# Optical theorem

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.19.

There is a simple relationship between the total cross section and the scattering amplitude in three dimensional scattering. The formulas are (in terms of phase shifts):

 $\displaystyle f\left(\theta\right)$ $\displaystyle =$ $\displaystyle \frac{1}{k}\sum_{l=0}^{\infty}\left(2l+1\right)e^{i\delta_{l}}\sin\delta_{l}P_{l}\left(\cos\theta\right)\ \ \ \ \ (1)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{4\pi}{k^{2}}\sum_{l=0}^{\infty}\left(2l+1\right)\sin^{2}\delta_{l} \ \ \ \ \ (2)$

If ${\theta=0}$, we can use the fact that ${P_{l}\left(1\right)=1}$ for all ${l}$ (from the definition) so taking the imaginary part of ${f\left(0\right)}$ we get

$\displaystyle \Im f\left(0\right)=\frac{1}{k}\sum_{l=0}^{\infty}\left(2l+1\right)\sin^{2}\delta_{l} \ \ \ \ \ (3)$

from which we get the optical theorem:

$\displaystyle \sigma=\frac{4\pi}{k}\Im f\left(0\right) \ \ \ \ \ (4)$

# Born approximation of delta function well and finite square well

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.18.

We can use the one-dimensional Born approximation to get approximate values for transmission through a couple of test potentials. The Born approximation gives a reflection coefficient of

$\displaystyle R=\left(\frac{m}{\hbar^{2}k}\right)^{2}\left|\int_{-\infty}^{\infty}e^{2ikx}V\left(x\right)dx\right|^{2} \ \ \ \ \ (1)$

with

$\displaystyle k=\frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (2)$

Example 1 The delta function well. Here

$\displaystyle V=-\alpha\delta\left(x\right) \ \ \ \ \ (3)$

so

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \left(\frac{m}{\hbar^{2}k}\right)^{2}\left|-\alpha e^{2ik\times0}\right|^{2}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m\alpha^{2}}{2\hbar^{2}E}\ \ \ \ \ (5)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle 1-R=1-\frac{m\alpha^{2}}{2\hbar^{2}E} \ \ \ \ \ (6)$

The exact values are

 $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{1}{1+2\hbar^{2}E/m\alpha^{2}}\ \ \ \ \ (7)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{1}{1+m\alpha^{2}/2\hbar^{2}E} \ \ \ \ \ (8)$

For a weak potential, ${\alpha}$ is small, so ${2\hbar^{2}E/m\alpha^{2}\gg1}$ so we can approximate the exact formulas by

 $\displaystyle R$ $\displaystyle \approx$ $\displaystyle \frac{1}{2\hbar^{2}E/m\alpha^{2}}=\frac{m\alpha^{2}}{2\hbar^{2}E}\ \ \ \ \ (9)$ $\displaystyle T$ $\displaystyle \approx$ $\displaystyle 1-\frac{m\alpha^{2}}{2\hbar^{2}E} \ \ \ \ \ (10)$

in agreement with the Born approximation.

Example 2 The finite square well. Here

$\displaystyle V=\begin{cases} -V_{0} & -a

We get from 1

 $\displaystyle \int_{-\infty}^{\infty}e^{2ikx}V\left(x\right)dx$ $\displaystyle =$ $\displaystyle -V_{0}\int_{-a}^{a}e^{2ikx}dx\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -V_{0}\frac{e^{2ika}-e^{-2ika}}{2ik}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{V_{0}}{k}\sin\left(2ka\right)\ \ \ \ \ (14)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle \left(\frac{m}{\hbar^{2}k}\right)^{2}\left|-\frac{V_{0}}{k}\sin\left(2ka\right)\right|^{2}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V_{0}^{2}}{4E^{2}}\sin^{2}\left(\frac{2a\sqrt{2mE}}{\hbar}\right)\ \ \ \ \ (16)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle 1-\frac{V_{0}^{2}}{4E^{2}}\sin^{2}\left(\frac{2a\sqrt{2mE}}{\hbar}\right) \ \ \ \ \ (17)$

The exact formula is

$\displaystyle T^{-1}=1+\frac{V_{0}^{2}}{4E\left(E+V_{0}\right)}\sin^{2}\left(\frac{2a}{\hbar}\sqrt{2m\left(E+V_{0}\right)}\right) \ \ \ \ \ (18)$

For a weak potential (or, equivalently, a high incident energy), ${E\gg V_{0}}$ and we get

 $\displaystyle T^{-1}$ $\displaystyle \approx$ $\displaystyle 1+\frac{V_{0}^{2}}{4E^{2}}\sin^{2}\left(\frac{2a\sqrt{2mE}}{\hbar}\right)\ \ \ \ \ (19)$ $\displaystyle T$ $\displaystyle \approx$ $\displaystyle 1-\frac{V_{0}^{2}}{4E^{2}}\sin^{2}\left(\frac{2a\sqrt{2mE}}{\hbar}\right) \ \ \ \ \ (20)$

so again, the Born approximation gives a good result.

# Born approximation in one dimension

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.17.

Using Griffiths’s Green’s function for the one-dimensional Schrödinger equation:

$\displaystyle \psi\left(x\right)=\psi_{0}\left(x\right)-\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}e^{ik\left|x-x_{0}\right|}V\left(x_{0}\right)\psi\left(x_{0}\right)dx_{0} \ \ \ \ \ (1)$

we can work out the Born approximation in the one-dimensional case. The idea is we replace ${\psi\left(x_{0}\right)}$ inside the integral by the incident plane wave form ${\psi_{0}\left(x_{0}\right)}$. Assuming that the potential is zero outside a finite distance from the origin, we want the wave function in two regions: ${x\ll0}$ and ${x\gg0}$. The former will give the reflected wave and the latter the transmitted wave.

For ${x\ll0}$ (and at a distance where ${V\left(x\right)=0}$) we have

 $\displaystyle e^{ik\left|x-x_{0}\right|}$ $\displaystyle =$ $\displaystyle e^{-ikx}e^{ikx_{0}}\ \ \ \ \ (2)$ $\displaystyle \psi_{0}\left(x_{0}\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx_{0}} \ \ \ \ \ (3)$

where ${A}$ is the normalization constant. The Born approximation for this region is

 $\displaystyle \psi\left(x\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx}-\frac{im}{\hbar^{2}k}e^{-ikx}\int_{-\infty}^{\infty}e^{ikx_{0}}V\left(x_{0}\right)Ae^{ikx_{0}}dx_{0}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left[e^{ikx}-\frac{im}{\hbar^{2}k}e^{-ikx}\int_{-\infty}^{\infty}e^{2ikx_{0}}V\left(x_{0}\right)dx_{0}\right] \ \ \ \ \ (5)$

The scattering amplitude and reflection coefficient for the reflected particle are therefore

 $\displaystyle f_{R}$ $\displaystyle =$ $\displaystyle -\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}e^{2ikx_{0}}V\left(x_{0}\right)dx_{0}\ \ \ \ \ (6)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle \left|f_{R}\right|^{2}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m}{\hbar^{2}k}\right)^{2}\left|\int_{-\infty}^{\infty}e^{2ikx_{0}}V\left(x_{0}\right)dx_{0}\right|^{2} \ \ \ \ \ (8)$

[Note that we’re assuming that ${x so although the limits of the integral are infinite, we’re implicitly assuming that ${V=0}$ for all ${x_{0}>x}$ so the integral isn’t really over an infinite range.]

For ${x\gg0}$ we have

 $\displaystyle e^{ik\left|x-x_{0}\right|}$ $\displaystyle =$ $\displaystyle e^{ikx}e^{-ikx_{0}}\ \ \ \ \ (9)$ $\displaystyle \psi_{0}\left(x_{0}\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx_{0}} \ \ \ \ \ (10)$

The Born approximation here is

 $\displaystyle \psi\left(x\right)$ $\displaystyle =$ $\displaystyle Ae^{ikx}-\frac{im}{\hbar^{2}k}e^{ikx}\int_{-\infty}^{\infty}e^{-ikx_{0}}V\left(x_{0}\right)Ae^{ikx_{0}}dx_{0}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{ikx}\left[1-\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}V\left(x_{0}\right)dx_{0}\right] \ \ \ \ \ (12)$

The scattering amplitude and transmission coefficient are therefore

 $\displaystyle f_{T}$ $\displaystyle =$ $\displaystyle 1-\frac{im}{\hbar^{2}k}\int_{-\infty}^{\infty}V\left(x_{0}\right)dx_{0}\ \ \ \ \ (13)$ $\displaystyle T$ $\displaystyle =$ $\displaystyle \left|f_{T}\right|^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+\left(\frac{m}{\hbar^{2}k}\right)^{2}\left|\int_{-\infty}^{\infty}V\left(x_{0}\right)dx_{0}\right|^{2} \ \ \ \ \ (15)$

The Born approximation fails for transmission in this case, since ${T>1}$ which is impossible. We can still get an estimate of ${T}$ from ${T=1-R}$.

# Impulse approximation in scattering theory

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.14.

In classical scattering theory, the simplest approximation is the impulse approximation, in which a particle’s path is assumed to be a straight line right through the scattering region, and the total impulse resulting from the component of force perpendicular to the particle’s trajectory is calculated. This impulse ${I}$ is assumed to be a small fraction of the particle’s incoming horizontal momentum ${p}$, so the scattering angle should be small, and given approximately by

$\displaystyle \theta\approx\arctan\frac{I}{p} \ \ \ \ \ (1)$

As an example, we’ll apply the impulse approximation to Rutherford scattering of a charge ${q_{1}}$ travelling with kinetic energy ${E}$ from another charge ${q_{2}}$ at rest. We assume ${q_{1}}$ has an impact parameter ${b}$ (that is, if the particle didn’t interact with the target, it would pass by with a closest approach distance of ${b}$).

The impulse is the change in momentum that a force produces over a given time, so the required impulse from the perpendicular component of a force is

$\displaystyle I=\int F_{\perp}dt \ \ \ \ \ (2)$

We’ll take ${q_{1}}$‘s straight line trajectory to be along the ${x}$ axis and place ${q_{2}}$ at ${y=b}$ on the ${y}$ axis. Then the Coulomb force between ${q_{1}}$ and ${q_{2}}$ is

$\displaystyle F=\frac{q_{1}q_{2}}{4\pi\epsilon_{0}r^{2}} \ \ \ \ \ (3)$

with a perpendicular component of

$\displaystyle F_{\perp}=F\frac{b}{r}=\frac{q_{1}q_{2}b}{4\pi\epsilon_{0}r^{3}} \ \ \ \ \ (4)$

The impulse is

$\displaystyle I=\frac{q_{1}q_{2}b}{4\pi\epsilon_{0}}\int_{-\infty}^{\infty}\frac{dt}{r^{3}} \ \ \ \ \ (5)$

To convert this to an integral over ${r}$ we can use the fact that ${q_{1}}$‘s speed is constant at

$\displaystyle v=\frac{dx}{dt}=\sqrt{\frac{2E}{m}} \ \ \ \ \ (6)$

So we have

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \sqrt{r^{2}-b^{2}}\ \ \ \ \ (7)$ $\displaystyle dx$ $\displaystyle =$ $\displaystyle \frac{r\; dr}{\sqrt{r^{2}-b^{2}}}=\sqrt{\frac{2E}{m}}dt\ \ \ \ \ (8)$ $\displaystyle I$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}b}{4\pi\epsilon_{0}\sqrt{2E/m}}\times2\int_{b}^{\infty}\frac{dr}{r^{2}\sqrt{r^{2}-b^{2}}} \ \ \ \ \ (9)$

where the ${r}$ integral is over the range of ${r}$ from its closest approach when ${r=b}$ out to infinity. We’ve doubled the integral to account for the incoming (${-\infty) and outgoing (${0) legs of the journey.

The integral evaluates to

 $\displaystyle \int_{b}^{\infty}\frac{dr}{r^{2}\sqrt{r^{2}-b^{2}}}$ $\displaystyle =$ $\displaystyle \left.\frac{\sqrt{r^{2}-b^{2}}}{b^{2}r}\right|_{b}^{\infty}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{b^{2}}\ \ \ \ \ (11)$ $\displaystyle I$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}}{2\pi\epsilon_{0}b\sqrt{2E/m}} \ \ \ \ \ (12)$

The incoming momentum is

$\displaystyle p=m\frac{dx}{dt}=\sqrt{2mE} \ \ \ \ \ (13)$

so

 $\displaystyle \theta$ $\displaystyle \approx$ $\displaystyle \arctan\frac{I}{p}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\left[\frac{q_{1}q_{2}}{4\pi\epsilon_{0}bE}\right]\ \ \ \ \ (15)$ $\displaystyle b$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{4\pi\epsilon_{0}E}\cot\theta\ \ \ \ \ (16)$ $\displaystyle \tan\theta$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{4\pi\epsilon_{0}bE} \ \ \ \ \ (17)$

 $\displaystyle b$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}}{8\pi\epsilon_{0}E}\cot\frac{\theta}{2}\ \ \ \ \ (18)$ $\displaystyle \tan\frac{\theta}{2}$ $\displaystyle =$ $\displaystyle \frac{q_{1}q_{2}}{8\pi\epsilon_{0}bE} \ \ \ \ \ (19)$
For small ${\theta}$, ${\tan\frac{\theta}{2}\approx\frac{\theta}{2}}$ so
 $\displaystyle \frac{\theta}{2}$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{8\pi\epsilon_{0}bE}\ \ \ \ \ (20)$ $\displaystyle \theta$ $\displaystyle \approx$ $\displaystyle \frac{q_{1}q_{2}}{4\pi\epsilon_{0}bE} \ \ \ \ \ (21)$