# Spinless $e^{-}\gamma\rightarrow e^{-}$ Cross section

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?

• Normally when you square the matrix element you use the polarization sum rule for the photo polarization vectors $\sum_{\lambda \lambda'} \epsilon_{\mu}(\lambda)\epsilon_{\nu}^*(\lambda') = -g_{\mu \nu}$. May 17 '19 at 17:58
• @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? May 19 '19 at 9:23

Your expression for $$\mathcal{M}^2$$ is wrong. Inside $$\mathcal{M}$$ polarisation vectors are contracted with the momenta so for example $$\left|(P + P')^\mu \epsilon_\mu\right|^2 =(P + P')^\mu \epsilon_\mu \, (P + P')^\nu \epsilon_\nu =(P + P') \cdot \epsilon \, \,(P + P') \cdot \epsilon$$ It seems that you incorrectly contracted the $$(P + P')$$ factors with themselves and were left with $$\epsilon$$ vectors you didn't know what to do with.

The expected probability density for Compton scattering is

$$\begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega} +\left(\frac{m}{\omega}-\frac{m}{\omega'}+1\right)^2-1 \right) \end{equation*}$$

where $$\omega$$ is the angular frequency of the incident photon, $$\omega'$$ is the angular frequency of the scattered photon, and $$m$$ is electron mass.

The Compton formula is $$\begin{equation*} \frac{1}{\omega'}-\frac{1}{\omega}=\frac{1-\cos\theta}{m} \end{equation*}$$

It follows that $$\begin{equation*} \cos\theta=\frac{m}{\omega}-\frac{m}{\omega'}+1 \end{equation*}$$

Then by substitution $$\begin{equation*} \langle|\mathcal{M}|^2\rangle= 2e^4\left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega}+\cos^2\theta-1 \right) \end{equation*}$$

The differential cross section is $$\begin{equation*} \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{2m^2} \left(\frac{\omega'}{\omega}\right)^2 \left( \frac{\omega}{\omega'}+\frac{\omega'}{\omega}+\cos^2\theta-1 \right) \end{equation*}$$

For $$\omega\approx\omega'$$ $$\begin{equation*} \frac{d\sigma}{d\Omega} \approx \frac{\alpha^2}{2m^2} \left( \cos^2\theta+1 \right) \end{equation*}$$

See the following link for more on $$\langle|\mathcal{M}|^2\rangle$$ for Compton scattering.

http://www.eigenmath.org/compton-scattering.pdf

• OP is asking about interaction involving one photon, the fundamental vertex, not the two photon Compton amplitude
– lux
Jan 16 '20 at 0:51