TITLE: Noether's theorem in quantum mechanics
QUESTION [17 upvotes]: In classical mechanics:
If a Lagrangian $\mathcal{L}$ is preserved by an infinitesimal change in the state space variables $q_i \to q_i + \varepsilon K_i(q)$, this leads to only second order change in the Lagrangian:
$$ 0 = \frac{d\mathcal{L}}{d\varepsilon} = \sum_i \left( \frac{\partial \mathcal{L}}{\partial q_i}K_i + \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \dot{K}_i \right) = \frac{d}{dt}\left(\sum_i \frac{\partial \mathcal{L}}{\partial \dot{q}_i} K_i \right). $$
Then we get our conserved momentum because the rate of change on the right side is $0$.
In quantum mechanics, an observable $A$ commuting with the Hamiltonian, i.e. with $[\hat{H},A] = 0$, corresponds to a symmetry of the time-independent Schrödinger equation $\hat{H}\Psi = E \Psi$.  How do we compute the conserved quantity related to $A$?  In particular, what is the conserved quantity associated with the identity operator?

REPLY [4 votes]: (The laws of physics are supposedly invariant under time translations, so I hope the lateness of this answer can be forgiven.)
Both the question and the previous answers seem to have overlooked the fact that, in quantum mechanics, symmetries correspond to unitary operators on Hilbert space while observables (such as conserved charges) correspond to self-adjoint operators.
Supposing the Hilbert space to be finite-dimensional for simplicity, given any unitary matrix $A$, there exists an hermitian matrix $Q$ such that $A = e^{iQ}$, but $[H,A]=0$ (i.e. that $A$ is a symmetry of the hamiltonian) does not necessarily imply that $[H,Q]=0$ (i.e. that $Q$ is a conserved charge). Counterexamples may readily be found by considering $Q = \begin{pmatrix} 0 & -i\pi \\ i\pi & 0 \end{pmatrix}$, for which $A = -\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ commutes with any $H$.
But suppose, as for Noether's theorem in classical mechanics, that we have a smooth family $\theta \mapsto A_\theta$ of symmetry transformations such that $A_0$ is the identity transformation. Then the exponential map furnishes us with a $Q$ such that $A_\theta = e^{i\theta Q}$ and the Baker-Campbell-Hausdorff lemma in the form $[H,A_\theta] = i\theta [H,Q] + O(\theta^2)$ guarantees that $Q$ is conserved.