Reading through David Tong lecture notes on QFT.

On pages 58-60, he calculates the amplitude of nucleon scattering in scalar Yukawa theory using Dyson time ordering of operators and Wick's theorem. See below link:

QFT notes by Tong

He first uses $i\epsilon$ prescription for the Feynman propagator. However then drops $i\epsilon$ terms to arrive at the final result (3.52).

To show this we first go to the center of mass frame and conclude this below which is not clear to me why it is so and why it allows dropping the $i\epsilon$ terms in the final result:

"...This ensures that the 4-momentum of the meson is $$k=(0, \vec{p}-\vec{p}^{\prime}) $$ so $k^2<0$..."

When we do the integral of (3.51) using the delta functions we cancel $k$ but it is not obvious to me how a center of mass frame brings it back and let us drop the $i\epsilon$.


1 Answer 1


In the COM you have $k^2=-(\vec p-\vec p')^2<0$. But $k^2$ is a scalar, and thus independent of the frame of reference. Therefore, we conclude that $k^2<0$ in any frame of reference. In particular, it is non-zero and you can safety drop the $i\epsilon$ term.

In a more general context, you are always supposed to drop the $i\epsilon$ terms: by definition, there is always a limit implicit: $$ \int\mathrm dk\ \text{something with $i\epsilon$}\equiv \lim_{\epsilon\to 0^+}\int\mathrm dk\ \text{that thing} $$

In other words, after you integrate over momenta you must drop the epsilon terms. Note that in tree amplitudes you have already calculated all the momenta integrals, which means that you need not include the $i\epsilon$ terms in tree diagrams.

  • $\begingroup$ Sorry but the question is about why or how in COM 4-momentum of meson is given as above. The non zero part already does make sense. $\endgroup$
    – user56963
    Nov 12, 2016 at 18:32
  • $\begingroup$ @VictorVMotti because $k=p_1'-p_1$, and therefore $k^0=E_1'-E=0$ (where $E=\sqrt{|\vec p|^2+m^2}=\sqrt{|\vec p'|^2+m^2}=E'$, because $\vec p=\vec p'$) $\endgroup$ Nov 12, 2016 at 18:35
  • $\begingroup$ @VictorVMotti no problem. If you have any more questions, feel free to ask $\endgroup$ Nov 12, 2016 at 18:39
  • $\begingroup$ So we note that the meson is destroyed and the momentum is carried by one of the nucleons after scattering. What happens to the second unclean in the process in terms of its 4-momentum? $\endgroup$
    – user56963
    Nov 12, 2016 at 18:46
  • $\begingroup$ @VictorVMotti sorry but I don't seem to understand your question... maybe you can rephrase it? $\endgroup$ Nov 13, 2016 at 12:43

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