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# Posts Tagged ‘PHY456’

## Updated notes compilations for phy356 and phy456 (QM I & II)

Posted by peeterjoot on July 1, 2012

Here’s two updates of class notes compilations for Quantum Mechanics

The QM I notes updates are strictly cosmetic (the book template is updated to that of classicthesis since it was originally posted). The chapters in QM II are reorganized a bit, grouping things by topic instead by lecture dates.

## Second form of adiabatic approximation.

Posted by peeterjoot on December 11, 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.)]

# Motivation.

In class we were shown an adiabatic approximation where we started with (or worked our way towards) a representation of the form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k c_k(t) e^{-i \int_0^t (\omega_k(t') - \Gamma_k(t')) dt' } {\left\lvert {\psi_k(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.1)

where ${\left\lvert {\psi_k(t)} \right\rangle}$ were normalized energy eigenkets for the (slowly) evolving Hamiltonian

\begin{aligned}H(t) {\left\lvert {\psi_k(t)} \right\rangle} = E_k(t) {\left\lvert {\psi_k(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.2)

In the problem sets we were shown a different adiabatic approximation, where are starting point is

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_k c_k(t) {\left\lvert {\psi_k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(1.3)

For completeness, here’s a walk through of the general amplitude derivation that’s been used.

# Guts

We operate with our energy identity once again

\begin{aligned}0 &=\left(H - i \hbar \frac{d{{}}}{dt} \right) \sum_k c_k {\left\lvert {k} \right\rangle} \\ &=\sum_k c_k E_k {\left\lvert {k} \right\rangle} - i \hbar c_k' {\left\lvert {k} \right\rangle} - i \hbar c_k {\left\lvert {k'} \right\rangle} ,\end{aligned}

where

\begin{aligned}{\left\lvert {k'} \right\rangle} = \frac{d{{}}}{dt} {\left\lvert {k} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.4)

Bra’ing with ${\left\langle {m} \right\rvert}$, and split the sum into $k = m$ and $k \ne m$ parts

\begin{aligned}0 =c_m E_m - i \hbar c_m' - i \hbar c_m \left\langle{{m}} \vert {{m'}}\right\rangle - i \hbar \sum_{k \ne m} c_k \left\langle{{m}} \vert {{k'}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.5)

Again writing

\begin{aligned}\Gamma_m = i \left\langle{{m}} \vert {{m'}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.6)

We have

\begin{aligned}c_m' = \frac{1}{{i \hbar}} c_m (E_m - \hbar \Gamma_m) - \sum_{k \ne m} c_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.7)

In this form we can make an “Adiabatic” approximation, dropping the $k \ne m$ terms, and integrate

\begin{aligned}\int \frac{d c_m'}{c_m} = \frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \end{aligned} \hspace{\stretch{1}}(2.8)

or

\begin{aligned}c_m(t) = A \exp\left(\frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \right).\end{aligned} \hspace{\stretch{1}}(2.9)

Evaluating at $t = 0$, fixes the integration constant for

\begin{aligned}c_m(t) = c_m(0) \exp\left(\frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \right).\end{aligned} \hspace{\stretch{1}}(2.10)

Observe that this is very close to the starting point of the adiabatic approximation we performed in class since we end up with

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k c_k(0) e^{-i \int_0^t (\omega_k(t') - \Gamma_k(t')) dt' } {\left\lvert {k(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.11)

So, to perform the more detailed approximation, that started with 1.1, where we ended up with all the cross terms that had both $\omega_k$ and Berry phase $\Gamma_k$ dependence, we have only to generalize by replacing $c_k(0)$ with $c_k(t)$.

## Evaluating the squared sinc integral.

Posted by peeterjoot on December 10, 2011

# Motivation

In the Fermi’s golden rule lecture we used the result for the integral of the squared $\text{sinc}$ function. Here is a reminder of the contours required to perform this integral.

# Guts

We want to evaluate

\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx\end{aligned} \hspace{\stretch{1}}(1.2.1)

We make a few change of variables

\begin{aligned}\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx &= \left\lvert {\mu} \right\rvert \int_{-\infty}^\infty \frac{\sin^2 (y)}{y^2} dy \\ &= -i \left\lvert {\mu} \right\rvert \int_{-\infty}^\infty \frac{(e^{iy} - e^{-iy})^2}{(2 i y)^2} i dy \\ &=-\frac{i \left\lvert {\mu} \right\rvert}{4} \int_{-i\infty}^{i\infty} \frac{e^{2z} + e^{-2z} - 2}{z^2} dz\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.2)

Now we pick a contour that is distorted to one side of the origin as in fig. 1.1

Fig 1.1: Contour distorted to one side of the double pole at the origin

We employ Jordan’s theorem (section 8.12 [1]) now to pick the contours for each of the integrals since we need to ensure the $e^{\pm z}$ terms converges as $R \rightarrow \infty$ for the $z = R e^{i\theta}$ part of the contour. We can write

\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx=-\frac{i \left\lvert {\mu} \right\rvert}{4} \left(\int_{C_0 + C_2} \frac{e^{2z}}{z^2} dz-\int_{C_0 + C_1} \frac{e^{-2z}}{z^2} dz-\int_{C_0 + C_1} \frac{2}{z^2} dz\right)\end{aligned} \hspace{\stretch{1}}(1.2.3)

The second two integrals both surround no poles, so we have only the first to deal with

\begin{aligned}\begin{aligned}\int_{C_0 + C_2} \frac{e^{2z}}{z^2} dz &= 2 \pi i \frac{1}{{1!}} {\left.{{ \frac{d}{dz} e^{2z}}}\right\vert}_{{z=0}} \\ &= 4 \pi i \end{aligned}\end{aligned} \hspace{\stretch{1}}(1.2.4)

Putting everything back together we have

\begin{aligned}\int_{-\infty}^\infty \frac{\sin^2 (x\left\lvert {\mu} \right\rvert)}{x^2} dx &= -\frac{i \left\lvert {\mu} \right\rvert}{4} 4 \pi i \\ &= \pi \left\lvert {\mu} \right\rvert\end{aligned} \hspace{\stretch{1}}(1.2.5)

## On the cavalier choice of contours

The choice of which contours to pick above may seem pretty arbitrary, but they are for good reason. Suppose you picked $C_0 + C_1$ for the first integral. On the big $C_1$ arc, then with a $z = R e^{i \theta}$ substitution we have

\begin{aligned}\left\lvert {\int_{C_1} \frac{e^{2 z}}{z^2} dz} \right\rvert &= \left\lvert {\int_{\theta = \pi/2}^{-\pi/2} \frac{e^{ 2 R (\cos\theta + i \sin\theta) }}{R^2 e^{ 2 i \theta}}R i e^{i \theta} d\theta} \right\rvert \\ &= \frac{1}{R}\left\lvert {\int_{\theta = \pi/2}^{-\pi/2} e^{ 2 R (\cos\theta + i \sin\theta) }e^{-i \theta} d\theta} \right\rvert \\ &\le \frac{1}{R}\int_{\theta = -\pi/2}^{\pi/2} \left\lvert {e^{ 2 R \cos\theta }} \right\rvert d\theta \\ &\le \frac{\pi e^{2 R}}{R}\end{aligned} \hspace{\stretch{1}}(1.2.6)

This clearly doesn’t have the zero convergence property that we desire. We need to pick the $C_2$ contour for the first (positive exponent) integral since in that $[\pi/2, 3\pi/2]$ range, $\cos\theta$ is always negative. We can however, use the $C_1$ contour for the second (negative exponent) integral. Explicitly, again by example, using $C_2$ contour for the first integral, over that portion of the arc we have

\begin{aligned}\left\lvert {\int_{C_2} \frac{e^{2 z}}{z^2} dz} \right\rvert &= \left\lvert {\int_{\theta = \pi/2}^{3 \pi/2} \frac{e^{ 2 R (\cos\theta + i \sin\theta) }}{R^2 e^{ 2 i \theta}}R i e^{i \theta} d\theta} \right\rvert \\ &= \frac{1}{R}\left\lvert {\int_{\theta = \pi/2}^{3 \pi/2} e^{ 2 R (\cos\theta + i \sin\theta) }e^{-i \theta} d\theta} \right\rvert \\ &\le \frac{1}{R}\int_{\theta = \pi/2}^{3 \pi/2} \left\lvert {e^{ 2 R \cos\theta }d\theta} \right\rvert \\ &\approx \frac{1}{R}\int_{\theta = \pi/2}^{3 \pi/2} \left\lvert {e^{ -2 R }d\theta} \right\rvert \\ &= \frac{\pi e^{-2 R} }{R}\end{aligned} \hspace{\stretch{1}}(1.2.7)

# References

[1] W.R. Le Page and W.R. LePage. Complex Variables and the Laplace Transform for Engineers. Courier Dover Publications, 1980.

Posted in Math and Physics Learning. | Tagged: , , , | 5 Comments »

## Verifying the Helmholtz Green’s function.

Posted by peeterjoot on December 9, 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.)]

# Motivation.

In class this week, looking at an instance of the Helmholtz equation

\begin{aligned}\left( \boldsymbol{\nabla}^2 + \mathbf{k}^2\right) \psi_\mathbf{k}(\mathbf{r}) = s(\mathbf{r}).\end{aligned} \hspace{\stretch{1}}(1.1)

We were told that the Green’s 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}}(1.2)

that can be used to solve for a particular solution this differential equation via convolution

\begin{aligned}\psi_\mathbf{k}(\mathbf{r}) = \int G^0(\mathbf{r}, \mathbf{r}') s(\mathbf{r}') d^3 \mathbf{r}',\end{aligned} \hspace{\stretch{1}}(1.3)

\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}}(1.4)

Let’s try to verify this.

# Guts

Application of the Helmholtz differential operator $\boldsymbol{\nabla}^2 + \mathbf{k}^2$ on the presumed solution gives

\begin{aligned}(\boldsymbol{\nabla}^2 + \mathbf{k}^2) \psi_\mathbf{k}(\mathbf{r}) = - \frac{1}{{4 \pi}} \int (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \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}}(2.5)

## When $\mathbf{r} \ne \mathbf{r}'$.

To proceed we’ll need to evaluate

\begin{aligned}\boldsymbol{\nabla}^2 \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}}(2.6)

Writing $\mu = {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}$ we start with the computation of

\begin{aligned}\frac{\partial {}}{\partial {x}} \frac{e^{i k \mu} }{\mu}&=\frac{\partial {\mu}}{\partial {x}} \left( \frac{i k}{\mu} - \frac{1}{{\mu^2}} \right) e^{i k \mu} \\ &=\frac{\partial {\mu}}{\partial {x}} \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu}\end{aligned}

We see that we’ll have

\begin{aligned}\boldsymbol{\nabla} \frac{e^{i k \mu} }{\mu} = \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu} \boldsymbol{\nabla} \mu.\end{aligned} \hspace{\stretch{1}}(2.7)

Taking second derivatives with respect to $x$ we find

\begin{aligned}\frac{\partial^2 {{}}}{\partial {{x}}^2} \frac{e^{i k \mu} }{\mu}&=\frac{\partial^2 {{\mu}}}{\partial {{x}}^2} \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu}+\frac{\partial {\mu}}{\partial {x}} \frac{\partial {\mu}}{\partial {x}} \frac{1}{{\mu^2}} \frac{e^{i k \mu}}{\mu}+\left( \frac{\partial {\mu}}{\partial {x}} \right)^2 \left( i k - \frac{1}{{\mu}} \right)^2 \frac{e^{i k \mu}}{\mu} \\ &=\frac{\partial^2 {{\mu}}}{\partial {{x}}^2} \left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu}+\left( \frac{\partial {\mu}}{\partial {x}} \right)^2 \left( -k^2 - \frac{ 2 i k }{\mu} + \frac{2}{\mu^2} \right)\frac{e^{i k \mu}}{\mu}.\end{aligned}

Our Laplacian is then

\begin{aligned}\boldsymbol{\nabla}^2\frac{e^{i k \mu} }{\mu} =\left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu} \boldsymbol{\nabla}^2 \mu+\left( -k^2 - \frac{ 2 i k }{\mu} + \frac{2}{\mu^2} \right)\frac{e^{i k \mu}}{\mu} (\boldsymbol{\nabla} \mu)^2.\end{aligned} \hspace{\stretch{1}}(2.8)

Now lets calculate the derivatives of $\mu$. Working on $x$ again, we have

\begin{aligned}\frac{\partial {}}{\partial {x}} \mu&=\frac{\partial {}}{\partial {x}} \sqrt{ (x - x')^2 +(y - y')^2 +(z - z')^2 } \\ &=\frac{1}{{2}} 2 (x - x')\frac{1}{{\sqrt{ (x - x')^2 +(y - y')^2 +(z - z')^2 }}} \\ &=\frac{x - x'}{\mu}.\end{aligned}

So we have

\begin{aligned}\boldsymbol{\nabla} \mu &= \frac{\mathbf{r} - \mathbf{r}'}{\mu} \\ (\boldsymbol{\nabla} \mu)^2 &= 1 \end{aligned} \hspace{\stretch{1}}(2.9)

Taking second derivatives with respect to $x$ we find

\begin{aligned}\frac{\partial^2 {{}}}{\partial {{x}}^2} \mu&= \frac{\partial {}}{\partial {x}}\frac{x - x'}{\mu} \\ &= \frac{1}{\mu} - (x - x') \frac{\partial {\mu}}{\partial {x}} \frac{1}{{\mu^2}}\\ &=\frac{1}{\mu} - (x - x') \frac{x - x'}{\mu} \frac{1}{{\mu^2}}\\ &=\frac{1}{\mu} - (x - x')^2 \frac{1}{{\mu^3}}.\end{aligned}

So we find

\begin{aligned}\boldsymbol{\nabla}^2 \mu = \frac{3}{\mu} - \frac{1}{{\mu}},\end{aligned} \hspace{\stretch{1}}(2.11)

or

\begin{aligned}\boldsymbol{\nabla}^2 \mu = \frac{2}{\mu}.\end{aligned} \hspace{\stretch{1}}(2.12)

Inserting this and $(\boldsymbol{\nabla} \mu)^2$ into 2.8 we find

\begin{aligned}\begin{aligned}\boldsymbol{\nabla}^2\frac{e^{i k \mu} }{\mu} &=\left( i k - \frac{1}{{\mu}} \right) \frac{e^{i k \mu}}{\mu} \frac{2}{\mu}+\left( -k^2 - \frac{ 2 i k }{\mu} + \frac{2}{\mu^2} \right)\frac{e^{i k \mu}}{\mu}&=-k^2 \frac{e^{i k \mu}}{\mu}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.13)

This shows us that provided $\mathbf{r} \ne \mathbf{r}'$ we have

\begin{aligned}(\boldsymbol{\nabla}^2 + \mathbf{k}^2) G^0(\mathbf{r}, \mathbf{r}') = 0.\end{aligned} \hspace{\stretch{1}}(2.14)

## In the neighborhood of ${\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} < \epsilon$.

Having shown that we end up with zero everywhere that $\mathbf{r} \ne \mathbf{r}'$ we are left to consider a neighborhood of the volume surrounding the point $\mathbf{r}$ in our integral. Following the Coulomb treatment in section 2.2 of [1] we use a spherical volume element centered around $\mathbf{r}$ of radius $\epsilon$, and then convert a divergence to a surface area to evaluate the integral away from the problematic point

\begin{aligned}-\frac{1}{{4\pi}} \int_{\text{all space}} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \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}'=-\frac{1}{{4\pi}} \int_{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} < \epsilon} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \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}}(2.15)

We make the change of variables $\mathbf{r}' = \mathbf{r} + \mathbf{a}$. We add an explicit $\mathbf{r}$ suffix to our Laplacian at the same time to remind us that it is taking derivatives with respect to the coordinates of $\mathbf{r} = (x, y, z)$, and not the coordinates of our integration variable $\mathbf{a} = (a_x, a_y, a_z)$. Assuming sufficient continuity and “well behavedness” of $s(\mathbf{r}')$ we’ll be able to pull it out of the integral, giving

\begin{aligned}-\frac{1}{{4\pi}} \int_{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{r}^2 + \mathbf{k}^2) \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}'&= -\frac{1}{4\pi} \int_{{\left\lvert{\mathbf{a}}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{r}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} s(\mathbf{r} + \mathbf{a}) d^3 \mathbf{a} \\ &= -\frac{s(\mathbf{r})}{4\pi} \int_{{\left\lvert{\mathbf{a}}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{r}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} \end{aligned}

Recalling the dependencies on the derivatives of ${\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}$ in our previous gradient evaluations, we note that we have

\begin{aligned}\boldsymbol{\nabla}_\mathbf{r} {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} &= -\boldsymbol{\nabla}_\mathbf{a} {\left\lvert{\mathbf{a}}\right\rvert} \\ (\boldsymbol{\nabla}_\mathbf{r} {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert})^2 &= (\boldsymbol{\nabla}_\mathbf{a} {\left\lvert{\mathbf{a}}\right\rvert})^2 \\ \boldsymbol{\nabla}_\mathbf{r}^2 {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert} &= \boldsymbol{\nabla}_\mathbf{a}^2 {\left\lvert{\mathbf{a}}\right\rvert},\end{aligned} \hspace{\stretch{1}}(2.16)

so with $\mathbf{a} = \mathbf{r} - \mathbf{r}'$, we can rewrite our Laplacian as

\begin{aligned}\boldsymbol{\nabla}_\mathbf{r}^2 \frac{e^{i k {\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}}}{{\left\lvert{\mathbf{r} - \mathbf{r}'}\right\rvert}} = \boldsymbol{\nabla}_\mathbf{a}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} = \boldsymbol{\nabla}_\mathbf{a} \cdot \left(\boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right)\end{aligned} \hspace{\stretch{1}}(2.19)

This gives us

\begin{aligned}-\frac{s(\mathbf{r})}{4\pi} \int_{{\left\lvert{\mathbf{a}}\right\rvert} < \epsilon} (\boldsymbol{\nabla}_\mathbf{a}^2 + \mathbf{k}^2) \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} &=-\frac{s(\mathbf{r})}{4\pi} \int_{dV} \boldsymbol{\nabla}_\mathbf{a} \cdot \left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) d^3 \mathbf{a} -\frac{s(\mathbf{r})}{4\pi} \int_{dV}\mathbf{k}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} \\ &=-\frac{s(\mathbf{r})}{4\pi} \int_{dA} \left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) \cdot \hat{\mathbf{a}} d^2 \mathbf{a} -\frac{s(\mathbf{r})}{4\pi} \int_{dV}\mathbf{k}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} \end{aligned}

To complete these evaluations, we can now employ a spherical coordinate change of variables. Let’s do the $\mathbf{k}^2$ volume integral first. We have

\begin{aligned}\int_{dV}\mathbf{k}^2 \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} d^3 \mathbf{a} &=\int_{a = 0}^\epsilon \int_{\theta = 0}^\pi \int_{\phi=0}^{2\pi}\mathbf{k}^2 \frac{e^{i k a}}{a} a^2 da \sin\theta d\theta d\phi \\ &=4\pi k^2\int_{a = 0}^\epsilon a e^{i k a} da \\ &=4\pi \int_{u = 0}^{k\epsilon}u e^{i u} du \\ &=4\pi {\left.(-i u + 1) e^{i u} \right\vert}_0^{k \epsilon} \\ &=4 \pi \left( (-i k \epsilon + 1)e^{i k \epsilon} - 1 \right)\end{aligned}

To evaluate the surface integral we note that we’ll require only the radial portion of the gradient, so have

\begin{aligned}\left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) \cdot \hat{\mathbf{a}}&=\left( \hat{\mathbf{a}} \frac{\partial {}}{\partial {a}} \frac{e^{i k a}}{a} \right) \cdot \hat{\mathbf{a}} \\ &=\frac{\partial {}}{\partial {a}} \frac{e^{i k a}}{a} \\ &=\left( i k \frac{1}{{a}} - \frac{1}{{a^2}} \right)e^{i k a} \\ &=\left( i k a - 1 \right)\frac{e^{i k a}}{a^2}\end{aligned}

Our area element is $a^2 \sin\theta d\theta d\phi$, so we are left with

\begin{aligned}\begin{aligned}\int_{dA} \left( \boldsymbol{\nabla}_\mathbf{a} \frac{e^{i k {\left\lvert{\mathbf{a}}\right\rvert}}}{{\left\lvert{\mathbf{a}}\right\rvert}} \right) \cdot \hat{\mathbf{a}} d^2 \mathbf{a} &={\left.{{\int_{\theta = 0}^\pi \int_{\phi=0}^{2\pi}\left( i k a - 1 \right)\frac{e^{i k a}}{a^2}a^2 \sin\theta d\theta d\phi }}\right\vert}_{{a = \epsilon}}\\ &=4 \pi\left( i k \epsilon - 1 \right) e^{i k \epsilon}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.20)

Putting everything back together we have

\begin{aligned}-\frac{1}{{4\pi}} \int_{\text{all space}} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \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}'&=-s(\mathbf{r})\left((-i k \epsilon + 1)e^{i k \epsilon} - 1 +\left( i k \epsilon - 1 \right) e^{i k \epsilon}\right) \\ &=-s(\mathbf{r})\left((-i k \epsilon + 1 + i k \epsilon - 1 )e^{i k \epsilon} - 1 \right) \end{aligned}

But this is just

\begin{aligned}-\frac{1}{{4\pi}} \int_{\text{all space}} (\boldsymbol{\nabla}^2 + \mathbf{k}^2) \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}' = s(\mathbf{r}).\end{aligned} \hspace{\stretch{1}}(2.21)

This completes the desired verification of the Green’s function for the Helmholtz operator. Observe the perfect cancellation here, so the limit of $\epsilon \rightarrow 0$ can be independent of how large $k$ is made. You have to complete the integrals for both the Laplacian and the $\mathbf{k}^2$ portions of the integrals and add them, before taking any limits, or else you’ll get into trouble (as I did in my first attempt).

# References

[1] M. Schwartz. Principles of Electrodynamics. Dover Publications, 1987.

## A short derivation of the time dependent pertubation result.

Posted by peeterjoot on December 9, 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.)]

# Guts

A super short derivation of the time dependent pertubation result. With

\begin{aligned}{\left\lvert {\psi{t}} \right\rangle} = \sum_k c_k(t) e^{-i\omega_k t} {\left\lvert {k} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.1)

\begin{aligned}0&=\left( H_0 + H' - i\hbar \frac{d}{dt} \right){\left\lvert {\psi{t}} \right\rangle} \\ &=\left( H_0 + H' - i\hbar \frac{d}{dt} \right)\sum_k c_k e^{-i\omega_k t} {\left\lvert {k} \right\rangle} \\ &=\sum_k e^{-i\omega_k t} \left(\not{{c_k E_k}} + H' c_k - \not{{i\hbar (-i \omega_k) c_k}} -i\hbar c_k'\right){\left\lvert {k} \right\rangle}\end{aligned}

Bra with ${\left\langle {m} \right\rvert}$

\begin{aligned}\sum_k e^{-i\omega_k t} H'_{mk} c_k =i\hbar e^{-i\omega_m t} c_m',\end{aligned} \hspace{\stretch{1}}(1.2)

or

\begin{aligned}c_m'=\frac{1}{{i\hbar}}\sum_k e^{-i\omega_{km} t} H'_{mk} c_k \end{aligned} \hspace{\stretch{1}}(1.3)

Now we can make the assumptions about the initial state and away we go.

## One more adiabatic pertubation derivation.

Posted by peeterjoot on December 8, 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.)]

# Motivation.

I liked one of the adiabatic pertubation derivations that I did to review the material, and am recording it for reference.

# Build up.

In time dependent pertubation we started after noting that our ket in the interaction picture, for a Hamiltonian $H = H_0 + H'(t)$, took the form

\begin{aligned}{\left\lvert {\alpha_S(t)} \right\rangle} = e^{-i H_0 t/\hbar} {\left\lvert {\alpha_I(t)} \right\rangle} = e^{-i H_0 t/\hbar} U_I(t) {\left\lvert {\alpha_I(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.1)

Here we have basically assumed that the time evolution can be factored into a portion dependent on only the static portion of the Hamiltonian, with some other operator $U_I(t)$, providing the remainder of the time evolution. From 2.1 that operator $U_I(t)$ is found to behave according to

\begin{aligned}i \hbar \frac{d{{U_I}}}{dt} = e^{i H_0 t/\hbar} H'(t) e^{-i H_0 t/\hbar} U_I,\end{aligned} \hspace{\stretch{1}}(2.2)

but for our purposes we just assumed it existed, and used this for motivation. With the assumption that the interaction picture kets can be written in terms of the basis kets for the system at $t=0$ we write our Schr\”{o}dinger ket as

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i H_0 t/\hbar} a_k(t) {\left\lvert {k} \right\rangle}= \sum_k e^{-i \omega_k t/\hbar} a_k(t) {\left\lvert {k} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.3)

where ${\left\lvert {k} \right\rangle}$ are the energy eigenkets for the initial time equation problem

\begin{aligned}H_0 {\left\lvert {k} \right\rangle} = E_k^0 {\left\lvert {k} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.4)

For the adiabatic problem, we assume the system is changing very slowly, as described by the instantanious energy eigenkets

\begin{aligned}H(t) {\left\lvert {k(t)} \right\rangle} = E_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.5)

Can we assume a similar representation to 2.3 above, but allow ${\left\lvert {k} \right\rangle}$ to vary in time? This doesn’t quite work since ${\left\lvert {k(t)} \right\rangle}$ are no longer eigenkets of $H_0$

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i H_0 t/\hbar} a_k(t) {\left\lvert {k(t)} \right\rangle}\ne \sum_k e^{-i \omega_k t} a_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.6)

Operating with $e^{i H_0 t/\hbar}$ does not give the proper time evolution of ${\left\lvert {k(t)} \right\rangle}$, and we will in general have a more complex functional dependence in our evolution operator for each ${\left\lvert {k(t)} \right\rangle}$. Instead of an $\omega_k t$ dependence in this time evolution operator let’s assume we have some function $\alpha_k(t)$ to be determined, and can write our ket as

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i \alpha_k(t)} a_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.7)

Operating on this with our energy operator equation we have

\begin{aligned}0 &=\left(H - i \hbar \frac{d}{dt} \right) {\left\lvert {\psi} \right\rangle} \\ &=\left(H - i \hbar \frac{d}{dt} \right) \sum_k e^{-i \alpha_k} a_k {\left\lvert {k} \right\rangle} \\ &=\sum_k e^{-i \alpha_k(t)} \left( \left( E_k a_k-i \hbar (-i \alpha_k' a_k + a_k')\right) {\left\lvert {k} \right\rangle}-i \hbar a_k {\left\lvert {k'} \right\rangle}\right) \\ \end{aligned}

Here I’ve written ${\left\lvert {k'} \right\rangle} = d{\left\lvert {k} \right\rangle}/dt$. In our original time dependent pertubaton the $-i \alpha_k'$ term was $-i \omega_k$, so this killed off the $E_k$. If we assume this still kills off the $E_k$, we must have

\begin{aligned}\alpha_k = \frac{1}{{\hbar}} \int_0^t E_k(t') dt',\end{aligned} \hspace{\stretch{1}}(3.8)

and are left with

\begin{aligned}0=\sum_k e^{-i \alpha_k(t)} \left( a_k' {\left\lvert {k} \right\rangle}+a_k {\left\lvert {k'} \right\rangle}\right).\end{aligned} \hspace{\stretch{1}}(3.9)

Bra’ing with ${\left\langle {m} \right\rvert}$ we have

\begin{aligned}0=e^{-i \alpha_m(t)} a_m' +e^{-i \alpha_m(t)} a_m \left\langle{{m}} \vert {{m'}}\right\rangle+\sum_{k \ne m} e^{-i \alpha_k(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.10)

or

\begin{aligned}a_m' +a_m \left\langle{{m}} \vert {{m'}}\right\rangle=-\sum_{k \ne m} e^{-i \alpha_k(t)} e^{i \alpha_m(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.11)

The LHS is a perfect differential if we introduce an integration factor $e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle}$, so we can write

\begin{aligned}e^{-\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle} ( a_m e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } )'=-\sum_{k \ne m} e^{-i \alpha_k(t)} e^{i \alpha_m(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.12)

This suggests that we want to form a new function

\begin{aligned}b_m = a_m e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } \end{aligned} \hspace{\stretch{1}}(3.13)

or

\begin{aligned}a_m = b_m e^{-\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } \end{aligned} \hspace{\stretch{1}}(3.14)

Plugging this into our assumed representation we have a more concrete form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- \int_0^t dt' ( i \omega_k + \left\langle{{k}} \vert {{k'}}\right\rangle ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.15)

Writing

\begin{aligned}\Gamma_k = i \left\langle{{k}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.16)

this becomes

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.17)

## A final pass.

Now that we have what appears to be a good representation for any given state if we wish to examine the time evolution, let’s start over, reapplying our instantaneous energy operator equality

\begin{aligned}0 &=\left(H - i \hbar \frac{d}{dt} \right){\left\lvert {\psi} \right\rangle} \\ &=\left(H - i \hbar \frac{d}{dt} \right)\sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k {\left\lvert {k} \right\rangle} \\ &=- i \hbar \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } \left(i \Gamma_kb_k {\left\lvert {k} \right\rangle} +b_k' {\left\lvert {k} \right\rangle} +b_k {\left\lvert {k'} \right\rangle} \right).\end{aligned}

Bra’ing with ${\left\langle {m} \right\rvert}$ we find

\begin{aligned}0&=e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } i \Gamma_mb_m +e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } b_m' \\ &+e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } b_m \left\langle{{m}} \vert {{m'}}\right\rangle +\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle \end{aligned}

Since $i \Gamma_m = \left\langle{{m}} \vert {{m'}}\right\rangle$ the first and third terms cancel leaving us just

\begin{aligned}b_m'=-\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_{km} - \Gamma_{km} ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.18)

where $\omega_{km} = \omega_k - \omega_m$ and $\Gamma_{km} = \Gamma_k - \Gamma_m$.

# Summary

We assumed that a ket for the system has a representation in the form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i \alpha_k(t) } a_k(t) {\left\lvert {k(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.20)

where $a_k(t)$ and $\alpha_k(t)$ are given or to be determined. Application of our energy operator identity provides us with an alternate representation that simplifes the results

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(4.20)

With

\begin{aligned}{\left\lvert {m'} \right\rangle} &= \frac{d}{dt} {\left\lvert {m} \right\rangle} \\ \Gamma_k &= i \left\langle{{m}} \vert {{m'}}\right\rangle \\ \omega_{km} &= \omega_k - \omega_m \\ \Gamma_{km} &= \Gamma_k - \Gamma_m\end{aligned} \hspace{\stretch{1}}(4.21)

we find that our dynamics of the coefficients are related by

\begin{aligned}b_m'=-\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_{km} - \Gamma_{km} ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(4.25)

## 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.

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 L24 (Taught by Prof J.E. Sipe). 3D Scattering cross sections (cont.)

Posted by peeterjoot on December 5, 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 cross sections.

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})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{qmTwoL22fig5}
\caption{Bounded potential.}
\end{figure}

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)

Using

\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)

when

\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})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{qmTwoL24fig3}
\caption{plane wave front incident on particle}
\end{figure}

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

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{qmTwoL24fig4}
\caption{k space localized wave packet}
\end{figure}

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})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{qmTwoL24fig5}
\caption{point of measurement of scattering cross section}
\end{figure}

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})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.3\textheight]{qmTwoL24fig6}
\caption{Plane wave vs packet wave front}
\end{figure}

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)

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

where

\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)

and

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

# References

[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.

## 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{forlatex 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{forlatex 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{forlatex 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}
\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.

## Believed to be typos in Desai’s QM Text

Posted by peeterjoot on November 25, 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.)]

Here’s a few more typos, in addition to those noted previously, spotted in our QM text [1] from chapters we’ve covered in this term’s class.

# Chapter 17.

\begin{itemize}
\item Page 297. (17.38). appears to be off by a factor of $2$ since $(\sin^2 2x)' = 2 \sin x \cos x = \sin(2 x)$.
\item Page 311. (17.134). $d/dt'$ missing on the $H_{ss}'$ term in the integral.
\item Page 311. (17.136). First term (non-integral part) should be negated.
\item Page 312. (17.144). Sign on $\lambda$ before sum positive instead of negative.
\item Page 313. (17.149). $1/\hbar$ missing.
\item Page 313. (17.152,17.154). extra bra around the bra.
\item Page 313. (17.153). bra missing on $\phi_n$
\end{itemize}

# Chapter 24.

\begin{itemize}
\item Page 450. (24.6). $k^2(x)$ should be $k^2(x) u$.
\item Page 452. (24.18). In the $E < V$ case should be $1/\sqrt{\kappa}$ instead of $1/\sqrt{k}$ (although what's in the text is strictly still correct since it only changes the phase of the wavefunction).

\item Page 455. (24.40). RHD should be multiplied by $\hbar$.

\item Page 460. (24.71). $\psi$ should be $\phi$.
\item Page 460. Third paragraph. $(\mathbf{r}_1 - \mathbf{r})$ should be $(\mathbf{r}_1 - \mathbf{r}_2)$.
\item Page 460. (24.76). The integral should be $1/3$, not $5 \pi/32$. This messes up some of the subsequent stuff, unless there is also another compensating error. Note that one can check this easily since the derivative of $-1/(3 (1+x)^3)$ is $(1 + x)^{-4}$.
\end{itemize}

# Chapter 25.

\begin{itemize}
\item Page 471. (25.18). $i$ subscripts missing on $\mathcal{L}$ and $\dot{y}^2$.
\end{itemize}

# Chapter 26.

\begin{itemize}
\item Page 486. (26.60). $\mathbf{n} \times \mathbf{r} \cdot \boldsymbol{\nabla}$ ought to have braces and read $(\mathbf{n} \times \mathbf{r}) \cdot \boldsymbol{\nabla}$.
\item Page 487. (26.67). $0$ in the $3,3$ position should be $1$.
\item Page 489. before (26.84). For rotations about the imaginary axis was probably meant to be the i’th axis.
\item Page 495. (26.149,26.150). Looks like $\hbar$‘s are missing (esp. compared to 26.144-145).
\item Page 495. (26.150). $J_y$ off by $-1$. ($J_x + iJ_y \ne J_{+}$) as is.
\item Page 496. (26.154). An extra $Y_{l'm}$ in the integral, in between $Y_{l' m'}$ and the $(\theta, \phi)$.
\item Page 498. (26.175). $e^{i\phi}$ should be $e^{-i\phi}$ in the first line.
\item Page 498. (26.178). An $\hbar$ factor has been lost in either (26.178) or (26.179).
\item Page 499. (26.190). minor: $\mathbf{j}$ should be $\mathbf{J}$.
\item Page 450. (26.192). minor: $\mathbf{j}$ should be $\mathbf{J}$, and $R$ should be $\hbar$.
\end{itemize}

# Chapter 27.

\begin{itemize}
\item Page 503. (27.8). minor: bold $\chi$. $R(\theta, \phi)$ probably meant to be $R(\chi)$.
\item Page 504. (27.20). Missing $\hbar E_n$ factor on LHS.
\item Page 507. before (27.53). minor: Velocity $v$ missing bold.
\item Page 510. before (27.78). minor: periods in the two kets should be commas.
\item Page 510. (27.80). $\sigma_y$ and $\sigma_z$ should be interchanged (if $\alpha$ is the polar angle then $\hat{\mathbf{n}} = \hat{\mathbf{y}}$ for that rotation, and $\hat{\mathbf{n}} = \hat{\mathbf{z}}$ for the rotation in the $x,y$ plane). The $\hbar$‘s here should also be dropped.
\item Page 511. (27.81). Same as 27.80.
\item Page 511. (27.83). $\hbar$‘s should be dropped.
\item Page 514. (27.109). minor: dot instead of cdot.
\item Page 515. (27.117). Same error as in (26.149-150). $\hbar$‘s missing, and wrong sign on $J_y$.
\end{itemize}

# Chapter 28.

This chapter written as if $\hbar = 1$, without a statement that this is being done.
\begin{itemize}
\item Page 518. (28.4). $\hbar$ missing. Also in text following, eigenvalue should be $m \hbar = \hbar (m_1 + m_2)$.
\item Page 519. (28.9). $\hbar$ missing LHS.
\item Page 519. (28.11). $\hbar$ missing (after each equality). Text following $m$ and $j(j+1)$ eigenvalues should be multipled by $\hbar$ and $\hbar^2$ respectively.
\item Page 519. following (28.14). $\hbar$ missing in $J_{-}$ equality.
\item Page 520. (28.15). $\hbar$ missing for two factors after last $=$.
\item Page 520. (28.16). $\hbar$ missing LHS.
\item Page 520. (28.21). $\downarrow \downarrow$ should be $\uparrow \downarrow$.
\item Page 521. following (28.25). Chapter 2 should read Chapter 5.
\item Page 522. (28.31). Notational inconsistency. ${\left\lvert {j_1 j_2 jm} \right\rangle}$ should read ${\left\lvert {j_1 j_2, j m} \right\rangle}$
\item Page 522. (28.31). Extra $\vert$ between braket and ket.
\item Page 523. (28.36). Notational inconsistency. $\left\langle{{m_1, m_2}} \vert {{jm -1}}\right\rangle$ should read $\left\langle{{m_1, m_2}} \vert {{j,m -1}}\right\rangle$
\item Page 525. following (28.52). $l(l+1)$, $s(s+1)$, $j(j+1)$ eigenvalues all missing $\hbar^2$.
\item Page 525. (28.53). LHS missing $\hbar^2$.
\item Page 525. (28.54). LHS missing $\hbar^2$.
\item Page 525. (28.57). RHS missing $\hbar$.
\item Page 525. (28.58). RHS missing $\hbar$. $m \pm \frac{1}{{2}}$ should be $m \pm 1$.
\item Page 526. (28.60). $\hbar^2$ missing from both terms.
\item Page 526. (28.61). In first term $\sqrt{l + m_1 + 1}$ should be $\sqrt{(l + m_1 + 1)(l - m_1)}$.
\end{itemize}

# Chapter 29.

This chapter written as if $\hbar = 1$, without a statement that this is being done.
\begin{itemize}
\item Page 531. (29.23). $\hbar$ missing from second two lines.
\item Page 533. (29.25). $\hbar$ should multiply all.
\item Page 533. (29.26). $\hbar$ should multiply all (RHS).
\item Page 533. (29.29). $\hbar$ should multiply all (RHS).
\item Page 533. (29.30). $\hbar$ should multiply all (RHS).
\item Page 533. (29.31). $\hbar$ should multiply all (RHS).
\item Page 536. (29.59). $\hbar$ should multiply RHS.
\item Page 536. (29.60). $\hbar$ should multiply RHS. ${\left\lvert {j m + 1} \right\rangle}$ should be ${\left\lvert {j, m+1} \right\rangle}$.
\item Page 536. (29.61). $\hbar$ should multiply RHS. ${\left\lvert {j'm' - 1} \right\rangle}$ should be ${\left\lvert {j', m'-1} \right\rangle}$.
\item Page 537. (29.65). $\left\langle{{j'm' - 1}} \vert {{m, q}}\right\rangle$ should be $\left\langle{{j', m'-1}} \vert {{m, q}}\right\rangle$.
\end{itemize}

# References

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