QED Commutation Relations Implications

Question

In Brian Hatfield's book on QFT and Strings there is the following quote:

In particular $$ [A_i (x,t), E_j(y,t)] = -i \delta_{ij}\delta(x-y) $$ implies that $$ [A_i(x,t),\nabla \cdot E(y,t)] = -i\partial _i \delta(x-y).$$

I'm not sure how to get between those lines. If I take the partial of the fist line I get $$ [\partial_j A_i(x,t),E_j(y,t)] +[A_i(x,t),\partial_jE_j(y,t)] = -i\partial_i \delta(x-y) $$ So perhaps my question turns into: "Why is $[\partial_j A_i(x,t),E_j(y,t)] = 0$ ?" Thanks.

Answer 1

There's nothing strange going on. $\partial_i$ is shorthand for $$\frac{\partial}{\partial X^i},$$ where some coordinate set $X^i$ is implied. Since $E_j = E_j(y,t)$ you 'obviously' need to derive with respect to $y$ (as twistor59 notes), i.e. $$\nabla \cdot E = \frac{\partial}{\partial y^i} E^i(y,t).$$ The derivative doesn't act on $A_i(x,t)$, so you're done.