# Deriving the Sommerfeld expansion by contour integration (Le Bellac p. 277)

In Le Bellac's statistical physics book he derives the Sommerfeld expansion by a contour integral.

The idea is to expand integrals of the type $I(\beta)\equiv \int_{0}^{\infty}d\epsilon\, \frac{\varphi'(\epsilon)}{e^{\beta(\epsilon-\mu)}+1}$. Integrate by parts and expand $\varphi(\epsilon)$

$$\varphi(\epsilon)=\sum_{m=0}^{\infty}\frac{(\epsilon-\mu)^{m}}{m!}\left(\frac{d ^{m}\varphi}{d \epsilon}\Bigg |_{\epsilon=\mu}\right)$$

Make a change of variables to write $I(\beta)=\sum_{m=0}^{\infty}\frac{\beta^{-m}\varphi^{(m)(\mu)}}{m!}I_{m}$ where $I_{m}=\int_{-\infty}^{\infty}dx\, \frac{x^{m}e^{x}}{(e^{x}+1)^{2}}$.

The idea then to evaluate the $I_m$ is to consider the following integral

$$J(p)=\int_{-\infty}^{\infty}dx\, \frac{e^{ipx}}{(e^{x}+1)(e^{-x}+1)}=\sum_{m=0}^{\infty}\frac{(ip)^{m}}{m!}\int_{-\infty}^{\infty}\, dx \frac{x^{m}}{(e^{x}+1)(e^{-x}+1)}= \sum_{m=0}^{\infty}\frac{(ip)^{m}}{m!}I_{m}$$

Integrating $J(p)$ gives $J(p)=\frac{2\pi pe^{-\pi p}}{1-e^{-2\pi p}}=1-\frac{1}{6}(\pi p)^2+\ldots$ hence,

$$I(\beta)= \varphi(\mu) + \frac{\pi^{2}}{6}(k_{B}T)^{2}\varphi^{(2)}(\mu)+O(k_{B}T^{4})$$

So far so good, my problem comes when he writes the expansion of the Fermi-Dirac distribution itself as

$$\frac{1}{e^{\beta(\epsilon-\mu)}+1}\simeq \Theta_{H}(\mu-\epsilon)-\frac{\pi^{2}}{6}(k_{B}T)^{2}\delta'(\epsilon-\mu)+O(k_{B}T^{4})$$

which is meant to be valid under an integral. It appears to mimic the expansion of the generating function, at least where the signs are concerned. If we consider the definition of the Sommerfeld expansion from Wikipedia, we have a formula in the form

$$\int_{-\infty}^\infty \frac{H(\varepsilon)}{e^{\beta(\varepsilon - \mu)} + 1}\,\mathrm{d}\varepsilon = \int_{-\infty}^\mu H(\varepsilon)\,\mathrm{d}\varepsilon + \frac{\pi^2}{6}\left(\frac{1}{\beta}\right)^2H^\prime(\mu) + \mathrm{O} \left(\frac{1}{\beta\mu}\right)^4$$

If we give Le Bellac's form a try,

$$I(\beta)\simeq\int_{0}^{\infty}d\epsilon\, \varphi'(\epsilon)\left[\Theta_{H}(\mu-\epsilon)-\frac{\pi^{2}}{6}(k_{B}T)^{2}\delta'(\epsilon-\mu)+O(k_{B}T^{4}) \right] \\ =\varphi(\mu)-\frac{\pi^{2}}{6}(k_{B}T)^{2}\underbrace{\int_{0}^{\infty}d\epsilon\, \varphi'(\epsilon)\delta'(\epsilon-\mu)}_{K}$$

but integrating $K$ by parts gives

$$K=\left[\delta(\epsilon-\mu)\varphi'(\epsilon) \right]_{0}^{\infty}- \int_{0}^{\infty}d\epsilon\, \varphi''(\epsilon)\delta(\epsilon-\mu)=-\varphi''(\mu)$$

So the signs work out, and the negative sign is probably not a typo. On the other hand, the next term in the expansion would be $+\frac{7\pi^4}{360}(k_{B}T)^{4}\delta^{(3)}(\epsilon-\mu)$ which would take a negative sign when integrated.

The alternating behavior of the signs, as far as I can tell, is due to the imaginary unit that was included in the exponential for the contour integral. A very similar approach is done on p. 255 of the following notes except there is no imaginary unit in the exponential. Also, the expansion is meant to operate only on $\varphi$ instead of actually correcting the Fermi-Dirac distribution for small $T$.

My questions are as follows:

1. Are the corrections to the Fermi-Dirac distribution given by Le Bellac correct to all orders? They seem to work for the 0th and 1st order terms.

2. Assuming (1) is correct, how do we "know", or what's the reasoning, that the series expansion of the generating function would give the corrections to the Fermi-Dirac distribution? Because it seemed to me that the generating function was only to determine the $I_m$ coefficients.

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I) Preliminaries. First let us state explicitly the Sommerfeld expansion to all orders. To this end, let

$$\tag{1} B(x)~:=~\frac{x}{e^x-1}~=~\sum_{m=0}^{\infty}\frac{B_m}{m!}x^m~=~1-\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720}+\frac{x^6}{30240}+{\cal O}(x^8)\qquad$$

be the generating function for the Bernoulli numbers$^1$. Let

$$\widehat{A}(x)~:=~\frac{x/2}{\sinh\frac{x}{2}}~=~2B(\frac{x}{2})-B(x)~=~\sum_{m=0}^{\infty}\frac{\widehat{A}_m}{m!}x^m = \widehat{A}(-x)$$ $$\tag{2} ~=~1-\frac{x^2}{24}+\frac{7x^4}{5760}-\frac{31x^6}{967680}+{\cal O}(x^8)$$

be the so-called $A$-roof function with the coefficients

$$\tag{3} \widehat{A}_m~:=~(2^{1-m}-1)B_m, \qquad m~\in~\mathbb{N}_0,$$

given in terms of the Bernoulli numbers. Let

$$\tag{4} f(\varepsilon)~:=~\frac{1}{e^{\beta\varepsilon}+1}$$

be the bare Fermi-Dirac distribution without the chemical potential $\mu$. In this answer, we will for technical reasons work in terms of (minus) the derivative

$$\tag{5} -f^{\prime}(\varepsilon) ~=~\frac{\beta}{4\cosh^2 \frac{\beta\varepsilon}{2}}~>~0,$$

because contrary to the Fermi-Dirac distribution $f(\varepsilon)$ itself, the derivative $f^{\prime}(\varepsilon)$ is exponentially suppressed for $\varepsilon \to -\infty$. (As an added bonus the derivative $f^{\prime}(\varepsilon)=f^{\prime}(-\varepsilon)$ happens to be an even function.)

II) The Sommerfeld functional. Define the Sommerfeld functional as

$$\tag{6} I[\Phi] ~:=~-\int_{\mathbb{R}} \! d\varepsilon~f^{\prime}(\varepsilon-\mu) \Phi(\varepsilon).$$

Energies $\varepsilon$ with $|\varepsilon-\mu|\gg 1/\beta$ do effectively not contribute to the integral (6). Here $\Phi:\mathbb{R}\to\mathbb{C}$ is a real analytic test function

$$\tag{7} \Phi(\varepsilon)~=~\sum_{m=0}^{\infty}\frac{\Phi^{(m)}(\mu)}{m!}(\varepsilon-\mu)^m.$$

Beware that analyticity (7) is a mathematical abstraction that is almost never justifiable in actual physics applications. Let us furthermore assume that the integrand (6) has the function

$$\tag{8}\varepsilon~\mapsto~ -f^{\prime}(\varepsilon) \sum_{m=0}^{\infty}\frac{|\Phi^{(m)}(\mu)|}{m!}|\varepsilon-\mu|^m~\geq~0$$

as a Lebesgue integrable majorant, such that (according to Lebesgue dominated convergence theorem) the order of integration and summation in eq. (6) can be interchanged

$$\tag{9} I[\Phi]~=~-\sum_{m=0}^{\infty}\frac{\Phi^{(m)}(\mu)}{m!}\int_{\mathbb{R}} \! d\varepsilon~ (\varepsilon-\mu)^m f^{\prime}(\varepsilon-\mu) .$$

Existence of a Lebesgue integrable majorant (8) is a relatively mild technical assumption that is almost always satisfied in actual physics applications.

III) The Sommerfeld expansion. The integrals in eq. (9) are well-defined and can be computed by either real or complex methods, cf. Refs. 1, 2, and 3. The Sommerfeld expansion to all orders becomes

$$\tag{10} I[\Phi]~=~\sum_{m=0}^{\infty}\left(\frac{2\pi i}{\beta}\right)^{m} \frac{\widehat{A}_m}{m!}\Phi^{(m)}(\mu),$$

We will view eq. (10) as the main result.

IV) It is tempting to formally rewrite eq. (10) as

$$\tag{11}I[\Phi] ~=~ \widehat{A}\left(\frac{2\pi i}{\beta} \frac{d}{d\mu}\right) \Phi(\mu),$$

or as

$$\tag{12} -f^{\prime}(\varepsilon-\mu) ~=~ \sum_{m=0}^{\infty}\left(\frac{2\pi }{i\beta}\right)^{m} \frac{\widehat{A}_m}{m!}\delta^{(m)}(\varepsilon-\mu).$$

A formal integration of eq. (12) wrt. $\varepsilon$ yields

$$\tag{13} f(\varepsilon-\mu) ~=~\theta(\mu-\varepsilon)- \sum_{m=1}^{\infty}\left(\frac{2\pi}{i\beta}\right)^{m} \frac{\widehat{A}_m}{m!}\delta^{(m-1)}(\varepsilon-\mu).$$

Eqs. (12) and (13) are strictly speaking mathematical nonsense. They should only be thought of as a mnemonic to recall eq. (10).

V) Finally let us address OP's questions. Although on one hand, the Sommerfeld expansion (10) is mathematically accurate to all orders (when applied to above real analytic test function $\Phi$), on the other hand in physical applications, the test function $\Phi$ often fails to be real analytic.

Example: $I[\phi]$ could be the number of particles (per volume) of a system, and the derivative $\Phi^{\prime}(\varepsilon)$ could be the density $g(\varepsilon)$ of energy levels (per volume). There are no physical reasons to expect $\Phi$ to be real analytic.

Hence in physical applications, we only trust the first few terms in the Taylor expansion (7) with $|\varepsilon-\mu|\ll \mu$. Therefore the effective support of $f^{\prime}(\varepsilon-\mu)$ in the integral (6) should be restricted to this interval. This is guaranteed in the low temperature regime $1/\beta\ll\mu$. In this low temperature limit $1/\beta\ll\mu$, the first few terms in the Taylor expansion (7) effectively lead to the first few terms in the Sommerfeld expansion (10).

References:

1. N.W. Ashcroft and N.D. Mermin, Solid State Physics, 1976, Appendix C, p.760-761.

2. M. Le Bellac, F. Mortessagne and G.G. Batrouni, Equilibrium and Non-Equilibrium Statistical Thermodynamics, 2004, Section 5.2.2, p. 276-279.

3. Daniel Arovas, Lecture Notes on Thermodynamics and Statistical Mechanics, 2012, Section 5.8.5, p. 255-257. The pdf file is available here.

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$^1$ Beware of slightly different definitions of Bernoulli numbers in the literature.

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Thanks for the reply. I had just a question before I mark yours as the answer. How did you get equation 12? I could have probably deduced the form by guessing and checking, but I was wondering if there was more to it. – k-selectride Feb 17 '13 at 23:17
Eq. (12) follows formally by rewriting eq. (10) with delta functions, integrate by part, and finally removing the test function $\Phi$ on both sides. Are you really asking about eq. (10)? – Qmechanic Feb 17 '13 at 23:23
No my difficulty is with the step you just described, the rewriting with delta functions. Is it making a substitution of the type $\phi^{(m)}(\mu) \to \int \phi^{(m)}(\epsilon) \delta(\epsilon- \mu$? – k-selectride Feb 18 '13 at 7:41
Yes. Specifically, we have to assume that the test function $\Phi\in{\cal S}(\mathbb{R})$ belongs to the Schwartz space ${\cal S}(\mathbb{R})$. – Qmechanic Feb 18 '13 at 9:26
Last question, in the integration from 12 to 13, are the limits $\epsilon \to \infty$? On my end I get $f(\epsilon-\mu)=-\int_{\epsilon}^{\infty}d\epsilon \delta(\epsilon-\mu)-\sum...=-\theta(\mu-\epsilon)-\sum...$. I'm not sure where my error is to reconcile with the signs in 13. – k-selectride Feb 19 '13 at 7:09