Phase shift of elastics scattering

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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^-

but that is not so natural at all, 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^+)


\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

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In a scattering problem, the main objective is solving the Schrödinger equation


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.