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# Posts Tagged ‘scattering’

## PHY456H1F: Quantum Mechanics II. Lecture 25 (Taught by Prof J.E. Sipe). Born approximation.

Posted by peeterjoot on December 7, 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.

# Born approximation.

READING: section 20 [1]

We’ve been arguing that we can write the stationary equation

\begin{aligned}\left( \boldsymbol{\nabla}^2 + \mathbf{k}^2\right) \psi_\mathbf{k}(\mathbf{r}) = s(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.1)

with

\begin{aligned}s(\mathbf{r}) = \frac{2\mu}{\hbar^2} V(\mathbf{r}) \psi_\mathbf{k}(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.2)

\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}}(2.3)

Introduce Green function

\begin{aligned}\left( \boldsymbol{\nabla}^2 + \mathbf{k}^2\right) G^0(\mathbf{r}, \mathbf{r}') = \delta(\mathbf{r}- \mathbf{r}')\end{aligned} \hspace{\stretch{1}}(2.4)

Suppose that I can find $G^0(\mathbf{r}, \mathbf{r}')$, then

\begin{aligned}\psi_\mathbf{k}^{\text{particular}}(\mathbf{r}) = \int G^0(\mathbf{r}, \mathbf{r}') s(\mathbf{r}') d^3 \mathbf{r}'\end{aligned} \hspace{\stretch{1}}(2.5)

It turns out that finding the Green’s function $G^0(\mathbf{r}, \mathbf{r}')$ is not so hard. Note the following, for $k = 0$, we have

\begin{aligned}\boldsymbol{\nabla}^2 G^0_0(\mathbf{r}, \mathbf{r}') = \delta(\mathbf{r} - \mathbf{r}')\end{aligned} \hspace{\stretch{1}}(2.6)

(where a zero subscript is used to mark the $k = 0$ case). We know this Green’s function from electrostatics, and conclude that

\begin{aligned}G^0_0(\mathbf{r}, \mathbf{r}') = - \frac{1}{{4 \pi}} \frac{1}{{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}\end{aligned} \hspace{\stretch{1}}(2.7)

For $\mathbf{r} \ne \mathbf{r}'$ we can easily show that

\begin{aligned}G^0(\mathbf{r}, \mathbf{r}') = - \frac{1}{{4 \pi}} \frac{e^{i k{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(2.8)

This is correct for all $\mathbf{r}$ because it also gives the right limit as $\mathbf{r} \rightarrow \mathbf{r}'$. This argument was first given by Lorentz. We can now write our particular solution

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{r}}- \frac{1}{{4 \pi}} \int \frac{e^{i k{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} s(\mathbf{r}') d^3 \mathbf{r}'\end{aligned} \hspace{\stretch{1}}(2.9)

This is of no immediate help since we don’t know $\psi_\mathbf{k}(\mathbf{r})$ and that is embedded in $s(\mathbf{r})$.

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) = e^{i \mathbf{k} \cdot \mathbf{r}}- \frac{2 \mu}{4 \pi \hbar^2} \int \frac{e^{i k{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} V(\mathbf{r}') \psi_\mathbf{k}(\mathbf{r}') d^3 \mathbf{r}'\end{aligned} \hspace{\stretch{1}}(2.10)

Now look at this for $\mathbf{r} \gg \mathbf{r}'$

\begin{aligned}{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} &= \left( \mathbf{r}^2 + (\mathbf{r}')^2 - 2 \mathbf{r} \cdot \mathbf{r}'\right)^{1/2} \\ &=r \left( 1 + \frac{(\mathbf{r}')^2}{\mathbf{r}^2} - 2 \frac{1}{{\mathbf{r}^2}} \mathbf{r} \cdot \mathbf{r}'\right)^{1/2} \\ &=r \left( 1 - \frac{1}{{2}} \frac{2}{\mathbf{r}^2} \mathbf{r} \cdot \mathbf{r}'+ O\left(\frac{r'}{r}\right)^2\right)^{1/2} \\ &=r - \hat{\mathbf{r}} \cdot \mathbf{r}'+ O\left(\frac{{r'}^2}{r}\right)\end{aligned}

We get

\begin{aligned}\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) &\rightarrow e^{i \mathbf{k} \cdot \mathbf{r}} - \frac{2 \mu}{4 \pi \hbar^2} \frac{ e^{i k r}}{r} \int e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}'} V(\mathbf{r}') \psi_\mathbf{k}(\mathbf{r}') d^3 \mathbf{r}' \\ &=e^{i \mathbf{k} \cdot \mathbf{r}} + f_\mathbf{k}(\theta, \phi) \frac{ e^{i k r}}{r},\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.11)

where

\begin{aligned}f_\mathbf{k}(\theta, \phi) =- \frac{\mu}{2 \pi \hbar^2} \int e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}'} V(\mathbf{r}') \psi_\mathbf{k}(\mathbf{r}') d^3 \mathbf{r}' \end{aligned} \hspace{\stretch{1}}(2.12)

If the scattering is weak we have the Born approximation

\begin{aligned}f_\mathbf{k}(\theta, \phi) =- \frac{\mu}{2 \pi \hbar^2} \int e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}'} V(\mathbf{r}') e^{i \mathbf{k} \cdot \mathbf{r}'} d^3 \mathbf{r}',\end{aligned} \hspace{\stretch{1}}(2.13)

or

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) =e^{i \mathbf{k} \cdot \mathbf{r}} - \frac{\mu}{2 \pi \hbar^2} \frac{ e^{i k r}}{r} \int e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}'} V(\mathbf{r}') e^{i \mathbf{k} \cdot \mathbf{r}'} d^3 \mathbf{r}'.\end{aligned} \hspace{\stretch{1}}(2.14)

Should we wish to make a further approximation, we can take the wave function resulting from application of the Born approximation, and use that a second time. This gives us the “Born again” approximation of

\begin{aligned}\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) &=e^{i \mathbf{k} \cdot \mathbf{r}} - \frac{\mu}{2 \pi \hbar^2} \frac{ e^{i k r}}{r} \int e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}'} V(\mathbf{r}') \left( e^{i \mathbf{k} \cdot \mathbf{r}'} - \frac{\mu}{2 \pi \hbar^2} \frac{ e^{i k r'}}{r'} \int e^{-i k \hat{\mathbf{r}}' \cdot \mathbf{r}''} V(\mathbf{r}'') e^{i \mathbf{k} \cdot \mathbf{r}''} d^3 \mathbf{r}''\right) d^3 \mathbf{r}' \\ &=e^{i \mathbf{k} \cdot \mathbf{r}} - \frac{\mu}{2 \pi \hbar^2} \frac{ e^{i k r}}{r} \int e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}'} V(\mathbf{r}') e^{i \mathbf{k} \cdot \mathbf{r}'} d^3 \mathbf{r}' \\ &\quad +\frac{\mu^2}{(2 \pi)^2 \hbar^4}\frac{ e^{i k r}}{r} \int e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}'} V(\mathbf{r}') \frac{ e^{i k r'}}{r'} \int e^{-i k \hat{\mathbf{r}}' \cdot \mathbf{r}''} V(\mathbf{r}'') e^{i \mathbf{k} \cdot \mathbf{r}''} d^3 \mathbf{r}'' d^3 \mathbf{r}'.\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.15)

# References

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

## PHY456H1F: Quantum Mechanics II. Lecture 21 (Taught by Prof J.E. Sipe). Scattering theory

Posted by peeterjoot on November 24, 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 theory.

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

Here’s (\ref{fig:qmTwoL21:qmTwoL21Fig1}) a simple classical picture of a two particle scattering collision

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig1}
\caption{classical collision of particles.}
\end{figure}

We will focus on point particle elastic collisions (no energy lost in the collision). With particles of mass $m_1$ and $m_2$ we write for the total and reduced mass respectively

\begin{aligned}M = m_1 + m_2\end{aligned} \hspace{\stretch{1}}(2.1)

\begin{aligned}\frac{1}{{\mu}} = \frac{1}{{m_1}} + \frac{1}{{m_2}},\end{aligned} \hspace{\stretch{1}}(2.2)

so that interaction due to a potential $V(\mathbf{r}_1 - \mathbf{r}_2)$ that depends on the difference in position $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}$ has, in the center of mass frame, the Hamiltonian

\begin{aligned}H = \frac{\mathbf{p}^2}{2 \mu} + V(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.3)

In the classical picture we would investigate the scattering radius $r_0$ associated with the impact parameter $\rho$ as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig2}
\caption{Classical scattering radius and impact parameter.}
\end{figure}

## 1D QM scattering. No potential wave packet time evolution.

Now lets move to the QM picture where we assume that we have a particle that can be represented as a wave packet as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig3}
\caption{Wave packet for a particle wavefunction $\Re(\psi(x,0))$}
\end{figure}

First without any potential $V(x) = 0$, lets consider the evolution. Our position and momentum space representations are related by

\begin{aligned}\int {\left\lvert{\psi(x, t)}\right\rvert}^2 dx = 1 = \int {\left\lvert{\psi(p, t)}\right\rvert}^2 dp,\end{aligned} \hspace{\stretch{1}}(2.4)

and by Fourier transform

\begin{aligned}\psi(x, t) = \int \frac{dp}{\sqrt{2 \pi \hbar}} \overline{\psi}(p, t) e^{i p x/\hbar}.\end{aligned} \hspace{\stretch{1}}(2.5)

Schr\”{o}dinger’s equation takes the form

\begin{aligned}i \hbar \frac{\partial {\psi(x,t)}}{\partial {t}} = - \frac{\hbar^2}{2 \mu} \frac{\partial^2 {{\psi(x, t)}}}{\partial {{x}}^2},\end{aligned} \hspace{\stretch{1}}(2.6)

or more simply in momentum space

\begin{aligned}i \hbar \frac{\partial {\overline{\psi}(p,t)}}{\partial {t}} = \frac{p^2}{2 \mu} \frac{\partial^2 {{\overline{\psi}(p, t)}}}{\partial {{x}}^2}.\end{aligned} \hspace{\stretch{1}}(2.7)

Rearranging to integrate we have

\begin{aligned}\frac{\partial {\overline{\psi}}}{\partial {t}} = -\frac{i p^2}{2 \mu \hbar} \overline{\psi},\end{aligned} \hspace{\stretch{1}}(2.8)

and integrating

\begin{aligned}\ln \overline{\psi} = -\frac{i p^2 t}{2 \mu \hbar} + \ln C,\end{aligned} \hspace{\stretch{1}}(2.9)

or

\begin{aligned}\overline{\psi} = C e^{-\frac{i p^2 t}{2 \mu \hbar}} = \overline{\psi}(p, 0) e^{-\frac{i p^2 t}{2 \mu \hbar}}.\end{aligned} \hspace{\stretch{1}}(2.10)

Time evolution in momentum space for the free particle changes only the phase of the wavefunction, the momentum probability density of that particle.

Fourier transforming, we find our position space wavefunction to be

\begin{aligned}\psi(x, t) = \int \frac{dp}{\sqrt{2 \pi \hbar}} \overline{\psi}(p, 0) e^{i p x/\hbar} e^{-i p^2 t/2 \mu \hbar}.\end{aligned} \hspace{\stretch{1}}(2.11)

To clean things up, write

\begin{aligned}p = \hbar k,\end{aligned} \hspace{\stretch{1}}(2.12)

for

\begin{aligned}\psi(x, t) = \int \frac{dk}{\sqrt{2 \pi}} a(k, 0) ) e^{i k x} e^{-i \hbar k^2 t/2 \mu},\end{aligned} \hspace{\stretch{1}}(2.13)

where

\begin{aligned}a(k, 0) = \sqrt{\hbar} \overline{\psi}(p, 0).\end{aligned} \hspace{\stretch{1}}(2.14)

Putting

\begin{aligned}a(k, t) = a(k, 0) e^{ -i \hbar k^2/2 \mu},\end{aligned} \hspace{\stretch{1}}(2.15)

we have

\begin{aligned}\psi(x, t) = \int \frac{dk}{\sqrt{2 \pi}} a(k, t) ) e^{i k x} \end{aligned} \hspace{\stretch{1}}(2.16)

Observe that we have

\begin{aligned}\int dk {\left\lvert{ a(k, t)}\right\rvert}^2 = \int dp {\left\lvert{ \overline{\psi}(p, t)}\right\rvert}^2 = 1.\end{aligned} \hspace{\stretch{1}}(2.17)

## A Gaussian wave packet

Suppose that we have, as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig4}
\caption{Gaussian wave packet.}
\end{figure}

a Gaussian wave packet of the form

\begin{aligned}\psi(x, 0) = \frac{ (\pi \Delta^2)^{1/4}} e^{i k_0 x} e^{- x^2/2 \Delta^2}.\end{aligned} \hspace{\stretch{1}}(2.18)

This is actually a minimum uncertainty packet with

\begin{aligned}\Delta x &= \frac{\Delta}{\sqrt{2}} \\ \Delta p &= \frac{\hbar}{\Delta \sqrt{2}}.\end{aligned} \hspace{\stretch{1}}(2.19)

Taking Fourier transforms we have

\begin{aligned}a(k, 0) &= \left(\frac{\Delta^2}{\pi}\right)^{1/4} e^{-(k - k_0)^2 \Delta^2/2} \\ a(k, t) &= \left(\frac{\Delta^2}{\pi}\right)^{1/4} e^{-(k - k_0)^2 \Delta^2/2} e^{ -i \hbar k^2 t/ 2\mu} \equiv \alpha(k, t)\end{aligned} \hspace{\stretch{1}}(2.21)

For $t > 0$ our wave packet will start moving and spreading as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig5}
\caption{moving spreading Gaussian packet.}
\end{figure}

## With a potential.

Now “switch on” a potential, still assuming a wave packet representation for the particle. With a positive (repulsive) potential as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig6}), at a time long before the interaction of the wave packet with the potential we can visualize the packet as heading towards the barrier.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig6}
\caption{QM wave packet prior to interaction with repulsive potential.}
\end{figure}

After some time long after the interaction, classically for this sort of potential where the particle kinetic energy is less than the barrier “height”, we would have total reflection. In the QM case, we’ve seen before that we will have a reflected and a transmitted portion of the wave packet as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig7})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig7}
\caption{QM wave packet long after interaction with repulsive potential.}
\end{figure}

Even if the particle kinetic energy is greater than the barrier height, as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig8}), we can still have a reflected component.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig8}
\caption{Kinetic energy greater than potential energy.}
\end{figure}

This is even true for a negative potential as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig9})!

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig9}
\caption{qmTwoL21Fig9}
\end{figure}

Consider the probability for the particle to be found anywhere long after the interaction, summing over the transmitted and reflected wave functions, we have

\begin{aligned}1 &= \int {\left\lvert{\psi_r + \psi_t}\right\rvert}^2 \\ &= \int {\left\lvert{\psi_r}\right\rvert}^2 + \int {\left\lvert{\psi_t}\right\rvert}^2 + 2 \Re \int \psi_r^{*} \psi_t\end{aligned}

Observe that long after the interaction the cross terms in the probabilities will vanish because they are non-overlapping, leaving just the probably densities for the transmitted and reflected probably densities independently.

We define

\begin{aligned}T &= \int {\left\lvert{\psi_t(x, t)}\right\rvert}^2 dx \\ R &= \int {\left\lvert{\psi_r(x, t)}\right\rvert}^2 dx.\end{aligned} \hspace{\stretch{1}}(2.23)

The objective of most of our scattering problems will be the calculation of these probabilities and the comparisons of their ratios.

Question. Can we have more than one wave packet reflect off. Yes, we could have multiple wave packets for both the reflected and the transmitted portions. For example, if the potential has some internal structure there could be internal reflections before anything emerges on either side and things could get quite messy.

# Considering the time independent case temporarily.

We are going to work through something that is going to seem at first to be completely unrelated. We will (eventually) see that this can be applied to this problem, so a bit of patience will be required.

We will be using the time independent Schr\”{o}dinger equation

\begin{aligned}- \frac{\hbar^2}{2 \mu} \psi_k''(x) = V(x) \psi_k(x) = E \psi_k(x),\end{aligned} \hspace{\stretch{1}}(3.25)

where we have added a subscript $k$ to our wave function with the intention (later) of allowing this to vary. For “future use” we define for $k > 0$

\begin{aligned}E = \frac{\hbar^2 k^2}{2 \mu}.\end{aligned} \hspace{\stretch{1}}(3.26)

Consider a potential as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig10}), where $V(x) = 0$ for $x > x_2$ and $x < x_1$.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig10}
\caption{potential zero outside of a specific region.}
\end{figure}

We won't have bound states here (repulsive potential). There will be many possible solutions, but we want to look for a solution that is of the form

\begin{aligned}\psi_k(x) = C e^{i k x}, \qquad x > x_2\end{aligned} \hspace{\stretch{1}}(3.27)

Suppose $x = x_3 > x_2$, we have

\begin{aligned}\psi_k(x_3) = C e^{i k x_3}\end{aligned} \hspace{\stretch{1}}(3.28)

\begin{aligned}{\left.{{\frac{d\psi_k}{dx}}}\right\vert}_{{x = x_3}} = i k C e^{i k x_3} \equiv \phi_k(x_3)\end{aligned} \hspace{\stretch{1}}(3.29)

\begin{aligned}{\left.{{\frac{d^2\psi_k}{dx^2}}}\right\vert}_{{x = x_3}} = -k^2 C e^{i k x_3} \end{aligned} \hspace{\stretch{1}}(3.30)

Defining

\begin{aligned}\phi_k(x) = \frac{d\psi_k}{dx},\end{aligned} \hspace{\stretch{1}}(3.31)

we write Schr\”{o}dinger’s equation as a pair of coupled first order equations

\begin{aligned}\frac{d\psi_k}{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} \psi_k(x).\end{aligned} \hspace{\stretch{1}}(3.32)

At this $x = x_3$ specifically, we “know” both $\phi_k(x_3)$ and $\psi_k(x_3)$ and have

\begin{aligned}{\left.{{\frac{d\psi_k}{dx}}}\right\vert}_{{x_3}} &= \phi_k(x) \\ -\frac{\hbar^2}{2 \mu} {\left.{{\frac{d\phi_k(x)}{dx}}}\right\vert}_{{x_3}} = - V(x_3) \psi_k(x_3) + \frac{\hbar^2 k^2}{2\mu} \psi_k(x_3),\end{aligned} \hspace{\stretch{1}}(3.34)

This allows us to find both

\begin{aligned}{dx}}}\right\vert}_{{x_3}} \\ {dx}}}\right\vert}_{{x_3}} \end{aligned} \hspace{\stretch{1}}(3.36)

then proceed to numerically calculate $\phi_k(x)$ and $\psi_k(x)$ at neighboring points $x = x_3 + \epsilon$. Essentially, this allows us to numerically integrate backwards from $x_3$ to find the wave function at previous points for any sort of potential.

# References

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