## Absolute polarization measurement by elastic scattering

The magnitude of proton polarization can be measured by NMR technique with a reference. Because the NMR gives the free-induction decay signal, which is a voltage or current. For Boltzmann polarization using strong magnetic field and low temperature, the polarization can be calculated. However, when a reference point is not available, the absolute magnitude of proton polarization can be measured using proton-proton elastic scattering. The principle is the nuclear spin-orbital coupling. That creates left-right asymmetry on the scattering cross section.

Because of spin-orbital interaction:

$V_{ls}(r) = f(r) \vec{l} \cdot \vec{s} ,$

where $f(r)$ is the distance function, $\vec{l}$ is the relative angular momentum, $\vec{s}$ is the spin of the incident proton. In the following picture, the spin of the incident proton can be either out of the plane ($\uparrow$ ) or into the plan ($\downarrow$). When the proton coming above, the angular momentum is into the plane ($\downarrow$). The 4 possible sign of the spin-orbital interaction is shown. We can see, when the spin is up, the spin-orbital force repulses the proton above and attracts the proton below. That creates an asymmetry in the scattering cross section.

The cross section is distorted and characterized using analysing power $A_y$. Analyzing power is proportional to the difference between left-right cross-section. By symmetry (parity, time-reversal) consideration, $A_y = 1 + P sin(2\theta)$ (why?), in center of mass frame. In past post, the transformation between difference Lorentz frame. The angle in the $A_y$ has to be expressed in lab angle. The cross section and $A_y$ can be obtained from http://gwdac.phys.gwu.edu/ .

In scattering experiment, the number of proton (yield) is counted in left and right detectors. The yield should be difference when either proton is polarized. The yield is

$Y(\theta, \phi) = L \epsilon \sigma_0 (1 + cos(\phi)A_y(\theta) P) ,$

where $L$ is the luminosity, $\epsilon$ is the detector efficiency, $\sigma_0$ is the integrated cross-section of un-polarized beam and target of the detector, $P$ is the polarization of either the target or beam. When both target and the beam are polarized, the cross section is

$\sigma = \sigma_0 (1 + (P + P_T)A_y + P P_T C_yy),$

where $C_yy$ is spin-spin correlation due to spin-spin interaction of nuclear force.

Using the left-right yield difference, the absolute polarization of the target or the beam can be found using,

$\displaystyle A_y P = \frac{Y_L - Y_R}{Y_L + Y_R} ,$

where $Y_L = Y(\phi =0)$ and $Y_R = Y(\phi=\pi)$.

## Cross section IV

A more microscopic view of the cross section is

$\displaystyle \frac{dN}{dS} \sigma \rho dt dA= dn$,

where $N$ is the number of incident particle, $S$ is the area of the beam, $\sigma$ is the total cross section, $\rho$ is the particle density in the target, $A$ is the area of the target, $t$ is the thickness of the target, and $n$ is the particle detected.

The cross section is a constant, so, after integration on the target area and thickness,

$\displaystyle \sigma = \frac{n}{N \rho t}$.

If the beam is not uniform, $dN/dS = f(S)$ is a function of $S$. The integration has to be careful.

## Cross seciton of Coulomb Scattering

The coulomb scattering looks very easy, the formula of the differential cross section in CM frame is,

$\frac{d\sigma}{d\Omega} = (\frac{Z1 Z2 e}{4 E_{cm}})^2 \frac{1}{sin^4(\theta_{cm}/2)}$,

where $e = 1.44 MeV fm$ and $1 fm^2 = 10 mb$. The tricky point is, in most experiment, we are working in Laboratory frame that require frame transformation.

The relationship of the energy in CM frame and the energy in the Lab frame can be found by Lorentz transform, and use the total kinematic energy (both particle 1 and particle 2). In the CM frame, we can image we have a fixed virtual target on the center of mass, and there is only 1 object moving at energy of the total kinematic energy.

For example, we have a target of mass $m_2$, a projectile with mass $m_1$ and energy $T_1$, a classical energy in CM frame is

$E_{cm} = \frac{m_2}{m_1+m_2} T_1$

In fact the $E_{cm}$ has only 5% difference between relativistic and non-relativistic even up to 500 MeV

When calculating the integrated cross section, we can do it in the CM frame, but it is more intuitive to do it in the Lab frame. In this case, we need to transform the differential cross section from the CM frame to the Lab frame, mathematically, the transformation is done by a factor called Jacobian.

We can compare the result using the kinematic calculator in LISE++.

In the above plot, the blue line is the d.s.c. in CM frame, and the red line is d.s.c. of the 9Be in Lab frame. Jacobian was added, therefore, the zero degree d.s.c. of 9Be is larger than the 180 degree d.s.c. in the CM frame.

The grazing angle of the scattering, can be calculated by the shorted distance between the target and the projectile. In the Lab frame, the target is not fixed, so it is not easy to know the shortest distance. But in the CM frame, the virtual target is fixed, and we can calculate the distance using the $E_{cm}$ and $\theta_{cm}$.

## 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.

## a GEANT4 Simulation

The GEANT4 program structure was borrow from the example B5. I found that the most confusing part is the Action. Before that, let me start with the main().

GEANT4 is a set of library and toolkit, so, to use it, basically, you add a alot GEANT4 header files on your c++ programs. And, every c++ program started with main(). I write the tree diagram for my simplified exampleB5,

main()
+- DetectorConstruction.hh
+- Construct()
+- HodoscopeSD.hh     // SD = sensitive detector
+- HodoscopeHit.hh //information for a hit
+- ProcessHits()   //save the hit information
+- ConstructSDandField() //define which solid is SD
+- ConstructMaterials()  //define material
+- PhysicsList.hh  // use FTFP_BERT, a default physics for high energy physics
+- ActionInitialization.hh
+- PrimaryGeneratorAction.hh // define the particle gun
+- RunAction.hh // define what to do when a Run start, like define root tree
+- Analysis.h  // call for g4root.h
+- EventAction.hh //fill the tree and show some information during an event

A GEANT4 program contains 3 parts: Detector Construction, Physics, and Action. The detector construction is very straight forward. GEANT4 manual is very good place to start. The physics is a kind of mystery for me. The Action part can be complicated, because there could be many things to do, like getting the 2ndary beam, the particle type, the reaction channel, energy deposit, etc…

Anyway, I managed to do a simple scattering simulation, 6Li(2mm) + 22Ne(60A MeV) scattering in vacuum. A 100 events were drawn. The detector is a 2 layers plastic hodoscope, 1 mm for dE detector, 5 mm for E detector. I generated 1 million events. The default color code is Blue for positive charge, Green for neutral, Red for negative charge. So, the green rays could be gamma or neutron. The red rays could be positron, because it passed through the dE detector.

The histogram for the dE-TOF is

We can see a tiny spot on (3.15,140), this is the elastics scattered beam, which is 20Ne. We can see 11 loci, started from Na on the top, to H at the very bottom.

The histogram of dE-E

For Mass identification, I follow the paper T. Shimoda et al., Nucl. Instrum. Methods 165, 261 (1979).

I counted the 20Na from 0.1 billion beam, the cross section is 2.15 barn.

## Mean free path of a nucleon inside a nucleus

The mean free path is the average distance between 2 collisions. We simply copy the things in Bohr and Mottelson, Bertulani and Banielewicz, and John Lilley in here.

Since a nucleus has finite size, the mean free path can be view as transparency of nucleon scattering. Since the total cross section is smaller for higher energy, the mean free path is proportional to energy.

The wave number $K$ under a complex optical potential $V+iW$ is

$K=\sqrt{\frac{2m}{\hbar^2}(E-V-iW)}=k_r+ik_i$.

There are two solutions, one has imaginary $k_r$,

$k_r=\frac{\sqrt{m}}{\hbar}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}$

$k_i=\frac{m}{\hbar^2}\frac{W}{k_r}=\frac{1}{2\lambda}$

where $m$ is the nucleon mass, $E$ is the incident energy, $v$ is the velocity inside the nucleus, and $\lambda$ is the mean free path,

$\lambda=\frac{\hbar}{2W\sqrt{m}}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}\approx\frac{\hbar}{W\sqrt{2m}}\sqrt{(E-V)}$

Since the imaginary potential $W\approx-15-0.07E$, $V\approx-39+0.11E$ from 150 MeV to 400 MeV. The calculated mean free path for proton is shown in here. (Take $\hbar = 197.33 [MeV\cdot fm/c]$, proton mass $m=938.272 [MeV/c]$, the $\lambda$ is in fm.)

The blue line on the plot is exact calculation, the purple line is approximation $W << E-V$.

This plot is taken from S.S.M. Wong, showing the radial shapes of the volume term (simialr to central term) of proton-nucleus optical potential.

————–

in J. Killey, the $k_r$ is taken as,

$k_r = \frac{\sqrt{2m(E-V)}}{\hbar}$,

which is the wave number without $W$. The result only for weak $W << E-V$.

## example on spin scattering

From previous post, the scattering amplitude matrix is :

$F(p' \leftarrow p ) = a I + i b \hat{n} \cdot \sigma$

let the in- density matrix be for our spin ½ ensemble is :

$\rho_{in} = \frac{1}{2} ( I + \vec{\pi}_{in} \cdot \sigma )$

then the out-matrix is:

$\rho_{out} = F \rho_{in} F^\dagger$

using:

$(u\cdot \sigma ) (v \cdot \sigma ) = u\cdot v I + i u \times v \cdot \sigma$

with some messy algebra, and focus out pull out the identity matrix and the Pauli’s matrix, we will get:

$\rho_{out} = \frac{1}{2} ( |a|^2 + |b|^2 +2 Im( a b^*) \hat{n} \cdot \vec{\pi}_{in}) ( I + \vec{\pi}_{out}\cdot \sigma)$

the form of out-polarization vector takes very complicate form. however, this is not bother in the differential cross section, coz it only care the trace.

$\frac{d\sigma}{d\Omega} ( p' \leftarrow p, \vec{\pi}_{in} ) = ( |a|^2 +|b|^2 ) ( 1+ v(\theta) \hat{n}\cdot \vec{\pi}_{in} )$

$v(\theta) = 2 \frac{ Im( a b^*)}{|a|^2+|b|^2}$