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On page 29 of Peskin and Schroeder's An Introduction to Quantum Field Theory, the authors write that the propagator is given by: $$\begin{align} \langle 0|[\phi(x),\phi(y)]|0\rangle&=\int{d^3p\over (2\pi)^3}\bigg\{{1\over 2E_p}e^{-ip\cdot(x-y)}\bigg|_{p^0=E_p}+{1\over -2E_p}e^{-ip\cdot(x-y)}\bigg|_{p^0=-E_p}\bigg\}\cr&=\int{d^3p\over(2\pi)^3}\int{dp^0\over 2\pi i}{-1\over p^2-m^2}e^{-ip\cdot(x-y)} \quad(eq.2.54). \end{align}$$ How does the author derive this expression (eq.2.54)? Apparently, the author did not arrive at the expression via Fourier transformation of the inhomogeneous Klein-Gordon with a delta function source, because this technique is alluded to later with the claim that the expression in question (2.54) had already been derived earlier. To my knowledge, the expression is written for the first time on page 30 without any derivation or proof. So how does the author derive the expression?

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    $\begingroup$ This is simply the reverse of the Residue Theorem. $\endgroup$ Commented May 15 at 16:37
  • $\begingroup$ @JeanbaptisteRoux Thank you! That makes a good deal of sense. $\endgroup$ Commented May 15 at 16:41
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    $\begingroup$ @JeanbaptisteRoux Consider to add some details and write an answer. $\endgroup$ Commented May 15 at 16:46
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    $\begingroup$ @TobiasFünke This is very strange, I was sure I've answered a similar question in the past but I can't find it... I will try to write an answer. $\endgroup$ Commented May 15 at 16:58
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    $\begingroup$ @TobiasFünke This is not exactly the same because the one you linked is for the Feynman propagator, and not the retarded one. Furthermore, it was for the position space expression while OP's question is about the momentum space only. $\endgroup$ Commented May 15 at 17:08

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There are several ways ways to do the calculation, with the Residue Theorem. Either you foresee that this is an instance of this theorem, or you know the solution and do reverse reasoning to find back the first line. I will adopt the first method.

You have the equality \begin{equation} \langle 0 |[\phi(x),\phi(y)]| 0 \rangle = \int \frac{d^3p}{(2\pi)^3}\left\{ \frac{1}{2E_p}\left.e^{-ip\cdot(x-y)}\right|_{p^0=E_p} + \frac{1}{-2E_p}\left.e^{-ip\cdot(x-y)}\right|_{p^0=-E_p}\right\}. \end{equation} P&S wrote the minus sign in the denominator of the factor of the second exponential to emphasize that this equality really is: \begin{equation} \langle 0 |[\phi(x),\phi(y)]| 0 \rangle = \int \frac{d^3p}{(2\pi)^3}\left\{ \left.\left[\frac{1}{2p^0}e^{-ip\cdot(x-y)}\right]\right|_{p^0=E_p} + \left. \left[\frac{1}{2p^0}e^{-ip\cdot(x-y)}\right]\right|_{p^0=-E_p}\right\} \end{equation} From here, and applying the Residue theorem, one had: \begin{align} \langle 0 |[\phi(x),\phi(y)]| 0 \rangle =& \int \frac{d^3p}{(2\pi)^3}\sum_{s=\pm 1} \lim_{p^0 \rightarrow s E_p}\left[\frac{(p^0-sE_p)}{(p^0+E_p)(p^0-E_p)}e^{-ip\cdot(x-y)}\right] \\ =& \int \frac{d^3p}{(2\pi)^3}\sum_{s=\pm 1} \lim_{p^0 \rightarrow s E_p}(p^0-sE_p)\left[\frac{1}{(p^0)^2-E_p^2}e^{-ip\cdot(x-y)}\right] \\ \stackrel{\text{Res. Th.}}{=}& -\int \frac{d^3p}{(2\pi)^3}\int_C \frac{dp^0}{2i\pi}\left[\frac{1}{(p^0)^2-E_p^2}e^{-ip\cdot(x-y)}\right] \end{align} Where the last line is the reverse residue theorem, and the contour $C$ encloses the two poles $p^0=\pm E_p$ by passing along the inferior part complex plane (this is where the overall minus sign comes from). But, by the estimation lemma, the integral over $p^0$ equal to the integral over $C \leadsto \mathbb{R}$. So we indeed have: \begin{align} \langle 0 |[\phi(x),\phi(y)]| 0 \rangle =& -\int \frac{d^3p}{(2\pi)^3}\int \frac{dp^0}{2i\pi}\left[\frac{1}{(p^0)^2-p^2-m^2}e^{-ip\cdot(x-y)}\right] \end{align}

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  • $\begingroup$ You have written one of the equalities as $\stackrel{!}{=}$, which I think is supposed to indicate a surprising equality? ...But, I think this $\stackrel{!}{=}$ symbol could be misunderstood as meaning $\neq$, since $!$ is used in some contexts to indicate NOT. $\endgroup$
    – hft
    Commented May 15 at 17:30
  • $\begingroup$ @hft The only meaning of this symbol I know is "must be equal to" $\endgroup$ Commented May 15 at 17:33
  • $\begingroup$ Merci beaucoup pour l'aide. $\endgroup$ Commented May 15 at 17:35
  • $\begingroup$ @TobiasFünke So, why not just use the symbol "$=$" $\endgroup$
    – hft
    Commented May 15 at 17:40
  • $\begingroup$ From Microsoft API documentation: "The not-equal-to operator (!=) returns true if the operands don't have the same value; otherwise, it returns false." $\endgroup$
    – hft
    Commented May 15 at 17:42

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