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

[aaaa202]

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

 

[Douasing]

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.

 

[t!m]

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."

 

[Douasing]

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.