1
$\begingroup$

For the Free electron model, we can easily describe the Fermi surface at T=0 due to the uniqueness of the Fermi-Dirac Distribution at T=0; below the Fermi-level, a state is definitely filled, above, the state is not filled. So due to the this, the Energy contours are just regular spheres right?:

$k_{F}$ fixed depending on how many electrons are present to fill the states. Hence: $E = \frac{\hbar^2}{2m}k_{F}^2$, which is a sphere in reciprocal space.

enter image description here

However, when the temperature is greater than T=0, then what happens? The states are now not in a filled or not filled position but follow instead a probability distribution: $n = \large \frac{1}{1 + e^{(E-E_{f})/kT} }$. So now what happens to the Fermi-sphere?

Could it be that now the temperature is non-zero, Bragg diffraction has an effect on the sphere at the edges of the Brillouin Zone where Bragg Diffraction is most prominent. If this were the case, in what way does this change the Fermi sphere anyway?

$\endgroup$
  • $\begingroup$ Your question first mentions free electron model and the Fermi sphere, and later Brillouin Zone and Bragg diffraction. These are mutually exclusive concepts. If you have one, you can't have the other. Non-zero temperature causes the Fermi surface gets fuzzy; it's not an abrupt, discrete surface. If you allow for a lattice and have Bragg diffraction, then you get Brillouin Zones, and the Fermi surface is no longer spherical. $\endgroup$ – garyp Jun 25 '16 at 20:42
1
$\begingroup$

The Fermi surface is a $T = 0$ property. But as the Fermi energies in metals are on the order of a few electron volts (that is a few $10^4\,\mathrm{K}$), it is a good concept for approximately discussing matters at $T > 0$ (where the sharp surface will not be present, but there will only be excitations states close to the Fermi surface).

Furthermore for interacting electrons there will not even at $T = 0$ be a sharp Fermi surface in the sense, that the occupation drops to zero above the Fermi energy. Rather the quasi-particle density will be smeared out around the Fermi surface (which survives as a jump in the occupation density).

The sphere is just an approximation to the form of the Fermi surface. To handle this correctly you have to consider the dispersion relation of the electrons in the lattice, which is determined by the lattice periodic potential, and will certainly be anisotropic. So the Fermi surface will be a deformed sphere anyway. Often the Fermi surface in real metals even intersects the Brillouion zone boundary! (While I heard an explanation for some qualitative features in terms of Bragg diffraction, I don't think it is a very fruitful concept to understand this.)

$\endgroup$
  • $\begingroup$ Would it be possible to explain these qualitative features in terms of Bragg diffraction as I'm not looking for a rigorous understanding, more of just a qualitative one. (More specifically, my lecturer keeps explaining things (such as dispersion relation cut-off's in weak binding model) in terms of Bragg diffraction so I feel this is the explanation I ought to know for now at least.) $\endgroup$ – SomePhysicsStudent May 31 '15 at 20:18
  • $\begingroup$ Someone certainly can, I can't because I've thrown it out of my head at some point (and do not have the time to recover it), sorry. $\endgroup$ – Sebastian Riese May 31 '15 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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