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In hopping model, we can get the Hamitonian as $H_0=-t\sum a^\dagger_ia_{i'}$. Then we take the fourier transform and put the operator which are in momentum space in the Hamitonian above. However, I found that there will be terms like $blabla~a^\dagger_ka_{k'}$, which is not remained at last when you finally got the energy spectrum like $E_ka^\dagger_ka_k$. So my question is how do these non-diagonal terms vanish.

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  • $\begingroup$ Is that the tight-binding model? If so you're missing the h.c. Your $H_0$ is not Hermitian. $\endgroup$ – Bubble Oct 4 '14 at 15:23
  • $\begingroup$ Yeah, it's the tight-binding model. It seems that if $i$ and $i'$ can run all of the lattice, there is no need to add the hc term, and it is Hermitian. $\endgroup$ – Simon Oct 4 '14 at 15:26
  • $\begingroup$ Aha, so the summation takes care of h.c. and it is Hermitian hence. But its not nearest neighbor if $i$ and $i'$ run over the entire lattice? If so in that case I don't think you can diagonalize the model with Fourier transform, though I'm unsure. $\endgroup$ – Bubble Oct 4 '14 at 15:32
  • $\begingroup$ Well, I think $i$ can run all of the lattice, and keep $i$ and $i'$ to be the nearest ones at the same time. That's what the sum means. $\endgroup$ – Simon Oct 4 '14 at 15:36
  • $\begingroup$ Then you're making a mistake in your calculation. Without seeing the details we can't know what it is. $\endgroup$ – Bubble Oct 4 '14 at 15:37
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When you Fourier transform the tight-binding Hamiltonian, $$c_{j,s}=\sum_{k,s} a_{k,s} e^{i R_j k},$$ with periodic boundary conditions, you will be left with a diagonalized Hamiltonian in the desired form. For details see section 2.3 of these lectures notes, starting on p.18: http://manybody.skku.edu/Lecture%20notes/Solid%20State%20Physics.pdf

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  • $\begingroup$ Yeah, I think I find where I made the mistake. I forget to use the anti-commutation relations with $a^\dagger_k$ and $a_{k'}$, this will help me cancel those crossing terms. Thank you very much. $\endgroup$ – Simon Oct 4 '14 at 15:52
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    $\begingroup$ @Simon.Z If you found Bubble's answer helpful, perhaps you might consider upvoting it. $\endgroup$ – WetSavannaAnimal Nov 5 '14 at 13:29

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