# Fermi Surface squashed by potentials

## Main Question or Discussion Point

Taken from my textbook: My understanding is that:

• One valence electron, 2 spin states -> Half-filled Brillouin zone
• Seeking inspiration from "Nearly Free Electron Model": gaps open up at zone boundaries
• States nearer to zone boundaries get pushed down in energy further

Since a fermi surface is a surface of constant energy in k-space, shouldn't the surfaces nearer to the zone boundaries that get pushed down in energies get repelled even more? It seems that surfaces nearer to the boundaries get closer even! Why are the fermi seas like these below? Related Atomic and Condensed Matter News on Phys.org
DrDu
Take a simple potential with this symmetry as an example: ##V(x,y)=A \cos(x)\cos(y)##. Can you find a direction along which the potential is constant? To which lines do these directions in reciprocal space? What happens at the cell boundary?

Take a simple potential with this symmetry as an example: ##V(x,y)=A \cos(x)\cos(y)##. Can you find a direction along which the potential is constant? To which lines do these directions in reciprocal space? What happens at the cell boundary? The surfaces of constant potential are the circles* about (0,0). Given this is the fermi surface, these surfaces are the reciprocal space. Still doesn't answer my question about "pushed down in energy -> decrease in radius, squashed inwards"?

bumpp

DrDu