Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.