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This is from a book by Srednicki:

\begin{eqnarray} [\Pi(x),H] &=& \frac{1}{2}\int{\rm d}^3y~[\Pi(x),\nabla^i\varphi(y)\nabla^i\varphi(y) + m^2\varphi^2(y)] \\ &=& \frac{1}{2}\int{\rm d}^3y~ \left( \nabla^i_y[\Pi(x),\varphi(y)]\nabla^i\varphi(y) + \nabla^i\varphi(y)\nabla^i_y[\Pi(x),\varphi(y)]\right. \\ &&~~~~~ + \left.m^2[\Pi(x),\varphi(y)]\varphi(y) + m^2\phi(y)[\Pi(x),\varphi(y)]\right) \\ &=& -i\int{\rm d}^3y~(\nabla^i_y\delta^3({\bf x}-{\bf y})\nabla^i\varphi(y)) + m^2\delta^3({\bf x}-{\bf y})\varphi(y)) \\ &=& -i\int{\rm d}^3y~ (-\delta^3({\bf x}-{\bf y})\nabla^2\varphi(y) + m^2\delta^3({\bf x}-{\bf y})\varphi(y))\\ &=& -i(-\nabla^2 +m^2)\varphi(x). \end{eqnarray}

The third equality is boggling me: it seems that he assumes that $\nabla^i(\delta(x-y))$ and $\nabla^i\phi(y)$ commute, but seems false to me. In fact, in the fourth equality he seems to use $\nabla^i\delta(x-y)=-\delta(x-y)\nabla^i$ (which seems to be in accordance to my notes on functional analysis), but if we apply this to the expression of the second equality, we get that the first term in the sum is $$ -\delta(x-y)(\nabla^i)^2\phi(y) $$ and the second one is $$ -\nabla^i\phi(y)\delta(x-y)\nabla^i $$ and these are not equal, as I suspected. So, what's going on here?

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    $\begingroup$ Mark Srednicki was my instructor. Nice guy! $\endgroup$ – amcalde Oct 10 '17 at 22:24
  • $\begingroup$ Which book? Which page? $\endgroup$ – Qmechanic Oct 14 '17 at 13:59
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$\hat{\varphi}$ is an operator, whereas $\delta(x - y)$ is not, neither is its derivative. So

$$ \color{red}{\nabla \delta(x-y)}\color{blue}{\nabla\hat{\varphi}(y)} = \color{blue}{\nabla\hat{\varphi}(y)}\color{red}{\nabla \delta(x-y)} $$

In other words, they commute! So does $\hat{\phi}$ in the potential term

$$ \color{red}{\delta(x-y)}\color{blue}{\hat{\varphi}(y)} = \color{blue}{\hat{\varphi}(y)}\color{red}{ \delta(x-y)} $$

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  • $\begingroup$ But what I said above is that I thought that the derivative of $\delta$ is an operator (as explained in the text). And, he seems to use what I wrote about the derivative of $\delta$, like I described in my question. $\endgroup$ – Soap Oct 10 '17 at 22:44
  • $\begingroup$ In other words, what in my text is incorrect? $\endgroup$ – Soap Oct 10 '17 at 22:44
  • $\begingroup$ @Soap You can go from the third to the fourth equation by integration by parts, will not need to assume that $\nabla \delta$ is an operator $\endgroup$ – caverac Oct 10 '17 at 22:47
  • $\begingroup$ What about the term $\nabla(\delta(x-y)\nabla\phi)$? $\endgroup$ – Soap Oct 11 '17 at 9:25
  • $\begingroup$ @Soap You can assume it vanishes at the boundary of the volume. This integral is carried out over the whole space, it is natural to assume that a infinity both the field and its gradient are zero $\endgroup$ – caverac Oct 11 '17 at 9:57

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