What is the Fermi Surface? I hope this question is not too elementary for this forum, and apologise in advance in case it is.

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:


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:

What is Fermi surface and why is this concept so useful in metals research?

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.)

  • $\begingroup$ The Fermi surface is defined at a temperature of absolute zero, so you take the ground-state solutions $E_0(k)=\mu$... $\endgroup$
    – lemon
    Oct 14, 2016 at 12:26
  • $\begingroup$ And in a solid, you look at the states within a (Wigner-Seitz) unit cell. $\endgroup$
    – Jon Custer
    Oct 14, 2016 at 12:39
  • $\begingroup$ Lemon: I find that also quite confusing. So your statement would be 'The Fermi surface is the set of $k$ such that $E_0(k)=\mu$,' where $E_0(k)$ is the lowest energy with momentum $k$. But then, in a solid where many of the lower energy bands are filled, there would be many electrons above the Fermi level. This seems not to agree with the usual picture. $\endgroup$ Oct 14, 2016 at 12:49
  • $\begingroup$ Jon Custer: I guess you're referring to the fact that each of the $u_{kn}$ are determined by their values in a cell. That's true. But there are no states that are just concentrated in a cell. (The $u_{kn}$ are periodic.) In any case, I don't see how this answers the question. The way you phrase it, you make it sound like ' for each $k$, there is a unique $\psi_{kn}$ concentrated in a cell, and its energy is what we use to define the Fermi surface.' This doesn't sound right for a variety of reasons. $\endgroup$ Oct 14, 2016 at 12:54

2 Answers 2


Everything you say is correct. The Fermi surface is defined to be the set of points $k$ such that $E_n(k) = \mu$ for any band $n$. However, typically the bands are spaced relatively far apart and don't overlap in energy, like this:

enter image description here

As we can see, the bands 1 and 3 lie completely above or completely below the chemical potential $\mu$ and so are irrelevant for determining the Fermi surface (in fact, at low temperatures, those bands are pretty much irrelevant for any physical phenomena - only bands near the chemical potential are physically important). That's why in practice, you can get away with just considering one or two bands and completely ignoring all the others - and when there's a Fermi surface (i.e. the chemical potential intersects a band(s)), one band is almost always enough.

In more complicated/unusual systems though, you do need to keep track of multiple bands. For example, sometimes bands can touch or cross, and funny things can happen if you tune the chemical potential exactly to the crossing point. Even more unusually, two bands can share a whole finite range of energy - e.g. two cosine curves shifted vertically by a tiny amount. But these cases are very rare - for most everyday materials, $\mu$ sits in at most one band and you don't need to worry about this. (In fact, professional physicists like to find/create unusual materials where the chemical potential does sit right at a band crossing, precisely because such systems aren't as theoretically well-understood, so there's more to learn.)

BTW, in 1-D, like the plot above, the Fermi "surface" just consists of isolated values of $k$, but in 2-D it's usually a closed curve in the $k_x$-$k_y$ plane, and in 3-D it's usually a closed surface, like a sphere. Sometimes the Fermi surface can actually consist of two (or more) spheres, with one inside the other, and the filled "Fermi sea" for the relavant band lies in between them. This phenomenon is called "Fermi surface nesting." But if you're just learning about Fermi surfaces, then you won't have to worry about these complicated situations for a long time.

  • $\begingroup$ Thanks for the clear answer. By the way, I've gather now that the word 'band' is used in two distinct ways in solid state physics. The word you use here just refers to an energy level. But there is also the notion of a band as an essentially continuous distribution of energy levels, between which are 'gaps.' I think this was a major part of my confusion. Correct me if I'm wrong about this. $\endgroup$ Oct 22, 2016 at 17:31
  • $\begingroup$ @MinhyongKim A "band" is defined to be a single curve $E_n(k)$ for a given value of $n$. (I think it's somewhat misleading to call that an "energy level" because the function is generally not constant, so it takes values over a whole finite interval of energies.) People occasionally abuse terminology and also use the word "band" to refer to the interval of energy over which the function ranges - i.e. collapsing the momentum dependence. You're right that this is what people are thinking about when they talk about "band gaps." But the two senses of "band" are really almost identical ... $\endgroup$
    – tparker
    Oct 22, 2016 at 18:49
  • $\begingroup$ ... the only difference is whether you keep track of the dependence on $k$ or just consider the function's range. $\endgroup$
    – tparker
    Oct 22, 2016 at 18:50
  • 1
    $\begingroup$ Thanks for the further explanation. But it seems to me somewhat important to distinguish the two senses. If the word 'band' were being used in the sense of electronic band structure, then the equation $E_n(k)=\mu$ wouldn't be well-defined even for a fixed value of $n$. This was one of the very confusing things for a novice like me. In any case, thanks again! $\endgroup$ Oct 22, 2016 at 22:52

The Fermi surface is the surface in the reciprocal space (the dual of the real space you live in) delimitating the fermionic occupied states from the fermionic unoccupied ones at zero temperature. So it is a momentum ($k$) construction rather than an energy construction.

The logic is the following : try to put all together a given number of fermions. Since they follow the Pauli exclusion principle, you can not pack these fermions the way you want. Each time there is some room for a state in the momentum space, only one fermion can occupy this empty room. So you must start to pile up the fermions. It has a complete analogy with filling up a bookcase with books : you must use the next row when the previous one is full. You may use smaller intervals between raws, enlarge the size of each raw, ..., if you have too many books, you might use the next raw, which is nothing but using the next momentum branch in your dispersion relation (what you call $k_n(E)$). When you put the last fermion in your fermionic bookcase, the corresponding momentum state is called the Fermi momentum, the corresponding energy is called the Fermi energy, ..., and the surface of iso-$k$ at the Fermi momentum is called the Fermi surface.

Few remarks now

  • There will never be an infinite number of branches used to fill up a finite number of fermions in the dispersion relations (the band structure of the material if you prefer).

  • There is no contradiction in supposing the Fermi surface has several sheets. Even on Wikipedia you already have example of Fermi surface with electron and hole pockets

  • The concept of Fermi surface comes from the notion of (Fermi-Dirac) statistics, when you have a finite number of particles to deal with (in an ancient terminology, it is a second quantised problem), whereas the band structure is the complete spectrum of available states for one particle (in the ancient terminology, it is a first quantised problem) in a periodic potential. The easy way to pass from one to the other is the use of the chemical potential, which fixes the number of particles per energy state (more precisely, the amount of energy required to add a particle to the thermodynamic system).

  • The Fermi surface is a particularly useful concept to understand a few transport properties (electric, heat, ... transports) for materials with simple band structures, like pure metals and doped semi-conductors. When the Fermi surface becomes too complicated, it becomes difficult to get any intuition from it. I think this is at the heart of the misunderstanding of the concept in your question.


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