Scattering in a Central Potential

I hope that most of this is just a reminder to you of work covered in other classes, hopefully the introductory quantum lectures. We are considering the situation of a wavepacket, localized at -ve $\infty$ at time $t~=~-\infty$. We wish to form the (differential) cross section for this wave packet to scatter off a scattering centre. This amounts to determining the probability for the particle in the wave packet to end up travelling at some angle with respect to its initial direction. Wave packets are inherently messy objects to deal with and so it is more convenient to determine the scattering amplitudes of incident plane waves, and the convolve with the wavepacket amplitudes at the end. In reality we never do this as the momentum spread of the wavepacket, is small compared to other scales in the problem.

When we are far from the scattering potential for a finite range potential, then we know that the incident wavefunction is just a plane wave $e^{i k z}$, where $k~=~\vert{\bf k}\vert~=~\sqrt{2 \mu E}$ where $\mu$ is the reduced mass of the system. As we are going to be dealing with scattering in states of angular momentum $l$ it is convenient to expand the incident plane wave in the z-direction into angular momentum states

 

$$e^{i k z } \ =\ e^{i k r \cos\theta} = \sum_{l=0}\ (2 l+1) \ i^l \ j_l(k r) \ P_l(\cos\theta) \ \ \ .$$ (1)

It is obvious that this plane wave is a solution to the schrodinger equation with no potential. In order to determine the amplitude for scattering of this incident plane wave we need to solve the schrodinger equation with the complete potential $V(r)$ (we are assuming a central potential). We have that

 

$$H\psi \ =\ E\psi$$

$$\left[ -{1\over 2 \mu} \nabla^2 + V(r) \right] \psi({\bf r}) \ =\ - E\psi({\bf r}) \ \ \ .$$ (2)

We use the separability of $\psi$ and define

 

$$\psi({\bf r}) \ =\ {1\over r} u(r)_l Y_l^m (\Omega) \ \ ,$$ (3)

to write

 

$$\left[ -{d^2\over dr^2} + {l (l+1)\over r^2} + 2\mu V(r) - k^2\right] u_l(r)\ =\ 0 \ \ \ ,$$ (4)

where we have used that the radial part of the laplacian is

 

$$\nabla^2 = {1\over r^2}{d\over dr}\left( r^2 {d\over dr} \right) + {\rm angular.....} \ \ \ .$$ (5)

We have a boundary condition on $u_l(r)$ arising from the fact that the wavefunction $\psi$ must be finite everywhere. This gives that $u$ must vanish at the origin, $u_l(0)~=~0$. The slope of $u_l$ at the origin is determined by the normalization of the wavefunction, and can be chosen to be a real number, making $u_l$ a real valued function.

If there was no potential $V(r)=0$ everywhere, then the solution to the schrodinger equation would be $u_l(r) \sim r j_l(kr)$, where $j_l(z)$ is a spherical Bessel function. However, when there is a non-zero potential, we can still have $u_l$ vanish at the origin, while at long distances (much larger than the range of the potential) the solution has the form

 

$$u_l(r) \ =\ r \left( A_l j_l(kr) + B_l n_l(kr) \right)$$

$$\rightarrow {1\over k}\left( A_l \sin(kr-l{\pi\over 2}) - B_l \cos(kr- l{\pi\over 2} ) \right) \ \ \ ,$$ (6)

as $r\rightarrow\infty$. The eigenfunctions are $j_l(z)$ ( the spherical Bessel functions) which are regular at the origin, and $n_l(z)$ (the Neumann functions) which are singular at the origin. Our statements about boundary conditions lead us to $B_l=0$when $V=0$.

It is useful to have the asymptotic forms of the Bessel and Neumann functions explicitly to understand the behaviour of the amplitudes

 

$$j_0(z) \rightarrow {\sin z\over z}$$

$$j_1(z) \rightarrow {\sin z\over z^2} - {\cos z\over z}$$

$$j_2 (z) \rightarrow \left({3\over z^3}-{1\over z}\right)\sin z - {3\over z^2}\cos z$$

$$n_0 (z) \rightarrow -{\cos z\over z}$$

$$n_1 (z) \rightarrow -{\cos z\over z^2} - {\sin z\over z}$$

$$n_2 (z) \rightarrow -\left({3\over z^3}-{1\over z}\right)\cos z - {3\over z^2}\sin z \ \ \ .$$ (7)

For a non-vanishing potential, we are free to define the ratio of coefficients $A_l$ and $B_l$ to be

 

$${B_l\over A_l} = -\tan\delta_l \ \ \ ,$$ (8)

where $\delta_l$ is a energy dependent phase. Further, for convenience we normalize the states to have a coefficient of unity,

 

$$u_l(r)\rightarrow {1\over k} \sin (kr-l{\pi\over 2} + \delta_l ) \ \ .$$ (9)

The complete wavefunction (denoted by $\psi^+$) for the scattering of a plane wave is the sum of the incident plane wave and the outgoing wave (outgoing means $e^{ikr}$ and incoming means $e^{-ikr}$) is

 

$$\psi^+ = e^{ikr\cos\theta} + f(\theta) {e^{ikr}\over r}$$

$$= \sum_{l=0} (2l+1) i^l {\sin (kr-l{\pi\over 2})\over kr} P_l(\cos\theta) + f(\theta) {e^{ikr}\over r} \ \ \ ,$$ (10)

and we know that this can be written in terms of eigenstates of the hamiltonian, including the potential,

 

$$\psi^+ = \sum_{l=0} a_l P_l(\cos\theta) {u_l\over r}$$

$$\rightarrow \sum_{l=0} a_l P_l (\cos\theta) {\sin (kr-l{\pi\over 2} + \delta_l )\over kr} \ \ \ ,$$ (11)

for unknown coefficients $a_l$. By comparing the asymptotic form of the incoming waves ($e^{-ikr}$) of these expressions, we find that the $a_l$'s are $a_l = (2l+1) i^l e^{+i \delta_l}$leading to

 

$$\psi^+ = \sum_{l=0} (2l+1) i^l e^{i\delta_l} P_l (\cos\theta) {\sin (kr-l{\pi\over 2} + \delta_l )\over kr} \ \ \ .$$ (12)

A comparison now, between the outgoing waves ($e^{+ikr}$) yields

 

$$f(\cos\theta) = {1\over 2 i k} \sum_{l=0} (2l+1) (e^{2 i \delta_l} - 1) P_l(\cos\theta)$$

$$= {1\over k} \sum_{l=0} (2l+1) e^{i\delta_l} \sin\delta_l \ P_l(\cos\theta) \ \ \ .$$ (13)

We are now in a position to determine the cross section for a given process. We start by computing the flux of particles associated with $\psi^+$. From NR quantum mechanics we have that the current density is

$${\bf j}\ =\ {1\over \mu} {\rm Re}\left( \psi^*\nabla\psi\right) \ \ \ ,$$ (14)

from which we have that

$${\bf j}_{\rm inc}\ =\ v\ =\ {k\over\mu}$$

$${\bf j}_{\rm scatt}\ =\ v {1\over r^2} \vert f(\Omega)\vert^2 \ \ \ .$$ (15)

The differential cross section is defined to be

$${d\sigma\over d\Omega}\ =\ \vert f(\Omega)\vert^2 \ \ \ ,$$ (16)

and the probability of scattering from a beam incident upon a target with $N$ scatterers per unit volume and thickness $T$ into a solid angle $d\Omega$ is

$${\rm prob} \ =\ {d\sigma\over d\Omega} d\Omega N T \ \ \ ,$$ (17)

assuming that the beam is negligibly attentuated in passing through the target.

Hence the cross section is

 

$${d\sigma\over d\Omega} = \vert f(\cos\theta)\vert^2$$

$$= {1\over k^2} \vert \sum_{l=0} (2l+1) e^{i\delta_l} \sin\delta_l P_l(\cos\theta) \vert^2$$

$$\sigma = {4\pi\over k^2} \sum_{l=0} (2l+1) \sin^2\delta_l \ \ \ ,$$ (18)

where we have used the orthogonality of Legendre polynomials

 

$$\int \ dx\ P_l(x) P_{l^\prime}(x) = {2\over 2l +1 } \delta_{l l^\prime} \ \ \ .$$ (19)

We see from this that the unitarity of the scattering amplitude ( $\vert\sin\delta_l\vert\le 1$) leads to the contribution from each partial wave being bounded from above by

 

$$\sigma_l \le {4\pi\over k^2} (2l+1) \ \ .$$ (20)