# QED Commutation Relations Implications

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.

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3
Maybe the partial is with respect to the y coordinate? – twistor59 Dec 22 '12 at 23:07
Lol, of course. Thanks. – kηives Dec 23 '12 at 2:20

## 1 Answer

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.

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