Commutator with gradient operator (nabla)

[Replusz]

TL;DR

I don't understand what is happening from line 1 to 2.
It looks like he is taking out the derivative operator from the commutator.
Which I presume is not allowed?

Or he could be using product rule on commutator and then neglecting the 2nd part (maybe because it is trivially 0? I don't see why it is 0 tho: it is [nabla, Pi]Phi )

 

[Gaussian97]

I think you can take the gradient outside because $\pi(t,\vec{x})$ doesn't depend on $\vec{x}'$

 

[Replusz]

I think you can take the gradient outside because $\pi(t,\vec{x})$ doesn't depend on $\vec{x}'$

And the Nabla acts on x' right?
Then it makes sense.

 

[Gaussian97]

Yes, normally one makes the distinction and calls it $\nabla'$ or $\nabla_{\vec{x}'}$, but in this case, you simply must deduce it from the context. Look that they change
$$\nabla_i = \frac{\partial}{\partial x' ^i}$$

 

[Replusz]

Nice!
Thanks so much and all the best! :)