# What "goes wrong" in a Sommerfeld expansion?

Let $$f$$ be the Fermi function and $$H$$ be a function which which vanishes as $$\epsilon \to -\infty$$ and which diverges at $$\infty$$ no worse than some power of $$\epsilon$$. In the Sommerfeld expansion of solid state physics (see e.g. Ashcroft and Mermin Appendix C), one writes an integral of the form $$\int_{-\infty}^\infty H(\epsilon)f(\epsilon)\, d\epsilon$$ instead as, for some $$\mu \in \mathbb{R}$$, $$\int_{-\infty}^\infty H(\epsilon)f(\epsilon)\, d\epsilon = \int_{-\infty}^\mu H(\epsilon)f(\epsilon)\, d\epsilon + \sum_{n=1}^{\infty}\int_{-\infty}^\infty \frac{(\epsilon - \mu)^{2n}}{2n!}(-\partial f/\partial \epsilon)\, d\epsilon \left(\frac{d^{2n-1}H}{d\epsilon^{2n-1}}\right)(\mu).$$ The key step is doing a Taylor expansion of $$K(\epsilon) := \int_{-\infty}^\epsilon H({\epsilon'})\,d\epsilon'$$ about $$\epsilon = \mu$$ and then exchanging the infinite sum and integral.

My question is, when is this exchange justified? In particular, $$H$$ usually has some sort of lack of smoothness at $$\epsilon = 0$$ (e.g. in 2D, $$H$$ can be a positive constant for $$\epsilon>0$$ and $$0$$ for negative numbers). I would have thought that the expansion would be rigorously valid as long as we restrict ourselves to $$\mu >0$$ (since usually $$H = 0$$ for $$\epsilon<0$$ so that $$\int_{-\infty}^\infty H(\epsilon)f(\epsilon)\, d\epsilon = \int_{0}^\infty H(\epsilon)f(\epsilon)\, d\epsilon$$ anyway), and yet I have just worked a problem which shows that this is not so; that somehow the exchange "knows about" the singularity at $$0$$ in spite of the fact that $$\epsilon < 0$$ "doesn't matter". I'm hoping someone can help me understand what's going on.

Further, some of my confusion lies in that, in 3D (where $$H$$ is not differentiable) the expansion works whereas in 2D (where $$H$$ is not continuous) it does not work. Does the fact that the expansion is valid to one extra term in the 3D case really change things that materially?

• From wikipedia: "The expansion is only valid if $H(\varepsilon )$ vanishes as $\varepsilon \rightarrow -\infty$ and goes no faster than polynomially in $\varepsilon$ as $\varepsilon \rightarrow \infty$. If the integral is from zero to infinity, then the integral in the first term of the expansion is from zero to $\mu$ and the second term is unchanged." Does this answer your question? Commented Apr 19, 2023 at 2:24
• Also just a small note, but this statement is somewhat confusing: "The key step is doing a Taylor expansion of $K(\varepsilon) = \int_{-\infty}^\infty H(\varepsilon)$ about $\varepsilon=\mu$," since $K(\varepsilon)$ depends on $\varepsilon$ but the integral on the right hand side (the integration measure is presumably $d\varepsilon$?) doesn't depend on $\varepsilon$. Is it possible you intended $K(\varepsilon)=f(\varepsilon)H(\varepsilon)$? Commented Apr 19, 2023 at 2:29
• I'm not sure I follow your first sentence (those are conditions I asserted in my question)? As for your second, I had an error and have fixed it now. @Andrew
– EE18
Commented Apr 19, 2023 at 2:56
• @Andrew no, that form of K is standard to an analysis of the Sommerfeld expansion. Commented Apr 19, 2023 at 3:12
• Crossposted from math.stackexchange.com/q/4681863/11127 Commented Apr 23, 2023 at 18:17

## 1 Answer

For a while I kept thinking that you were asking something about Sommerfeld expansion. Then I realised that you are actually asking something about mathematics, really.

https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged

I think it is important to point out an important fact about the Sommerfeld expansion. It is known that the Sommerfeld expansion is an asymptotic expansion. It has a radius of convergence of zero. It behaves a little like the renormalisation problem as in QFT. But you can easily sidestep this problem because you can do a variant of this that would not have the problem---manipulate the integrals analytically exactly until you have the same initial terms and the final small remainder term is an integral, and just evaluate this last integral numerically.

Note, however, that just because it is looks mathematically nonsense to do such asymptotic expansions, it is actually perfectly fine. These approximations are highly motivated by physics, and so the maths would work hard to give us something sensible and incredibly accurate, even though the expressions may be formally silly.

• I find that this doesn't answer the question the OP intended to make, if a bit misworded. Let $H(\epsilon)$ be an heaviside function, as the OP proposes. The Sommerfeld expansion for the density of states gives that the fermi energy is the same as the chemical potential for all temperatures, which is not true if one calculates the integral analytically. Why the expansion fails for this particular discontinuity I am not sure mathematicaly, but it upholds the restrictions on the growth of the density of states, and the derivatives of the expansion are taken on positive values... Commented May 23, 2023 at 16:15
• Oh, it's because the chemical potential can be both positive and negative, my mistake, so the derivatives can be taken on zero. Commented May 23, 2023 at 16:35
• Nice to see that you have figured out a thing, @RicardoMM, so I can focus on another issue: "The Sommerfeld expansion for the density of states gives that the Fermi energy is the same as the chemical potential for all temperatures" is not the case. Fermi energy is by definition the zero temperature limit of chemical potential, and what we do is to basically always start the expansion from the chemical potential for the temperature of concern. If we want to figure out the result of the chemical potential itself changing, that is a totally different problem requiring a different computation. Commented May 24, 2023 at 2:05
• Yeah, it is not, but for this particular example I gave one gets that idea after looking at the Sommerfeld expansion, where all the terms except the first one disappear. Commented May 24, 2023 at 3:57
• @RicardoMM no, the underlying idea in Sommerfeld expansion is really just a rearrangement of the integration, and it is really only used when $\mu>0$ and at low $T$ so that when the Taylor expansion steps are unphysical, we could just go back to the integral forms. And when $\mu<0$ the thing to do is to just approximate with Maxwell-Boltzmann. Commented May 24, 2023 at 4:09