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Say I have some Hamiltonian $\mathcal{H}$ that describes a 1-D spin chain (i.e. Heisenberg model) which has a number of compatible symmetries $\{\mathcal{O}^i\}$ (i.e. total spin...etc). For all $\mathcal{O}^i$ we have $[\mathcal{H},\mathcal{O}^i]=0$ and thus for every $\mathcal{O}^i$ there exist a basis $\{o_j^i\}$ which spans both $\mathcal{O}_i$ and $\mathcal{H}$. As it seems standard in the lore, in order to make numerics more efficient, one for instance claims that the magnetisation is zero and proceeds to work with $\mathcal{H}$ in a “basis” (smaller than $\{o_j^i\}$) where this is true.

My questions are:

  1. This basis is not a basis of the whole Hilbert space, which means we can’t fully decompose $\mathcal{H}$ in that basis (no resolution of the id), what is going on here? (In this basis, are the missing matrix elements of $\mathcal{H}$ simply zero?)

  2. Say I have no problems with 1) and that there is another symmetry I can use, i.e. translational symmetry and picking $k=0$. To import both symmetries in my problem, would I have to choose the intersection between the two basis?

Thank you in advance!

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  1. Yes, if you choose some magnetization sector you are by definition projecting over some subspace (one of the many subspaces of the total magnetization). You can however resolve the identity for that subspace (the magnetization sectors are block diagonals of the total problem, i.e., akin to irreducible representations).

  2. You have to guarantee mutual commutativity in order to simultaneously diagonalize two or more matrices.

If you are dealing with specific ranges of temperature, then most likely such an approximation will suffice. Treating dynamics however, is whole another beast, since arbitrary excited states may come into play.

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    $\begingroup$ Oh yes. But, are you familiar with Bethe Ansatz? XXX chain (in fact XYZ generally) is integrable. Could help you gauge the numerics... $\endgroup$
    – daydreamer
    Oct 14, 2020 at 10:47
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    $\begingroup$ 1) actually, I guess you mean that I act with the projector of that sub space on $\mathcal{H}$ where only its matrix elements that lie in that sub space survive, leaving us with actually a smaller matrix, as opposed to a more sparse matrix, right? We can do this as if we did some measurements and knew that we were in that sub system where we would collapse our system into the smaller one just like with state vectors in a superposition after a measurement, right? $\endgroup$ Oct 14, 2020 at 10:49
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    $\begingroup$ Precisely that, my friend. $\endgroup$
    – daydreamer
    Oct 14, 2020 at 10:51
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    $\begingroup$ No problem! Good luck with your hamiltonian (I'm curious to ge to know him lol) $\endgroup$
    – daydreamer
    Oct 14, 2020 at 10:54
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    $\begingroup$ Its a 1d Heisenberg Hamiltonian with an $S^z_iS^z_{i+2}$ term which is responsible of the non integrability of the model. But since my problem was on general lines I decided not to post the whole problem but rather to sketch it :). $\endgroup$ Oct 14, 2020 at 10:57

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