I Noether's Theorem in the Presence of a Charged Operator

[thatboi]

I am trying to understand the following idea that I found from some notes: Generally, a system with U(1) symmetry will have a conserved current: $\partial_{\mu}j^{\mu} = 0$. The notes then state that in the presence of a local operator $\mathcal{O}(x)$ with charge $q\in \mathbb{Z}$ under U(1), the continuity equation becomes: $\mathcal{O}(x)\partial_{\mu}j^{\mu}(x') = q\delta(x-x')\mathcal{O}(x)$. I just wanted to better understand the intuition behind this equation. How can I derive this?

 

[samalkhaiat]

\mathcal{O}(x)\partial_{\mu}j^{\mu}(x') = q\delta(x-x')\mathcal{O}(x)

It is not clear what meaning one can associate with the product of operators on the left hand side.

I you have an exact symmetry, then the Noether current is conserved, i.e., \partial_{\mu}j^{\mu}(x) = 0, and its associated charge, Q = \int d^{3}x \ j^{0}(x), generates the correct infinitesimal symmetry transformation of local operators: \left[Q , \mathcal{O}(y)\right] = \delta \mathcal{O}(y) = - i q \mathcal{O}(y) , or \left[ j^{0}(x) , \mathcal{O}(y) \right] = -i q \delta^{3}(\vec{x} - \vec{y}) \mathcal{O}(y) . \ \ \ \ (1)

Now consider the following time-ordered product T\left( j^{\mu}(x)\mathcal{O}(y)\right) \equiv j^{\mu}(x)\mathcal{O}(y)\theta (x^{0} - y^{0}) + \mathcal{O}(y)j^{\mu}(x) \theta (y^{0} - x^{0}) . Differentiation gives you \frac{\partial}{\partial x^{\mu}} T\left(j^{\mu}(x)\mathcal{O}(y) \right) = T\left( \partial_{\mu}j^{\mu}(x) \mathcal{O}(y)\right) + \delta (x^{0} - y^{0}) \left[ j^{0}(x) , \mathcal{O}(y)\right] . If the symmetry is exact, then current conservation and eq(1) give you the following (Ward identity):

\partial_{\mu}^{(x)} \left( T\left( j^{\mu}(x)\mathcal{O}(y)\right)\right) = - i q \delta^{4}(x - y) \mathcal{O}(y).