# Analyzing electron-positron scattering cross section in detail

It can be found in any quantum electrodynamics book that the cross section for electron-positron scattering in the high energy limit is $$\dfrac{\mathrm{d} \sigma}{\sin \theta \, \mathrm{d} \theta} = \frac{k}{E^2} \left( \dfrac{1+\cos^{4}\theta/2}{\sin^{4} \theta/2} + \dfrac{1+\cos^{2} \theta/2}{2} - 2 \dfrac{\cos^{4} \theta/2}{\sin^{2} \theta/2} \right)$$ where $k$ is a constant. The first term is due to the $t$-channel scattering, second from the $s$-channel annihilation, and the third is the interference of the two. Below is a Mathematica plot of three terms plotted separately.

Blue: 1st term ($s$-channel annihilation), Pink: 2nd term ($t$-channel scattering), Green: 3rd term (interference)

My questions are:

1. Why is the $s$-channel contribution so low?

2. If you see graph carefully, the $s$-channel curve is not monotonously increasing/decreasing as you can see below:

1. What exactly does interference mean?
• seeking_infinity, please think twice before "correcting" s-channel and t-channel in the question again. The original question had it all upside-down. – adavid Dec 30 '15 at 23:18

So, first make sure to look through the basics of Bhabha scattering and then let's start from the end:

In quantum mechanics, amplitudes add linearly. In this case, the $t$-channel scattering and $s$-channel amplitudes, that can be neatly described by the corresponding Feynman diagrams, are added to give the full amplitude for the $e^+e^-\to e^+e^-$ process. In this case the amplitudes have opposite signs (see basics), such that $\mathcal{M}=\mathcal{M}_t-\mathcal{M}_s$.

Since observables are proportional to the modulo of the amplitude squared, you will have something like $|\mathcal{M}|^2=|\mathcal{M}_t-\mathcal{M}_s|^2=|\mathcal{M}_t|^2+|\mathcal{M}_s|^2-2|\mathcal{M}_t\mathcal{M}_s|$.

So that last piece is the result of the (quantum mechanical) interference of the amplitudes.

For a more visual explanation have a look at pages 35-38 of this lecture.

## About the size of the $s$-channel annihilation contribution

First of all, it is not small. The issue is that since what we labelled the "$t$-channel scattering contribution" is divergent at low angles (more on that below,) it looks smaller. But note that only the sum of all three pieces is observable, so this sort of per-piece analysis is flawed to some extent.

## About the shape of the $s$-channel annihilation contribution

Regarding the shape, let me first discuss the $t$-channel scattering shape: it is monotonically changing, rapidly increasing to small scattering angles as you correctly observed. That is intuitively understandable since the probability of not scattering at all should be huge: why scatter when you can just do nothing? Of course, as hinted above, looking at a single amplitude-squared is bogus.

Now, in the case of $s$-channel annihilation, the situation is very different. You are not scattering, but you actually have an intermediate product of the annihilation, typically a virtual photon (that can go through the Z boson pole). It is then this virtual photon (a spin-1 "particle") that decays. That decay has the well-known $(1+\cos^2\theta)$ dependency. Now, I am not sure about your calculations, in particular the factors $1/2$ in all the $\theta/2$ in the right-hand-side but that may be the reason you get what you get and not something more familiar. Experimentalists usually plot $x=\cos\theta$, between $x=-1$ and $x=+1$ and you can see the result in page 37 of this lecture.

• I know the relative minus sign in $\mathcal{M_{t}}$ and $\mathcal{M_{s}}$ comes because of exchange in two fermions. But I am not comfortable with the idea that we are subtracting the amplitude of two possible processes. – seeking_infinity Jan 1 '16 at 7:38
• There is nothing being subtracted; there is only the amplitudes having opposite signs. (Destructive) interference is a very basic property even of classical wave mechanics: en.wikipedia.org/wiki/Interference_(wave_propagation) – adavid Jan 1 '16 at 15:56