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Just want to make sure I am setting this up correct.

I am trying to find the eigen energies for a tight binding Hamiltonian, let's say with nearest neighbor hopping term $t$. We can solve for the energies with a FT, which gives us (in a 2D cubic system): $$ \epsilon(\vec{k})=-2t(cos(k_x)+cos(k_y)) $$ I set the lattice constant to 1 and the k-vectors live in the Brillouin zone. I want to check this with a computer program I am making. I use the lattice model: $$ \hat{H}=-t\sum_{\langle r,r'\rangle}\hat{c}^\dagger_{r}\hat{c}_{r'} $$ The r's above just refer to the lattice sites, so I label an N by N lattice by calling any given site $j+N(i-1)$ where j is row and i column, so the top left corner is 1, for instance. I then try and rewrite the Hamiltonian in the form: $$ \hat{H}=\vec{c}^\dagger H \vec{c} $$

Here $\vec{c}=\begin{pmatrix}\hat{c}_1\\ \hat{c}_2\\ .\\.\\.\\\hat{c}_{N^2} \end{pmatrix}$ etc. We build $H$ in such a way that we get the sum above after doing our matrix multiplication. For a $3x3$ lattice it would be ($t=1$):

 0    -1    -1    -1     0     0    -1     0     0
-1     0    -1     0    -1     0     0    -1     0
-1    -1     0     0     0    -1     0     0    -1
-1     0     0     0    -1    -1    -1     0     0
 0    -1     0    -1     0    -1     0    -1     0
 0     0    -1    -1    -1     0     0     0    -1
-1     0     0    -1     0     0     0    -1    -1
 0    -1     0     0    -1     0    -1     0    -1
 0     0    -1     0     0    -1    -1    -1     0

The first row has an entry in the second column so that we have a transition from lattice site 2 to 1 and so on. I also put in PBCs.

Question: If I diagonalize this H using this method, should I get out ALL the energies of the analytic expression (restricting k to Brillouin zone)? I think this should be yes, but it seems I do not: certain energies in the analytic solution do not show up.

Maybe something is wrong with my logic, so I want to see if someone can tell me if I am going the wrong direction here, or if this should be right. I need to adapt this to a unconventional superconductor so this needs to be right!!

Thanks for any help!

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  • $\begingroup$ I think you are doing it correctly. I don't understand why do you claim not to get the correct result for the Hamiltonian matrix you gave here. When plugging it into my computer I got the correct energy spectrum, corresponding to $-2[\cos(k_x)+\cos(k_y)]$ when $k_{x,y} = 2\pi n_{x,y}/3$ for $n_{x,y} = 0,1,2$ (as these are the available momenta for this Hamiltonian). Did you get a different result? $\endgroup$ – yu-v Dec 2 '19 at 14:06

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