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Usually, we define the momentum $k$-space Brillouin zone (by Fourier transformed from the real space $x$ with a wavefunction $\psi(x)$ to the momentum $k$-space) for:

(1) quadratic non-interacting (free) systems (such as those can be written in terms of BdG equation.)

and

(2) translational invariant systems (so one can define the conjugate momentum $k$ as a good quantum number).

Question: Could we define the momentum $k$-space Brillouin zone for

non-quadratic and interacting systems

but translational invariant systems? (Namely can we modify (1) to interacting, but keep (2)?)

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The Brillouin zone is typically defined as the Wigner-Seitz cell of the reciprocal lattice. It's definition does not have any physics content to it; you give me a direct lattice, and I transform it to its reciprocal lattice and I can tell you the Brillouin zone. So, the answer to your question technically is yes, in a trivial sense.

As a side comment, the inverse of quadratic non-interacting isn't non-quadratic interacting. Typically free systems are by default quadratic, since kinetic energy is quadratic, so I think what you meant are non-interacting systems in general. In this case, there are plenty. The tight-binding model for crystals in standard band structure theory, for example, is an example of interacting systems where the concept of Brillouin zone is useful.

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  • $\begingroup$ 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.. $\endgroup$
    – user32229
    Commented Nov 1, 2016 at 14:55
  • $\begingroup$ although you did not precisely address my question, I vote you up still +1. Please do think more precisely for the interacting cases. $\endgroup$
    – user32229
    Commented Nov 1, 2016 at 15:10
  • $\begingroup$ 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? $\endgroup$
    – Aaron
    Commented Nov 1, 2016 at 15:31
  • $\begingroup$ 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. $\endgroup$
    – user32229
    Commented Nov 1, 2016 at 16:25
  • $\begingroup$ 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? $\endgroup$
    – Aaron
    Commented Nov 1, 2016 at 16:40

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