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I can understand that band diagram is a diagram that relates the Energy and Momentum, E-k relationship.

But I still have vague understanding about the Fermi Surface. I cannot see the difference between Fermi Surface and Band Diagram.

  1. can I consider Fermi Surface as a Band Diagram drawn in 3D?

  2. For one element, we can draw different kind of Fermi Surface because it depends on which energy level we want to draw from its "origin" point i.e. $\Gamma$ point. Is it correct?

Cu Band Diagram

enter image description here

Silicon Band Diagram.

enter image description here

Codelearner777
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  • The band diagram is a 3D entity (the standard presentation is cuts along high-symmetry directions). The Fermi surface is a 3D surface (potentially unconnected) representing the highest filled states in the band structure. This surface is where conduction takes place. – Jon Custer Mar 04 '19 at 16:38
  • Hi, I added a picture in my post. which energy level we see in the Fermi Surface, the blue line (= Fermi level) or the yellow line (the highest EV level = the highest filled state) or my picture is not relevant? – Codelearner777 Mar 04 '19 at 16:53
  • Interesting, but perhaps philosophical question. The Fermi surface is usually presented for metals, for things like understanding de Haas-van Alphen effects, where the shape and connectivity of the surface determine the oscillation frequency. I’ve not seen it used often for semiconductors. – Jon Custer Mar 04 '19 at 16:59
  • "The Fermi surface is usually presented for metals" aah..ok. I updated my post again. – Codelearner777 Mar 04 '19 at 18:36
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    The Fermi surface in the top plot would be a contour of the highest filled states in the metal, thus the yellow curve (in its full 3D glory). Again, it may not be a simply connected surface (for example, iron has a highly disconnected Fermi surface with bits and blobs here and there, at least partly because of the different spin states - this is a major reason that the conductivity of Fe is low compared with, say, Cu). – Jon Custer Mar 04 '19 at 18:49
  • Oh, and the added difficulty that the Fermi surface is taken as real space, not $k$ space.. – Jon Custer Mar 04 '19 at 18:50
  • from wiki, it is drawn in k space.

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

     – Codelearner777 Mar 04 '19 at 18:51
  • Fair enough - I blame not enough coffee. But, it is still the surface of the last-filled states. – Jon Custer Mar 04 '19 at 18:55
  • from this dicussion https://physics.stackexchange.com/questions/286301/what-is-the-fermi-surface "The Fermi surface is defined to be the set of points k such that En(k)=μ for any band n" If Fermi surface refers to the yellow line, then it should be like min(highest filled state,Fermi Level) and En(k)=μ is now always fulfilled.

    I think I get the general idea now, but most of the time I will doubt my self again after that, if I am not precise from the first time. lol.

     – Codelearner777 Mar 04 '19 at 19:00
  • hi, would you like to write your comment in the answer section? – Codelearner777 Mar 06 '19 at 00:11
  • I’ll try to find time tomorrow once I’m back in my office. – Jon Custer Mar 06 '19 at 00:17

1 Answers1

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My usual go-to book for this kind of a thing is Ashcroft & Mermin's excellent 'Solid State Physics' text, at least as a starting point.

So, in Chapter 8 where they introduce the Bloch formalism, one finds a section on the Fermi surface. So,

The ground state of $N$ free electrons is constructed by occupying all one-electron levels $k$ with energies $\epsilon (k) = {\hbar^{2}k^{2} \over 2m}$ less than $\epsilon_{F}$, where $\epsilon_{F}$ is determined by requiring the total number of one-electron leves with energies less than $\epsilon_{F}$ to be equal to the total number of electrons.

Fine - for free electrons, you get a sphere of states up to the Fermi level. Next,

The ground state of $N$ Bloch electrons is similarly constructed, except that the one-electron levels are now labeled by the quantum numbers $n$ and $k$, $\epsilon_{n}(k)$ does not have the simple explicit free electron form, and $k$ must be confined to a single primitive cell of the reciprocal lattice if each level is to be counted only once. When the lowest of these levels are filled by a specified number of electrons, two quite distinct types of configuration can result...

OK, the top surface could be weird, may not be free-electron like. So far so good. Well, the first configuration described is for semiconductors and insulators, where there are full bands and empty bands. Note quite what we want here, so on to type number 2:

A number of bands may be partially filled. When this occurs, the energy of the highest occupied level, the Fermi energy $\epsilon_{F}$, lies within the energy range of one of more bands. 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, as 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 branches of the Fermi surface.

Now, if one looks at An Experimental Determination of the Fermi Surface in Copper, AB Pippard, Phil Tran A, Volume 250, Issue 979, pp 325-357 (1957) one sees a pretty spherical Fermi surface (the black areas are contact areas, these are easily mapped out with various experimental techniques):

Fermi surface of copper

In contrast, a bcc metal such as molybdenum or tungsten might look more like this from R. Herrmann:

enter image description here

Iron is even worse, with disconnected pockets of electrons.

So, in summary, the Fermi surface is the set of highest occupied electron states in $k$ space. By definition, these states have energies at or below the Fermi energy, but the surface does not have to be, and often is not, at the Fermi energy. The behavior of electrons at the Fermi surface determines the electrical conductivity behavior of metals.

Jon Custer
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