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PHY456H1F: Quantum Mechanics II. Lecture L24 (Taught by Prof J.E. Sipe). 3D Scattering cross sections (cont.)

Posted by peeterjoot on December 5, 2011

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Peeter’s lecture notes from class. May not be entirely coherent.

Scattering cross sections.

READING: section 20 [1]

Recall that we are studing the case of a potential that is zero outside of a fixed bound, V(\mathbf{r}) = 0 for r > r_0, as in figure (\ref{fig:qmTwoL24:qmTwoL22fig5})

\caption{Bounded potential.}

and were looking for solutions to Schr\”{o}dinger’s equation

\begin{aligned}-\frac{\hbar^2}{2\mu} \boldsymbol{\nabla}^2\psi_\mathbf{k}(\mathbf{r})+ V(\mathbf{r})\psi_\mathbf{k}(\mathbf{r})=\frac{\hbar^2 \mathbf{k}^2}{2 \mu}\psi_\mathbf{k}(\mathbf{r}),\end{aligned} \hspace{\stretch{1}}(2.1)

in regions of space, where r > r_0 is very large. We found

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) \sim e^{i \mathbf{k} \cdot \mathbf{r}} + \frac{e^{i k r}}{r} f_\mathbf{k}(\theta, \phi).\end{aligned} \hspace{\stretch{1}}(2.2)

For r \le r_0 this will be something much more complicated.

To study scattering we’ll use the concept of probability flux as in electromagnetism

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{j} + \dot{\rho} = 0\end{aligned} \hspace{\stretch{1}}(2.3)


\begin{aligned}\psi(\mathbf{r}, t) =\psi_\mathbf{k}(\mathbf{r})^{*}\psi_\mathbf{k}(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.4)

we find

\begin{aligned}\mathbf{j}(\mathbf{r}, t) = \frac{\hbar}{2 \mu i} \Bigl(\psi_\mathbf{k}(\mathbf{r})^{*} \boldsymbol{\nabla} \psi_\mathbf{k}(\mathbf{r})- (\boldsymbol{\nabla} \psi_\mathbf{k}^{*}(\mathbf{r})) \psi_\mathbf{k}(\mathbf{r})\Bigr)\end{aligned} \hspace{\stretch{1}}(2.5)


\begin{aligned}-\frac{\hbar^2}{2\mu} \boldsymbol{\nabla}^2\psi_\mathbf{k}(\mathbf{r})+ V(\mathbf{r})\psi_\mathbf{k}(\mathbf{r})=i \hbar \frac{\partial {\psi_\mathbf{k}(\mathbf{r})}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(2.6)

In a fashion similar to what we did in the 1D case, let’s suppose that we can write our wave function

\begin{aligned}\psi(\mathbf{r}, t_{\text{initial}}) = \int d^3k \alpha(\mathbf{k}, t_{\text{initial}}) \psi_\mathbf{k}(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.7)

and treat the scattering as the scattering of a plane wave front (idealizing a set of wave packets) off of the object of interest as depicted in figure (\ref{fig:qmTwoL24:qmTwoL24fig3})

\caption{plane wave front incident on particle}

We assume that our incoming particles are sufficiently localized in k space as depicted in the idealized representation of figure (\ref{fig:qmTwoL24:qmTwoL24fig4})

\caption{k space localized wave packet}

we assume that \alpha(\mathbf{k}, t_{\text{initial}}) is localized.

\begin{aligned}\psi(\mathbf{r}, t_{\text{initial}}) =\int d^3k\left(\alpha(\mathbf{k}, t_{\text{initial}})e^{i k_z z}+\alpha(\mathbf{k}, t_{\text{initial}}) \frac{e^{i k r}}{r} f_\mathbf{k}(\theta, \phi)\right)\end{aligned} \hspace{\stretch{1}}(2.8)

We suppose that

\begin{aligned}\alpha(\mathbf{k}, t_{\text{initial}}) = \alpha(\mathbf{k}) e^{-i \hbar k^2 t_{\text{initial}}/ 2\mu}\end{aligned} \hspace{\stretch{1}}(2.9)

where this is chosen (\alpha(\mathbf{k}, t_{\text{initial}}) is built in this fashion) so that this is non-zero for z large in magnitude and negative.

This last integral can be approximated

\begin{aligned}\begin{aligned}\int d^3k\alpha(\mathbf{k}, t_{\text{initial}}) \frac{e^{i k r}}{r} f_\mathbf{k}(\theta, \phi)&\approx\frac{f_{\mathbf{k}_0}(\theta, \phi)}{r}\int d^3k\alpha(\mathbf{k}, t_{\text{initial}}) e^{i k r} \\ &\rightarrow 0\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.10)

This is very much like the 1D case where we found no reflected component for our initial time.

We’ll normally look in a locality well away from the wave front as indicted in figure (\ref{fig:qmTwoL24:qmTwoL24fig5})

\caption{point of measurement of scattering cross section}

There are situations where we do look in the locality of the wave front that has been scattered.

Our income wave is of the form

\begin{aligned}\psi_i = A e^{i k z} e^{-i \hbar k^2 t/2 \mu}\end{aligned} \hspace{\stretch{1}}(2.11)

Here we’ve made the approximation that k = {\left\lvert{\mathbf{k}}\right\rvert} \sim k_z. We can calculate the probability current

\begin{aligned}\mathbf{j} = \hat{\mathbf{z}} \frac{\hbar k}{\mu} A\end{aligned} \hspace{\stretch{1}}(2.12)

(notice the v = p/m like term above, with p = \hbar k).

For the scattered wave (dropping A factor)

\begin{aligned}\mathbf{j} &=\frac{\hbar}{2 \mu i}\left(f_\mathbf{k}^{*}(\theta, \phi) \frac{e^{-i k r}}{r} \boldsymbol{\nabla} \left(f_\mathbf{k}(\theta, \phi) \frac{e^{i k r}}{r}\right)-\boldsymbol{\nabla} \left(f_\mathbf{k}^{*}(\theta, \phi) \frac{e^{-i k r}}{r}\right)f_\mathbf{k}(\theta, \phi) \frac{e^{i k r}}{r}\right)\\ &\approx\frac{\hbar}{2 \mu i}\left(f_\mathbf{k}^{*}(\theta, \phi) \frac{e^{-i k r}}{r} i k \hat{\mathbf{r}} f_\mathbf{k}(\theta, \phi)\frac{e^{i k r}}{r}-f_\mathbf{k}^{*}(\theta, \phi) \frac{e^{-i k r}}{r} (-i k \hat{\mathbf{r}}) f_\mathbf{k}(\theta, \phi)\frac{e^{i k r}}{r}\right)\end{aligned}

We find that the radial portion of the current density is

\begin{aligned}\hat{\mathbf{r}} \cdot \mathbf{j}&= \frac{\hbar}{2 \mu i} {\left\lvert{f}\right\rvert}^2 \frac{ 2 i k }{r^2} \\ &= \frac{\hbar k}{\mu} \frac{1}{{r^2}} {\left\lvert{f}\right\rvert}^2,\end{aligned}

and the flux through our element of solid angle is

\begin{aligned}\hat{\mathbf{r}} dA \cdot \mathbf{j}&=\frac{\text{probability}}{\text{unit area per time}} \times \text{area}  \\ &= \frac{\text{probability}}{\text{unit time}} \\ &=\frac{\hbar k}{\mu} \frac{{\left\lvert{f_\mathbf{k}(\theta, \phi)}\right\rvert}^2}{r^2} r^2 d\Omega \\ &=\frac{\hbar k }{\mu}{\left\lvert{f_\mathbf{k}(\theta, \phi)}\right\rvert}^2 d\Omega \\ &=j_{\text{incoming}}\underbrace{{\left\lvert{f_\mathbf{k}(\theta, \phi)}\right\rvert}^2}_{d\sigma/d\Omega} d\Omega.\end{aligned}

We identify the scattering cross section above

\begin{aligned}\frac{d\sigma}{d\Omega}={\left\lvert{f_\mathbf{k}(\theta, \phi)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.13)

\begin{aligned}\sigma = \int {\left\lvert{f_\mathbf{k}(\theta, \phi)}\right\rvert}^2 d\Omega\end{aligned} \hspace{\stretch{1}}(2.14)

We’ve been somewhat unrealistic here since we’ve used a plane wave approximation, and can as in figure (\ref{fig:qmTwoL24:qmTwoL24fig6})

\caption{Plane wave vs packet wave front}

will actually produce the same answer. For details we are referred to [2] and [3].

Working towards a solution

We’ve done a bunch of stuff here but are not much closer to a real solution because we don’t actually know what f_\mathbf{k} is.

Let’s write Schr\”{o}dinger

\begin{aligned}-\frac{\hbar^2}{2\mu} \boldsymbol{\nabla}^2\psi_\mathbf{k}(\mathbf{r})+ V(\mathbf{r})\psi_\mathbf{k}(\mathbf{r})=\frac{\hbar^2 \mathbf{k}^2}{2 \mu}\psi_\mathbf{k}(\mathbf{r}),\end{aligned} \hspace{\stretch{1}}(2.15)

instead as

\begin{aligned}(\boldsymbol{\nabla}^2 + \mathbf{k}^2)\psi_\mathbf{k}(\mathbf{r})= s(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.16)


\begin{aligned}s(\mathbf{r}) = \frac{2\mu}{\hbar} V(\mathbf{r}) \psi_\mathbf{k}(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.17)

where s(\mathbf{r}) is really the particular solution to this differential problem. We want

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) =\psi_\mathbf{k}^{\text{homogeneous}}(\mathbf{r})+ \psi_\mathbf{k}^{\text{particular}}(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.18)


\begin{aligned}\psi_\mathbf{k}^{\text{homogeneous}}(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{r}}\end{aligned} \hspace{\stretch{1}}(2.19)


[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] A. Messiah, G.M. Temmer, and J. Potter. Quantum mechanics: two volumes bound as one. Dover Publications New York, 1999.

[3] JR Taylor. {\em Scattering Theory: the Quantum Theory of Nonrelativistic Scattering}, volume 1. 1972.


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