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I'm reviewing page 59 of the QFT notes here and am a little confused by a reference frame argument. You can compute the second order probability amplitude term for nucleon-nucleon scattering to be

$$-ig^2\left[\frac{1}{(p_1-p_1')^2-m^2+i\epsilon}+\frac{1}{(p_1-p_2')^2-m^2+i\epsilon}\right](2\pi)^4\delta(p_1+p_2-p_1'-p_2')$$

in a scalar field theory approximation. Now the author argues that we may remove the $i\epsilon$ terms by moving to the centre of mass frame.

Here he says that $p_1=-p_2$ in this frame and that $|\vec{p_1}|=|\vec{p_1'}|$ by conservation of momentum. He continues to claim that the four-momentum of the meson is hence $k=(0,\vec{p}-\vec{p'})$ so $k^2<0$. Quite what $p,p'$ are I don't know exactly.

I don't understand this argument at all. Surely in the centre of mass frame, the sum of all momenta $p_i$ and $p_i'$ is zero (.)? Also where has the second constraint come from? I don't see how morally you could get more than my claim (.).

Could someone explain this argument to me? Very many thanks in advance.

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Be careful about distinguishing three- and four-vectors. Using boldface letters for 3-vectors, we have in the CM frame $\mathbf{p_1} + \mathbf{p_2} = 0$ (as 3-vectors), but $p_1^0, p_2^0 > 0$ (you can compute these using the on-shell condition), so $p_1 \neq -p_2.$ If you write $p := |\mathbf{p_1}| = |\mathbf{p_2}|,$ $p' := |\mathbf{p'_1}| = |\mathbf{p'_2}|$, then you can show that $p = p'$ due to conservation of energy (please do this carefully). – Vibert Dec 27 '12 at 14:02

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The transferred 4-momentum $k = p - p'$ is a difference (not a sum) and is a momentum of the meson. In the CM reference frame it has only space coordinates, so its square is negative: $k^2=0^2-(\vec{p}-\vec{p}')^2=-(2\vec{p})^2<0$.

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I think my problem is that I don't know how $p$ is related to $p_i$ and $p_i'$. Could you expand on that? And why does it have only space coordinates in the centre of mass frame? I understand if we're talking about the centre of mass for the meson itself, but aren't we interested in the centre of mass for the whole system? Sorry if I'm missing something simple! Many thanks. – Edward Hughes Dec 27 '12 at 13:56
The system is a couple of nucleons with the same mass. 4-momentum conservation law reads: $p_1+p_2=p'_1+p'_2$. The transfered momentum is by definition $k=p_1-p'_1$. In the CMRF $\vec{p}_1=-\vec{p}_2$. – Vladimir Kalitvianski Dec 27 '12 at 14:03
So in his exposition and your answer there is a subscript 1 missing from the $p$ then? Or have I missed something. Thanks a lot. – Edward Hughes Dec 27 '12 at 14:06
Yes, you can add a subscript 1 to $p$. Omitting the subscript is due to fact that nearly the same expression occurs to $p_2$ too because of 4-momentum conservation. – Vladimir Kalitvianski Dec 27 '12 at 14:10
Ah right that makes much more sense now. I still don't see why the first components of $p_1$ and $p_1'$ are necessarily the same though. I presume it's energy conservation, but why are you allowed to just do that locally? – Edward Hughes Dec 27 '12 at 14:13
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