I am now reading the David Tong's lecture note on Quantum Field theory.

I have some questions about the contour used in the integral \begin{equation} \int \frac{d^{4}p}{(2\pi)^{4}} \frac{i}{(p^{0})^{2} - {\bf p}^{2}} e^{-ip(x-y)} \end{equation} where $p(x-y) = p^{0}(x^{0}-y^{0}) - {\bf p} \cdot \left({\bf x}- {\bf y}\right)$.

In order to do the integral we can actually choose 4 different paths. And the contour we will choose for Feynman propagator is

enter image description here

depending on whether $x^{0}$ is larger than $y^{0}$.

I now want to work out the integral \begin{equation} \int \frac{d{p^{0}}}{2\pi} \frac{i}{(p^{0})^2 - E_{\bf p}^{2}} e^{-ip^{0}(x^{0}-y^{0})} \end{equation}

But I have some questions on the path and the answer I am looking for.

  1. Do we ultimately let the radius of half circle in the integral path $C_{1}$ and $C_{2}$ $\to 0$ ?

  2. For $x^{0} < y^{0}$ The integral of the total contour can be decomposed into \begin{equation} \int_{L_1} + \int_{C_{1}} + \int_{L_2} + \int_{C_{2}} + \int_{L_{3}} + \int_{C_{3}} = 2\pi i Res(z=+E_{\bf p}) \end{equation} And since the $C_{3}$ path is chosen in such a way that when we make the radius of $C_{3} \to \infty$ then $\int_{C_{3}} \to 0$ we will ignore it hereafter.

Now I am wondering what is the value we want

(1) \begin{align} \int \frac{d{p^{0}}}{2\pi} \frac{i}{(p^{0})^2 - E_{\bf p}^{2}} e^{-ip^{0}(x^{0}-y^{0})} & = \int_{L_1} + \int_{C_{1}} + \int_{L_2} + \int_{C_{2}} + \int_{L_{3}} \\ & = 2\pi i Res(z=+E_{\bf p}) \end{align} or

(2) \begin{align} \int \frac{d{p^{0}}}{2\pi} \frac{i}{(p^{0})^2 - E_{\bf p}^{2}} e^{-ip^{0}(x^{0}-y^{0})} &= \int_{L_1} + \int_{L_2} + \int_{L_{3}} \\ & = 2\pi i Res(z=+E_{\bf p}) - \int_{C_{1}} - \int_{C_{2}} \\ & \ with\ \rho \to 0 \end{align}

The reason that I am confused is that if we choose to evaluate the (2) with $\rho \to 0$ then the $\int_{C_{1}}$ and $\int_{C_{2}}$ do have some nonzero value (because of the simple pole at $E_{\bf p}$ and $-E_{\bf p}$) and can not cancel each other. And this will make the answer I get different from the case (1).

Therefore, I am wondering which one should I calculate, (1) or (2)?


3 Answers 3


It is much easier to let the poles go off the real axis with Feynman's $i\epsilon$. Then the integration can stay straight on the real line and you will not need to draw C1 and C2. If you do draw them like you did, then yes, you will need to shrink them and calculate the contributions of the half circles, since they do not cancel. They are half circles, going in opposite directions around poles of opposite residue, so each contribute $+i\pi$ for $+2i\pi$ in total, and obviously the same in both cases. The line integrals L1, L2 and L3 are zero!

In the Feynman's way of doing it, and that is how most physists get to the answer quickly without calculating six segments every time, we just remember that the answer is this: the residue of whichever pole is inside the contour is equal to the integral we want. For $x0<y0$ you capture the pole at z=-Ep, going counterclockwise, and for $x0>y0$ you capture the pole at z=+Ep going clockwise. The value of the residue is opposite at z=-Ep and +Ep, therefore the answer is the same.

By the way, one very old QFT book claims there are not four but seven inequivalent ways to go around two poles...

  • $\begingroup$ Which old QFT book are you talking about? $\endgroup$
    – Sanjana
    May 14 at 17:49
  • $\begingroup$ @Sanjana , you are supposed to know just from the phrase "one very old QFT book" :-) Of course that is the book of Bogoliubov and Shirkov... $\endgroup$
    – Kphysics
    May 16 at 9:31

You have to take (1) and let $\rho \to 0$, because in (2), in the limit, you get the integral over the real line minus the two points $\pm E_p$, which is not what you want (and as you have observed the two integrals after the residue do not disappear because they carry valuable information). In (1) the contour is continuous (uninterrupted) and $\rho \to 0$ is roughly speaking a homotopy (a continuous deformation of loops/contours) from the original contour (on your pictures) to the whole real line. This is what you need because the values of the integrals of holomorphic (complex analytic) functions do not change under homotopies, i.e. the value of the integral stays the same when $\rho \to 0$ while the contours deform.


This also caused me a deal of confusion because I too was trying to calculate the contributions $\int_{C_1}$, $\int_{C_2}$ in order to determine the integral along the real axis (as is so often the target of exercises in complex variable courses).

The key realisation is that the propagator $\Delta_F(x-y)$ is defined (by Tong) as the entire integral \begin{equation} \int \frac{d^{4}p}{(2\pi)^{4}} \frac{i}{(p^{0})^{2} - {\bf p}^{2}} e^{-ip(x-y)} \end{equation} In particular, the integral over $p^0$ is not the (principle value of the) integral along the real axis (which would be (2)) but entire contour, thus in the case $x^0<y^0$ always takes the value $2\pi i \text{Res}\left(p^0=+E_p\right)$, by the residue theorem. You do not need to consider the indented parts of the contour at all (i.e. (1) is what you want).


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