# Probability of spin-flip due to scattering

devd

## Homework Statement

A beam of spin-1/2 particles scatters off of a target consisting of spin-1/2 heavy nuclei. The interaction between the particle and nucleus is given by $$V(\vec{ r})= V_0~\delta (\vec{r})~ \vec{S}_1. \vec{S}_2$$

1) Averaging over initial spin states, find the differential scattering cross section.
2) Assuming the incident particles are polarized along ##+\hat{z}## and target nuclei are polarized along ##-\hat{z}##, find the probability that the spins of the incoming will have flipped.

## The Attempt at a Solution

I can do the first part, by constructing the triplet and singlet states and using the expression for scattering cross section in Born approximation and using the fact that the probabilities of triplet and singlet are 3/4 and 1/4.

I can begin by writing ##\vec{S}_1 . \vec{S}_2=S_{1z}S_{2z}+\frac{1}{2} S_{1+}S_{2-}+S_{1-}S_{2+}##. So, the potential couples the initial ##|\uparrow_{1}>|\downarrow_{2}>## state and the final ## |\downarrow_{1}>|\uparrow_{2}>## state.

For the probability I thought of using Fermi's Golden rule for the rate of transition and from there the probability. But, the whole setup of the scattering problem is essentially time independent, so this makes a bit hesitant. How do I go about it?

Mentor
I read the problem as asking to compare the probability of detecting the scattered particle as flipped with that of detecting it in its initial spin state.

I can begin by writing ##\vec{S}_1 . \vec{S}_2=S_{1z}S_{2z}+\frac{1}{2} S_{1+}S_{2-}+S_{1-}S_{2+}##.
There is a little mistake here.