## Momentum Matching in Transfer Reaction

In transfer reaction, the kinematics determines the scattering angles and momenta, but it does not tell the probability of the reaction. Not all angle have same cross section, because when a particle is being transfer to or pickup from a nucleus, the momentum has to be matched. It is like a spaceship need a proper angle and speed in order to orbit around the earth, it the incident angle or speed are not good, the spaceship will just go out or crash into the earth.

Suppose our reaction is $A(B,1)2$, $A$ is incident particle, $B$ is the target, $1, 2$ are the outgoing particles. The momentum transfer from $A$ via $2$ to $B$ or $1$ is

$\vec{q} = \vec{p}_2 - \vec{p}_A = \vec{p}_f - \vec{p}_i$

where $\vec{p}_i$ is initial momentum and $\vec{p}_f$ is final momentum.

Assume the reaction take place at the surface of nucleus $B$. The maximum angular momentum it can create is

$\vec{L} = \vec{r} \times \vec{q} \rightarrow L = r q$

In QM, the angular momentum must be $L = \sqrt{l(l+1)} \hbar$. Here we pause for unit conversion. In this blog, we usually using nuclear unit, that momentum in [MeV/c] and $\hbar = 197.327 \textrm{MeV fm/c}$, thus, to calculate the angular momentum, simple multiple the momentum in MeV/c and nuclear radius in fm. The radius of a nucleus is roughly $r = 1.25 A^{1/3} \textrm{fm}$.

So, if the momentum $\sqrt{(l+1)(l+2)} \hbar > r|q| > \sqrt{l(l+1)} \hbar$, the possible angular momentum is $\sqrt{l(l+1)} \hbar$ or the l-orbit.

In the transfer reaction, the equation for $\vec{p}_k$ and $\vec{p}_i$ are known. It is straight forward to calculate $r q / \hbar$, plot against scattering angle, and we look for the value for $\sqrt{l(l+1)}$.

Or, we can scale the y-axis by

$\displaystyle y' = \sqrt{\left(q\frac{r}{\hbar}\right)^2 + \frac{1}{4}} - \frac{1}{2}$

The new y-axis is in unit of $\sqrt{l(l+1)}$.

We can also plot the contour of momentum matching on the $p_k$ versus scatering angle. The momentum matching is

$\displaystyle q^2 = \frac{l(l+1)\hbar^2}{r^2} = p_f^2 + p_i^2 - 2 p_f p_i \cos(\theta)$

where $\theta$ is the scattering angle of particle $1$. Solve for $p_f$

$\displaystyle p_f = p_i \cos(\theta) \pm \sqrt{ \left( \frac{l(l+1) \hbar^2}{r^2} \right) - k_i^2 \sin^2(\theta) }$

We can plot the contour plot for difference $l$. and then overlap with the momentum and the scattering angle of the outgoing particle $2$.

a correction, the y-axis should be $p_2$, momentum of particle 2.

## Kinematics of Transfer Reaction

Another surprise that no post on the kinematics of transfer reaction!

Suppose the reaction is $A(B, 1)2$, where $A$ is the incident particle, $B$ is the target nucleus that is at rest, $1, 2$ are the outgoing particle.

The Q-value for the reaction is the difference between initial mass and final mass

$Q = m_A+ m_B - m_1-m_2$

When the incident energy is too low, lower then the Q-value (we will see the relationship later) , the reaction is impossible, because the nuclei-1 and 2 cannot be created.

The equation for the reaction is

$P_A + P_B = P_1 + P_2$

where $P_i$ are 4-momenta. The number of freedom is 2. The total unknown is 8 from $P_1, P_2$, the masses of particle-1, and -2 give 2 constrains, the equations give 4 constrains.

$P_A = ( E_A , 0, 0, k_A) = ( m_A + T_A, 0, 0, \sqrt{2 m_A T_A + T_A^2})\\ P_B = (m_B, 0, 0, 0)$

where $T_A$ is the incident energy. We first transform the system into center of momentum (CM) frame. Since the total momentum is zero in the CM frame, we construct a CM 4-momentum

$P_{cm} = (P_A+P_B)/2 = ( m_A+ m_B + T_A, 0, 0, k_A)$

The Lorentz beta is $\beta = k_A/(m_A+m_B+T_A)$.

The invariance mass of the CM 4-momentum is unchanged, which is equal to the total energy of the particles in the CM frame. We can check

$P_A' = ( \gamma E_A - \gamma \beta k_A, 0, 0, -\gamma \beta E_A + \gamma k_A) \\ P_B' = ( \gamma m_B , 0, 0, -\gamma \beta m_B )$

The total energy in CM frame is

$E_{tot} = \gamma( m_A+m_B + T_A) - \gamma \beta k_A = \sqrt{(m_A+m_B+T_A)^2 - k_A^2}$

After the reaction, the momenta of the particle-1 and -2 are the same an opposite direction in the CM frame.Balancing the energy and momentum, the momentum is

$\displaystyle k = \frac{1}{2 E_{tot}} \sqrt{(E_{tot}^2 - (m_1+m_2)^2)(E_{tot}^2 - (m_1-m_2)^2)}$

In the equation, if either of the bracket is negative, the momentum cannot be formed. The smaller term is

$E_{tot}^2 - (m_1+m_2)^2 = (m_A+m_B+T_A)^2 - k_A^2 - (m_1+m_2)^2 \\ =(m_A+m_B+T_A)^2 - k_A^2 - (m_A+m_B-Q)^2 \\ = 2m_BT_A + 2(m_A + m_B)Q - Q^2$

If we set the line for this quantity be zero,

$\displaystyle 2m_BT_A + 2(m_A + m_B)Q - Q^2 = 0 \\ T_A = \frac{Q^2 - 2(m_A+m_B)}{2m_B} \\ = \frac{1}{2m_B} ( (Q-m_A-m_B)^2 - (m_A+m_B)^2) \\ = \frac{1}{2m_B} ( (m_1+m_2)^2 - (m_A+m_B)^2)$

Thus, if the $T_A$ is smaller then that value, the reaction is impossible.

The 4-momenta of the outgoing particles can be constructed.

$P_1' = ( \sqrt{m_1^2 - k^2} , k \cos(\phi) \sin(\theta) , k \sin(\phi) \sin(\theta) , k \cos(\theta)) \\ P_2' = ( \sqrt{m_2^2 - k^2} , -k \cos(\phi) \sin(\theta) ,- k \sin(\phi) \sin(\theta) , -k \cos(\theta))$

Back to Lab frame,

$P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 + k^2} + \gamma \beta k \cos(\theta) \\ k \cos(\phi) \sin(\theta) \\ k \sin(\phi) \sin(\theta) \\ \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta)\end{pmatrix}$

For simplicity, set $\phi = 0$,

$P_1 = \begin{pmatrix} \gamma \sqrt{m_1^2 + k^2} + \gamma \beta k \cos(\theta) \\ k \cos(\phi) \sin(\theta) \\ 0 \\ \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta) \end{pmatrix}$

We can see, the locus in the momentum space is

$( k cos(\phi) \sin(\theta) , 0, \gamma \beta \sqrt{m_1^2 - k^2} + \gamma k \cos(\theta))$

This is an ellipse with length $k (1, \gamma)$, and centered at $(0, \gamma \beta \sqrt{m_1^2 + k^2})$. If the relativistic effect is small, it is a circle.

When the ellipse long axis $\gamma k$ is longer then the center position $\gamma \beta \sqrt{m_1^2 + k^2}$, the scattering angle can be to whole circle, i.e.

$\displaystyle \gamma \beta \sqrt{m_1^2 + k^2} < \gamma k \rightarrow \gamma \beta m_1 < k$

in here, we can treat $\gamma \beta m_1 = k_\beta(1)$, this is the momentum “gain” from shifting from CM frame to Lab frame. By compare $k_\beta(1)$ or $k_beta(2)$ to $k$ we can know the span of scattering angle.

Here is an example,

## 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.