A valuable math result for quantum mechanics is that if two hermitian operators (physical observables) commute, then a simultaneous basis of eigenvectors exists. Nevertheless, there are cases in which two operators commute without being both hermitians -- a really common one is when one operator is hermitian (such as an hamiltonian) and the other is unitary (say, a spatial translation operator which we'll consider a symmetry of the hamiltonian).(adsbygoogle = window.adsbygoogle || []).push({});

On QM books I didn't find any math results about what happens when an hermitian operator commutes with an unitary one, but this rarely matters. In fact, in the case above, I shall expand the unitary operator on small parameter and then consider the generator of the symmetry, which is hermitian, rather than the unitary operator itself.

But what happens when I couldn't invoke the generator? Say that a quantum system is invariant only under discrete translation (a crystal, for instance). So its hamiltonian won't commute with the momentum, but it'll surely do with the unitary spatial translation operator when the proper parameter is set to crystal period. Usually on physics books there's written "since they commute an orthonormal basis of eigenvectors exists" (see, Ashcroft and Mermin - Solid State Physcis. Bloch Theorem, first demonstration). But which math theorem am I using ? Does a basis of ort. eigenvectors exists even if one operator is hermitian and the other is unitary?

I'm sorry if the question sounds like homework, but I hope it won't.

Ll.

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# Simultaneous diagonalisation of non-hermitian operator (Bloch theorem)

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