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PHY456H1F: Quantum Mechanics II. Lecture 22 (Taught by Prof J.E. Sipe). Scattering (cont.)

Posted by peeterjoot on November 30, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer.

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

Scattering. Recap

READING: section 19, section 20 of the text [1].

We used a positive potential of the form of figure (\ref{fig:qmTwoL22:qmTwoL22fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL22fig1}
\caption{A bounded positive potential.}
\end{figure}

\begin{aligned}-\frac{\hbar^2}{2 \mu} \frac{\partial^2 {{\psi_k(x)}}}{\partial {{x}}^2} + V(x) \psi_k(x) = \frac{\hbar^2 k^2}{2 \mu}\end{aligned} \hspace{\stretch{1}}(2.1)

for x \ge x_3

\begin{aligned}\psi_k(x) = C e^{i k x}\end{aligned} \hspace{\stretch{1}}(2.2)

\begin{aligned}\phi_k(x) = \frac{d{{\psi_k(x)}}}{dx}\end{aligned} \hspace{\stretch{1}}(2.3)

for x \ge x_3

\begin{aligned}\phi_k(x) = i k C e^{i k x}\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}\frac{d{{\psi_k(x)}}}{dx} &= \phi_k(x) \\ -\frac{\hbar^2}{2 \mu} \frac{d{{\phi_k(x)}}}{dx} &= - V(x) \psi_k(x) + \frac{\hbar^2 k^2}{2 \mu}\end{aligned} \hspace{\stretch{1}}(2.5)

integrate these equations back to x_1.

For x \le x_1

\begin{aligned}\psi_k(x) = A e^{i k x} + B e^{-i k x},\end{aligned} \hspace{\stretch{1}}(2.7)

where both A and B are proportional to C, dependent on k.

There are cases where we can solve this analytically (one of these is on our problem set).

Alternatively, write as (so long as A \ne 0)

\begin{aligned}\begin{array}{l l l}\psi_k(x)&\rightarrow e^{i k x} + \beta_k e^{-i k x} & \quad \mbox{for latex x x_2$}\end{array}\end{aligned} \hspace{\stretch{1}}(2.8)$

Now want to consider the problem of no potential in the interval of interest, and our window bounded potential as in figure (\ref{fig:qmTwoL22:qmTwoL22fig3})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL22fig3}
\caption{Wave packet in free space and with positive potential.}
\end{figure}

where we model our particle as a wave packet as we found can have the fourier transform description, for t_{\text{initial}} < 0, of

\begin{aligned}\psi(x, t_{\text{initial}}) = \int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{initial}}) e^{i k x}\end{aligned} \hspace{\stretch{1}}(2.9)

Returning to the same coefficients, the solution of the Schr\”{o}dinger eqn for problem with the potential 2.8

For x \le x_1,

\begin{aligned}\psi(x, t) = \psi_i(x, t) + \psi_r(x, t)\end{aligned} \hspace{\stretch{1}}(2.10)

where as illustrated in figure (\ref{fig:qmTwoL22:qmTwoL22fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL22fig4}
\caption{Reflection and transmission of wave packet.}
\end{figure}

\begin{aligned}\psi_i(x, t) &= \int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{initial}}) e^{i k x} \\ \psi_r(x, t) &= \int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{initial}}) \beta_k e^{-i k x}.\end{aligned} \hspace{\stretch{1}}(2.11)

For x > x_2

\begin{aligned}\psi(x, t) = \psi_t(x, t)\end{aligned} \hspace{\stretch{1}}(2.13)

and

\begin{aligned}\psi_t(x, t) = \int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{initial}}) \gamma_k e^{i k x}\end{aligned} \hspace{\stretch{1}}(2.14)

Look at

\begin{aligned}\psi_r(x, t) = \chi(-x, t)\end{aligned} \hspace{\stretch{1}}(2.15)

where

\begin{aligned}\begin{aligned}\chi(x, t)&= \int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{initial}}) \beta_k e^{i k x} \\ &\approx\beta_{k_0} \int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{initial}}) e^{i k x}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.16)

for t = t_{\text{initial}}, this is nonzero for x < x_1.

so for x < x_1

\begin{aligned}\psi_r(x, t_{\text{initial}}) = 0\end{aligned} \hspace{\stretch{1}}(2.17)

In the same way, for x > x_2

\begin{aligned}\psi_t(x, t_{\text{initial}}) = 0.\end{aligned} \hspace{\stretch{1}}(2.18)

What hasn’t been proved is that the wavefunction is also zero in the [x_1, x_2] interval.

Summarizing

For t = t_{\text{initial}}

\begin{aligned}\psi(x, t_{\text{initial}})=\left\{\begin{array}{l l}\int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{initial}}) e^{i k x} &\quad \mbox{for latex x x_2$ (and actually also for x > x_1 (unproven))}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.19)$

for t = t_{\text{final}}

\begin{aligned}\psi(x, t_{\text{final}})\rightarrow\left\{\begin{array}{l l}\int \frac{dk}{\sqrt{2 \pi}} \beta_k \alpha(k, t_{\text{final}}) e^{-i k x} &\quad \mbox{for latex x x_2$ }\end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.20)$

Probability of reflection is

\begin{aligned}\int {\left\lvert{\psi_r(x, t_{\text{final}})}\right\rvert}^2 dx\end{aligned} \hspace{\stretch{1}}(2.21)

If we have a sufficiently localized packet, we can form a first order approximation around the peak of \beta_k (FIXME: or is this a sufficiently localized responce to the potential on reflection?)

\begin{aligned}\psi_r(x, t_{\text{final}}) \approx \beta_{k_0}\int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{final}}) e^{-i k x},\end{aligned} \hspace{\stretch{1}}(2.22)

so

\begin{aligned}\int {\left\lvert{\psi_r(x, t_{\text{final}})}\right\rvert}^2 dx\approx {\left\lvert{\beta_{k_0}}\right\rvert}^2 \equiv R\end{aligned} \hspace{\stretch{1}}(2.23)

Probability of transmission is

\begin{aligned}\int {\left\lvert{\psi_t(x, t_{\text{final}})}\right\rvert}^2 dx\end{aligned} \hspace{\stretch{1}}(2.24)

Again, assuming a small spread in \gamma_k, with \gamma_k \approx \gamma_{k_0} for some k_0

\begin{aligned}\psi_t(x, t_{\text{final}}) \approx \gamma_{k_0}\int \frac{dk}{\sqrt{2 \pi}} \alpha(k, t_{\text{final}}) e^{i k x},\end{aligned} \hspace{\stretch{1}}(2.25)

we have for x > x_2

\begin{aligned}\int {\left\lvert{\psi_t(x, t_{\text{final}})}\right\rvert}^2 dx\approx {\left\lvert{\gamma_{k_0}}\right\rvert}^2 \equiv T.\end{aligned} \hspace{\stretch{1}}(2.26)

By constructing the wave packets in this fashion we get as a side effect the solution of the scattering problem.

The

\begin{aligned}\psi_k(x) \rightarrow & e^{i k x} + \beta_k e^{-i k x} \\ & \gamma_k e^{i k x}\end{aligned}

are called asymptotic in states. Their physical applicability is only once we have built wave packets out of them.

Moving to 3D

For a potential V(\mathbf{r}) \approx 0 for r > r_0 as in figure (\ref{fig:qmTwoL22:qmTwoL22fig5})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL22fig5}
\caption{Radially bounded spherical potential.}
\end{figure}

From 1D we’ve learned to build up solutions from time independent solutions (non normalizable). Consider an incident wave

\begin{aligned}e^{i \mathbf{k} \cdot \mathbf{r}} = e^{i k \hat{\mathbf{n}} \cdot \mathbf{r}}\end{aligned} \hspace{\stretch{1}}(3.27)

This is a solution of the time independent Schr\”{o}dinger equation

\begin{aligned}-\frac{\hbar^2}{2 \mu} \boldsymbol{\nabla}^2 e^{i \mathbf{k} \cdot \mathbf{r}} = Ee^{i \mathbf{k} \cdot \mathbf{r}},\end{aligned} \hspace{\stretch{1}}(3.28)

where

\begin{aligned}E = \frac{\hbar^2 \mathbf{k}^2}{2 \mu}.\end{aligned} \hspace{\stretch{1}}(3.29)

In the presence of a potential expect scattered waves. We’ll next be indentifying the nature of these solutions.

References

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

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