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:

Can someone please clarify whether

  • Non-unitary transfomrtaions of the nderlying hilbert space can be symmetries of an observable. Why or why not?
  • What the significance- physical or not- of such symmetries would be.
  • $\begingroup$ I'd suggest you to start by taking a look of section 2.2 in Weinberg's The Quantum Theory of Fields, Vol 1. It points to the relevance and motivation behind Wigner's result and other interesting facts. The short answer is that symmetry transformations must be unitary or anti-unitary. $\endgroup$ – secavara Dec 2 '18 at 14:27
  • $\begingroup$ AFAIK unitarity is in general necessary in the quantum regime because it is connected with the probability which has to integrate to 1. $\endgroup$ – anna v Dec 2 '18 at 17:36
  • $\begingroup$ I don't follow. Wigner's theorem implies that only unitary evolution is allowed as only it will preserve probability. But a unitary evolution is generated by an Hermitian operator, which we call the Hamiltonian. What is symmetric under this evolution is hence by definition the Hermitian (not unitary) operations that commute with this Hamiltonian, as they share the same (constant) eigenstates. $\endgroup$ – PhysicsTeacher Dec 2 '18 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.