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I'm trying to derive the eq.2.56 of P&S's QFT textbook step by step:

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

I have no problem with the first step:

$(\partial^2+m^2)D_R(x-y)=(\partial^2\theta(x^0-y^0))\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle+2(\partial_\mu\theta(x^0-y^0))(\partial^\mu\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle))+\theta(x^0-y^0)(\partial^2+m^2)\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle$ (*)

And here is what my derivation, please point out the mistakes I've made, I would be appreciated for your help:

1.for the third term in the RHS of the eq.(*):

I think it used the Klein-Gordon equation so that $(\partial^2+m^2)\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle=\langle0\rvert[(\partial^2+m^2)\phi(x),\phi(y)]\rvert0\rangle+\langle0\rvert[\phi(x),(\partial^2+m^2)\phi(y)]\rvert0\rangle=0$

2.for the first term in the RHS of the eq.(*):

Since $\theta(x^0-y^0)$ only contains the time variable, then only partial derivative wrt. time remained, then $\partial^2\theta(x^0-y^0)=\partial_t\delta(x^0-y^0)$, then

$\partial_t\delta(x^0-y^0)\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle=-\delta(x^0-y^0)\partial_t\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle$

in which the property of $\delta$ function $f(x)\delta^\prime(x)=-f^\prime(x)\delta(x)$ has been used.

3.for the second term in the RHS of the eq.(*):

it equals to $2\delta(x^0-y^0)\partial_t\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle$

Finally, combined the three terms we have obtained, we get:

$\delta(x^0-y^0)\partial_t\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle$

I think I have some troubles in dealing with $\partial_t\langle0\rvert[\phi(x),\phi(y)]\rvert0\rangle$, I think it should be:

$\langle0\rvert[\partial_t\phi(x),\phi(y)]\rvert0\rangle+\langle0\rvert[\phi(x),\partial_t\phi(y)]\rvert0\rangle=\langle0\rvert[\pi(x),\phi(y)]\rvert0\rangle+\langle0\rvert[\phi(x),\pi(y)]\rvert0\rangle$

but obviously, the authors do not really think so, where I am wrong? Thank you very much.

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1 Answer 1

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The derivatives in $\partial^2+m^2$ are only with respect to the $x$ argument, not the $y$ (clearly it can't be both). So in all your steps you have the right idea but only the term with $\pi(x)$ appears.

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    $\begingroup$ Thanks, got it! $\endgroup$
    – 白琢丛
    Commented May 26, 2022 at 6:02

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