Tight binding simulation issues

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!

[Vladimir Sovkov]

Anyway, the total number of the computed eigenenergies must be exactly equal to the size of the Hamiltonian matrix, i.e., N if the size of the matrix is NxN, no more no less. I do not clearly understand the physics behind your problem, but in many physical problems imposing symmetry conditions help to reduce the degeneracy, such as a definite state parity, Pauli principle, etc.

[Marcus Rosales]

I'm not confussed why you get 4N*N egien values since that's just basic linear algebra.

The analytic solution has half the number of a given energy (degeneracy) which I believe is due to a particle hole symmetry: in superconductors the excitation spectrum is built out of quasi particles which are linear combinations of particles and holes. So I believe you are correct in diagonosing the degeneracy; thanks for the sanity check.

The fact that there are energies in one spectrum not found in the other is what really worries me. Something seems to be semi correct since I am getting the right answers, but the extra ones are stumping me.

I asked on physics stack exchange which no one could answer so I thought I'd try here, but this is maybe too much on the physics side of things. Thanks for the repy!

[Vladimir Sovkov]

From elementary Physics courses, I have always thought that the superconductivity was produced by the Cooper pairs, i.e., couples of electrons exchanging phonons, but I have never been an expert in the subject. The idea that the same effect can be achieved by an electron-hole pair (can we call it "exciton"?) is new for me. It is interesting.

Is it possible that your extra states all possess a wrong symmetry, so that imposing the symmetry conditions would cancell them totally?

I think that there is nothing else I can advise.

[Marcus Rosales]

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.