a scattering amplitude with spin can be more specific if we impose some symmetry on it. for now, we like to impose the rotation and parity.

the rotation operator act on a state of momentum p and spinor χ should be in this way:

$R \left| p, \chi \right> = \left|p_R, \chi_R \right>$

with out lose of generality, we set the incoming momentum lay on z axis, out target is on the origin. thus, the out-going momentum is making some angle (θ, φ ) on the coordinate system. and we also set the spinor takes z-axis as spin up, thus we write a particular state with spin z-component as m, this specify the spin. we we rotate the system by -φ on z-axis, making the out-going momentum lay on x-z plane. thus the out-going state becomes:

$R \left| p', m' \right> = e^{i \phi m' } \left|p_R, m' \right>$

and the incoming state,

$R\left|p,m\right> = e^{ i \phi m} \left|p,m\right>$

the phase is come from the spin, with the rotation operator $Exp( - \frac{i}{\hbar} \phi J_z )$.

due to rotation symmetry, the scattering amplitude must be the same after rotation. and since conservation of energy, the amplitude must only be depended on energy, and scattering angle, thus, we have:

$\left = e^{i \phi (m -m') )}\left< p' , m'| S| p, m \right>$

$F_{m'm}(E, \theta, \phi) = e^{i \phi (m-m' )} F_{m'm}(E, \theta, 0)$

Thus, the off-diagonal element of the matrix has a phase. for spin half case:

$F(E,\theta, \phi) = \begin{pmatrix} f_{++} & f_{+-}e^{-i \phi} \\ f_{-+}e^{i\phi} & f_{--}\end{pmatrix}$

For parity invariant, the parity operator only change to space but not the spin. thus:

$P \left| p, m \right> = \left| - p , m \right>$

if we express the momentum in spherical coordinate, with its length is same as energy:

$P \left| E , \theta, \phi \right> = \left| E, \pi - \theta, \pi + \phi \right>$

which is not so nice and not easy to compare with the original scattering amplitude. thus, we rotate the system by π around y -axis. thus, the result of  rotating after parity is just a mirror of plane x-z, which is just a rotation. the rotation on y -axis is:

$R_y(\pi) \left|s, m\right>= (-1)^{s-m}\left|s, -m\right>$

where s is the length of the spinor. thus, the total transform for incoming state is:

$R_y(\pi) P \left|p, s , m \right> = (-1)^{s-m} \left|p, s, -m \right>$

for out-going state is

$R_y(\pi) \left|p', s , m' \right> = (-1)^{s-m'} \left| p_m , s, -m' \right>$

where $p_m$ is the mirror of $p'$. therefore, the scattering matrix is:

$\left = (-1)^{m'm} \left$

if we disregard the sign of azimuth angle, the scattering amplitude matrix is:

$F_{m'm}(E,\theta, \phi) = (-1)^{m-m'} F_{-m',-m}(E,\theta, \phi)$

for spin half case,

$F(E,\theta, \phi) = \begin{pmatrix} a & be^{-i \phi} \\ -be^{i\phi} & a \end{pmatrix}$

or, we can write in a simpler form:

$F = a I + i b \hat{n}\cdot \sigma$

where $\vec{n} = p \times p'$ and $\sigma$ is Pauli matrix vector.