PHY.F20 Molecular and Solid State Physics

Sommerfeld expansion

Arnold Sommerfeld described a way to perform integrals of the form,

\[ \begin{equation} \int \limits_{-\infty}^{\infty} H(E)f(E)~dE, \end{equation} \]

where $H(E)$ is any function that goes to zero for large negative energies, $H(-\infty)=0$, and $f(E)$ is the Fermi function,

\[ \begin{equation} f(E) = \frac{1}{ \exp\left(\frac{E-\mu}{k_B T}\right)+1}. \end{equation} \]

Integrating once by parts,

\[ \begin{equation} \int \limits_{-\infty}^{\infty} H(E)f(E)~dE=K(\infty)f(\infty)-K(-\infty)f(-\infty)-\int \limits_{-\infty}^{\infty} K(E)\frac{df}{dE}dE, \end{equation} \]

where $K(E)$ is,

\[ \begin{equation} K(E) = \int \limits_{-\infty}^{E} H(E')~dE'. \end{equation} \]

The boundary terms vanish because $K(-\infty) = 0$ and $f(\infty) = 0$. The integral can then be written as,

\[ \begin{equation} \int \limits_{-\infty}^{\infty} H(E)f(E)~dE=-\int \limits_{-\infty}^{\infty} K(E)\frac{df}{dE}dE, \end{equation} \]

This is convenient because the derivative of the Fermi function is only nonzero in a region a few $k_BT$ wide around the chemical potential $\mu$, \[ \begin{equation} -\frac{df(E)}{dE} = \frac{\exp\left(\frac{E-\mu}{k_B T}\right)}{k_B T \left( \exp\left(\frac{E-\mu}{k_B T}\right)+1 \right)^2}. \end{equation} \]

The original integral can then be approximated by an integral over a small energy range and this can be evaluated numerically.

\[ \begin{equation} \int \limits_{-\infty}^{\infty} H(E)f(E)~dE \approx -\int \limits_{-10\mu/k_BT}^{10\mu/k_BT} K(E)\frac{df}{dE}dE. \end{equation} \]

This relationship was used to write programs to numerically calculate the temperature dependence of the electronic contribution to the thermodynamic properties of metals such as the chemical potential, the internal energy, and the specific heat.

There were no computers when Sommerfeld was working on this problem so he expressed $K(E)$ as a Taylor expansion,

\[ \begin{equation} K(E) = K(\mu)+\frac{dK}{dE}\biggr\rvert_{E=\mu}(E-\mu)+\frac{1}{2}\frac{d^2K}{dE^2}\biggr\rvert_{E=\mu}(E-\mu)^2+\cdots = K(\mu)+H(\mu)(E-\mu)+\frac{1}{2}\frac{dH}{dE}\biggr\rvert_{E=\mu}(E-\mu)^2+\cdots, \end{equation} \]

and then integrated term by term. Using the definite integrals,

\[ \begin{equation} \int \limits_{-\infty}^{\infty} \frac{e^x}{(1+e^x)^2}=1 \hspace{1.5cm} \int \limits_{-\infty}^{\infty} \frac{xe^x}{(1+e^x)^2}=0 \end{equation} \] \[ \begin{equation} \int \limits_{-\infty}^{\infty} \frac{x^2e^x}{(1+e^x)^2}=\frac{\pi^2}{3} \hspace{1.5cm} \int \limits_{-\infty}^{\infty} \frac{x^3e^x}{(1+e^x)^2}=0 \end{equation} \] \[ \begin{equation} \int \limits_{-\infty}^{\infty} \frac{x^4e^x}{(1+e^x)^2}=\frac{7\pi^4}{15} \hspace{1.5cm} \int \limits_{-\infty}^{\infty} \frac{x^5e^x}{(1+e^x)^2}=0 \end{equation} \]

The Sommerfeld expansion is,

\[ \begin{equation} \int \limits_{-\infty}^{\infty} H(E)f(E)~dE = K(\mu)+\frac{\pi^2}{6}(k_BT)^2\frac{dH}{dE}\biggr\rvert_{E=\mu}+\frac{7\pi^4}{360}(k_BT)^4\frac{d^3H}{dE^3}\biggr\rvert_{E=\mu}+\cdots. \end{equation} \]

The first two terms of the Sommerfeld expansion can be used to approximate the temperature dependence the thermodynamic properties of the free electron model. This model has two parameters which can be taken to be the electron density $n$ and the effective mass $m$ or the two parameters can be the electron density of states at the Fermi energy $D(E_F)$, and the derivative of the electron density of states at the Fermi energy $\frac{dD(E_F)}{dE}=D'(E_F)$. The zero of energy can be chosen so it is possible to define $E_F=0$.

$n$ and $m$

$D(E_F)$ and $\frac{dD(E_F)}{dE}=D'(E_F)$

Chemical potential μ
Implicitly defined by
\[ \begin{equation} n =\int \limits_{-\infty}^{\infty} D(E)f(E)dE \end{equation} \]
\[ \begin{equation} \mu \approx \frac{\hbar^2}{2m}(3\pi^2n)^{2/3}-\frac{\pi^{2/3}m}{2\hbar^23^{10/3}n^{2/3}}(k_BT)^2\,[\text{J}] \end{equation} \] \[ \begin{equation} \mu \approx E_F-\frac{\pi^2}{6}(k_BT)^2\frac{D'(E_F)}{D(E_F)}\,[\text{J}] \end{equation} \]
Internal energy
\[ \begin{equation} u= \int \limits_{-\infty}^{\infty} ED(E)f(E)~dE \end{equation} \]
\[ \begin{equation} u\approx\frac{\hbar^2}{10m}(\pi^43^5n^5)^{1/3}+\frac{(3\pi^2n)^{1/3}m}{6\hbar^2}(k_BT)^2\,[\text{J m}^{-3}] \end{equation} \] \[ \begin{equation} u\approx\int\limits_{-\infty}^{E_F}ED(E)dE+\frac{\pi^2D(E_F)}{6}(k_BT)^2\,[\text{J m}^{-3}] \end{equation} \]
Specific heat
\[ \begin{equation} c_v=\left(\frac{du}{dT}\right)_{V=\text{const}} \end{equation} \]
\[ \begin{equation} c_v\approx\frac{(3\pi^2n)^{1/3}m}{3\hbar^2}k_B^2T\,\,[\text{J m}^{-3}\text{ K}^{-1}] \end{equation} \] \[ \begin{equation} c_v\approx\frac{\pi^2D(E_F)}{3}k_B^2T\,\,[\text{J m}^{-3}\text{ K}^{-1}] \end{equation} \]
Entropy
\[ \begin{equation} s=\int\frac{c_v}{T}dT \end{equation} \]
\[ \begin{equation} s\approx\frac{(3\pi^2n)^{1/3}m}{3\hbar^2}k_B^2T\,\,[\text{J m}^{-3}\text{ K}^{-1}] \end{equation} \] \[ \begin{equation} s\approx\frac{\pi^2D(E_F)}{3}k_B^2T\,\,[\text{J m}^{-3}\text{ K}^{-1}] \end{equation} \]
Helmholtz free energy
\[ \begin{equation} f=u-Ts \end{equation} \]
\[ \begin{equation} f\approx\frac{\hbar^2}{10m}(\pi^43^5n^5)^{1/3}-\frac{(3\pi^2n)^{1/3}m}{6\hbar^2}(k_BT)^2\,[\text{J m}^{-3}] \end{equation} \] \[ \begin{equation} f\approx\int\limits_{-\infty}^{E_F}ED(E)dE-\frac{\pi^2D(E_F)}{6}(k_BT)^2\,[\text{J m}^{-3}] \end{equation} \]

The density of states at the Fermi energy and the derivative of the density of states at the Fermi energy are given for a few materials in the table below.

$D(E_F)$ J-1 m-3

$D'(E_F)$ J-2 m-3

 Al 

1.3 × 1047

-5.8 × 1065

Cu

1.5 × 1047

-1.2 × 1065

K

7.4 × 1046

3.3 × 1065

Li

1.5 × 1047

7.0 × 1065

Na

8.4 × 1046

2.2 × 1065

V

9.9 × 1047

-8.9 × 1065