In the middle of reading a book chapter, I was somehow drawn to think about symmetries in Quantum Mechanics. It was rather uncalled for, but quite tantalizing. It was all triggered by this short passage: A Hamiltonian $H=H_1 + H_2$ is given; if a unitary tranformation $U$ is applied to it one gets $UHU^{\dagger} = - H_1 + H_2$. "So," the author concluded, "the sign of $H_1$ is not essential."

I was puzzled: $U$ was not a symmetry of the system, that is it did not leave the Hamiltonian unchanged. In fact, it changed the sign of the first term. Yet, still...?! I started thinking about a symmetry operator, and remmbered vaguely that at some point when I was learning the subject in school, a friend of mine and I had given ourselves this line of argument: "A symmetry operator should leave the observables of the system unchanged, including, say, its energy spectrum, so it should commute with the energy operator, i.e. Hamiltonian." Now, it's obvious that this is a faulty logic. First of all, any unitary transformation leaves the spectrum of observables unchanged: $UAU^{\dagger}$ and $A$ have the same set of eigenvalues. Secondly, if that was to be true a symmetry operator had to commute with all observables of the system, rendering it hardly different from unity! After a few minutes of puzzlement the correct picture dawned at me. In finding the right statement, I also looked at Weinberg's first volume on QFT, where in chapter 2 he gives a concise but quite elegant overview of (reletavistic) quantum mechanics.

1. Weinberg defines a symmetry transformation, as one that leaves the transition probabilities unchanged:
   
   
   (1)$\mathcal{P}(\psi_1 \to \psi_2) = \mathcal{P}(T \psi_1 \to T \psi_2)$,
   
   
   where $T$ is the transformation and the $\psi_{1,2}$ are two arbitrary states of the system. From this condition, it is possible to prove a symmetry representaion theorem, due to Wigner, that states $T$ could be represented on the Hilbert space by either a unitary and linear, or anti-unitary and anti-linear operator $U$: $U^{-1} = U^{\dagger}$.


2. From Eq. (1) alone it is not possible to draw any conclusion on the properties of $U$ as regards the equations of motion, or the Hamiltonian. To the first order in time the transition probabilty amplitude is given by $\langle\psi_2|\psi_1\rangle - \frac{i}{\hbar}\langle\psi_2|H|\psi_1\rangle$ where $H$ is the Hamiltonian. Now, if $U$ is any unitary transformation this amplitude does not change if we replace $\psi_{1,2}$ with $U \psi_{1,2}$ and $H$ with $UHU^{\dagger}$. This is nothing but the statement that the choice of the basis does not matter for the physics: If the $\psi_{1,2}$ are given by the columns $[\left(\psi_{1,2}\right)_i]$ given$U$ and $H$ are given by matrices $[U_{ij}]$ and $[H_{ij}]$ respectively in a basis $\hat{e}_i$, then in the basis $\hat{e}_i^\prime = \sum_j U_{ij}^\dagger\hat{e}_j$ they are given by $[\left(U\psi_{1,2}\right)_i]$ and $[\left(UHU^{\dagger}\right)_{ij}]$. There is no change of observer in this.


3. If, on the other hand, we change to a new observer, and demand that the equations of motion, or equivalently the Hamiltonian matrix, retain their original form under this change, we have established a symmetry of the system. It is in this definitive sense that symmetries are represented by (anti-)unitary transformations that commute with the Hamiltonian.