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(I'm following these notes by Vadim Kaplunovsky titled "Feynman Propagator of a Scalar Field," specifically asking about equation 10.)

When calculating the time derivative of the Feynman scalar propagator, we get a term $\delta(x^0-y^0)\times\langle 0|[\phi(x),\phi(y)]|0\rangle$. According to Kaplunovsky, the delta function forces the term to be zero at unequal times but the term is zero at equal times also because then the $\phi$'s commute. However, if one just naively plugs in $x^0=y^0$ into the equation, we get an indeterminate form $\infty\times0$. I realize this doesn't make sense because $\delta$ is a distribution, not a function, but is there a mathematical way to show that this term yields zero?

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    $\begingroup$ Here we are using Green's functions, so one must not get confused with what is happening here. We are dealing with distributions, not functions, and Green's function are defined as being solution to inhomogeneous equation where the inhomogeneous term is a Dirac delta. So here, you have to keep in mind that is all going to be under an integral when we are done dealing with them, hence there are no problems. $\endgroup$ – Soap312 Oct 31 '17 at 15:58
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Because you have a $\delta$-function, the expression you get only makes sense when integrated against a test function. So, consider a test function $g(x^0)$. We have \begin{align} \int dx^0\ g(x^0) \langle[\phi(x),\phi(y)]\rangle \delta(x^0-y^0) &= g(y^0)\langle[\phi(y^0,\vec{x}),\phi(y^0,\vec{y})]\rangle\\ &= g(y^0)\times 0\\ &=0\\ \end{align} where in the first step I used the definition of the delta function, in the second I used the commutation relation assuming $\vec{x}\neq\vec{y}$. You're right that naively plugging in $x^0=y^0$ gives an indeterminant form, but that's only because you're dealing with a distribution instead of a function. Since you have a distribution, you learn about it by integrating it against test functions instead of plugging in numbers. Once you accept that, it's pretty easy to see that the distribution is the zero distribution, since it integrates to zero against any test function.

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