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In Peskin/Schroeder there is an explicit calculation showing that the retarded Green's function of the real Klein-Gordon field

$$D_R(x-y) ~\equiv~ \theta(x^0 - y^0) \langle 0 | [\phi(x), \phi(y)] |0\rangle\tag{2.55} $$

fulfills the equation

$$(\partial^2 + m^2) D_R(x-y) = -i\delta^4(x-y).\tag{2.56}$$

I can't follow one specific step in the derivation: It seems like they are doing the substitution

$$(\partial_{x^0}\delta(x^0-y^0))\langle 0 | [\phi(x), \phi(y)] |0\rangle = -\delta(x^0-y^0)\partial_{x^0}\langle 0 | [\phi(x), \phi(y)] |0\rangle. \tag{A} $$

But I don't see how this is justified: If we interpret $(\partial_t \delta(t)) f(t)$ as a distribution and act with it on a test function $g(t)$, we get $$\int (\partial_t \delta(t)) f(t)g(t)dt = - \int \delta(t) \partial_t(f(t)g(t))dt = - \partial_t(f(t)g(t))|_{t=0}.\tag{B} $$

If we act instead with $-\delta(t) \partial_t f(t)$ on $g(t)$ we get $$-(g(t)\partial_t f(t))|_{t=0}.\tag{C} $$

Does somebody have an explanation?

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    $\begingroup$ If you just imagine an integral on each side of the equality then there's no problem, correct? $\endgroup$
    – Danu
    Jul 26, 2015 at 11:44
  • $\begingroup$ Yes, but I don't see how this helps me. $\endgroup$
    – LLang
    Jul 26, 2015 at 11:48
  • $\begingroup$ Why don't you just do the entire derivation under some integral signs and keep in mind that maybe P&S's derivation should be viewed as formal manipulations rather than a real proof? It's seriously non-rigorous throughout... $\endgroup$
    – Danu
    Jul 26, 2015 at 11:55
  • $\begingroup$ @LLang: Which pages in P&S? $\endgroup$
    – Qmechanic
    Jul 26, 2015 at 12:00
  • $\begingroup$ In my edition it is page 30, Eq. 2.56. $\endgroup$
    – LLang
    Jul 26, 2015 at 12:16

2 Answers 2

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Peskin & Schroeder, An Intro to QFT, are using that$^1$

$$i\Delta(x-y)~:=~\langle 0 | [\phi(x), \phi(y)] |0\rangle \tag{K} $$ vanishes for space-like vectors, see below eq. (2.53) on p. 28. In particular for equal times $x^0=y^0$, we have

$$i\Delta(0,{\bf x}-{\bf y})~=~0.\tag{L}$$

Therefore at the physics level of rigor

$$i\Delta(x-y)\delta(x^0-y^0)~=~0.\tag{M}$$

Differentiation of eq. (M) wrt. $x^0$ then yields OP's eq. (A).

Eq. (A) can alternatively be established using test functions.

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$^1$ The notation (K) is taken from Itzykson & Zuber, QFT, eq. (3-55).

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Yes, they are using the substitution of dirac delta $$ \delta' f(x)=-\delta(x)f'(x) $$ And the calculation follows as $$ (\partial^2+m^2)D_R(x-y)=(\partial^2\theta(x^0-y^0))\langle 0 |[\phi(x),\phi(y)]|0\rangle +\theta(x^0-y^0)(\partial^2+m^2)\langle 0 |[\phi(x),\phi(y)]|0\rangle + 2\partial_{\mu}\theta(x^0-y^0)\partial^{\mu}\langle0|[\phi(x),\phi(y)]|0\rangle $$ Since $\theta(x^0-y^0)$ just depends on time, last term both derivatives must be $\partial_t$. Using these equations below $$ \partial_{\mu}\theta(x^0-y^0)=\delta(x^0-y^0) $$ $$ \partial^2\theta(x^0-y^0)=\partial_t\delta(x^0-y^0) $$

Also commutation relation $[\Pi(x),\phi(x)]=-i\delta^3(x-y)$ and dirac notation, one can compute $$ (\partial^2+m^2)D(x-y)=-i\delta^4(x-y) $$

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