Confusion about conserved current in quantum field theory In classical mechanics, it is known from Noether's theorem every continuous symmetry gives a conserved current
\begin{equation}
\partial_{\mu}J^{\mu}=0,
\end{equation}
where $J^{\mu}$ (generally) can be written in terms of fields and their conjugate momentum. The conserved charge $Q$ is
\begin{equation}
Q=\int d^3x J^0
\end{equation}
In quantum case, our fields are promoted to operators. So $J^{\mu} \rightarrow \hat{J^{\mu}}$, $Q\rightarrow \hat{Q}$. Conservation of $\hat{Q}$ means $d\hat{Q}/dt=-i[\hat{Q},\hat{H}]=0$, where $\hat{H}$ is the Hamiltonian.
My question is: How to write $\partial_{\mu}J^{\mu}=0$ in quantum case? Do we have $\partial_{\mu}\hat{J^{\mu}}=0$ or $\langle\alpha|\partial_{\mu}\hat{J^{\mu}}|\alpha\rangle=0$ for any ket $|\alpha\rangle$?
 A: The quantum version of the Noether theorem is Ward-Takanashi identity - https://en.wikipedia.org/wiki/Ward%E2%80%93Takahashi_identity.
It gives relationships between correlation functions in the QFT :
$$
\langle \partial^{\mu} j_{\mu} (x) \mathcal{O}_1 (x_1) \ldots 
\mathcal{O}_N (x_N) \rangle = 
\sum_{n = 1}^{N} \delta(x - x_n)\langle \mathcal{O}_1 (x_1) \ldots \delta \mathcal{O}_n (x_N) \ldots 
\mathcal{O}_N (x_N) \rangle
$$
Where $\delta$ denotes the variation, induced by the conserved charge $Q$ :
$$
\delta \mathcal{O}_n (x_N) = i [Q, \mathcal{O}_n]
$$
This identity can be derived rather shortly via the path integral formalism.
A: We should have the operator equation $\partial_\mu \hat J^\mu=0$. This implies that $\langle a|\partial_\mu \hat J^\mu|b\rangle=0$, $\forall a,b$.
A: Other answers are all fine, but they don't give you the commutator formula you seem to be seeking.
Just like
$$
\frac{d}{dt} Q = - i \hbar [Q, H],
$$
the generators of the translation group (4-momentum) give you space-time derivatives in QFT:
$$
\partial_{\mu} J^{\mu} = - i \hbar [J^{\mu}, P_{\mu}],
$$
with $P_{\mu} = (E, -\vec{P})$.
So your quantum Noether equation is an operator equation
$$
\sum_{\mu = 0}^4 [J^{\mu}(x), P_{\mu}] = 0.
$$
