As we mentioned before, a density matrix with weight w for an ensemble  is in a from:

$\rho = \sum w_i \left|\psi_i \right> \left< \psi_i \right|$

For any Operator Q, the expectation value is:

$\left_{\rho} = Tr[ A \rho ] = Tr [ \rho A ]$

for a random polarization state:

$\rho = \sum \frac{1}{2s+1} \left|m\right>\left

where I is identity matrix. any density matrix can be expressed by the polarization vector π.

$\rho = \lambda ( I + \vec{\pi} \cdot \sigma )$

where λ is a normalization factor and σ is the Pauli’s matrix.

from this post, we knew that the out-spin and in-spin is related by the “scattering amplitude matrix” F.

$\xi_{out} = F(p' \leftarrow p) \xi_{in}$

the in-density matrix for an ensemble is:

$\rho_{in} = \sum w_i \left| \xi_{in}^i\right> \left< \xi_{in}^i\right|$

thus, the out-density matrix is:

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

which is just the usual transformation rule for density matrix or a general tensor.

For a single particle, we have :

$\frac{d \sigma}{d \Omega} (p' \leftarrow p, \xi_{in} ) = | \xi_{out} |^2$

$=\sum\left<\chi|\xi_{out}\right>\left<\xi_{out}|\chi\right>=Tr[ \left|\xi_{out} \right> \left< \xi_{out} \right|$

For an ensemble, we have:

$\frac{d \sigma}{d \Omega} (p' \leftarrow p,\rho_{in})=\sum w_i \frac{d\sigma}{d\Omega} (p' \leftarrow p, \xi_{in}^i )$

$=\sum w_i Tr[\left| \xi_{out}^i \right>\left< \xi_{out}^i \right| ] = Tr [\rho_{out}]$