Confusion regarding time ordering in the Ward Identity derivation

Question

This question does not follow from reading any text, but I was watching Shiraz Minwalla's CFT lectures on YouTube. At 51:36 into the lecture, he raises the question that, as an operator statement, we know $$\partial_\mu j^\mu = 0,$$ where $j^\mu$ is a conserved current of the theory. However, we know that the Ward Identities are

\begin{equation} \langle \partial_\mu j^\mu(x) O_1(x_1)\cdots O_n(x_n)\rangle + \sum_i \langle \cdots O_i(x_i)\delta(x-x_i)\cdots\rangle = 0, \end{equation}

where $\langle \cdots \rangle$ denote a path integral average. Now he says that this is consistent with the fact that $\partial_\mu j^\mu = 0$ as an operator statement simply because path integral calculates time ordered correlators. He continues to say that "having $\partial_\mu j^\mu$ inside a path integral really means \begin{equation} \partial_\mu\langle j^\mu O_1(x_1)\cdots O_n(x_n)\rangle \end{equation} i.e. computing the average at two infinitesimally close insertions and then taking the difference, further dividing by the infinitesimal distance between the insertions. Now what I do not understand is how having $\partial_\mu j^\mu$ 'inside' the path integral actually means the above, since if I just pull out the $\partial_\mu$ from the Ward Identities as well, I get something completely different. Is this really just some abuse of notation and I am unnecessarily confused? I hope I was able to explain my question.

Answer 1

The main point is that each correlator $\langle\cdots\rangle$ implicitly contains a covariant time-ordering $T_{\rm cov}$, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$ in the correlator. This is necessary in order to correctly translate between the operator formulation and the path integral formulation, cf. e.g. this related Phys.SE post.