Last time, we saw how symmetry fixed the form of scattering amplitude. Now, the rotation symmetry also impose the scattering amplitude as a series. to see this, we first need to know, the energy- angular momentum state is an eigen state of scattering matrix.

$S \left| E, l, m \right> = s_l(E) \left| E, l, m \right>$

because of the conservation of energy and conservation of angular momentum by a central potential. and a momentum state can be expressed as:

$\left< p| E, l, m\right> = \frac{1}{\sqrt{mp}} \delta(E_p-E)Y_{lm}(\hat{p})$

thus, the scattering amplitude is :

$\delta(E_{p'}-E_p) f( p' \leftarrow p) = \frac{2 \pi}{i} m \left$

$\left = \int dE \sum \left\left$

$= \frac{1}{mp} \int dE \sum \delta(E_{p'}-E) \delta(E_p-E) Y_{lm}(\hat{p'}) (s_l(E)-1) Y_{lm}^*(\hat{p})$

$= \frac{1}{mp} \sum \delta(E_{p'}-E_p) Y_{lm}(\hat{p'}) (s_l(E_p)-1) Y_{lm}^*(\hat{p})$

if we set the incoming momentum lay on z-axis. thus:

$Y_{lm}(\hat{p}) =0$ for $m\neq 0$

and using:

$\sum_m Y_{lm} (\hat{p'}) Y_{lm}^*(\hat{p}) = \frac{2l+1}{4\pi} P_l(cos(\theta))$

thus, we have:

$f(E_p, \theta) = f(p' \leftarrow p) = \frac{2\pi}{ip} \sum_l (2l+1) (s_l(E_p)-1) P_l(cos(\theta))$

if we define the Partial-wave amplitude:

$f_l(E_p,\theta) = \frac{s_l(E_p)-1}{2ip} = \frac{1}{p} sin(\delta_l)$

since the S is unitary, the eigen value should be simply $e^{i 2 \delta_l}$. thus, we have:

$f(E_p, \theta) = \sum (2l+1) f_l(E_p,\theta) P_l(cos(\theta))$

this result means the scattering amplitude can be de-composited into Legendre polynomial. the total scattering cross section is:

$\sigma = \int |f(E_p,\theta)|^2 d\Omega$

by using the orthogonal properties,

$\int_{-1}^{+1} P_l(x) P_l'(x) dx = \frac{2}{2l+1} \delta_{l'l}$

then, we have a partial cross section.

$\sigma = \sum_l \sigma_l$

$\sigma_l = 4\pi (2l+1)| f_l(E)|^2= 4\pi(2l+1) \frac{ sin^2(\delta_l)}{p}$

from the last equality, we have a boundary for partial cross section:

$\sigma_l \leq \frac{2\pi (2l+1)}{p}$

this inequality is called unitarity condition.