## Resonance Scattering

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

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

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

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.

## Coulomb wave function (II)

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

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

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

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

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

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

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.

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

## Coulomb wave function

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.

## Meaning of phase shift

For hard sphere with radius R, the phase shift is

$\displaystyle \tan \delta_l = - \frac{\hat{j}_l(kR)}{ \hat{n}_l(kR)}$

for $l = 0$, $\displaystyle \delta_0 = - k R$

In general, the phase shift has these properties

• The phase shift is use to calculate the cross section.
• At very large energy, the phase shift approach 0, $\lim_{k\rightarrow \infty} \delta_l = 0$
• When $\delta_l = (n+1) \frac{\pi}{2}$, it is where resonance take place.
• At resonance, the slope of the $\delta_l(E)$ is related to the resonance width.
• The absolute phase shift is defined at $\lim_{k\rightarrow \infty} \delta_l = 0$, thus, the absolute phase shift is always positive.
• The absolute phase shift at $k = 0$ is equal to $n \pi$, where $n$ is the number of bound state. ( this is also known as Levinson’s theorem )
• When $\delta_l = n \pi$, $S_l = 1$ and the scattering amplitude for that partial wave is vanished.

Suppose the outer wave take the form

$\displaystyle u_l^{II}(k,r) = a_l ( \hat{j}_l(kr) + b_l \hat{n}_l(kr) )$

The logarithmic derivative at $r = R$ is

$\displaystyle L^{II} = R \frac{1}{u_l^{II}(kR)} \left. \frac{du_l^{II}(kr)}{dr} \right|_{r=R} = kR \frac{\hat{j}_l'(kR) + b_l \hat{n}_l'(kR)}{\hat{j}_l(kR) + b_l \hat{n}_l(kR)}$

The inner function will be calculated by numerical method, and the logarithmic derivative at $r = R$,

$\displaystyle L_{I} = R \frac{(u_l^I)'}{u_l^I}$

matching the boundary condition,

$\displaystyle L^{I} = L^{II} = kR \frac{\hat{j}_l'(kR) + b_l \hat{n}_l'(kR)}{\hat{j}_l(kR) + b_l \hat{n}_l(kR)}$

solve for $b_l$

$\displaystyle b_l = - \frac{kR \hat{j}_l'(kR) - L^I \hat{j}_l(kR)}{kR \hat{n}_l'(kR) - L^I \hat{n}_l(kR)} = \tan (\delta_l)$

If Coulomb potential included, simply replace $\hat{j}_l \rightarrow F_l$ and $\hat{n}_l \rightarrow G_l$.

In more general, the phase shift could indicate the attractive/repulsive of the potential.

• positive phase shift = attractive potential
• negative phase shift = repulsive potential

When the potential is attractive ( repulsive), the wave-number $k$ will be larger (smaller) than free plane wave inside the potential, thus, the wave will “advance” more (less), resulting positive (negative) phase shift. For example, I use a Woods-Saxon potential with depth only -1 MeV. A proton s-wave radial wave function is

The blue curve is the radial wave function, the orange curve is $\hat{j}_0$. The red curve is the potential / 10, and the green line is the energy / 100.

The phase shift is 7 deg.

If I flip the potential to be positive

The phase shift is -7 deg.

However, there is an ambiguity of the phase shift, because -7 deg = +  353 deg. Thus, the phase shift need to be measure at various incident energies and infer from the data. For example, using high incident energy so that the wave length inside the potential will not be different from outside. That will give a general ideal on the attractive/repulsive nature of the potential. However, high energy also means the s-wave component is small.

But the S-matrix does not have ambiguity.

## S-matrix and scattering amplitude

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.

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

After some manipulation,

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

$\displaystyle S_l = \frac{a + i b}{ a - i b }$

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

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.

The time-independent Schrödinger equation is

$(-\frac{\hbar^2}{2m}\nabla^2 + V ) \Psi = E \Psi$

Using the Laplacian in spherical coordinate. and Set $\Psi = R Y$

$\nabla^2 R Y - \frac{2m}{\hbar^2}(V-E) R Y = 0$

$\nabla^2 = \frac{1}{r^2}\frac{d}{dr}(r^2 \frac{d}{dr}) - \frac{1}{r^2} L^2$

The angular part,

$L^2 Y = l(l+1) Y$

$\frac{d}{dr}(r^2\frac{dR}{dr}) - l(l+1)R - \frac{2mr^2}{\hbar^2}(V-E) R = 0$

To simplify the first term,

$R = \frac{u}{r}$

$\frac{d}{dr}(r^2 \frac{dR}{dr})= r \frac{d^2u}{dr^2}$

A more easy form of the radial function is,

$\frac{d^2u}{dr^2} + \frac{l(l+1)}{r^2} u - \frac{2m}{\hbar^2} (V-E) u = 0$

The effective potential $U$

$U = V + \frac{\hbar^2}{m} \frac{l(l+1)}{r^2}$

$\frac{d^2u}{dr^2} + \frac{2m}{\hbar^2} (E - U) u = 0$

We can use Rungu-Kutta method to numerically solve the equation.

The initial condition of $u$ has to be 0. (home work)

I used excel to calculate a scattered state of L = 0 of energy 30 MeV. The potential is a Wood-Saxon of depth 50 MeV, radius 3.5 fm, diffusiveness 0.8 fm.

Another example if bound state of L = 0. I have to search for the energy, so that the wavefunction is flat at large distance. The outermost eigen energy is -7.27 MeV. From the radial function, we know it is a 2s orbit.

## 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 $\theta_{cm}$.