As an alternative approach to the Sommerfeld-expansion, my lecturer tries to motivate properties of free fermions, such as temperature dependencies of the chemical potential $\mu(T)$, electron number $N_e(T)$, energy density $U(T)$, etc. by expanding the Fourier transform of the Fermi function for low temperatures:

$$\int d\epsilon g(\epsilon)f(\epsilon)=\int d\epsilon g(\epsilon)\int dt \tilde{f}(t)e^{-i\epsilon t}\\ \text{where}~ \tilde{f}(t)=\frac{e^{i\mu t}}{2\pi i}\left(\pi i \delta (t)+\frac{1}{t}\frac{\pi t /\beta}{\sinh(\pi t /\beta)}\right)\\ \text{and} ~ g(\epsilon) ~\text{some arbitrary well-behaved function}. $$ I have only come so far to calculate the Fourier transform $\tilde{f}(t)$ of $f(\epsilon)=\frac{1}{(e^{\beta(\epsilon-\mu)}+1)}$: $$\tilde{f}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} d\epsilon \frac{e^{i\epsilon t}}{e^{\beta(\epsilon-\mu)}+1}=\frac{e^{i\mu t}}{2\pi \beta}\int dx \frac{e^{ixt/\beta}}{e^x+1}=\dots= \frac{e^{i\mu t}}{2\pi i}\left(\pi i \delta (t)+\frac{1}{t}\frac{\pi t /\beta}{\sinh(\pi t /\beta)}\right)$$

I have used $x=\beta(\epsilon-\mu)$.

Can anybody give me hints for the calculation steps in between?


2 Answers 2


I have in my notes a related Laplace transform: $$ I=\int_{-\infty}^{\infty} \frac{d\epsilon}{2\pi} e^{\tau\epsilon/2\pi} \left\{ \frac{1}{1+e^{\beta(\epsilon-\mu)}}-\theta(-\epsilon)\right\}= \frac 1{\tau}\left\{ \frac{(\frac{\tau T}{2})}{\sin(\frac{\tau T}{2})} e^{\tau\mu/2\pi}-1\right\}, \quad 0<\tau T/2\pi< 1. $$ I evaluated it goes as follows: $$ I=\int_0^\infty \frac{d\epsilon}{2\pi} \left\{\frac{e^{\tau\epsilon/2\pi}}{1+e^{\beta(\epsilon-\mu)}}+\frac{e^{-\tau\epsilon/2\pi}}{1+e^{\beta(-\epsilon-\mu)}}- e^{-\tau\epsilon/2\pi}\right\}\nonumber\\ = \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \left\{\frac{e^{\tau\epsilon/2\pi}}{1+e^{\beta(\epsilon-\mu)}}\right\}-\frac 1\tau\\ = e^{\mu\tau/2\pi} \int_{-\infty}^\infty \frac{d\epsilon}{2\pi} \left\{\frac{e^{\tau(\epsilon-\mu)/2\pi}}{1+e^{\beta(\epsilon-\mu)}}\right\}-\frac 1\tau\\ = e^{\mu\tau/2\pi} T\int_{-\infty}^\infty \frac{d\xi }{2\pi} \left\{\frac{e^{\xi T\tau/2\pi}}{1+e^\xi}\right\}-\frac 1\tau\\ = e^{\mu\tau/2\pi} T \int_{0}^\infty \frac{dx }{2\pi}\frac{x^{T\tau/2\pi-1}}{1+x} -\frac 1\tau\\ = \frac 1{\tau}\left\{ \frac{(\frac{\tau T}{2})}{\sin(\frac{\tau T}{2})} e^{\tau\mu/2\pi}-1\right\}. $$ I set $x=\exp\{\beta(\epsilon-\mu)\}$ and at the last step used the standard integral $$ \int_0^\infty dx \frac{x^{\alpha-1}}{1+x}= \frac{\pi}{\sin\pi \alpha}, \quad 0<\alpha<1. $$ There may be an easier way!

The integral is interesting because it seems to be related to the generating function for the $\hat A$ genus that appears in the Dirac index theorem. I learned this from a paper of Loganayagam and Surówka: arXiv:1201.2812


You can write $f(\epsilon) = \frac{1}{2} (1 - \tanh[\beta(\epsilon-\mu)/2])$. The Fourier transform of the first term gives the Dirac delta, and the second term is more difficult. In this answer, the Fourier transform of $\tanh(x)$ is stated, although no proof is given.

Also see this related answer: https://math.stackexchange.com/questions/2569814/fourier-transform-of-sigmoid-function


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