Let $\phi$ denote the Klein-Gordon field. Then its propagator $\langle 0 \mid [\phi(x), \phi(y)] \mid 0 \rangle$ can be calculated as $$\int \frac{d^4}{(2\pi)^3} \frac{-e^{-ip(x-y)}}{p^2 -m ^2}. \tag{1}$$ Isolating just the $p^0$ part of this integral, which is where the problem is, we get $$\int \frac{dp^0}{2\pi i} \frac{-e^{-p^0(x^0-y^0)}}{(p^0 + E_p)(p^0 -E_p)} \tag{2}$$ and so there are poles at $\pm E_m$.

The standard way of dealing with such an integral is of course to use Cauchy's residue theorem and choose a contour across the real line with small semi-circles of radius $\epsilon$ around the poles and then sending $\epsilon$ to zero in the limit. Thus no matter what contour we choose we should get the same answer.

What I am confused about is why we do not send $\epsilon \rightarrow 0$ when calculating these propagators. Since we do not do this we get different values for the integral (which depend on the choice of contour and how whether we push the poles above or below the real axis) and hence different propagators (advanced, retarded, and Feynman). I understand that the different propagators correspond to different supports (in terms of time), but unless we take $\epsilon \rightarrow 0$ we're no longer solving for $\langle 0 \mid [\phi(x), \phi(y)] \mid 0 \rangle$ but rather a modified version of it. What is going on here?

EDIT: To make things clearer, my question is the following: shouldn't the answer to (2) give rise to the same propagator in the limit $\epsilon \rightarrow 0$ regardless of what contour is used?

  • 1
    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – ACuriousMind
    Commented Feb 9 at 19:04
  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/138217/50583 and its linked questions $\endgroup$
    – ACuriousMind
    Commented Feb 9 at 19:05
  • 1
    $\begingroup$ @CBBAM: My understanding is that you have a good point. In principle, the limit $\epsilon\rightarrow0$ must be taken before doing anything else in order to be faithful to the original expression, and cannot be iterated with the integration. The condition for iterating a limit and integration (dominated convergence) is not met in this case. $\endgroup$
    – Yair
    Commented Mar 3 at 9:50

1 Answer 1


Here is a cute toy example from What is a Quantum Field Theory? A First Introduction for Mathematicians. M. Talagrand. CUP. Appendix N: Feynman Propagator and Klein-Gordon Equation, which hopefully clears some confusion.

Consider the (linear) differential operator $D(u):=\partial_t^2 u+m^2u$. We now seek fundamental solutions of $D$.

Depending on the boundary conditions we employ, we find different fundamental solutions. For example, the following functions, corresponding to the retarded, advanced and Feynman prescription: $$u_1(t)=\begin{cases} \displaystyle 0 \quad &\text{for} \quad t\leq 0 \\ \frac{1}{m} \sin(mt)\quad &\text{for} \quad t\geq0 \end{cases}$$

$$u_2(t)=\begin{cases} \displaystyle 0 \quad &\text{for} \quad t\geq 0 \\ -\frac{1}{m} \sin(mt)\quad &\text{for} \quad t\leq 0 \end{cases}$$

$$u_3(t)= i\frac{e^{-i|t|m}}{2m}\quad .$$

It should be easy to verify that these are indeed fundamental solutions of $D$.

To find these fundamental solutions, we might take the Fourier transform of $D(u)=\delta$ which gives $$-\omega^2 \hat u(\omega)+m^2 \hat u(\omega)=1\quad .$$

Naively inverting the above equation would give $$\hat u(\omega)=\frac{1}{-\omega^2+m^2} \quad , $$

which, with the inverse Fourier transform, would yield

$$u(t)=\int_\mathbb R\frac{\mathrm d \omega}{2\pi}\frac{e^{it\omega}}{-\omega^2+m^2} \quad . $$

The integral however does not make sense, the function is not defined at $\omega=\pm m$. But we can compute slightly different (well-defined) integrals.

For example by e.g. bypassing both poles from below, i.e. replacing $\omega \to \omega -i\epsilon$, with $\epsilon >0$: If $t<0$, the integral vanishes, and for $t>0$ it equals $\sin(mt)/m$, i.e. we obtain $u_1$. Similarly, bypassing the poles both from above gives $u_2$. Finally, you might bypass the poles by computing $$\int_\mathbb R\frac{\mathrm d \omega}{2\pi}\frac{e^{it\omega}}{-\omega^2+m^2-i\epsilon} \quad ,$$ which gives $u_3$.

Note that we've implicitly let $\epsilon \to 0^+$ in all cases after the integration.

Regarding your specific case, which is very similar to the toy example above: What can be shown is that the Feynman propagator, as a tempered distribution, is given by $$ \Delta_F(x)=\lim\limits_{\epsilon \to 0^+}\int \frac{\mathrm d^4p}{(2\pi)^3}\frac{\exp-i(x,p)}{-p^2+m^2-i\epsilon}\quad .$$

This is Lemma 13.10.4. (p. 370) in the same book.

  • 3
    $\begingroup$ It is also interesting to emphasize that $u_3$ corresponds to bypassing one pole from above and one pole from below $\endgroup$ Commented Feb 8 at 22:46
  • $\begingroup$ Thank you very much! I think the big takeaway for me is if the integrand is ill defined then its value is contour dependent. I had incorrectly thought that by using the residue theorem the integral will give a unique value. $\endgroup$
    – CBBAM
    Commented Feb 9 at 0:33
  • 1
    $\begingroup$ @CBBAM It might be instructive, too, to consider an even easier example: Take the derivative operator and find the general form of its fundamental solutions by solving the differential equation (DE). Then, for comparison, take the Fourier transform (FT) of the DE and a) do the naive inversion and try to compute via the inverse FT the fundamental solution(s). Of course, it is ill-defined, so you might make the replacement $\omega\to \omega\pm i\epsilon$, compute the inverse FT and let $\epsilon \to 0^+$, which should give the retarded and advanced solutions. $\endgroup$ Commented Feb 9 at 18:31
  • 1
    $\begingroup$ b) Do the correct distributional inversion, e.g. use the fact that the general distributional solution of $xf(x)=1$ is $f(x)=\mathrm{p.v.} 1/x + c \delta(x)$ for a constant $c$. Then take the inverse FT of this to find the correct general form of the fundamental solution. If I am not mistaken, everything should be more or less straightforward if you recall some facts about the Heaviside function. Of course, and as usual, to make everything very rigorous is a completely different story--but this is not the root of your question, I think, so we are fine with the usual physics level of rigor. $\endgroup$ Commented Feb 9 at 18:40
  • 1
    $\begingroup$ @TobiasFünke Thank you very much for all your help! I now fully understand the problem, it is precisely because no meaning can be given to the ill-defined integral. However we can use it to create a sort of ansatz for what the Greens function might look like. I have edited my post to make my question more clear, please let me know if I should edit it further. $\endgroup$
    – CBBAM
    Commented Feb 9 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.