What is a Fermi surface?

Question

I posted this question on the physics site, but then received immediately the sense that I won't be able to understand answers even if they come. So I hope it's all right if I post it here, since there appears to be mathematical interest as evidenced by this question:

Topology and the 2016 Nobel Prize in Physics

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Allow me to explain my confusion. Given a solid, I believe I have some feeling for the Fermi level. I can understand it, for example, as the characteristic parameter $\mu$ in the Fermi-Dirac distribution of energy levels for the electrons in the system: $$f(\epsilon)=\frac{1}{e^{(\epsilon-\mu)/kT}+1}$$ ignoring for the moment other physical interpretations. Thus, it is the unique energy level that has probability 1/2 of being occupied.

The definition of the Fermi surface, on the other hand, is usually given as 'the iso-surface of states with energy equal to the Fermi level' in the three-dimensional space of wave-vectors $k$, for example in this Wikipedia article:

https://en.wikipedia.org/wiki/Electronic_band_structure

In other words, it is defined to be those $k$ such that $$E(k)=\mu.$$ So far, so good. The problem is, I don't quite understand what $E(k)$ is.

One situation seems to be straightforward, namely a Fermi gas of identical particles. Then $$E(k)=\frac{k^2}{2m}$$ and the Fermi surface is a sphere. However, if we are in an infinite periodic potential, the usual idealised model for Bloch theory, then the solutions to the Schroedinger equation come out in the form $$\psi_{kn}(r)=e^{ik\cdot r}u_{kn}(r),$$ where $u_{kn}$ is a periodic function and $n$ is a discrete index for energy levels. In other words, for each wave vector $k$,

there are many energy levels $E_n(k)$.

So the equation for the Fermi surface would actually look like $$E_n(k)=\mu.$$My question, therefore, is which energy level is the $E(k)$ that occurs in the definition of the Fermi surface? Perhaps there is one Fermi surface for every level $n$? (Assuming that the levels vary continuously over the momentum space, enabling us to consistently index the levels for varying $k$.)

If I could elaborate on my confusion a little bit more, I don't quite understand the definition in this answer to this question:

https://physics.stackexchange.com/q/5739/

It is stated that

'The Fermi surface is simply the surface in momentum space where, in the limit of zero interactions, all fermion states with (crystal) momentum $|k|<|k_F|$ are occupied, and all higher momentum states are empty. '

For one thing, as mentioned above, for any momentum $k$, there is an infinite sequence of fermion states. The other problem is that I'm not sure that the statement above defines a unique surface, even if I were able to somehow pick out a fermion state $\psi(k)$ for each $k$ that the statement refers to. (I would need to draw a picture to explain this point, which I don't have the competence to do.)

Answer 1

Quoting p. 142 of the trusty Ashcroft-Mermin (who write $\mathcal E_F$ for your $\mu$):

For each partially filled band there will be a surface in $k$-space separating the occupied from the unoccupied levels. The set of all such surfaces is known as the Fermi surface, and is the generalization to Bloch electrons of the free electron Fermi sphere. The parts of the Fermi surface arising from individual partially filled bands are known as the branches of the Fermi surface. We shall see (Chapter 12) that a solid has metallic properties provided that a Fermi surface exists.

Analytically, the branch of the Fermi surface in the $n$th band is that surface in $k$-space (if there is one) determined by $$\mathcal E_n(\mathbf k)=\mathcal E_F.$$

As to "what $\mathcal E_n(\mathbf k)$ is": it arises from looking for eigenfunctions of a "single-electron" hamiltonian $$ \left(-\frac{\hbar^2}{2m}\nabla^2+U(\mathbf r)\right)\psi(\mathbf r)=E\psi(\mathbf r) $$ in the "Bloch" form: $\psi(\mathbf r)=e^{i\mathbf k\cdot\mathbf r}u(\mathbf r)$ with (Bravais lattice) periodic boundary conditions on $u$, i.e. $$ \left(\frac{\hbar^2}{2m}\left(\frac1i\nabla + \mathbf k\right)^2+U(\mathbf r)\right)u_{\mathbf k}(\mathbf r)=\mathcal E(\mathbf k)u_{\mathbf k}(\mathbf r). $$

Because of the periodic boundary condition we can regard [this] as a Hermitian eigenvalue problem restricted to a single primitive cell of the crystal. Because the eigenvalue problem is set in a fixed finite volume, we expect on general grounds to find an infinite family of solutions with discretely spaced eigenvalues, which we label with the band index $n$.