1
$\begingroup$

I know the fermi level is the highest energy level in an atom for its electrons and the fermi surface is (in reciprocal space) a sphere of radius fermi level, if that makes sense. So when two different fermi surfaces meet do they create a band structure and because of the difference is this where band gaps arise? Thanks in advance

$\endgroup$
  • $\begingroup$ Fermi levels apply to solids, not individual atoms. It's also not the highest level but it's the energy of a hypothetical state that has a 50% chance of being occupied at a certain temperature. At 0K this would be the energy level of the first empty state in the solid's band structure. At higher temperatures it's a thermodynamic average. $\endgroup$ – CuriousOne Nov 1 '15 at 15:28
  • $\begingroup$ Interesting, So what happens when the two states meet? $\endgroup$ – AlwaysStuck Nov 1 '15 at 15:35
  • $\begingroup$ When two states "meet" they form a new set of states, but again, Fermi levels are not states, they are thermodynamic energy levels that depend on the temperature. Fermi levels are usually used to define chemical potentials for solids, i.e. one can calculate thermo-voltages and pn-junction bias-voltages with them. $\endgroup$ – CuriousOne Nov 1 '15 at 15:43
2
$\begingroup$

Fermi surface is not necessarily a sphere. It can have an arbitrary shape in the reciprocal space (momentum space) that respects the symmetry of the crystal. Because crystals are periodic arrangements of atoms in direct space, in the reciprocal space it implies that the Fermi surface must repeat itself periodically in every direction. Every such "period" is called a Brillouin zone. So the simplest example would be a sphere that repeats itself (this is the case for Na, K or Rb). If the diameter of the sphere becomes large enough (comparable to the Brillouin zone size), neighboring spheres would merge forming "necks" (e.g. in Cs or Ca). You can see this very nicely illustrated here: http://www.phys.ufl.edu/fermisurface/periodic_table.html

$\endgroup$

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.