What is the energy of a photon in an electron-muon scattering?

Question

Currently I am reading about this process in an Introduction to Quantum Field Theory by Peskin and Schroeder (pages 153-154). It should be mentioned that they are working in a center-of-mass (CM) frame and they make an approximation that the mass of the electron in equal to zero ($m_e = 0$). To understand my confusion let's look at the first diagram. For four-momentums $p_1$, $p'_1$, $p_2$, $p'_2$, $q$ we can write the following: $$p_1 = p'_1 + q$$ $$p_2 = p'_2 - q$$ Which means that for zero's (energy) components we have the following equations: $$E_{p_1} = E_{p'_1} + E_{q}$$ $$E_{p_2} = E_{p'_2} - E_{q}$$ This means that after the scattering process the energy of an electron ($E_{p'_1}$) must decrease and the energy of a muon ($E_{p'_2}$) must increase: $$E_{p'_1} = E_{p_1} - E_{q}$$ $$E_{p'_2} = E_{p_2} + E_{q}$$ However, if you look at diagram 2 you will see that the energy of the electron after the scattering remains the same (they denote it as k). The same is true for the energy of the muon.

Therefore, it appears that they can exchange only a photon with ZERO ENERGY!? Does it make sense or my logic is flawed somewhere?

Does it mean that as long as the scattering process is ELASTIC the energy of the exchanged particle(or particles) must be zero?!

My derivation of this result is the following:

1. Electron and muon before and after the scattering are on shell (meaning the energy-momentum relation is true: $E^2 = m^2 + \textbf{p}^2$ in the natural units)
2. We are working in the center of mass frame so $\textbf{p}_1 = - \textbf{p}_2$ and $\textbf{p}'_1 = - \textbf{p}'_2$
3. The total energy before and after the scattering must remain the same: $$E_{p_1} + E_{p_2} = E_{p'_1} + E_{p'_2}$$ Which means the following: $$ \sqrt{m_e^2 + \textbf{p}_1^2} + \sqrt{m_{\mu}^2 + \textbf{p}_1^2} = \sqrt{m_e^2 + \textbf{p}{'}_1^{2}} + \sqrt{m_{\mu}^2 + \textbf{p}{'}_1^2}$$ Thus, since $\textbf{p}_1^2$, $\textbf{p}{'}_1^2$ are nonnegative it must be true that $|\textbf{p}_1|=|\textbf{p}{'}_1|$ which gives us that the energy of the electron after the scattering is the same as before the scattering and the energy of the muon before and after the scattering also remains the same: $$E_{p_1} = E_{p'_1} $$ $$E_{p_2} = E_{p'_2} $$ If we go back to the previous equations: $$E_{p_1} = E_{p'_1} + E_{q}$$ $$E_{p_2} = E_{p'_2} - E_{q}$$ Therefore, it looks like $E_{q}$ must be ZERO. Which seems kind of weird.

Answer 1

Your reasoning is correct* and indeed \begin{equation} E_q=0 \,. \end{equation}

Note however that only the external legs of the diagram correspond to the asymptotic states that can be measured, while instead the internal photon of moment $q$ is not. Moreover, it is in general off-shell and neither does $E_q$ need to be positive. So there is no physical relevance in the energy of the virtual photon being zero nor you should think it is wierd, it comes simply from how you are instructed to perform a perturbative computation.

* Actually, since $E_q$ need not be positive, you are wrong in assuming $E_{p'_1}<E_{p_1}$ and $E_{p'_2}>E_{p_2}$. That does not affect your conclusion however.