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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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