# Why are we only interested in unitary/anti-unitary transformations of the underlying Hilbert space when considering symmetries in quantum mechanics?

Background to question:

We briefly looked at 'symmetries' in my quantum mechanics course. I was dissatisfied with the fact that we only considered unitary (touched on antiunitary) operators when looking at symmetries of the hamiltonian and other observables. It was mentioned at the start that a theorem due to Wigner stated that only unitary and antiunitary operators preserved the inner product squared: $$|\langle \phi | \psi \rangle|^2$$ for all kets $$|\psi \rangle$$ and $$|\phi\rangle$$ in the Hilbert space. However I did not see why this meant that only (anti-)unitary operators are significant in considering the symmetries of observables. I.e. whether the observable commutes with the transformation operator or no.

I am considering, for instance, a transformation of the Hilbert space which has non-unitary action on the subspaces corresponding to given eigenvalues of some observable $$\hat A$$, but then I think this would commute with $$\hat A$$ still.

I saw this post point out that, if $$\hat U$$ is not unitary/anti, but it was a symmetry of $$\hat A$$, then we would have $$\hat A = \hat U \dagger \hat A \hat U$$ which cannot be true because on the LHS we have a non-Hermitian operator, and on the RHS we have a Hermitian operator. Also, since $$\hat U \dagger \hat U \neq \hat I$$, it is not the commutator as defined that vanishes, but instead $$(\hat U \dagger)^{-1} \hat H - \hat H \hat U = 0$$. The vanishing of this object may stil have some sugnificance however, phsyical or purely mahematical.

So the question: