0
$\begingroup$

I was trying to figure out the cross section $\frac{d\sigma}{d\Omega}$ for spinless $e^{-}\gamma\rightarrow e^{-}$ scattering. First I wrote the terms associated with each component.

Vertex: $$ie(P_A+P_B)^{\mu}$$ External Boson: $1$

Photon: $\epsilon_{\mu}$

Multiplying these will give the invariant amplitude. $$i\mathcal{M} =ie(P_A+P_B)^{\mu}\epsilon_{\mu}$$ Now consider the momenta in high energy approximation $$P_A =(p,P)$$ $$P_B=(p,P')$$ Such that $|P|=|P'|=p$ Then $$P_A+P_B=(2p,P+P')$$ Now squaring $\mathcal{M}$ $$\mathcal{M}^2 = e²(6p^2+2p^2\cos\theta)\epsilon^2$$ The differential cross section will become: $$\frac{d\sigma}{d\Omega}=\frac{p^2e²}{32\pi^2s}(3+\cos\theta)\epsilon^2$$

Now I have two questions:

1) What have I done wrong? I couldn't find the answer anywhere online , is there something obvious that I am missing? I know I am wrong because $\epsilon^2$ is a $3\times 3$ matrix. A cross section can't be a matrix (As far as I know).

2) What will $s$ be? In the book Martin and Halzen the definition $s$ was simply $$s=(P_A+P_B)^2$$ But $s$ in Martin and Halzen was defined in the case of two vertex diagram. What will be the definition of $s$ in single vertex diagram?

$\endgroup$
  • $\begingroup$ Normally when you square the matrix element you use the polarization sum rule for the photo polarization vectors $\sum_{\lambda \lambda'} \epsilon_{\mu}(\lambda)\epsilon_{\nu}^*(\lambda') = -g_{\mu \nu}$. $\endgroup$ – Triatticus May 17 at 17:58
  • $\begingroup$ @Triatticus So you're saying I'll get a factor of 4 because $$g_{\mu\nu}g^{\mu\nu}=4$$ Is this what you're saying? $\endgroup$ – Manvendra Somvanshi May 19 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.