Let us consider a unitary or antiunitary operator $\hat{U}$, that associates with each quantum state $| \psi \rangle$ another state $\hat{U} | \psi \rangle$. I have read that for $\hat{U}$ to be a symmetry transformation it has to keep the Hamiltonian $\hat{H}$ invariant. It means that $\hat{U}^{\dagger} \hat{H} \hat{U} = \hat{H} \Rightarrow [\hat{H},\hat{U}] = 0$. But what does it mean physically?
I believe that a symmetry is a transformation that doesn't change the physics of the system, that is, this doesn't change neither the expection values of the physical observables nor the probabilities, right? So, how is this related to the invariance of the Hamiltonian?
Best Answer
Sometimes this is claimed without much explanation.
The time evolution operator is given by exponentiating the Hamiltonian: $$ U(t) = \exp(-i t\hat H / \hbar ). $$ For concreteness, when we think about a symmetry operation (what you called $U$) let's think about rotations around the $z$-axis. A rotation by $\theta$ degrees is given by $$ R(\theta) = \exp(-i\theta \hat J_z/\hbar) $$ where $\hat J_z$ is the angular momentum operator in the $z$-direction.
If our symmetry commutes with time translations, we have
$$ [U(t), R(\theta)] = 0 \implies U(t) R(\theta) = R(\theta) U(t). $$
This means that, for any $|\psi \rangle$,
$$ U(t) R(\theta) |\psi \rangle = R(\theta) U(t) |\psi \rangle. $$
In other words, if you rotate the state by $\theta$ degrees and then wait $t$ seconds, you will end up with the same state as if you first waited $t$ seconds before rotating $\theta$ degrees.
The "commutativity" of these operations is often what physicists mean when they say they have a symmetry.
By differentiating the equation $[U(t), R(\theta)] = 0$ by $t$, $\theta$, or both, we can see that this statement is actually equivalent to four closely related statements