Is the wavefunction constant in the tight-binding model?

[Niles]

Hi

I've been reading about the tight-binding model, and I have some questions. Let's say we have the Hamiltonian H for our lattice, and it satisfies

Hψ = Eψ,

where ψ is a vector containing the wavefunction for each atom in the lattice. When I solve the above equation e.g. numerically, I get the eigenvectors ψ. In my case the ψ's just contain numbers, but does this mean that the waverfuncion for each atom is constant?

 

[kanato]

No, it doesn't mean the wavefunction is a constant.
Say you have two atoms in the unit cell, and one orbital for each atom. Then your Hamiltonian is 2x2, and your wavefunction is a two component vector. The values of these components indicate the relative contribution of each atom. If $$\phi_i(r)$$ are your basis orbitals, then your wavefunction is

$$u_k(r) = c_1 \phi_1(r) + c_2 \phi_2(r)$$

Note that this is the solution for the periodic part of the Bloch wavefunction, so there is a long range exp(ik*r) phase factor.

 

[Niles]

Ok, so the two eigenvectors I get have the above form. Does each eigenvector (and corresponding eigenvalue) corresponding to a single particle state? I.e., eigenvector #1 is for the single particle state at atom #1 and eigenvector #2 is for the single particle state at atom #2?

I really appreciate this. Thanks.

 

[kanato]

Yes they refer to single particle states. But the second part is wrong.

If you have two eigenvectors, they will be of form (a, b) and (-b, a), as required by orthogonality. The first state (a,b) will be a single particle state that has a contribution a from atom 1 and a contribution b from atom 2. You will not get "eigenvector #1 is for the single particle state at atom #1 and eigenvector #2 is for the single particle state at atom #2?" unless b = 0, which will only happen if your Hamiltonian is diagonal. Both eigenvectors will have contributions on both atoms for realistic tight binding models.

 

[Niles]

Thank you. That clarified it for me.