From previous post, the scattering amplitude matrix is :

$F(p' \leftarrow p ) = a I + i b \hat{n} \cdot \sigma$

let the in- density matrix be for our spin ½ ensemble is :

$\rho_{in} = \frac{1}{2} ( I + \vec{\pi}_{in} \cdot \sigma )$

then the out-matrix is:

$\rho_{out} = F \rho_{in} F^\dagger$

using:

$(u\cdot \sigma ) (v \cdot \sigma ) = u\cdot v I + i u \times v \cdot \sigma$

with some messy algebra, and focus out pull out the identity matrix and the Pauli’s matrix, we will get:

$\rho_{out} = \frac{1}{2} ( |a|^2 + |b|^2 +2 Im( a b^*) \hat{n} \cdot \vec{\pi}_{in}) ( I + \vec{\pi}_{out}\cdot \sigma)$

the form of out-polarization vector takes very complicate form. however, this is not bother in the differential cross section, coz it only care the trace.

$\frac{d\sigma}{d\Omega} ( p' \leftarrow p, \vec{\pi}_{in} ) = ( |a|^2 +|b|^2 ) ( 1+ v(\theta) \hat{n}\cdot \vec{\pi}_{in} )$

$v(\theta) = 2 \frac{ Im( a b^*)}{|a|^2+|b|^2}$