DISCLAIMER: I edited the question to include the integrals and the picture

I'm stuyding Klein-Gordon field in Peskin & Schroeder An Introduction to Quantum Field Theory There is an integral that I don't see, related to the computation of the propagators, in particular for the K-G field. I'm refering to two steps

$$\langle 0| [\phi(x),\phi(y)] 0\rangle=\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}(e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}).\tag{2.54}$$

Now that integral is written as

$$\int \frac{d^3p}{(2\pi)^3} \left\{ \frac{1}{2E_p}e^{-ip\cdot(x-y)}\bigg|_{p_0=E_p}+\frac{1}{-2E_p}e^{-ip\cdot(x-y)}\bigg|_{p_0=-E_p}\right\} $$

Question 1) How can be fixed the vale of $p_0$ to be $\pm E_p$? It is only notation?

Now it says that it can be written as a $p_0$ integral in the following contour

enter image description here

If $x^0>y^0$ then

$$\int \frac{d^3p}{(2\pi)^3}\int\frac{dp^0}{2\pi i}\frac{-1}{p^2-m^2}e^{-ip\cdot(x-y)}$$

and the contour is closed below. If $x^0<y^0$ the contour is closed above, giving 0 (this i can see it)

Question 2) Why that particular choice of closed contours?

When I had to do any integral with residues theorem, usually they are integrals over $\mathbb{R}$ so that the contour of integration in the complex plane is exactly a semicircle of radius $R$ with $R\to \infty$. Now this is not the case, because the contour does not pass throught the entire real line.

Question 3) The semicircles centered at $\pm E_p$ affect the $p_0$ integration?

  • $\begingroup$ This is related to contour integrals. Do you what that is? $\endgroup$
    – Prahar
    Sep 22, 2013 at 18:17
  • $\begingroup$ Yes, I know this is related to residues theorem but I don't see those particular integrals $\endgroup$
    – J L
    Sep 22, 2013 at 18:39
  • $\begingroup$ OK. Often ordinary integrals over the real line can be extended to the complex plane where the residue theorem can then be used. Do you know how that is done? It is the same thing that applies here. $\endgroup$
    – Prahar
    Sep 22, 2013 at 19:20
  • $\begingroup$ Yes, my particular problem is with the countour of integration, being a semicircle but with two semicircles at $\pm E_p$. Is in page 30 of the book $\endgroup$
    – J L
    Sep 22, 2013 at 19:30

1 Answer 1


OK, I have Peskin & Schroeder open in front of me. They're not saying that $\langle 0|[\phi(x),\phi(y)]|0\rangle$ takes on two different values depending on how you close the contour. They're saying that the integral given on the last line of equation 2.54 $$\int \frac{d^3p}{(2 \pi)^3}\int \frac{dp^0}{2\pi i} \frac{-1}{p^2 - m^2} e^{-i p \cdot (x - y)}$$ is either equal to $\langle 0|[\phi(x),\phi(y)]|0\rangle$ or zero, depending on whether $x^0 > y^0$ or $x^0 < y^0$ (which in turn dictates which way you should close the contour).

You can evaluate the integral by adding an arc to it either above or below, making it a closed contour. But you have to choose the arc such that in the limit that its radius goes to infinity, the integral along the arc goes to zero. Otherwise, adding the arc would change the value of the integral. For $x^0 > y^0$ the right choice is to close the contour below. This encloses both poles, and it follows from the residue theorem that the integral is $\langle 0|[\phi(x),\phi(y)]|0\rangle$. For $x^0 < y^0$, the right choice is to close the contour above, enclosing no poles, from which it follows that the integral is zero.

Therefore in the general case, the integral is $\theta(x^0 - y^0)\langle 0|[\phi(x),\phi(y)]|0\rangle$, which is the retarded Green's function. (Here $\theta$ is the step function.)

  • $\begingroup$ Hi thanks. If the poles are displaced from the real axis some $i \epsilon$ then you always get one pole don't you? $\endgroup$
    – J L
    Sep 23, 2013 at 6:38
  • 2
    $\begingroup$ @Nivalth : See also Propagators, and look at the different $\epsilon$ prescriptions, for the different propagators. $\endgroup$
    – Trimok
    Sep 23, 2013 at 7:25
  • $\begingroup$ @Nivalth: You can do the displacement in several ways. With the choice P. & S. make on page 30, you enclose two or zero poles. With the Feynman prescription they give on the following page, you always enclose one pole. Trimok's link has good details. $\endgroup$ Sep 23, 2013 at 13:46
  • $\begingroup$ Hi Thanks. If the integral is for $p_0 \in \mathbb{R}$ is there any problem with the semicircles in both poles? Usually the countour is a semicircle stricly, this more or less the same but not exactly $\endgroup$
    – J L
    Sep 23, 2013 at 21:19

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