Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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

Particularly, I can somewhat appreciate the Fermi energy idea - the radius of Fermi surface which is a sphere. But is there any quantitative use of more complicated Fermi surfaces?

share|improve this question
add comment

4 Answers 4

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. Amazingly, Luttinger and Ward showed that the Fermi surface survives even with interactions to all orders in perturbation theory (Oshikawa later showed this nonperturbatively, see also the arXiv version).

The point of the Fermi surface is that this is where all of the low-lying excitations of the system live -- the Fermi energy is so much larger than room temperature that for room-temperature experiments, all of the thermodynamics is dominated by excitations right at the Fermi surface, and thus knowing its structure is very important.

A more advanced reason for the importance of the Fermi surface is that by knowing its structure (look up "Fermi surface nesting"), we can understand the instabilities of a metal, for example, for sufficiently low temperature, a normal metal settling into a charge density wave state.

share|improve this answer
    
hm, I see. Let me appologize for my hard head and restate the question in a different way - how do we use the anisotropy part of the Fermi surface knowledge? That is, what does the direction in momentum space tell us? Maybe I am missing something rather simple here... –  Michal Feb 23 '11 at 18:10
1  
I suppose I'm not sure what you're asking... the Fermi surface is anisotropic because the underlying material is anisotropic (i.e., the electrons are in a periodic lattice potential). It's not that there's an anisotropic "part"... the idea of a Fermi "sphere" only holds for electrons in free space. –  wsc Feb 23 '11 at 18:47
    
yes, Fermi "sphere" holds for free electrons. I agree completely. In a crystal structure - which is prety much by definition anisotropic - the Fermi surface is also anisotropic. So, what I am asking is what is the relation between these two - anisotropy of Fermi surface and anisotropy of underlying material - since one is in momentum space and another in real space. I am looking for a "visualization" something like this " if I see Fermi surface has long peaks in this direction, what does it tell me about the electrons velocities in real space?" –  Michal Feb 23 '11 at 18:57
    
Ah, the electron velocity is related to its $k$-space state through $\hbar\vec{v}=-\vec{\nabla}_{\vec{k}}E(\vec{k})$, so if you are interested in electron velocities at the Fermi surface, you need to find out how the energy varies in the Brillouin zone, and simply take the derivatives at $\vec{k}=\vec{k_F}$. –  wsc Feb 23 '11 at 19:09
    
Thank you very much for your patience and clear explanations. I dare to ask one more question - the example with velocity is clear to me since velocity has direction - so one can talk about relationship of this direction to direction of something (e.g. gradient) on Fermi surface. Is there anything similar? For example, if the Fermi surface touches Brillouin zone - that place is definitely special. In cubic structure the Fermi surface seems "cut" at those places...what is that? –  Michal Feb 23 '11 at 19:39
show 1 more comment

I think what you might be interested in are Van Hove singularities or the critical points of the Fermi surface, where the density of states as given by $dN/dK_{|k=k_f|}$ diverges. Now $dN/dk$ is proportional to the inverse of the gradient of the energy $ dN/dk \propto 1/\nabla E $. The locations with the greatest d.o.s. on the Fermi surface will exhibit singularities in various absorption and emission spectra. These are precisely those locations where the Fermi surface fails to be a smooth, differentiable surface.

Critical surfaces can be 0D (Fermi point), 1D (line), 2D (Fermi surface). Complicated substances (such as the High $T_c$ superconductors for e.g. which are composed of layers of cuprates) will in general have complicated Fermi surfaces. Because critical surfaces are topological entities they are robust with respect to small perturbations of the microscopic Hamiltonian of the system. In other words the critical surfaces determine the universality class to which the given surface belongs. The universality class determines whether a given material is a superconductor, ferromagnet, Mott insulator etc. Clearly to take a material from one universality class to another requires that one cross a phase boundary. Such a change also requires that the Fermi surface topology undergo a change. Consequently changes in Fermi surface topology can be understood as signs of phase transitions.

A Fermi surface is also important for reasons other than its topology, which describes the global characteristics of the material. Others will likely mention some of the complementary local aspects of the Fermi surface.

share|improve this answer
    
aha, I see. This is quite interesting, thank you. The basic motivation of my original question goes like this: many people talk about Fermi surfaces, Ok, I sit down and calculate one. Than what? After I see its complicated shape, how do I judge it? What you write is one example - if the F. surface has any discontinuity, I can immediately see there will be some interesting properties of the material. I can judge it in that sense. Is there anything similar? –  Michal Feb 23 '11 at 19:46
1  
You have to be careful with this answer: what Deepak is talking about is the density of states at the Fermi surface. As I said above, the thermodynamics is dominated by the Fermi surface, and so for example the specific heat scales as $C_V \propto g(E_F)$ (DOS at the Fermi surface). Obviously then, v.H. singularities signal that something is going awry. This doesn't change where the Fermi surface is, and in particular (with the exception of some controversial theoretical arguments) we never see "discontinuous" Fermi surfaces. –  wsc Feb 23 '11 at 21:31
    
@wsc to the contrary, we see observe such changes in 2D materials such as a graphene. At $0$T its dispersion relation takes the form of a cone in $(k_x,k_y,E(k))$ space symmetric around the $(k_x,k_y)$ plane. The Fermi level is $E_F = 0 $, the ground state is not gapped. If you turn on a parameter $t'$ which controls the next-to-nearest neighbor coupling in the Hamiltonian, the tips of the cones become blunt with extrema at $E_f = e_k,-e_k$, i.e. a gap develops between the "upper" and "lower" bands of the dispersion. This is an example where a Fermi point appears in the limit that $t'=0$ –  user346 Feb 23 '11 at 21:52
    
Probably the single best reference for the physics of Graphene in general is the RMP article from two years ago. It already has 996 citations. In two years. Whew! –  user346 Feb 23 '11 at 21:53
    
Of course, but what you're describing is quite fine-tuned. I'm not arguing that Fermi surface topology isn't indicative of states of matter, or that (Dirac) Fermi points don't exist (though I won't be coaxed to argue about "Fermi arcs")... But if you gate the graphene you have two continuous circles of zero-energy excitations around $K$ and $K'$. My point was simply that you can't just measure the Fermi surface, stare at it, and say useful things. –  wsc Feb 23 '11 at 22:19
add comment

One more thing about geometrical properties of the Fermi surface. Its structure defines material transport properties like conductivity. It is actually equal to the integral of the mean free path along those wave vectors that define a Fermi surface.

Knowing this is very important. How do you sample the Fermi surface of a given metal? By means of the de Haas-van Alphen effect.

share|improve this answer
add comment

A Fermi surface is one of constant energy in a k-space

share|improve this answer
    
While this statement is correct, you are encouraged to go into more detail. –  dj_mummy Oct 4 '13 at 10:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.