D.63 Fermi-Dirac integrals at low temperature

This note finds the basic Fermi-Dirac integrals for the free-electron gas at low temperature. To summarize the main text, the number of particles and total energy per unit volume are to be found from

\begin{displaymath}
\frac{I}{{\cal V}} = \int_0^\infty \iota^{\rm {f}}{\cal D}...
...{\rm p}\iota^{\rm {f}}{\cal D}{\,\rm d}{\vphantom' E}^{\rm p}
\end{displaymath}

where the Fermi-Dirac distribution and the density of states are:

\begin{displaymath}
\iota^{\rm {f}} = \frac{1}{e^{({\vphantom' E}^{\rm p}-\mu)...
...\frac{2m}{\hbar^2}\right)^{3/2} \sqrt{{\vphantom' E}^{\rm p}}
\end{displaymath}

and the number of spin states $n_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2s+1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 for systems of electrons. This may be rewritten in terms of the scaled energies

\begin{displaymath}
u = \frac{{\vphantom' E}^{\rm p}}{k_{\rm B}T} \qquad
u_0 = \frac{\mu}{k_{\rm B}T}
\end{displaymath}

to give

\begin{displaymath}
\frac{I}{{\cal V}} = \frac{n_s}{4\pi^2} \left(\frac{2m}{\h...
...u=0}^\infty \frac{(u/u_0)^{1/2}}{e^{u-u_0}+1}{\,\rm d}(u/u_0)
\end{displaymath}


\begin{displaymath}
\frac{E}{{\cal V}} = \frac{n_s}{4\pi^2} \left(\frac{2m}{\h...
...u=0}^\infty \frac{(u/u_0)^{3/2}}{e^{u-u_0}+1}{\,\rm d}(u/u_0)
\end{displaymath}

To find the number of particles per unit volume for small but nonzero temperature, in the final integral change integration variable to $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(u/u_0)-1$, then take the integral apart as

\begin{displaymath}
\int_{-1}^0 \sqrt{1+v}{\,\rm d}v
- \int_{-1}^0 \frac{\sq...
...}
+ \int_{0}^\infty \frac{\sqrt{1+v}{\,\rm d}v}{e^{u_0v}+1}
\end{displaymath}

and clean it up, by dividing top and bottom of the center integral by the exponential and then inverting the sign of $v$ in the integral, to give

\begin{displaymath}
\int_{-1}^0 \sqrt{1+v}{\,\rm d}v
+ \int_{0}^1 \frac{(\sq...
...}
+ \int_{1}^\infty \frac{\sqrt{1+v}{\,\rm d}v}{e^{u_0v}+1}
\end{displaymath}

In the second integral, the range that is not killed off by the exponential in the bottom is very small for large $u_0$ and you can therefore approximate $\sqrt{1+v}-\sqrt{1-v}$ as $v$, or using a Taylor series if still higher precision is required. (Note that the Taylor series only includes odd terms. That makes the final expansions proceed in powers of 1/$u_0^2$.) The range of integration can be extended to infinity, since the exponential in the bottom is exponentially large beyond $v$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. For the same reason, the third integral can be ignored completely. Note that $\int_0^{\infty}x{\,\rm d}{x}$$\raisebox{.5pt}{$/$}$$(e^x+1)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi^2$$\raisebox{.5pt}{$/$}$​12, see [40, 18.81-82, p. 132] for this and additional integrals.

Finding the number of particles per unit volume $I$$\raisebox{.5pt}{$/$}$${\cal V}$ this way and then solving the expression for the Fermi level $\mu$ gives

\begin{displaymath}
\mu = {\vphantom' E}^{\rm p}_{\rm {F}}
- \frac{\pi^2}{12...
...ht)^{2/3}
\frac{\hbar^2}{2m} \left(\frac{I}{V}\right)^{2/3}
\end{displaymath} (D.39)

This used the approximations that $\mu$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ ${\vphantom' E}^{\rm p}_{\rm {F}}$ and $u_0^{-2}$ is small, so

\begin{displaymath}
u_0^{-2} = \left(\frac{k_{\rm B}T}{\mu}\right)^2 \approx
...
...)^{-2/3}
\approx 1 - \frac{2}{3} \frac{\pi^2}{8\,} u_0^{-2}
\end{displaymath}

The integral in the expression for the total energy per unit volume goes exactly the same way. That gives the average energy per particle as

\begin{displaymath}
\frac{E}{I} = {\vphantom' E}^{\rm p}_{\rm ave} = \frac{3}{...
...rm {F}}}\right)^2 {\vphantom' E}^{\rm p}_{\rm {F}}
+ \ldots
\end{displaymath} (D.40)

To get the specific heat at constant volume, divide by $m$ and differentiate with respect to temperature:

\begin{displaymath}
C_v = \frac{\pi^2}{2\,} \frac{k_{\rm B}T}{{\vphantom' E}^{\rm p}_{\rm {F}}} \frac{k_{\rm B}}{m} + \ldots
\end{displaymath}