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This questions relates to equation (4.43) in Timo Weigand's QFT lecture note (page 108). Weigand makes the claim that $\partial_\mu A^\mu=0$, interpreted as an operator equation, causes issues due to the following:

$$0\overset{!}{=}[\partial_\mu A^\mu,A^\nu]\overset{?}{=}[\dot A^0,A^\nu]=i\eta^{0\nu}\delta^{(3)}(\vec x-\vec y) \tag{4.43}.$$

My question is why the second equality (with a $?$ over it) holds? This seems to imply that

$$[\partial_i A^i,A^\nu]=0,$$

however I cannot figure out why this should be true.

Qmechanic
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Charlie
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2 Answers2

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If you write out the functional dependencies in the equation you are referring to, you might see this more clearly. Note that $$ [\partial_i A^i(0,\mathbf{x}), A^\nu(0,\mathbf{y})] = \partial_{x,i} [A^i(0,\mathbf{x}), A^\nu(0,\mathbf{y})] = 0 $$ which follows by the fact that the derivative acts on $\mathbf{x}$ and can therefore be pulled out, and the canonical commutation relations.

sondre
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The Poisson bracket reads:

$$\{\partial_i A^i , A^\nu\} = \sum_{\mu}\left(\frac{\delta A^\nu}{\delta A^\mu}\frac{\delta(\partial_i A^i)}{\delta \pi_\mu} - \frac{\delta A^\nu}{\delta \pi^\mu}\frac{\delta(\partial_i A^i)}{\delta A_\mu}\right) = \frac{\delta(\partial_i A^i)}{\delta \pi_\nu} = \frac{\delta (\partial_i A^i)}{\delta\dot{A}_\nu} = 0,$$

since the off-shell spatial derivatives are not functionally dependent on the temporal derivatives.

ACuriousMind
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    That checks out, I am a little surprised that we can argue this from the classical Poisson bracket though. – Charlie Jul 17 '22 at 12:05
  • @Charlie Since we're doing canonical quantization here, the Poisson bracket directly corresponds to the quantum commutator. – ACuriousMind Jul 17 '22 at 13:39
  • I suppose my surprise here comes from the fact that I am not familiar with anything other than the "put hats on things" approach to quantization. It is correct to say that other more careful methods of quantization justify being able to carry results over from the classical Poisson bracket to the quantized theory? – Charlie Jul 17 '22 at 13:41
  • @Charlie "Putting hats on operators" is exactly what I mean by "canonical quantization" - how do you know what the commutators of the "hatted stuff" is if not by using the Poisson bracket and adding an $\mathrm{i}$? See e.g. this answer of mine for why careful quantization methods in general will not preserve all Poisson brackets (but that's irrelevant here since these problems generally only occur for polynomials of higher order in the canonical coordinates). – ACuriousMind Jul 17 '22 at 13:47
  • Ok I see what you mean, thank you :) – Charlie Jul 17 '22 at 14:05