Advanced Solid State Physics

7.1.3 Sommerfeld Expansion

When you calculate the thermodynamic quantities the normal way you get stuck analytically at the
density of statesD(E)of the dispersion relationship. From this point on you have to work numerically.
The Sommerfeld-Expansion allows you to calculate analytically the leading order term(s) as a function
of temperature, which are enough in a lot of cases. The number of electronsnand the internal energy
uare:

n=

∫∞

−∞

D(E)f(E)dE u=

∫∞

−∞

E·D(E)f(E)dE

withf(E)as the fermi function andEas the energy. We would like to perform integrals of the form:
∫∞

−∞

H(E)f(E)dE

H(E)stands for an arbitrary function which is multiplied by the fermi function. Such an integral
appears if you want to calculate the total number of states or for instanceμ, the chemical potential,
which is inside the fermi-function. We integrate by parts:
∫∞

−∞

H(E)f(E)dE=K(∞)f(∞)−K(−∞)f(−∞)−

∫∞

−∞

K(E)

df(E)

dE

dE

with

K(E) =

∫E

−∞

H(E′)dE′ i.e. H(E) =

dK(E)

dE

(47)

The fermi functionf(E)atT= 0K looks like a step function at the chemical potential. The last term
with the integral includes the derivation of the fermi function, which is a peak around the chemical
potential. The fermi function vanishes forE→∞,f(−∞) = 1andH(−∞) = 0. With eqn. (47):

∫∞

−∞

H(E)f(E)dE=−

∫∞

−∞

K(E)

df(E)

dE

dE

Then K(E) is expanded aroundE=μwith a Taylor series (everywhere else the derivation is zero):

K(E)≈K(μ) +

dK

dEE=μ

(E−μ) +

1

2

d^2 K

dE^2 E=μ

(E−μ)^2 +...

Solving the integral this way is similar that you say that just the states near the fermi surface con-
tribute to the physical properties of the electrons. Only states near the fermi sphere contribute to the
measurable quantities.

With

x=

E−μ

kBT

∫∞

−∞

H(E)f(E)dE=K(μ)

∫∞

−∞

ex

(1 +ex)^2

dx+kBT

dK

dEE=μ

∫∞

−∞

x·ex

(1 +ex)^2

+...