. (The only exception is when the resulting integrals over ɛ vanish by symmetry; in this case it is in general necessary to keep the term in ρ (ɛ) in the Taylor expansion 9.) Thus, for example, the expressions (33) and (34) for the compressibility K and the spin susceptibility χ remain valid at nonzero T to within terms of relative order (T/T F ) 2. In the case of χ, in particular, it is easy to see why: to this order the number of particles which have to migrate from states to is exactly the same as at T =. A quantity which is meaningful only at nonzero T is the specific heat C V (T ). The easiest way to calculate it is probably by writing down the expression for the mean energy E(T ): E(T ) = ρ(ɛ)(ɛ µ(t ))(exp β(ɛ µ) + 1) 1 dɛ + µ(t )N (42) where we used the fact that the integral without the ɛ is N (eqn. (35)) to add and subtract a term µn. Differentiating (42) with respect 1 to T : C V (T ) de dt = β V 4 ρ(ɛ)(ɛ µ) 2 sech 2 β(ɛ µ)/2 dɛ (43) As in the above argument concerning µ, in evaluating this integral we set µ = µ(), ρ(ɛ) = ρ(ɛ F ) and thereby obtain (cf. eqn. (4)) the well-known result C V (T ) = π2 3 k2 BT ρ(ɛ F ) (+O(T 3 )) (44) 9 An example is thermoelectic power (see second half of course). 1 It is of course not necessary to differentiate the µ(t ), since the result cancels in the two terms on the RHS. 11

The low-temperature specific heat of a free Fermi gas is therefore of order T/T F times its classical value 3Nk 2 B; this is intuitively reasonable since only a fraction of the particles of this order are thermally excited. Note: in comparing formulae (37)-(44) with those quoted in textbooks, you should beware of the fact that many of the latter assume a free-gas spectrum and thus ρ(ɛ) ɛ 1/2, and express results in terms of the ration k B T/ɛ F. 12