S-matrix, elastics and absorption cross section

April 18, 2020

GoLuckyRyan Basic, 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.

S-matrix and scattering amplitude

April 16, 2020

GoLuckyRyan Basic, Scattering , , , , Leave a comment

In this post and this post talk about the scattering and phase shift. But the relation is not shown clearly.

The central idea is solving the Schrödinger equation

\displaystyle  \left( -\frac{\hbar^2}{2m} \nabla^2 + V(r) \right) \Psi = E \Psi

rewrite,

\displaystyle \left(\frac{\hbar^2}{2m} \nabla^2 + E \right) \Psi = V(r) \Psi

The homogeneous solution ( i.e. V(r) = 0 ) is a plane wave e^{i\vec{k} \cdot \vec{r} } . Thus, the general solution is

\displaystyle \Psi = A \left(e^{i\vec{k} \cdot \vec{r}} + f(\theta) \frac{e^{ikr}}{r} \right)

Here is the outline:

  1. Solve the Schrödinger equation in partial wave
    1. the inner part is solved numerically
    2. the outer part is asymptotically approach to free plane wave with a phase shift ( or the S-Matrix)
  2. matching the boundary condition
  3. compare the outer part with the general form of the solution:
    \displaystyle \Psi = A \left(e^{i\vec{k} \cdot \vec{r}} + f(\theta) \frac{e^{ikr}}{r} \right)
    that give out the scattering amplitude f(\theta)
  4. The elastic cross section is |f(\theta)|^2

Another way to solve the Schrödinger equation is solve it in a partial wave,

\displaystyle \Psi_{lm} = \frac{u_l(kr)}{kr} Y_{lm}(\theta, \phi)

where the radial solution u_l satisfy,

\displaystyle \left( \frac{d^2}{d \rho^2} +1 - \frac{l(l+1)}{\rho^2} \right) u_l = U(\rho) u_l

with \rho = kr , U(\rho ) = V(r)/E . The total solution is

\displaystyle \Psi = \sum_l C_l Y_{lm}(\theta, \phi) \frac{u_l(kr)}{kr}

As very far away, (the arrow stands for very long distance)

\displaystyle u_l(kr) \rightarrow a \hat{j}_l + b \hat{n}_l

where \hat{j}_l(r) is Riccati-Bessel function. The spherical Bessel-function can be expressed into spherical Hankel function.

\displaystyle \hat{h}_l^{\pm}(r) = \hat{n}_l(r) \pm i \hat{j}_l(r)  \rightarrow  e^{\pm i (r - l \frac{\pi}{2}) }

Follow below manipulation,

\displaystyle u_l(kr) \rightarrow a \hat{j}_l + b \hat{n}_l = \frac{a}{2i}\left(\hat{h}_l^+ - \hat{h}_l^- \right) + \frac{b}{2}\left(\hat{h}_l^+ + \hat{h}_l^- \right) \\ ~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{a+ib}{2i} \hat{h}_l^+ - \frac{a-ib}{2i} \hat{h}_l^- \\~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{a-ib}{2i} \left( S_l \hat{h}^+ - \hat{h}^- \right), ~~S_l = \frac{a + i b}{ a - i b }

Drop the a-ib,

\displaystyle u_l(kr) \rightarrow \frac{1}{2i} \left(  S_l \hat{h}^+ - \hat{h}^- \right) = \frac{1}{2i} \left( S_l e^{i(kr - l \pi/2)} - e^{-i(kr - l\pi/2)} \right)

The total wave function is the sum of all partial waves,

\displaystyle \Psi = \sum_l C_l Y_{lm}(\theta, \phi) \frac{u_l(kr)}{kr}

At far distance, m = 0 and Y_{lm} \rightarrow P_l(cos\theta) .

\displaystyle \Psi \rightarrow \sum_l C_l P_l(\cos\theta)\frac{1}{2i} \frac{ S_l e^{i(kr - l \pi/2)} - e^{-i(kr - l\pi/2)} }{kr}

Separate out e^{\pm ikr}

\displaystyle \Psi \rightarrow  \frac{e^{-ikr}}{r} \sum_l P_l(\cos\theta) \frac{ -C_l  e^{i l \pi/2}}{2ik} +  \frac{e^{ikr}}{r} \sum_l P_l(\cos\theta) \frac{ C_l S_l e^{- i l \pi/2}}{2ik}

The plane wave solution can be expanded into partial wave,

\displaystyle e^{i\vec{k} \cdot \vec{r}} = \sum_l (2l+1) i^l P_l(\cos\theta) \frac{\hat{j}_l(kr)}{kr}

Thus,

\displaystyle \hat{j}_l = \frac{\hat{h}_l^+ - \hat{h}_l^-}{2i} \rightarrow \frac{e^{i(r-l\pi/2)} -  e^{-i(r-l\pi/2)}}{2i}

substitute into the plane wave

\displaystyle e^{i\vec{k} \cdot \vec{r}} \rightarrow  \sum_l (2l+1) i^l P_l(\cos\theta) \frac{e^{i(kr-l\pi/2)} -  e^{-i(kr-l\pi/2)}}{2ikr}

substitute into \Psi =A( e^{i\vec{k} \cdot \vec{r}} + f(\theta) \frac{e^{ikr}}{r} ) and collect e^{\pm ikr} .

\displaystyle \Psi \rightarrow A\left(\sum_l (2l+1) i^l P_l(\cos\theta) \frac{e^{i(kr-l\pi/2)} -  e^{-i(kr-l\pi/2)}}{2ikr} + f(\theta) \frac{e^{ikr}}{r} \right)

\displaystyle \Psi \rightarrow \frac{e^{- ikr}}{r} \sum_l (2l+1) i^l P_l(\cos\theta) \frac{-A e^{il\pi/2}}{2ik} + \\ \frac{e^{ikr}}{r} \left( A f(\theta) + \sum_l (2l+1) i^l P_l(\cos\theta) \frac{Ae^{-il\pi/2}}{2ik}\right)

Compare the coefficient of  \frac{e^{\pm ikr}}{r}

\displaystyle \frac{ -C_l  e^{i l \pi/2}}{2ik} = (2l+1) i^l \frac{-A e^{il\pi/2}}{2ik}

Thus,

\displaystyle \Rightarrow C_l = A (2l+1) i^l

And,

\displaystyle \sum_l P_l(\cos\theta) \frac{ C_l S_l e^{- i l \pi/2}}{2ik} = A f(\theta) + \sum_l (2l+1) i^l P_l(\cos\theta) \frac{Ae^{-il\pi/2}}{2ik}  

\displaystyle f(\theta)  = \sum_l P_l(\cos\theta) (2l+1) i^l (S_l -1)  \frac{e^{- i l \pi/2}}{2ik} 

And since e^{-i l \pi/2} = (-i)^l ,

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

And

\displaystyle S_l = \frac{a+i b}{a - ib} = e^{2i\delta_l }  \Rightarrow \frac{S_l -1 }{2 i} = e^{i \delta_l } \sin(\delta_l )

Finally,

\displaystyle f(\theta)  =   \frac{1}{k} \sum_l P_l(\cos\theta) (2l+1) e^{i \delta_l } \sin(\delta_l )

The elastics cross section ( because the energy is same as incoming and outgoing wave ) is

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

The total elastics cross section is

\displaystyle \sigma = 2\pi \int |f(\theta)|^2 d(\cos\theta)

Using

\displaystyle \int P_l(\cos\theta) P_{l'}(\cos\theta) d(\cos\theta) = \frac{2}{2l+1} \delta_{ll'}

\displaystyle \sigma = 2\pi  \int |f(\theta)|^2 d(\cos\theta) = \frac{4\pi}{k^2} \sum_l (2l+1) \sin^2(\delta_l ) 

At \theta = 0 , P_l(1) = 1, e^{i \delta_l } = \cos \delta_l + i \sin \delta_l ,

\displaystyle f(0)  =   \frac{1}{k} \sum_l (2l+1) ( \cos \delta_l \sin\delta_l + i   \sin^2\delta_l ) 

\displaystyle \sigma = \frac{4\pi}{k^2} \sum_l (2l+1) \sin^2(\delta_l ) = \frac{4\pi}{k} Im(f(0)) 

Phase shift of elastics scattering

September 29, 2016

GoLuckyRyan Basic, theory , , , , , Leave a comment

I found that the derivation of most “google result” is not clear enough. So here is my derivation. Before process, people may need to review the pervious post.

Most people start from,

u_l = A( \hat{h}_l^- - S_l \hat{h}_l^+ )

that annoying me because it is somehow not “natural”. Why there is a “minus” sign? Why the \hat{h}_l^- is the first term? For my self, a more natural way is,

u_l = a \hat{h}_l^+ + b \hat{h}_l^-

where a, b are complex numbers, but that is still not so natural, because in numerical calculation, for simplicity, there is no complex number, we only have,

u_l = \alpha \hat{j}_l + \beta \hat{n}_l

The first term is alway there as it is the free solution and bounded at r = 0. the second term is caused by the potential.

The goal is to find a solution take the form

\displaystyle \psi = A \left( e^{i \vec{k} \cdot \vec{r}} + f(\theta) \frac{e^{ikr}}{r} \right)

where the first term is free wave and the second term is scattered wave. The solution for elastics scattering is

\displaystyle \psi = \sum C_l P_l (\cos\theta) \frac{u_l}{kr} \rightarrow \sum C_l P_l(\cos\theta) (\alpha \hat{j}_l + \beta \hat{n}_l)

we used the substitution,

\displaystyle R_l(r) = \frac{u_l(\rho)}{\rho},  \rho = kr.

The radial function can be solved using Rungu-Kutta method on the equation,

\displaystyle \frac{d^2}{d\rho^2} u_l = \frac{2 m_\mu}{\hbar^2} (V-E) u_l + \frac{l(l+1)}{\rho^2}

and the solution of u_l at far away is,

u_l \rightarrow  \alpha \hat{j}_l + \beta \hat{n}_l.

the arrow means r \rightarrow \infty. So, the problem is how to rewrite the solution. In the way, we will see how the phase shift or the S-matrix was found.

The free solution is the spherical wave,

\displaystyle e^{i \vec{k} \cdot \vec{r}} = \sum_l (2l+1) i^l P_l(\cos\theta) j_l(kr)

The spherical Bessel function j_l(kr) cna be express as Heankel function

h_l^{\pm} = n_l \pm i j_l \rightarrow e^{\pm i (kr - l \frac{\pi}{2})}

The + sign is outgoing wave.

\displaystyle u_l \rightarrow (\alpha \hat{j}_l + \beta \hat{n}_l)

\displaystyle = \frac{\alpha}{2i} (\hat{h}_l^{+} - \hat{h}_l^{-}) + \frac{\beta}{2}(\hat{h}_l^{+} + \hat{h}_l^{-})

\displaystyle = \frac{\alpha - i \beta}{2i} (\hat{h}_l^{+} - \hat{h}_l^{-}) + \beta \hat{h}_l^{+}

\displaystyle = (\alpha - i \beta ) \left( \frac{\hat{h}_l^+ - \hat{h}_l^-}{2i} + \frac{\beta}{\alpha - i \beta} \hat{h}_l^+\right)

\displaystyle = (\alpha - i \beta ) \left( \hat{j}_l + \frac{\beta}{\alpha - i \beta} \hat{h}_l^+\right)

Since the u_l should be normalized, we can se \alpha = \cos \delta and \beta = \sin\delta.

\displaystyle \frac{\beta}{\alpha - i \beta } = \sin(\delta) e^{i\delta}

We put u_l back

\displaystyle \psi \rightarrow \sum_l C_l P_l (cos\theta)(\alpha - i \beta ) \left( \hat{j}_l + \sin(\delta) e^{i\delta} \hat{h}_l^+\right)

By setting

\displaystyle C_l = A i^l \frac{2l+1}{\alpha - i \beta} ,

we have the first term is the free wave function. In the second term, \hat{h}_l^+ \rightarrow e^{i(kr - l \frac{\pi}{2}}) / kr . Notice that

e^{i l \frac{\pi}{2}} = i^{-l}

That cancel the i^l term in C_l . And we have

 \displaystyle f(\theta) = \sum (2l+1) P_l (\cos\theta) \frac{\sin(\delta) e^{i\delta}}{k}

some people will write the u_l as \hat{h}_l^{\pm} and the S-matrix,

\displaystyle u_l = \frac{\alpha + i \beta} {2i} \hat{h}_l^+ - \frac{\alpha - i \beta}{2i} \hat{h}_l^-

\displaystyle = -\frac{\alpha - i \beta}{2i} \left( \hat{h}_l^- - \frac{\alpha + i \beta}{\alpha - i \beta} \hat{h}_l^+ \right)

\displaystyle = A' (\hat{h}_l^- - S_l \hat{h}_l^+)

where

\displaystyle S_l =\frac{\alpha + i \beta}{\alpha - i \beta} = e^{2i\delta} .

Remember that this is the S-matrix for elastics scattering.

 

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

February 26, 2016

GoLuckyRyan Basic, nuclear, Scattering, theory , , , , , , , , , , Leave a comment

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<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.

Scattering Matrix

May 8, 2011

GoLuckyRyan Scattering, theory , , Leave a comment

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<p | S| p' \right> \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<p'| [H^0,S] | p\right> = (E_{p'} -E_p ) \left< p'|S|p\right>

thus implies,

\left<p'|S|p\right> = \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<p'|S|p\right> =\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<p'| S|p \right> = \delta(p - p') + \frac{i}{2\pi m} \delta(E_{p'} - E ) f(p' \leftarrow p)