Daily Archives: Fri, 19 June 2015

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<x<a\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (11)

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<x_{0}} 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<x<0}) and outgoing ({0<x<\infty}) 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)

The exact answer is

\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)

which is consistent with 17.