# Reducing the Hamiltonian size by using a symmetry to speed computations

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?

Physics Asked by FriendlyLagrangian on December 27, 2020

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.

Correct answer by daydreamer on December 27, 2020

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