# 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 $$$$\partial_{\mu}J^{\mu}=0,$$$$ where $$J^{\mu}$$ (generally) can be written in terms of fields and their conjugate momentum. The conserved charge $$Q$$ is $$$$Q=\int d^3x J^0$$$$ In quantum case, our fields are promoted to operators. So $$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 $$|\alpha\rangle$$?

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.

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

• 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. Commented Sep 14, 2020 at 9:14
• 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. Commented Sep 14, 2020 at 14:33
• In words, he's saying that the vanishing of the diagonal in every basis implies the operator is zero. Commented Sep 14, 2020 at 16:59
• I agree with @Iván Mauricio Burbano. My original statement was wrong. I was thinking only of a single basis. Commented Sep 14, 2020 at 17:47
• @mikestone I would edit the answer to reflect that Commented Sep 14, 2020 at 22:18

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 $$\sum_{\mu = 0}^4 [J^{\mu}(x), P_{\mu}] = 0.$$