We now turn to examine the heat capacity of electrons in a metal. Analogous to Eq. 4.3, the total energy of our system of electrons is given now by
Etotal= 2V (2π)3 Z dk (k) nF(β((k) − µ)) = 2V (2π)3 Z ∞ 0 4πk2dk (k) nF(β((k) − µ))
where the chemical potential is defined as above by N = 2V (2π)3 Z dk nF(β((k) − µ)) = 2V (2π)3 Z ∞ 0 4πk2dk nF(β((k) − µ))
(Here we have changed to spherical coordinates to obtain a one dimensional integral and a factor of 4πk2 out front).
It is convenient to replace k in this equation by the energy E by using Eq. 4.2 or equivalently k = r 2m ~2 we then have dk = r m 2~2 d
We can then rewrite these expressions as Etotal = V Z ∞ 0 d g() nF(β( − µ)) (4.8) N = V Z ∞ 0 d g() nF(β( − µ)) (4.9)
4.2. ELECTRONIC HEAT CAPACITY 31 where g()d = 2 (2π)34πk 2dk = 2 (2π)34π 2m ~2 r m 2~2 d = (2m)3/2 2π2~3 1/2d (4.10)
is the density of states per unit volume. The definition7 of this quantity is such that g()d is the
total number of eigenstates (including both spin states) with energies between and + d. From Eq. 4.7 we can simply derive (2m)3/2/~3 = 3π2n/E3/2
F , thus we can simplify the
density of states expression to
g() = 3n 2EF EF 1/2 (4.11) which is a fair bit simpler. Note that the density of states has dimensions of a density (an inverse volume) divided by an energy. It is clear that this is the dimensions it must have given Eq. 4.9 for example.
Note that the expression Eq. 4.9 should be thought of as defining the chemical potential given the number of electrons in the system and the temperature. Once the chemical potential is fixed, then Eq. 4.8 gives us the total kinetic energy of the system. Differentiating that quantity would give us the heat capacity. Unfortunately there is no way to do this analytically in all generality. However, we can use to our advantage that T TF for any reasonable temperature, so that the
Fermi factors nF are close to a step function. Such an expansion was first used by Sommerfeld,
but it is algebraically rather complicated8 (See Ashcroft and Mermin chapter 2 to see how it is
done in detail). However, it is not hard to make an estimate of what such a calculation must give — which we shall now do.
When T = 0 the Fermi function is a step function and the chemical potential is (by definition) the Fermi energy. For small T , the step function is smeared out as we see in Fig. 4.1. Note, however, that in this smearing the number of states that are removed from below the chemical potential is almost exactly the same as the number of states that are added above the chemical potential9.
Thus, for small T , one does not have to move the chemical potential much from the Fermi energy in order to keep the number of particles fixed in Eq. 4.9. We conclude that µ ≈ EF for any low
temperature. (In fact, in more detail we find that µ(T ) = EF + O(T/TF)2, see Ashcroft and
Mermin chapter 2).
Thus we can focus on Eq. 4.8 with the assumption that µ = EF. At T = 0 let us call the
kinetic energy10 of the system E(T = 0). At finite temperature, instead of a step function in Eq.
4.8 the step is smeared out as in Fig. 4.1. We see in the figure that only electrons within an energy range of roughly kBT of the Fermi surface can be excited — in general they are excited above the
Fermi surface by an energy of about kBT . Thus we can approximately write
E(T ) = E(T = 0) + (γ/2)[V g(EF)(kBT )](kBT ) + . . .
Here V g(EF) is the density of states near the Fermi surface (Recall g is the density of states
per unit volume), so the number of particles close enough to the Fermi surface to be excited is V g(EF)(kBT ), and the final factor of (kBT ) is roughly the amount of energy that each one gets
7Compare the physical meaning of this definition to that of the density of states for sound waves given in Eq. 2.3
above.
8Such a calculation requires, among other things, the evaluation of some very nasty integrals which turn out to
be related to the Riemann Zeta function (see section 2.4 above).
9Since the Fermi function has a precise symmetry around µ given by n
F(β(E − µ)) = 1 − nF(β(µ − E)), this
equivalence of states removed from below the chemical potential and states inserted above would be an exact statement if the density of states in Eq. 4.9 were independent of energy.
10In fact E(T = 0) = (3/5)N E
32 CHAPTER 4. SOMMERFELD THEORY excited by. Here γ is some constant which we cannot get right by such an approximate argument (but it can be derived more carefully, and it turns out that γ = π2/3, see Ashcroft and Mermin).
We can then derive the heat capacity
C = ∂E/∂T = γkBg(EF)kBT V
which then using Eq. 4.11 we can rewrite as C = γ 3N kB 2 T TF
The first term in brackets is just the classical result for the heat capacity of a gas, but the final factor T /TF is tiny (0.01 or smaller!). This is the above promised linear T term in the specific
heat of electrons, which is far smaller than one would get for a classical gas.
This Sommerfeld prediction for the electronic (linear T ) contribution to the heat capacity of a metal is typically not far from being correct (The coefficient may be incorrect by factors of “order one”). A few metals, however, have specific heats that deviate from this prediction by as much as a factor of 10. Note that there are other measurements that indicate that these errors are associated with the electron mass being somehow changed in the metal. We will discover the reason for these deviations later when we study band theory (mainly in chapter 16).
Realizing now that the specific heat of the electron gas is reduced from that of the classical gas by a factor of T /TF . 0.01, we can return to the re-examine some of the above Drude
calculations of thermal transport. We had above found (See Eq. 3.3-3.5) that Drude theory predicts a thermopower S = Π/T = −cv/(3e) that is too large by a factor of 100. Now it is clear that
the reason for this error was that we used in this calculation (See Eq. 3.4) the specific heat per electron for a classical gas, which is too large by roughly TF/T ≈ 100. If we repeat the calculation
using the proper specific heat, we will now get a prediction for thermopower which is reasonably close to what is actually measured in experiment for most metals.
We also used the specific heat per particle in the Drude calculation of the thermal conduc- tivity κ = 1
3ncvhvi
2λ. In this case, the c
v that Drude used was too large by a factor of TF/T ,
but on the other hand the value of hvi2 that he used was too small by roughly the same factor
(Classically, one uses mv2/2 = k
BT whereas for the Sommerfeld model, one should use the Fermi
velocity mv2
F/2 = kBTF). Thus Drude’s prediction for thermal conductivity came out roughly
correct (and thus the Wiedemann-Franz law correctly holds).