Nucleon scattering and the meson 4-momentum in scalar Yukawa

Question

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 $i\epsilon$.

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.