If $A$ is a classical symmetry, it is possible that after quantization $A$ is no longer a symmetry. One of the ways to see this in the operator formulation of quantum mechanics is the following.
Let $H$ and $A$ be a self-adjoint operators on the Hilbert space $\mathcal{H}$. We consider $H$ as quantum hamiltonian and $A$ as its symmetry: $A$ and $H$ commute, meaning that operator-valued spectral measures commute.
If $\psi \in \mathcal{H}$ is a state, then $$ \frac{d}{dt} \langle A(t)\rangle= \frac{d}{dt}\langle \psi(t), A\psi(t)\rangle =\langle -i H \psi(t), A\psi(t)\rangle+ \langle\psi(t), A (-iH)\psi(t)\rangle\\ =i(\langle H \psi(t), A\psi(t)\rangle - \langle\psi(t), A H\psi(t)\rangle). \quad (\ast) $$ Quantum Hamiltonian $H$ is densely defined on $\mathcal{H}$ with domain $D(H)$ and if $A$ does not preserved $D(H)$ the first term $\langle H \psi(t), A\psi(t)\rangle$ is not equal to $\langle\psi(t), HA\psi(t)\rangle$ and $\frac{d}{dt}\langle A(t)\rangle \neq 0$. In other words even if $H=H^{\dagger}$ on $D(H)$ it is not necessarily so on $A(\mathcal{H})$$A(\mathcal{H})=\text{range}(A)$.
My question is how to derive formula $\ast$ using path integral formulation? Somehow the fact that Feynman measure is not invariant under symmetry $A$ has to play a role in such computation, but I don't see how to do it. If I start with $$ \int \mathcal Dp(s) \mathcal Dq(s) A(p(t),q(t))e^{iS(p(s),q(s))}, $$ with some boundary conditions at $s=0$ and $s=T$ and $0<t<T$, take derivative $\frac{d}{dt}$ I don't see how I can get anything equivalent to $H-H^{\dagger}$ on $A(\mathcal H)$.
