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When computing the electron density in metals, the usual crude result is computed for zero temperature. That is, we integrate

\begin{equation} n=\frac{8\sqrt{2}\pi m^{3/2}}{h^{3}} \int\limits_{0}^{E_{F}}\sqrt{E} dE = \frac{8\sqrt{2}\pi m^{3/2}}{h^{3}} \left ( \frac{2}{3}E_{F}^{3/2} \right ) . \end{equation}

I am looking for an electron density estimate for non zero temperatures, but I can't deal with following integration:

$$ n=\frac{8\sqrt{2}\pi m^{3/2}}{h^{3}} \int\limits_{0}^{\infty}\frac{\sqrt{E}}{1+\exp(\frac{E-E_{F}}{k_{B}T})} dE . $$

Can you help me with a link where this problem is solved? Or simply a term which describes (approximates) the electron density for higher temperatures. Is there some "rule of thumb" for the metals - e.g. some sort of exponential dependence. Does the derivation need more precise approach? Thank you :)

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The thing you are looking for is called the Sommerfeld Expansion. The integral you specify can be approximated quite well to calculate the chemical potential (different to $E_F$ when the electrons are not completely degenerate) and expressions for the number density and energy density of the electrons when the chemical potential (or $E_F$) is larger than $kT$, but the gas is not completely degenerate.

Often the expansion is limited to the first two terms; the second term is of order $(kT/E_F)^2$, while the third term is neglected as becoming small when $(kT/E_F)^4$ is small.

Full details can be found at the wikipedia link, but also a quick search turned up this student paper from a Graz Univ course on Advanced Solid State Physics. It seems to provide a very clear description, and derives the expressions you want for 1, 2 and 3D Fermi-Dirac gases.

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  • $\begingroup$ Thank you very much, after a short calculation the Sommerfeld expansion adds temperature dependent term to the electron density approximation, which I was looking for. Interesting point is, that I have always thought, that the thermal dependence would turn out linear rather than quadratic. $\endgroup$
    – Jan Hirschner
    Commented Sep 26, 2014 at 10:56
  • $\begingroup$ @Jan Hirschner It is the energy density that has the additional quadratic term. I suppose you could think of this as a fraction $\sim kT/E_F$ electrons are above the "roll-off" region in the distribution and each of these has $\sim kT$ of energy above $E_F$. Differentiating this gives a heat capacity that is linearly dependent on $T$. $\endgroup$
    – ProfRob
    Commented Sep 26, 2014 at 11:13

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