# Tight binding simulation issues

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Marcus Rosales on 2 Dec 2019
Commented: Marcus Rosales on 3 Dec 2019
Hello,
This may be too far on the physics side of things, but I figured I may try and ask anyways. My main issue comes from the eigenvalues I am recieving from the code when I diagonalize a matrix.
I am trying to diagonalize a 2D NxN square lattice Hamiltonian which contains a uniform d-wave super conducting order parameter and a nearest neighbor hoping term. We can solve this model analtically with a fourier transform due to translational invariance, so it gives me an opertunity to check my answer. This next part is not too important, but here is the explicit model: Here the delta is the d-wave superconductivity term which connects nearest neighbors on the lattice: it can be treated very similar to the first term. Basically we just need to throw in a negative sign for terms which are above a given lattice point (can go into more detail if needed). Doing the math we can find the following dispersion relation: Now for the simulation. We can rewrite the above summations in the following form: Where ; H is a matrix choosen to make the two definitions consitent. Finding the eigen energies here amounts to diagonalizing H.
My question is I get all of the energies I see with my analytic solution, but also quite a few extra I do not. Also, it seems like my degeneracy is way too high!
I am unsure why this could be the case. I am certain my Hamiltonian looks right. I can provide anyone who wants it with my code. Hopefully this is not too vague, and I can expand if needed.
Thanks!
Marcus Rosales on 3 Dec 2019
I'm thinking they are not idependent, so I can cut the degeneracy in half (for the energies which are actually in the analytic solution. idk about the others; seeing a grad student tomorrow to talk about it).
The excited states are a linear combinations of creation and annihilation operators, which we call quasi particles, so the holes and the electrons do behave as a single particle. Not going to say it is an exciton because we usually atribute these to a bound state of some sort (e.g. from the Coulomb interaction). Although, it might apply: I am only familiar with the term from a poster presentation given by an Engineering student years ago.
I think of it as we have a particle hole symmetric Hamiltonian, so any excitation should not favor one "particle" over the other; hence, we should expect some linear combination, intuitively. The specfic linear combinations are determined by a bogoliubov transformation, which is just a way of diagonalizing a matrix with operators.