How Does the Fermi Surface Form a Perfect Square in a 2D Tight Binding Model?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
Sign up Log in
3 replies · 7K views
aaaa202
Messages
1,144
Reaction score
2
On the attached file the tight binding dispersion for a 2d square lattice is described. It is then assumed that the fermi surface is a square. My question is: How can it ever be a perfect square when the dispersion looks as it does.
Also can someone explain:
Why does the half filled case correspond to:
-1 electron per lattice site (why will 2 electrons not occupy the ground state etc.)
- chemical potential μ=0
 

Attachments

  • fermisurface.png
    fermisurface.png
    13.6 KB · Views: 2,092
Physics news on Phys.org
aaaa202 said:
On the attached file the tight binding dispersion for a 2d square lattice is described. It is then assumed that the fermi surface is a square. My question is: How can it ever be a perfect square when the dispersion looks as it does.
Also can someone explain:
Why does the half filled case correspond to:
-1 electron per lattice site (why will 2 electrons not occupy the ground state etc.)
- chemical potential μ=0

I suppose that the formula (1) on the attachment file supplied by you has already answered your question,i.e., "How can it ever be a perfect square when the dispersion looks as it does."

Not noly tight binding can describe it,but also perturbation theory can support it.
 
You are exactly right that in principle 2 electrons could occupy each lattice site; this would be the "completely filled" case. If there is only 1 electron per lattice site, we call that "half filled."
 
t!m said:
You are exactly right that in principle 2 electrons could occupy each lattice site; this would be the "completely filled" case. If there is only 1 electron per lattice site, we call that "half filled."

Actually,I think the more important problem here is "how to caltulate the fermi level for actual computation",for example,how to make it come true in some DFT softwares ,and how about its effictiveness,and so on.