definitionhamiltonianhamiltonian-formalismquantum mechanicssymmetry
In Sakurai's Modern Quantum Mechanics, in Chapter 4, he effectively states that the operation of rotation or translation, represented by a unitary operator $U$, is customarily called a symmetry operator regardless of whether the physical system itself possesses the symmetry corresponding to $U$. It's a symmetry or invariance of the system only when $U^\dagger H U=H$.
Why are symmetries defined with respect to invariance of the Hamiltonian?
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
Best Answer
In classical mechanics, a conserved quantity has vanishing Poisson bracket with the Hamiltonian. Such quantities become "good quantum numbers" in QM: they commute with $H$, so simultaneous eigenstates from a complete basis. The evolution operator $e^{-iHt/\hbar}$ also commutes with good quantum numbers, so their probability distribution is unchanged. (Thus the connection from Noether's theorem of conservation laws to continuous symmetries survives in QM.) For unitary $U$ commuting with $H$, $U^\dagger=U^{-1}$ obtains your equation.