## Very short introduction to Partial-wave expansion of scattering wave function

In a scattering problem, the main objective is solving the Schrödinger equation

$H\psi=(K+V)\psi=E\psi$

where H is the total Hamiltonian of the scattering system in the center of momentum, K is the kinetic energy and V is the potential energy. We seek for a solution $\psi$,

$\displaystyle \psi_{k}^{+}(r)=e^{i\vec{k}\cdot \vec{r}}+f(\theta)\frac{e^{ikr}}{kr}$

The solution can be decomposed

$\displaystyle \psi_{k}^{+}(r)=R_{l}(k,r)Y_{lm}(\theta,\phi)=\frac{u_{l}(k,r)}{kr}Y_{lm}(\theta,\phi)$

The solution of $u_{l}(k,r)$ can be solve by Runge-Kutta method on the pdf

$\displaystyle \left(\frac{d^2}{d\rho^2} + 1 - \frac{l(l+1)}{\rho^2} \right)u_{l}(k,\rho)=U(\rho)u_{l}(k,\rho)$

where $\rho=kr, k=\sqrt{2\mu E}/\hbar, \mu=(m_1+m_2)/(m_1 m_2)$ and $U=V/E$.

For $U = 0$, the solution of $u_l$ is

$\displaystyle u_{l}(k,r)=\hat{j}_l(\rho) \xrightarrow{r\rightarrow \infty} \sin(r') = \frac{e^{ir'}-e^{-ir'}}{2i}$

where $r' = kr-l\pi/2$ and $\hat{j}_l$ is the Riccati-Bessel function. The free wave function is

$\displaystyle \phi_k(r)=e^{i\vec{k}\cdot\vec{r}}=\sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ikr}i^l (e^{ir'}-e^{-ir'})$

where $P_l(x)$ is the Legendre polynomial.

Note that, if we have Coulomb potential, we need to use the Coulomb wave instead of free wave, because the range of coulomb force is infinity.

For $U\neq 0$, the solution of $u_l(r can be found by Runge-Kutta method, where R is a sufficiency large that the potential $V$ is effectively equal to 0.  The solution of $u_l(r>R)$ is shifted

$\displaystyle u_{l}(k,r>R)=\hat{j}_l(\rho)+\beta_l \hat{n}_l(\rho) \xrightarrow{r\rightarrow \infty} \frac{1}{2i}(S_l e^{ir'}-e^{-ir'})$

where $S_l$ is the scattering matrix element, it is obtained by solving the boundary condition at $r = R$. The scattered wave function is

$\displaystyle \psi_k(r)=\sum\limits_{l=0} P_l(\cos(\theta)) (2l+1) i^l \frac{u_l(r)}{kr}$

put the scattered wave function and the free wave function back to the seeking solution, we have the $f(\theta)$

$\displaystyle f(\theta) = \sum\limits_{l=0} P_l(\cos(\theta)) \frac{2l+1}{2ik} (S_l - 1)$

and the differential cross section

$\displaystyle \frac{d\sigma}{d\Omega}=|f(\theta)|^2$.

In this very brief introduction, we can see

• How the scattering matrix $S_l$ is obtained
• How the scattering amplitude $f(\theta)$ relates to the scattering matrix

But what is scattering matrix? Although the page did not explained very well, especially how to use it.

## Angular distribution of emission that carry angular momentum

Before decay, the nucleus is in state with total angular momentum J and symmetry axis quantization M :

$\Phi_{JM}$

Say, the emitted radiation (can be EM wave or particle ) carries angular momentum  l and axis quantization m, its wavefunction is:

$\phi_{lm}$

then the daughter nucleus has angular momentum j and $m_j$, the wave function is

$\Psi_{j m_j}$

their relation is:

$\Phi_{JM} = \sum_{m, m_j}{\phi_{lm} \Psi_{j m_j} \left< l m j m_j | JM \right>}$

where the $Latex \left< l m j m_j | JM \right>$ is Clebsch-Gordan coefficient.

The wave function of the emitted radiation from a central interaction takes the form:

$\phi_{lm} = A_0 u_{nl}(r) Y_l^m(\theta,\phi)$

The angular distribution is:

$\int |\phi_{lm}|^2 dx^3 =A_0^2 \int |u_{nl}|^2 r^2 dr |Y_l^m|^2$

for a fixed distance detector, the radial part is a constant. Moreover, not every spherical harmonic contribute the same weight, there is weighting factor due to Clebsch-Gordan coefficient.   Thus, the angular distribution is proportional to:

$W(\theta) \propto \sum_{m_j=M-m}{|Y_l^m|^2 |\left|^2 }$

For example, JM=00, The possible (l, j) are (0,0), (1,1), (2,2) and so on, the $m=-m_j$. The C-G coefficient are,

$\langle 0000 | 00 \rangle = 1$

$\langle lml-m | 00 \rangle = \frac{1}{\sqrt{2l+1}}$

thus,

$Y_0^0 = \frac{1}{4\pi}$

$\displaystyle \sum_{m}{\left|Y_l^m\right|^2 \left|\langle l m l -m |0 0 \rangle \right|^2 } = \sum_m|Y_l^m|^2 \frac{1}{2l+1}= constant$

Thus, the angular distribution is isotropic.

## Density Matrix and Differential Cross Section

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:

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

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

## Rotation symmetry on scattering amplitude

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.

## rotation and parity symmetry on scattering amplitude

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.

## scattering with spin

from previous post, the scattering matrix for non-spin case is:

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

for a spin case, and notices that the real space and spin apace are not the same but a general state is wriiten by this:

$\left|\psi\right> = \left| x \right> \otimes \left|\chi\right> = \left| x, \chi \right>$

thus, the modification for the scattering matrix is:

$\left = \delta(p - p') \delta_{\chi' \chi} + \frac{i}{2\pi m} \delta(E_{p'} - E ) f(p' , \chi' \leftarrow p, \chi)$

and the differential cross section is:

$\frac{d\sigma}{d\Omega}=|f(p',\chi' \leftarrow p,\chi)|^2$

for a particular spin state, it can be expanded into a combination of the eigen spin state χ.

$\left| \xi \right> = \sum \xi_{\chi} \left| \chi \right>$

thus, from a particular spin state to another particular spin state will be:

$\frac{d\sigma}{d\Omega}=|\sum \xi'_{\chi'} f(p',\chi' \leftarrow p,\chi) \xi_{\chi}|^2$

the scattering amplitude now becomes a matrix, we define:

$F(p' \leftarrow p)=f_{\chi',\chi}(p' \leftarrow p)=f(p',\chi' \leftarrow p,\chi)$

and the d.c.s. becomes:

$\frac{d\sigma}{d\Omega}=|\xi'^\dagger F(p' \leftarrow p)\xi|^2$

for example, spin 1/2 cases. the scattering amplitude matrix is:

$F(p' \leftarrow p) = \begin{pmatrix} f_{++} & f_{+-} \\ f_{-+} & f_{--} \end{pmatrix}$

we can see that the diagonal terms are non-spin flip, while the off diagonal terms are spin flipped. it should be clear that:

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

$|\xi_{out}|^2 = \sum_{\chi} |\xi_{\chi' , out}|^2 = \sum | \sum f_{\chi', \chi} \xi_{\chi', out}|^2$

sub it into the d.s.c., we have:

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

for example, if we have an initial state is spin up, and we only measure the down state, thus:

$\frac{d\sigma}{d\Omega}(p', - \leftarrow p, +) =\frac{1}{2}|f_{+-}|^2$

the half is from the fact  that, each $\xi_{out} - \xi_{in}$ only contribute $\frac{1}{(2s_1+1) (2s_2+1)}$, where s is the spin of target of incident beam. in the example, target spin is 0, incident beam is 1/2 or vice via.

## Scattering Matrix

at the point of scattering ( t = 0 ), the wave function and the incoming and out-going wave function can be related as:

$\left| \psi_{in} \right> \rightarrow \left| \psi \right> \leftarrow \left| \psi_{out}\right>$

where the incoming and out-going wavefunction is very far away from the field of the scatter, and thus, they are free and we say they are asymptotic.

Let a time propagator with a full Hamiltonian be U(t), and a time propagator with free Hamiltonian be $U^0(t)$. thus the time behavior of the scattering wave functions can be related as:

$U(t) \left| \psi \right> \rightarrow U^0(t) \left| \psi_{in} \right>$ for $t \rightarrow - \infty$

$U(t) \left| \psi \right> \rightarrow U^0(t) \left| \psi_{out} \right>$ for $t \rightarrow + \infty$

in equation:

$\left| \psi \right> = lim_{t \rightarrow - \infty} U^\dagger (t) U^0(t) \left| \psi_{in} \right> = \Omega_+ \left| \psi_{in} \right>$

$\left| \psi \right> = lim_{t \rightarrow + \infty} U^\dagger (t) U^0(t) \left| \psi_{out} \right> = \Omega_- \left| \psi_{out} \right>$

Thus, we have the scattering matrix or the S-matrix.

$\left| \psi_{out} \right> = \Omega_-^\dagger \Omega_+ \left|\psi_{in} \right> = S\left| \psi_{in} \right>$

Now assume for a particular state generated by an accelerator is Φ, and a particular out-going asymptotic state is χ. we have:

$\left| \phi \right> = \Omega_+ \left| \phi_+ \right>$

$\left| \chi \right> = \Omega_+ \left| \chi_- \right>$

thus, the probability amplitude for the scattering between these 2 states is:

$\omega( \chi \leftarrow \phi) = | \left<\chi_- | \phi_+ \right> |^2 =| \left<\chi|\Omega_-^\dagger \Omega_+| \phi \right> |^2 = |\left< \chi | S | \phi \right>|^2$

if we expand a wave function  in momentum basis:

$\left| \psi \right> = \int d^3p \left| p \right> \left< p | \psi \right>$

$\psi_{out}(p) = \int d^3p' \left

\psi_{in}(p')$

now, we are going to show the energy conservation of the scattering operator or matrix, by showing that the scattering operator S commutes with the Hamiltonian. from

$Exp(\frac{i}{\hbar} H \tau) \Omega_\pm =\Omega_\pm Exp( \frac{i}{\hbar} H^0 \tau )$

differential it then we have :

$H \Omega_\pm = \Omega_\pm H^0$

thus,

$H^0 = \Omega_\pm^\dagger H \Omega_\pm$

$S H^0 = H^0 S$

together with the wavefunction:

$0 = \left = (E_{p'} -E_p ) \left< p'|S|p\right>$

thus implies,

$\left = \delta(E_{p'}-E_p) g( p' \leftarrow p)$

since, at the forward direction,  the change of momentum is zero, we can write S = 1 + R, then,

$\left =\delta(p-p') - 2 \pi i \delta(E_{p'}-E_p) t( p' \leftarrow p)$

the $t(p' \leftarrow p)$ is called on-shell T-matrix. since the energy must be equal, required by the delta function, thus, the momentum magnitude must be equal, therefore, the 2 momentums s on a shell. The T-matrix also related to the scattering amplitude by:

$f(p' \leftarrow p) = - (2 \pi)^2 m t(p' \leftarrow p)$

then the S-matrix becomes,

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