Last post on optical model, we did not include the spin. to introduced the spin, we just have to modify the wave function. For spin-½ case.

$\begin {pmatrix} \psi_i \\ \psi_2 \end {pmatrix} \rightarrow Exp( i k r ) \begin {pmatrix} a_1 \\ a_2 \end {pmatrix} + \frac { Exp ( i k r) }{r} M \begin {pmatrix} a_1 \\ a_2 \end {pmatrix}$

where the M is a matrix:

$M = f + g \vec{ \sigma } \cdot \vec{n}$

the f is for the spin-Independence part of the wave function. For the incident wave and the scattered plane wave.

$\begin {pmatrix} a_1 \\ a_2 \end {pmatrix} = \begin {pmatrix} Exp( - i \phi_s /2 ) cos ( \theta_s /2 ) \\ Exp( i \phi_s /2 ) sin ( \theta_s /2 ) \end {pmatrix}$

where $\theta_s$ and $\phi_s$ are the angle of spin . not the detector angle.

after calculation by routine algebra, we have the scattered spherical wave.

$\chi = M \cdot \begin {pmatrix} a_1 \\ a_2 \end {pmatrix} =$ $\begin {pmatrix} (f+g)Exp( - i \phi_s /2 ) cos ( \theta_s /2 ) \\ (f-g)Exp( i \phi_s /2 ) sin ( \theta_s /2 ) \end {pmatrix}$

The expectation of the wavefunction, or the intensity of the spherical part will be:

$I(\theta_s) = \chi^{\dagger} \chi = |f|^2 + |g|^2 + 2 Re( f^* g ) cos( \theta_s)$

the beam polarization should be equal the intensity and normalized polarization.

$I P_z = \chi^{\dagger} \sigma_z \chi = ( |f|^2+ |g|^2 ) cos ( \theta_s) + 2 Re(f^* g)$

Thus, we have the induced polarization when incident beam is unpolarized:

$P_z ( \theta_s = \pi /2 ) = \frac { 2 Re ( f^* g ) }{ |f|^2 + |g|^2 }$

for a beam of many particle and formed an ensemble, the $\theta_s$ is the average.

and Analyzing power, which is a short term for Polarization Analyzing Power , or the spin asymmetry, is given by

$A_y=\frac { I(\theta_s = 0 ) - I( \theta_s = \pi ) } { I ( \theta_s = 0 ) + I ( \theta_s = \pi ) } = \frac { 2 Re( f^*g) }{ |f|^2 + |g|^2 } =P_z$

Therefore, in order to get the spin asymmetry, we have to use 2 polarized beams, one is up-polarized, and another is down-polarized, to see the different between the scattering result.

However, to have 100% polarized beam is a luxury. in most cases, we only have certain polarization. thus, the spin-asymmetry is not equal to the analyzing power. the spin-asymmetry $\epsilon$ is from the yield measurement.

$\epsilon = \frac { I(\theta_s) - I(\theta_s) }{ I(\theta_s ) + I(\theta_s) }$

since f and g only depend on the detector angle. and we can assume they are symmetry. Thus

$\epsilon = \frac {2 Re( f^* g ) }{|f|^2 +|g|^2 } cos ( \theta_s) = A_y P$

the P is the polarization of the target.