Commutator with gradient operator (nabla)

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Discussion Overview

The discussion revolves around the application of the gradient operator (nabla) in the context of a function ##\pi(t,\vec{x})## and its dependence on the variable ##\vec{x}'##. Participants explore the implications of taking the gradient outside of certain expressions and the notation used for the gradient operator.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • Some participants propose that the gradient can be taken outside because the function ##\pi(t,\vec{x})## does not depend on ##\vec{x}'##.
  • Others clarify that the nabla operator acts on ##\vec{x}'##, suggesting that this understanding is crucial for the discussion.
  • A participant notes the distinction in notation, indicating that one might use ##\nabla'## or ##\nabla_{\vec{x}'}## to specify the gradient operator acting on ##\vec{x}'##, but context can help deduce the correct interpretation.
  • There is an acknowledgment of the mathematical representation of the gradient, specifically that ##\nabla_i = \frac{\partial}{\partial x' ^i}##.

Areas of Agreement / Disagreement

Participants generally agree on the notion that the gradient can be taken outside under certain conditions, but there is some discussion regarding the appropriate notation and context for the gradient operator.

Contextual Notes

The discussion does not resolve the nuances of notation and context fully, leaving some assumptions about the dependence of functions and the application of the gradient operator open to interpretation.

Replusz
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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 )
1586598000065.png
 
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I think you can take the gradient outside because ##\pi(t,\vec{x})## doesn't depend on ##\vec{x}'##
 
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Gaussian97 said:
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.
 
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
$$\nabla_i = \frac{\partial}{\partial x' ^i}$$
 
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Nice!
Thanks so much and all the best! :)
 

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