# Cross section in Coulomb scattering of relativistic electrons

i'm currently reading about electron scattering and i cant understand the following statement of the book im reading. I quote the book(translated by me):

Many scattering experiments are done with unpolarized beams, and the polarization of the scattered particles are not measured. It is required that we take the mean value of the absolute square of the Matrix element, summed over both the spin states of the incoming electrons and the outgoing electrons.

$$\frac{d\sigma}{d\Omega}\sim \frac{1}{2}\sum_{s_i,s_f}|\bar{u_f}\gamma^0u_i|^2$$

I can't understand the factor of $$\frac{1}{2}$$ in from of the summation. The way im thinking it you need to replace it with $$\frac{1}{4}$$ because of the four different combinations of the initial and the final spin, which are:

$$+ \rightarrow +$$,

$$+ \rightarrow -$$,

$$- \rightarrow +$$,

$$- \rightarrow -$$

where $$+$$ is spin up and $$-$$ is spin down.

• does ++ or -- make a dipole? hyperphysics.phy-astr.gsu.edu/hbase/electric/dipole.html, note, "opposite charges" Oct 19, 2020 at 11:07
• no it's the spin of the electron before and after the scattering Oct 19, 2020 at 11:32
• missfead, sorry, Maybe Pauli exclusion does not allow ++ and --spin? just guessing. Oct 19, 2020 at 16:34
• I don't think so because the logic would still hold for a single electron of which we don't know the initial spin. Read the answer of doublefelix he made it very clear how this relation is derived! Oct 19, 2020 at 17:23

## 1 Answer

The usual way to do this is to average over the possible incoming states, but sum over the outgoing states. In that case the 1/2 is correct as there are 2 possible incoming states. This would be correct for the scattering cross section, since you don't know what particle you started with (thus the average) and you end up with all possible combinations (thus the sum). That is likely to be the reason for the 1/2.

• Your answers clears things up a lot, but im still confused as to why we don't divide by one half when we sum over the final states. To my mind the final states should be averaged as well, since either one or the other will occur Oct 19, 2020 at 12:00
• In the differential cross section, we want to divide the outgoing flux (per unit solid angle) by the ingoing flux (per unit area). Thinking about how this is counted, let's first consider the outgoing flux. We will get flux of spin 1/2 states and spin -1/2 states, but we won't know which is which. The total outgoing flux we get is the sum of the spin1/2 flux and the spin -1/2 flux. That is why we do not divide by anything for the final states. Oct 19, 2020 at 12:56
• As for the incoming flux, compared to a polarized beam of (for example) only spin 1/2 electrons, we do not have more electrons. We simply don't know what the polarization is. Looking at the contribution of an individual incoming electron, we expect it to be equally likely spin 1/2 and -1/2. The effective outgoing flux we see over many particles is therefore an average of the outgoing flux which would come from a spin1/2 particle, and the outgoing flux which would come from a spin-1/2 particle. Oct 19, 2020 at 12:58
• thank you, these two comments made it very clear Oct 19, 2020 at 13:52
• No we do not, we just get the final momentum but don't know the spin. Oct 19, 2020 at 19:01