Timeline for Momentum $k$-space Brillouin zone for non-quadratic and interacting systems?
Current License: CC BY-SA 3.0
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| Nov 1, 2016 at 18:27 | comment | added | Aaron | No, band theory requires the quasiparticles to not interact with one another. The quadratic nature of the Hamiltonian is intimately tied with this statement, since given a quadratic, hermitian hamiltonian one can always diagonalize it. In other words, there is a basis such that it can be viewed as a free theory. As you note, by adding higher order interaction terms, generally speaking there is no basis you can transform to such that you can view the quasiparticles as independent of one another. Of course, if you choose interactions that merely modify energy then you can still use band theory. | |
| Nov 1, 2016 at 17:49 | comment | added | user32229 | My question means to ask: whether there is a band theory for interacting and non-quadratic Hamiltonian in the lattice systems? If yes, can you give an example? I doubt, thought the answer is no. | |
| Nov 1, 2016 at 16:40 | comment | added | Aaron | In those cases, the concept of a Brillouin zone is still useful. The point of the Brillouin zone is to give you a reduced set of k-vectors that are non-redundant. In these cases when you have higher degree interactions, you think of different points of k-space interacting with each other. Again, the concept of Brillouin zone is tied to the lattice, not to any physics. However, this concept becomes useful because of Bloch's theorem, which ties lattice periodicity in your Hamiltonian to a form of the wavefunction solution. Perhaps you're confusing the ideas of k-space and Brillouin zone? | |
| Nov 1, 2016 at 16:25 | comment | added | user32229 | well, any interactions like $c_i^\dagger c_j^\dagger c_k c_l$ are even more terms are interacting which cannot be written as the BdG equation. Usually people do mean-field approximation $\Delta_{ij}^\dagger c_k c_l$. More generally, if we have $c_i^\dagger c_j^\dagger \dots c_k c_l \dots$, then it seems impossible to consider the theory in Brillouin zone due to non-BdG form. | |
| Nov 1, 2016 at 15:31 | comment | added | Aaron | What do you mean by non-quadratic? Are you thinking about a non-quadratic kinetic term? Also, tight-binding is definitely interacting. You have a hopping potential between electrons on different atomic sites, so you cannot treat each electron independently; rather you're morally doing a first order approximation to a free picture of independent electrons. Do you mean a strongly interacting case where you cannot expand about some free theory? | |
| Nov 1, 2016 at 15:10 | comment | added | user32229 | although you did not precisely address my question, I vote you up still +1. Please do think more precisely for the interacting cases. | |
| Nov 1, 2016 at 14:55 | comment | added | user32229 | Actually what I meant is non-quadratic and interacting system. Tight binding model is quadratic and non-interacting system in most cases when you consider its band in k space.. | |
| Nov 1, 2016 at 6:52 | history | answered | Aaron | CC BY-SA 3.0 |