scattering | Nuclear Physics 101

Various cross sections

November 24, 2021

GoLuckyRyan Basic scattering Leave a comment

I have been confused by so many different cross sections for so long. What are the differences among them?

total cross section
- sum of all kind of cross sections. Any scattering due to the present of target.
elastic cross section
- both incident and target nuclei do not change, only change in direction. For example, Coulomb scattering.
inelastic cross section
- both incident and target nuclei do not change, only change in direction and energy. I am not sure, is Resonance reaction belong to inelastic cross section?
reaction cross section
- any thing not elastic. so, inelastic cross section is part of reaction cross section.
interaction cross section
- the incident or target nuclei change, direction and energy are also changed. For example, transfer reaction, knockout reaction.
total reaction cross section (?)
- I still have no idea what this is.

From the above, we have

$\displaystyle \sigma_{tot} = \sigma_{el} + \sigma_{R},~~~~\sigma_{R}= \sigma_{inel} + \sigma_{I}$

When the incident energy is above Coulomb barrier, the inelastic cross section is much smaller than the interaction cross section, because both incident and target nuclei are certainly “touch” and interaction with each other. So, in those cases, people usually assume that $\sigma_{R} \approx \sigma_{I}$ .

In scattering theory, the wave function of the incident beam is

$\displaystyle \Psi_i = e^{+i\vec{k}\cdot \vec{r}}$

After scattering, the wave function becomes

$\displaystyle \Psi = \begin{cases} e^{+i\vec{k}\cdot \vec{r}} + f_e(\theta) \frac{e^{ikr}}{r}, & \textrm{elastic} \\ f_{R}(\theta) \frac{e^{ik'r}}{r}, & \textrm{reaction} \end{cases}$

Note that the momentum is $k' \neq k$ in the reaction channel.

Rutherford scattering in CM and Lab frame

September 14, 2021

GoLuckyRyan Basic Rutherford, scattering Leave a comment

I spent sometime to complete and clear this and this posts. Now it is times to complete the Rutherford scattering in CM frame and Lab frame. We are restrict ourselves to low energy.

Recall the result, the Rutherford scattering in target-fixed frame is

$\displaystyle \frac{d\sigma}{d\Omega} = \frac{\eta'}{4 \sin^4(\theta/2)}, ~~\eta' = \frac{q_1 q_2 e^2}{2 T_{cm}}$

The relation between the target-fixed frame and the CM frame is

$\displaystyle \vec{r}_{cm} = \frac{m_1}{m_1+m_2} \vec{r}$

So, in the CM frame, there is a mass-correction,

$\displaystyle \frac{d\sigma}{d\Omega_{cm}} = \left(\frac{m_2}{m_1+m_2}\right)^2\frac{\eta'}{4 \sin^4(\theta/2)}$

From the CM frame and the Lab frame, the relation is a bit complicated, in the classical limit,

$\displaystyle \vec{v}_{cm} + \vec{v}_T = \vec{v}_{Lab}, ~~ \vec{v}_T = \frac{m_1}{m_1+m_2} \vec{v}$

The differential cross section in Lab frame, for low energy is

$\displaystyle \frac{d\sigma}{d\Omega_{Lab}} = \frac{d\sigma}{d\Omega_{cm}} \frac{d \Omega_{cm}}{d\Omega_{lab}} \\~~~~= \left(\frac{m_2}{m_1+m_2}\right)^2 \frac{\eta'^2}{4 \sin^4(\theta/2)} \frac{(1+\tau^2 + 2\tau \cos(\theta))^{3/2}}{ 1 + \tau \cos(\theta)}, ~~~ \tau = \frac{m_1}{m_2}$

The left-hand side still depends on the CM frame angle. It is better to express it in the Lab angle.

The geometrical relation between Lab frame angle and CM frame angle for low energy is

$\displaystyle \tan(\theta_L) = \frac{\sin(\theta)}{\tau + \cos(\theta)},~~~\tau = \frac{m_1}{m_2}$

Solve

$\displaystyle \cos(\theta) = -\tau \sin^2(\theta_L) \pm \cos(\theta_L) \sqrt{1-\tau^2 \sin^2(\theta_L)}$

When $\tau < 1$ , i.e. the projectile is lighter than the target. The angle in Lab and CM frame are 1-1 corresponding. But when $\tau > 1$ , there are 2 Lab angles for 1 CM angle.

differentiate

$\displaystyle \frac{d \cos(\theta)}{d \cos(\theta_L) } = \frac{\left(\pm \tau \cos(\theta_L) + \sqrt{1- \tau^2 \sin^2(\theta_L)}\right)^2}{\sqrt{1 - \tau^2 \sin^2(\theta_L)}}$

We can check that

$\displaystyle \frac{\left(\pm \tau \cos(\theta_L) + \sqrt{1- \tau^2 \sin^2(\theta_L)}\right)^2}{\sqrt{1 - \tau^2 \sin^2(\theta_L)}} = \frac{(1+\tau^2 + 2\tau \cos(\theta))^{3/2}}{ 1 + \tau \cos(\theta)}$

Now, the rest is convert $\sin^4(\theta/2)$ into Lab angle.

Well…. using Excel or Mathematica, we don’t need that.

proton-proton scattering at 5 MeV Lab energy.

Calculate phase shift of square well

September 7, 2021

GoLuckyRyan Basic phase shift, Riccati-Bessel, scattering, Woods-Saxon 4 Comments

In the book, Introduction to Nuclear Reaction, by C. A. Bertulani, Chapter 2, there is a section that shows an example of phase shift from a square well. I follow the receipt and able to reproduce the result.

The actual method I did is little different. I use Mathematica to solve the radial Schrodinger equation in $u(r) = R(r)/r$ , the differential equation I solve is

$\displaystyle \frac{d^2}{dr^2} u(r) = \begin{cases} 0 & r = 0 \\ \frac{2m}{\hbar^2} W(r) u(r) + \frac{l(l+1)}{r^2} u(r) & r > 0 \end{cases}$

where $W(r)$ is the Woods-Saxon function, set the diffusiveness parameter to 0.001, the Woods-Saxon shape becomes square well.

In solving the differential equation, I set the right-hand side to be zero when $r=0$ . That solved to problem when $l > 0$ , it has to deal with the $1/r^2$ . In fact, since $u(0) = 0$ , the right-hand side is always zero when $r = 0$ .

For the solving range, I set

$\displaystyle r_{max} = 5 \frac{2\pi}{k} + \frac{\sqrt{l(l+1)}}{k}$

This is because the “first” peak of the Riccati-Bessel function appear after $\sqrt{l(l+1)}/k$ . This will make sure the range has 5 oscillations in the solution.

Next, I table the solution for the last 2 oscillations, find the maximum, normalize the solution, and then fit with

$\displaystyle u(r) \xrightarrow{r \rightarrow \infty} A \hat{j}_l(kr) + B \hat{n}_l(kr)$

and extract $A$ , and $B$ .

The phase shift is

$\displaystyle \tan(\phi) = \frac{A}{-B}$ .

The negative is due to the $\hat{n}_l$ is negatively defined in Mathematica.

The elastics cross section is

$\displaystyle \sigma_{el} = \frac{4\pi}{k^2} \sum_{l} \sqrt{2l+1} \sin^2(\phi_l)$

From the calculation, we found that, at very small energy, only s-wave scattering cross section is non-zero, which is agree with theory.

[20220531] update

The Mathematical notebook can be downloaded here.

Two-body Coulomb scattering

July 19, 2021

GoLuckyRyan Basic Coulomb, Rutherford, scattering Leave a comment

This is a classical problem that is well know. This is more like my personal note. well, this blog is always my personal note. anyway,

The setting of the two-body problem is simple, there are two particles at position $\vec{r}_1, \vec{r}_2$ with velocity $\vec{v}_1 = v \hat{x}, \vec{v}_2 = 0$ , mass $m_1, m_2$ , and charge $q_1, q_2$ . There is a Coulomb force acting between the two particle and no external force.

The equations of motion for each particle are

$\displaystyle m_1 \frac{d^2 \vec{r}_1}{dt^2} = \frac{q_1 q_2 e^2}{|\vec{r}_1 - \vec{r}_2|^{3/2}} \left( \vec{r}_1 - \vec{r}_2 \right)$

$\displaystyle m_2 \frac{d^2 \vec{r}_2}{dt^2} = \frac{q_1 q_2 e^2}{|\vec{r}_1 - \vec{r}_2|^{3/2}} \left( \vec{r}_2 - \vec{r}_1 \right)$

Define the center of mass

$\displaystyle \vec{R} = \frac{m_1 \vec{r}_1+ m_2 \vec{r}_2 }{m_1+m_2}, ~~~\displaystyle \frac{d^2\vec{R}}{dt^2} = 0$

The particle-1 is moving at velocity $v \hat{x}$ at impact parameter $b$ . Shift to the center of mass frame,

$\displaystyle \vec{\rho}_1 = \vec{r}_1 - \vec{R} = \frac{m_2}{m_1+m_2} (\vec{r}_1 - \vec{r}_2)$

$\displaystyle \vec{\rho}_2 = \vec{r}_2 - \vec{R} = \frac{m_1}{m_1+m_2} (\vec{r}_2 - \vec{r}_1)$

The initial velocity at the center of mass frame is

$\displaystyle \vec{u}_1 = \frac{m_2}{m_1+m_2} v \hat{x},~~ \vec{u}_2 = - \frac{m_1}{m_1+m_2} v \hat{x}$

The total angular momentum with respect to the center of mass $\vec{R}$ is

$\displaystyle L = m_1 \vec{y}_1 \times \vec{u}_1 + m_2 \vec{y}_2 \times \vec{u}_2 = \frac{m_1 m_2^2}{(m_1+m_2)^2} vb + \frac{m_2 m_1^2}{(m_1+m_2)^2} vb = \mu v b$

where $\mu = m_1 m_2 / (m_1+m_2 )$ is the reduced mass.

The equation of motion for the relative position is

$\displaystyle m_1 m_2 \frac{d^2 (\vec{r}_1 - \vec{r}_2) }{dt^2} = (m_1+m_2) \frac{q_1 q_2 e^2}{ |\vec{r}_1 - \vec{r}_2|^{3/2}} \left( \vec{r}_1 - \vec{r}_2 \right)$

$\displaystyle \mu \frac{d^2 \vec{r} }{dt^2} = \frac{Z}{r^2} \hat{r}$

using polar coordinate,

$\displaystyle \vec{r} = r \hat{r} \\ \ddot{\vec{r}} =\dot{r} \hat{r} + r \dot{\theta} \hat{\theta} \\ \ddot{\vec{r}} = (\ddot{r} - r \dot{\theta}^2)\hat{r} + (2\dot{r}\dot{\theta} + r \ddot{\theta}) \hat{\theta}$

We have

$\displaystyle \ddot{r} - r \dot{\theta}^2 = \frac{Z}{\mu r^2}, ~~~ 2\dot{r}\dot{\theta} + r \ddot{\theta} = 0$

From the angular equation,

$2\dot{r}\dot{\theta} + r \ddot{\theta} = \frac{1}{r} \frac{r^2 \dot{\theta}}{dt}= 0$

And we know that $\mu r^2 \dot{\theta } = L$ , the angular momentum. Define, $L = \mu l$ . Replace $\dot{\theta} = l /r^2$ , we have

$\displaystyle \ddot{r} = \frac{l^2}{r^3} + \frac{Z}{\mu r^2}$

The energy equation is

$\displaystyle \frac{1}{2}\mu (\dot{r}^2 + (r\dot{\theta})^2 ) + \frac{Z}{r} = T_{cm} = \frac{1}{2}\mu v^2$

replace the $\dot{\theta} = l/r^2$ ,

$\displaystyle \dot{r}^2 + \frac{l^2}{r^2} + 2 \frac{Z}{\mu r} =2 \frac{T_{cm}}{\mu}$

Differentiate the above equation will get back the equation of motion.

At the closest distance between the two particle $r = b'$ , $\dot{r} = 0$ , $l = vb$

$\displaystyle \frac{v^2 b^2}{b'^2} + 2 \frac{Z}{\mu b'} = v^2$

Solve for $b'$ , we get back the Coulomb correction for Glauber model

$\displaystyle b' = \eta' + \sqrt{\eta'^2 + b^2}, ~~ \eta' = \frac{Z}{2 T_{cm}}$

where $\eta'$ is the half distance of closest approach.

Back to the equation of motion. it may be not possible to solve it analytically in term of time. But it can be solve it in term of angle by replacing $r = 1/u$ .

$\displaystyle \frac{dr}{dt} = - \frac{1}{u^2} \frac{du}{d\theta} \frac{d\theta}{dt} = - r^2 \dot{\theta} u'(\theta) = - l u'$

$\displaystyle \frac{d^2 r}{dt} = - l u'' \dot{\theta} = - l^2 u^2 u''$

The equation of motion becomes,

$\displaystyle - l u^2 u'' = l u^3+ \frac{Z}{\mu} u^2 \rightarrow u'' + u = - \frac{Z}{\mu l^2}$

The solution for this type of differential equation is

$\displaystyle \frac{1}{r(\theta)} = A \cos\theta + B \sin\theta - \frac{Z}{\mu l^2}$

$\displaystyle \frac{1}{r} = \frac{Z}{\mu l^2 } \left( e \cos(\theta - \phi) - 1 \right) = \kappa \left( e \cos(\theta - \phi) - 1 \right)$

in order to solve the equation of motion completely, the Initial condition has to be specified. Let our locus be like this:

The initial condition is $\theta \rightarrow \pi, r \rightarrow \infty, r \sin(\theta) = b, dr/dt = v$

The last condition can be rewrite as

$\displaystyle \frac{dr}{dt} = - l \frac{du}{d\theta} = v \Rightarrow \frac{du}{d\theta} = -\frac{1}{b}$

So, we solve the coupled equations:

$\displaystyle ~~~~~0 = \kappa ( e \cos(\theta - \phi) -1 ) \\ - \frac{1}{b} = - \kappa e \sin(\theta-\phi)$

$\displaystyle \frac{1}{b'} = \frac{Z}{\mu l^2 } \left( e \cos(- \phi) - 1 \right)$

The solution is

$\displaystyle e = \frac{\sqrt{1 + b^2 \kappa^2}}{ b \kappa } = \sqrt{\frac{b^2}{\eta'^2}+1}$

$\displaystyle \phi = \frac{\pi}{2} + \tan^{-1}(\kappa b) = \tan^{-1}(-\kappa b, 1)$

proton-proton Coulomb scattering in the relative position. The energies of the upper locus variate from 1, 2, 3, …, 9 MeV and the impact parameters are fixed at 1 fm. The energies of the lower locus are fixed at 5 MeV, and the impact parameters variates from 0.1, 0.2, … 0.9 fm.

From the above calculation. We can see there is a “shadow” behind the target. I generated many locus with different impact parameters with fixed energy of 5 MeV for proton-proton Coulomb scattering.

How to calculate the shadow???

With the equation of locus

$\displaystyle \frac{1}{r} = \kappa \left( e \cos(\theta- \phi) -1 \right), \\~ \kappa = \frac{Z}{\mu l^2} = \frac{\eta'}{b^2}, ~~ e = \sqrt{\frac{b^2}{\eta'^2}+1}, ~ \phi = \frac{\pi}{2} + \tan^{-1}(\frac{\eta'}{b})$ .

We can derived the Rutherford scattering.

The deflection angle is

$\displaystyle \Theta = 2 \phi - \pi =2 \tan^{-1}\left( \frac{Z}{\mu v^2 b} \right)$

express in term of $b$

$\displaystyle \frac{Z}{\mu v^2 b} = \tan\left(\frac{\Theta}{2} \right)$

$\displaystyle b = \frac{Z}{\mu v^2} \cot\left( \frac{\Theta}{2} \right) = \eta' \cot\left( \frac{\Theta}{2} \right)$

The last step used $2T_{cm} = \mu v^2$ . The differential cross section is the function of a uniform ring with impact parameter.

$\displaystyle 2\pi b db = - \frac{d\sigma}{d\Omega} d\Omega$

with the cylindrical symmetry, $d\Omega = 2\pi \sin \Theta d\Theta$ . The minus sign in the above equation reflect the fact that the change of impact parameter result smaller deflection angle.

$\displaystyle \frac{d\sigma}{d\Omega} = - \frac{b}{\sin \Theta} \frac{db}{d\Theta} = \frac{b}{\sin\Theta} \frac{\eta'}{2 \sin^2(\Theta/2)}$

Using $\sin(x) = 2 \sin(x/2) \cos(x/2)$

$\displaystyle \frac{d\sigma}{d\Omega} = \frac{\eta'^2}{4 \sin^4 (\Theta/2)}$

Using the trajectory equation, the nearest distance happens at $\theta = \phi$ .

$\displaystyle \frac{1}{b' } = \kappa ( e - 1) = \frac{\eta'}{b^2} \left( \sqrt{\frac{b^2}{\eta'^2}+1} -1 \right)$

rearrange, we get

$\displaystyle b' = \eta' + \sqrt{\eta'^2 +b^2}$

The only remining thing are the trajectory in CM frame and Lab frame. Recall that, in the CM frame,

$\displaystyle \vec{\rho}_1 = \frac{m_2}{m_1+m_2} (\vec{r}_1 - \vec{r}_2) , ~ \vec{\rho}_2 = \frac{m_1}{m_1+m_2} (\vec{r}_2 - \vec{r}_1)$ .

Thus, simply multiple a factor, we can get the CM frame trajectory.

$\displaystyle \rho_1 = \frac{m_2}{m_1+m_2} r$

Also, the impact parameter with respect to the origin in the CM frame will be smaller,

$\displaystyle b \rightarrow \frac{m_2}{m_1+m_2} b, ~~ \kappa = \frac{Z}{\mu l^2} \rightarrow \frac{Z}{\mu l^2} \left(\frac{m_1+m_2}{m_2} \right)^2 = \kappa \left(\frac{m_1+m_2}{m_2} \right)^2$

The over all effect is $\kappa \rightarrow \frac{m_1+m_2}{m_2} \kappa$ , the trajectory is

$\displaystyle \frac{1}{\rho} = \left(\frac{m_1+m_2}{m_2} \right) \kappa \left( e \cos(\theta -\phi) -1 \right)$

so the differential cross section is

$\displaystyle \frac{d\sigma_{cm}}{d\Omega} = \left(\frac{m_2}{m_1+m_2}\right)^2 \frac{d\sigma}{d\Omega}$

We can see, if the target mass is much bigger than the projectile mass, the correction is small. And we can see that the target will not recoil from the center, so $\vec{r} = \vec{r}_1 - \vec{r}_2 \approx \vec{r}_1$ . And for the proton-proton Coulomb scattering, in the CM frame, the cross section becomes quarter smaller. This is because the target will recoil and move, and that reduced the cross section, i.e. the field from the target is not rigid and becomes soft.

In order to see/verify the locus or the mass correction in CM frame. I solve the coupled equations using Mathematica,

$\displaystyle m_1 \frac{dx_1^2}{dt^2} = \frac{Z}{ \left((x_1-x_2)^2 + (y_1-y_2)^2\right)^{3/2}} (x_1 - x_2)$

$\displaystyle m_1 \frac{dy_1^2}{dt^2} = \frac{Z}{ \left((x_1-x_2)^2 + (y_1-y_2)^2\right)^{3/2}} (y_1 - y_2)$

$\displaystyle m_2 \frac{dx_2^2}{dt^2} = \frac{Z}{ \left((x_1-x_2)^2 + (y_1-y_2)^2\right)^{3/2}} (x_2 - x_1)$

$\displaystyle m_2 \frac{dy_2^2}{dt^2} = \frac{Z}{ \left((x_1-x_2)^2 + (y_1-y_2)^2\right)^{3/2}} (y_2 - y_1)$

These set of equations is same in all frame. Different frame has different initial conditions.

The CM frame initial condition are:

$\displaystyle (x_1, y_1)(0) = \left(-\infty, \frac{m_2}{m_1+m_2} b \right), (x_1', y_1')(0) = \left(\frac{m_2}{m_1+m_2} v, 0 \right)$

$\displaystyle (x_2, y_2)(0) = \left(+\infty, -\frac{m_1}{m_1+m_2} b \right), (x_2', y_2')(0) = \left(-\frac{m_1}{m_1+m_2} v, 0 \right)$

In Mathematica, it is, we cannot put infinity in the numerical solve, so, I set the initial position very far away.

In the below calculation, $x_1(0) = -150, x_1(0) = 150, v = 0.0462127$ . Since we use the nuclear unit, the unit of the speed is speed-of-light. The value corresponding to 1, 3, and 5 MeV proton energy. The impact parameter is 2 fm.

The solid locus are from the numerical solution. The dashed line is from the equation:

$\displaystyle \frac{1}{\rho} = \frac{m_1+m_2}{m_2} \kappa ( e \cos(\theta-\phi) - 1)$

I compare the target-fixed frame (upper), and CM frame (lower) trajectory in below:

For the Lab frame, the initial conditions are

$\displaystyle (x_1, y_1)(0) = \left(-\infty, b \right), (x_1', y_1')(0) = \left( v, 0 \right)$

$\displaystyle (x_2, y_2)(0) = \left(0, 0 \right), (x_2', y_2')(0) = \left(0, 0 \right)$

In the below simulation, the energy of proton is 1 MeV, the initial position of the projectile locates at -2000 fm.

Since the Coulomb force has infinite range, even at -2000 fm away, it still push the target proton away.

I guess it is almost most of the things about Two-body Coulomb scattering. We studied the locus in 3 frames: The target-fixed frame, the CM frame, and the Lab frame. There are few things to add into the calculation, e.g. lattice effect of the target that the target is bounded by a quadratic well, and screening of charge by electron. I will do the Rutherford differential cross section in Lab frame in other post. The idea is already in this post.

Scattering and Reaction cross sections

July 1, 2021

GoLuckyRyan Basic scattering Leave a comment

I found that there is no clear explanation on the reaction cross section but scattering cross section.

With the scattering matrix element $S_l$

The scattering cross section is

$\displaystyle \sigma_S = \frac{\pi}{k^2} \sum_{l=0} (2l+1) |1-S_l|^2$

The reaction cross section is

$\displaystyle \sigma_R = \frac{\pi}{k^2} \sum_{l=0} (2l+1) (1-|S_l|^2)$

Lets see how.

A plane wave toward z-direction is

$\displaystyle e^{+ikz} = \frac{2\sqrt{\pi}}{kr} \sum_{l=0} \sqrt{2l+1} i^l \hat{j}_l(kr) Y_{l0}(\theta, \phi)$

at long range

$\displaystyle \hat{j}_l(kr) \rightarrow \sin\left(kr - \frac{l\pi}{2}\right) = \frac{1}{2i}\left(e^{+i(kr-l\pi/2)} - e^{-i(kr-l\pi/2)} \right)$

The plane wave becomes:

$\displaystyle e^{+ikz} = \frac{\sqrt{\pi}}{kr} \sum_{l=0} \sqrt{2l+1} i^{l+1} \left(-e^{+i(kr-l\pi/2)} + e^{-i(kr-l\pi/2)} \right) Y_{l0}(\theta, \phi)$

If a target is presented, the target will modified the outgoing part will be modified and the complete wave function is

$\displaystyle \Psi = \frac{\sqrt{\pi}}{kr} \sum_{l=0} \sqrt{2l+1} i^{l+1} \left( - S_l e^{+i(kr-l\pi/2)} + e^{-i(kr-l\pi/2)} \right) Y_{l0}(\theta,\phi)$

With some calculations, and $Y_{l0} = \sqrt{\frac{2l+1}{4\pi}} P_l(\cos\theta)$ , we can convert it into a more common form

$\displaystyle \Psi = e^{+ikz} + f(\theta) \frac{e^{ikr}}{r}$

$\displaystyle f(\theta) = \frac{1}{k}\sum_{l=0}(2l+1) \frac{S_l-1}{2i} P_{l}(\cos\theta)$

which is the same in this post. Thus, the scattering cross section is derived.

Here we use another approach to calculate the scattering cross section. The scattering wave is then the modified wave subtract the incoming plane wave.

$\displaystyle \Psi_S = \Psi - e^{+ikz} = \frac{\sqrt{\pi}}{kr}\sum_{l=0}\sqrt{2l+1} i^{l+1} (1- S_l) e^{+i(kr-l\pi/2)}Y_{l0}(\theta, \phi)$

$\displaystyle \Psi_S = f(\theta) \frac{e^{ikr}}{r}$

The flux of the scattering wave is

$\displaystyle F_S =\frac{\hbar}{2mi}\left( \frac{\partial \Psi_S}{\partial r} \Psi_S^* - \frac{\partial \Psi_S^*}{\partial r} \Psi_S \right)$

Since only the radial part is the $e^{ikr}/r$

Evaluate

$\displaystyle \frac{d}{dr}\left( \frac{e^{ikr}}{r} \right) \frac{e^{-ikr}}{r} - \frac{d}{d r} \left( \frac{e^{-ikr}}{r} \right) \frac{e^{ikr}}{r} = \frac{2ik}{r^2}$

Thus,

$\displaystyle F_S =\frac{\hbar k}{mr^2} |f(\theta)|^2$

Integrate the surface of a sphere of the flux, which is the number of particle coming out.

$\displaystyle N_S = \int F_S r^2 d\Omega = \frac{\hbar k}{m} \frac{\pi}{k^2} \sum_{l=0} (2l+1)^2 |1-S_l|^2$

for non-relativistic, $k = mv/\hbar$ , where $v$ is the velocity of the outgoing particle

$\displaystyle N_S = v \frac{\pi}{k^2} \sum_{l=0} (2l+1) |1-S_l|^2$

And the scattering cross section is the number of outgoing particle divided by the velocity ( why? )

$\displaystyle \sigma_S = \frac{\pi}{k^2} \sum_{l=0} (2l+1) |1-S_l|^2$

Following the same receipt, we can repeat with the total wave function $\Psi$ . We get,

$\displaystyle \sigma_R = \frac{\pi}{k^2} \sum_{l=0} (2l+1) (1-|S_l|^2)$

In summary,

$\displaystyle \Phi_{Free} = e^{+ikz}$

with target presents,

$\displaystyle \Phi_{Target} = \Phi_{Free} + \Phi_{Scat} = e^{+ikz} + f(\theta) \frac{e^{ikr}}{r}$

The scattering cross section is from the $\Phi_{Scat}$ , and the reaction cross section is from the $\Phi_{Target}$ .

Although look at this post, we can somehow knowing what is the meaning of the scattering matrix element. But still, for me, it is not intuitive.

Resonance Scattering

May 21, 2020

GoLuckyRyan Basic, Scattering resonance, scattering Leave a comment

We will talk about the basics theory of resonance scattering or s-wave in this post.

The asymptotic behaviour of the scattered wave is (this post or this post), ignore th noramlization factor.

$\displaystyle u_l(r \rightarrow \infty) = \hat{h}^{-}(kr) - S_l \hat{h}^+(kr)$

For the s-wave,

$\displaystyle u_0(r \rightarrow \infty) = e^{-ikr} - S_0 e^{+ikr}$

The logarithmic derivative at the boundary $r=R$ is

$\displaystyle L^{II} = R \frac{u_0'(kR)}{u_0(kR)} = -ikR \frac{e^{-ikR} + S_0 e^{ikR}}{e^{-ikR} - S_0 e^{+ikR}} = -ikR \frac{1+ S_0 e^{2ikR}}{1 - S_0 e^{2ikR}}$

This must be matched with the logarithmic derivative of the inner wave,

$\displaystyle L^{I} = -ikR \frac{1+ S_0 e^{2ikR}}{1 - S_0 e^{2ikR}}$

Solve for $S_0$

$\displaystyle S_0 = e^{-2ikR} \frac{L^{I} + ik R}{L^I - ikR}$

For elastics scattering, $L^I$ is real, but in general, it can be a complex when a complex potential is used, that corresponding to an absorption, that other reaction channels are populated. Suppose

$L^I = a- i b$

$\displaystyle S_0 = e^{-2ikR} \frac{a + i(kR-b)}{a - i(kR + b) }$

Recall that the scattering amplitude

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

Assume only s-wave exist,

$\displaystyle f(\theta) = \frac{1}{2ik} (S_0 -1) = \frac{1}{2ik} \left( e^{-2ikR} \frac{a + i(kR-b)}{a - i(kR + b) } - 1\right)$

$\displaystyle f(\theta) =\frac{1}{2ik} \left( e^{-2ikR} \frac{a + i(kR-b)}{a - i(kR + b) } - e^{-2ikR} + e^{-2ikR} - 1\right) = f_R + f_S$

where

$\displaystyle f_S=\frac{1}{2ik} \left( e^{-2ikR} - 1\right)$

$\displaystyle f_R=\frac{e^{-2ikR}}{2ik} \left( \frac{a + i(kR-b)}{a - i(kR + b) } - 1\right) = \frac{e^{-2ikR}}{k} \left( \frac{kR}{a - i(kR + b) }\right)$

We can see that the phase shift of $f_S$ is $kR$ . It is like a scattering of a hard sphere.

The term $f_R$ is respond to resonance when $a = 0$ . We can imagine that $a$ is variate with the beam energy $E$ . Express $a(E)$ in Taylor series, the 1st order around the resonance energy $E_R$ .

$\displaystyle a(E) = \frac{da(E)}{dE} (E-E_R) = a'(E_R) (E-E_R)$

And define,

$\displaystyle \Gamma_e = - \frac{2kR}{a'(E_R)}$

$\displaystyle \Gamma_a = - \frac{2b}{a'(E_R)}$

$\displaystyle \Gamma = \Gamma_e + \Gamma_a$

where $\Gamma_e$ is scattering width, $\Gamma_a$ is absorption width, and $\Gamma$ is total width.

$\displaystyle f_R(E) \approx \frac{e^{-2ikR}}{k} \left( -\frac{\Gamma_e / 2}{(E-E_R) + i \Gamma/2 }\right)$

The cross section for s-wave

$\displaystyle \sigma = 4\pi |f_S+f_R|^2$

Near the resonance energy, $|f_r| >> |f_S|$ ,

$\displaystyle \sigma \approx 4\pi |f_R|^2 = \frac{\pi}{k^2} \frac{\Gamma_e^2}{(E-E_R)^2 + \Gamma^2/4 }$

At resonance energy $\displaystyle \sigma \approx \frac{4\pi}{k^2} \frac{\Gamma_e^2}{\Gamma^2 }$

For real potential, $b = 0$ and $\Gamma_e = \Gamma$ .

Scattering with Coulomb potential

April 21, 2020

GoLuckyRyan Basic, Scattering Coulomb, scattering Leave a comment

I only stay the result in here. The result is very similar to the case without Coulomb potential.

The general solution for Coulomb potential is

$\displaystyle \Phi_c( r \rightarrow \infty )= e^{i \vec{k} \cdot \vec{r}} e^{-i \eta \ln(k(r-z))} + f_c(\theta) \frac{e^{ikr}}{r} e^{-i \eta \ln(k(r-z))}$

$\displaystyle f_c(\theta) = - \frac{\eta}{ 2k \sin^2(\theta/2)} e^{-i \eta \ln(\sin^2(\theta/2))} e^{2i \sigma_0 }$ ,

$\displaystyle \sigma_l = \arg( \Gamma(1+l+i\eta) )$

In partial wave,

$\displaystyle \Phi_c = A \frac{1}{2kr} \sum_{l=0}^\infty (2l+1) i^{l+1} P_l(\cos\theta) e^{i\sigma_l} \left( H_l^{-} - H_l^+\right)$

Scattering with Coulomb and short-range potential, the potential is

$\displaystyle V(r) = V_c(r) + V_N(r)$

The solution takes the form

$\Psi(r) = \Phi_c + \Psi_N$

The asymptotic form of $\Psi_N$ is

$\displaystyle \Psi_N \rightarrow A f_N(\theta) \frac{e^{ikr}}{r} e^{-i\eta \ln(2kr)}$

$\displaystyle f_N(\theta) = \frac{1}{2ik} \sum_{l=0}^\infty (2l+1) P_l(\cos\theta) e^{2i\sigma_l} (S_l -1 )$

$S_l = e^{2i \delta_l}$

$\displaystyle \tan \delta_l = - \frac{kR F_l'(kR) - F_l(kR) L^I}{kR G_l'(kR) -G_l(kR) L^I}$

The cross section is

$\displaystyle \frac{d\sigma_c}{d\Omega} = |f_C(\theta) + f_N(\theta)|^2$

S-matrix, elastics and absorption cross section

April 18, 2020

GoLuckyRyan Basic, Scattering S-matrix, scattering Leave a comment

In this post, we shown that the elastics cross section is given by

$\displaystyle \sigma_{el} = \frac{\pi}{k^2} \sum_l (2l+1) |1- S_l|^2$

For elastics scattering $S_l = e^{2 i \delta_l}$ , so that

$\displaystyle |1-S_l|^2 = 4 \sin^2(\delta_l)$

Before we continuous, lets introduce the scattering operator $S$

$\displaystyle \left< \Psi_{k'}^- | \Psi_{k}^+ \right> = \left< \phi_{k'} | S | \phi_{k} \right> = S_{k'k}$

where $\Psi_k^+$ is the outgoing wave of the total Hamiltonian. And $\phi_k$ is the free plane wave. At very far way, a free plane wave is scattered by the scattering operator $S$ and result in different $k'$ . The nuclear potential is included in the operator.

$\displaystyle S_{k'k} = \delta(E_{k'} - E_k) \frac{\hbar^2}{4\pi m k} \sum_l (2l+1) P_l(\cos\theta) S_l(E_k)$

and the scattering amplitude is

$\displaystyle f(\theta) = \frac{1}{k} \sum_l (2l+1) P_l(\cos\theta) \frac{S_l(E_k) - 1}{2i}$

For absorption, $S_l = a e^{2 i \delta_l}, 0 < a < 1$

$\displaystyle \sigma_{a} = \frac{\pi}{k^2} \sum_l (2l+1) (1-|S_l|^2)$

The absorption cross section also called reaction cross section. The elastic cross section is from the real-part of the potential, and the absorption (reaction) cross section is from the imaginary-part of the potential. That means, part of the flux will go to the absorption (reaction) channels from the elastics channel.

For elastic scattering, $|S_l| = 1$ , thus, no absorption. Maximum elastic cross section when $S_1 = -1$ , and no elastic cross section when $S_l = 1$ .

The relation between $1-|S_l|^2$ and $|1-S_l|^2$ is plot below, where $S_l = a e^{2 i \delta_l}$ .

Annotation 2020-04-17 164051.png

When $a = 0$ , both elastics and absorption cross section is equal. The horizontal line is the step for $a$ from 0, to 1. The curved lines are step for $\delta_l$ from 0, 90 deg.

We can see that, the elastic scattering can be 4 times more, this is when the scattering is constructive interfered, i.e. $|1+1|^2 = 4$ . And the elastic scattering can be 0. time is when the scattering is destructive interfered. $|1-1|^2 = 0$

There can be elastics scattering but not reaction scattering (or absorption). But whenever there is reaction scattering, elastics scattering is there.

Coulomb wave function (II)

April 17, 2020

GoLuckyRyan Basic Coulomb, Gamma Function, hypergeometric, phase shift, scattering 5 Comments

In the previous post, we tried to derived to Coulomb wave function, and the regular Coulomb wave function $F_l(x)$ is derived, except for the normalization constant, and the mysterious Coulomb phase shift.

From this arxiv article, the Coulomb “Hankel” function is

$\displaystyle H_L^{pm}(x) = D_L^{\pm} x^{L+1} e^{\pm i x} U(L+1 \pm i \eta, 2L+2, - \pm 2 i x)$

$\displaystyle D_L^{\pm} = (-\pm 2i)^{2L+1} \frac{\Gamma(L+1\pm i \eta)}{ C_L \Gamma(2L+2)}$

$\displaystyle C_L = 2^L \frac{\sqrt{\Gamma(L+1+ i \eta) \Gamma(L+1- i \eta)}}{ e^{\eta \pi/2}\Gamma(2L+2)}$

where $U(a,b,z)$ is the confluent hypergeometric function of the second kind.

The regular Coulomb wave function is

$\displaystyle F_L(x) = C_L x^{L+1} e^{\pm i x} ~_1F_1(L+1 \pm i \eta, 2L+2, - \pm 2 i x) = \frac{1}{2i} \left( H_L^+ - H_L^- \right)$

I am fail to prove the last equality. The irregular Coulomb wave function is

$\displaystyle G_L(x) = \frac{1}{2} \left( H_L^+ + H_L^- \right)$

The form of the coefficience $C_L$ is different from the form give in this post, but they are the same.

$\displaystyle \frac{\sqrt{\Gamma(L+1+ i \eta) \Gamma(L+1- i \eta)}}{ \Gamma(2L+2)} = \frac{\left| \Gamma(L+1+i \eta) \right|}{(2L+1)!}$

In Mathematica 11.2, the $U(a,b,z)$ is fail to evaluate when $\eta > 2$ for $l = 0$ , and $\eta > 0.1$ for $l = 5$ .

Update 20200420, settign WorkingPrecision to 45 can solve the problem. Here are some plots for the Coulomb wave functions with $\eta = 10$ .

Annotation 2020-04-20 223816.png

Annotation 2020-04-20 223849.png

Annotation 2020-04-20 224003.png

When $\eta$ getting larger, the wave function pushed further.

Annotation 2020-04-20 224017.png

Coulomb wave function

April 17, 2020

GoLuckyRyan Basic Coulomb, Gamma Function, hypergeometric, Legendre polynomial, phase shift, Riccati-Bessel, scattering Leave a comment

The Coulomb wave function is the solution of a REPULSIVE Coulomb potential. The Schrödinger equation is

$\displaystyle \left( -\frac{\hbar^2}{2m} \nabla^2 + \frac{Q_1Q_2 e^2}{r} \right) \Phi = E \Phi$

where $e^2 = 1.44~\textrm{MeV.fm}$ , separate the radial and angular part as usual,

$\displaystyle \left( -\frac{\hbar^2}{2m} \frac{1}{r^2} \left( \frac{d}{dr} r^2 \frac{d}{dr} \right) + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2} + \frac{Q_1Q_2 e^2}{r} \right) R(r) = E R(r)$

setting

$\displaystyle k^2 = \frac{2mE}{\hbar^2}, \eta =\frac{k}{2E} Q_1Q_2 e^2$

$\displaystyle \left( \frac{1}{r^2} \left( \frac{d}{dr} r^2 \frac{d}{dr} \right) + k^2 - \frac{l(l+1)}{r^2} - \frac{2\eta k}{r} \right) R(r) =0$

using the usual substitution $R = u / r$

$\displaystyle \frac{1}{r^2} r \frac{d^2u(r)}{dr^2} + \left( k^2 - \frac{l(l+1)}{r^2} - \frac{2\eta k}{r} \right) \frac{u(r)}{r} =0$

$\displaystyle \frac{d^2u(r)}{dr^2} + \left( k^2 - \frac{l(l+1)}{r^2} - \frac{2\eta k}{r} \right) u(r) =0$

Setting $x = kr$

$\displaystyle \frac{d^2u(x)}{dx^2} + \left( 1 - \frac{l(l+1)}{x^2} - \frac{2\eta}{x} \right) u(x) =0$

The short range behaviour is approximated as

$\displaystyle \frac{d^2u(x)}{dx^2} = \frac{l(l+1)}{x^2}u(x) \Rightarrow u(x) \approx x^{l+1}~\textrm{or}~ x^{-l}$

and for long range behaviour, it should approach Riccati–Bessel functions $\hat{j}_l, \hat{n}_l$ with phase shift.

Set $u(x) = x^{l+1} \exp(i x) y(x)$ , the equation of $u(x)$ becomes,

$\displaystyle x \frac{d^2y}{dx^2} + (2L+2 + 2ix) \frac{dy}{dx} + (2i(L+1) - 2\eta) y = 0$

Change of variable $z = - 2i x$ ,

$\displaystyle z \frac{d^2y}{dz^2} + (2L+2 - z ) \frac{dy}{dx} - (L+1 + i \eta) y = 0$

This is our friend Laguerre polynomial again!!! with $\alpha = 2L+1, n = -(L+1+i\eta)$ . Since the $n$ is not an integer anymore, we go to a more general case, that is the Kummer’s equation,

$\displaystyle z \frac{d^2 w}{dz} + (b-z) \frac{dw}{dz} - a w = 0$

The solution of Kummer’s equation is the confluent hypergeometric function

$w(z) = _1F_1(a, b, z)$

Thus, the solution for the radial function is

$\displaystyle u_l(x) =A x^{l+1} e^{ix} _1F_1(L+1+i \eta, 2L+2, -2 i x )$

where $A$ is a normalization constant by compare the long range behaviour with Riccati-Bessel function. The full solution is,

$\displaystyle u_l(x) = \\ F_l(x) = \frac{2^l e^{-\pi \eta/2} |\Gamma(l+1+i\eta)| }{(2l+1)!} x^{l+1} e^{ix}~_1F_1(L+1+i \eta, 2L+2, -2 i x )$

At long range,

$\displaystyle F_l(x \rightarrow \infty) = \sin \left( x - l \frac{\pi}{2} - \eta \log(2x) + \sigma_l \right)$

where $\sigma_l = \arg( \Gamma(l+1+i\eta) )$ is the Coulomb phase shift.

Using Kummer’s transform

$\displaystyle _1F_1(a,b,z) = e^z ~_1F_1(b-a, b, -z)$

The solution can be written as

$\displaystyle u_l(x) = \\ F_l(x) = \frac{2^l e^{-\pi \eta/2} |\Gamma(l+1+i\eta)| }{(2l+1)!} x^{l+1} e^{-ix}~_1F_1(L+1-i \eta, 2L+2, 2 i x )$

Another solution should behave like $\cos$ and unbound at $x = 0$ .

Set $u(x) = x^{-l} \exp(i x) y(x)$ , the equation of $u(x)$ becomes,

$\displaystyle x \frac{d^2y}{dx^2} + (-2L + 2ix) \frac{dy}{dx} + (-2i L - 2\eta) y = 0$

Change of variable $z = -2ix$ ,

$\displaystyle z \frac{d^2y}{dz^2} + (-2L -z ) \frac{dy}{dx} - ( - L + i \eta) y = 0$

$\displaystyle u_l(x) =A x^{-l} e^{ix}~ _1F_1(-L+i \eta, -2L, -2 i x )$

However, $_1F_1(a, b, z)$ does not exist for non-position b.

We can also transform into Whittaker’s equation by change of variable $z = 2ix$

$\displaystyle -4 \frac{d^2u(x)}{dz^2} + \left( 1 - \frac{-4l(l+1)}{z^2} - \frac{4i\eta}{z} \right) u(x) =0$

$\displaystyle \frac{d^2u(x)}{dz^2} + \left( -\frac{1}{4} +\frac{i\eta}{z}- \frac{l(l+1)}{z^2} \right) u(x) =0$

using $\displaystyle l(l+1) = (l+1/2)^2 - 1/4$

$\displaystyle \frac{d^2u(x)}{dz^2} + \left( -\frac{1}{4} +\frac{i\eta}{z} + \frac{ 1/4 - (l+1/2)^2}{z^2} \right) u(x) =0$

This is the Whittaker’s equation with $\kappa = i \eta, \mu = l+1/2$

The solutions are

$u_l(x) = e^{-ix} (2ix)^{l+1} ~_1F_1(l+1-i \eta, 2l+2, 2ix)$

$u_l(x) = e^{-ix} (2ix)^{l+1} U(l+1-i \eta, 2l+2, 2ix)$

I still cannot get the second solution, the $G_l$ . According to Wolfram, https://mathworld.wolfram.com/CoulombWaveFunction.html

$\displaystyle G_l(x) = \frac{2}{\eta C_0^2(\eta)} F_l(x) \left( \log(2x) + \frac{q_l(\eta)}{p_l(\eta)} \right) + \frac{x^{-l}}{(2l+1) C_l(\eta)} \sum_{K=-l}^\infty a_k^l(\eta) x^{K+l}$ ,

where $q_l, p_l, a_k^l$ are defined inAbramowitz, M. and Stegun, I. A. (Eds.). “Coulomb Wave Functions.” Ch. 14 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 537-544, 1972.

Nuclear Physics 101

Various cross sections

Rutherford scattering in CM and Lab frame

Calculate phase shift of square well

Two-body Coulomb scattering

Scattering and Reaction cross sections

Resonance Scattering

Scattering with Coulomb potential

S-matrix, elastics and absorption cross section

Coulomb wave function (II)

Coulomb wave function

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