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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 $|\alpha\rangle$?

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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.

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We should have the operator equation $\partial_\mu \hat J^\mu=0$. This implies that $\langle a|\partial_\mu \hat J^\mu|b\rangle=0$, $\forall a,b$.

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    $\begingroup$ The condition $\langle a|\partial_\mu \hat{J}^\mu|b\rangle=0,\quad\forall a,b$ is not stronger that $\langle a|\partial_\mu\hat{J}^\mu|a\rangle=0,\quad\forall a$. They are equivalent. Indeed, if you assume the second one is true, then playing around with both $0=\langle a+b|\partial_\mu\hat{J}^\mu|a+b\rangle$ and $0=\langle a-b|\partial_\mu\hat{J}^\mu|a-b\rangle$ one can show the first. $\endgroup$
    – Ivan Burbano
    Commented Sep 14, 2020 at 9:14
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    $\begingroup$ No, I am not. The statement is $0=\langle a|\partial_\mu\hat{J}^\mu|a\rangle$ for all kets $|a\rangle$ in the Hilbert space is equivalent to $0=\langle a|\partial_\mu\hat{J}^\mu|b\rangle$ for all kets $|a\rangle,|b\rangle$ in the Hilbert space. What you are saying would be: $0=\langle a|\partial_\mu\hat{J}^\mu|a\rangle$ for all kets $|a\rangle$ in a basis of the Hilbert space is equivalent to $0=\langle a|\partial_\mu\hat{J}^\mu|b\rangle$ for all kets $|a\rangle,|b\rangle$ in the Hilbert space. $\endgroup$
    – Ivan Burbano
    Commented Sep 14, 2020 at 14:33
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    $\begingroup$ In words, he's saying that the vanishing of the diagonal in every basis implies the operator is zero. $\endgroup$
    – user1504
    Commented Sep 14, 2020 at 16:59
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    $\begingroup$ I agree with @Iván Mauricio Burbano. My original statement was wrong. I was thinking only of a single basis. $\endgroup$
    – mike stone
    Commented Sep 14, 2020 at 17:47
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    $\begingroup$ @mikestone I would edit the answer to reflect that $\endgroup$
    – Prof. Legolasov
    Commented Sep 14, 2020 at 22:18
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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 $$ \sum_{\mu = 0}^4 [J^{\mu}(x), P_{\mu}] = 0. $$

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