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# Posts Tagged ‘group velocity’

## Pre-final-exam update of notes for PHY487 (condensed matter physics)

Posted by peeterjoot on December 2, 2013

Here’s an update of my class notes from Winter 2013, University of Toronto Condensed Matter Physics course (PHY487H1F), taught by Prof. Stephen Julian.  This includes notes for all the examinable lectures (i.e. excluding superconductivity).  I’ll post at least one more update later, probably after the exam, including notes from the final lecture, and my problem set 10 solution.

NOTE: This v.4 update of these notes is still really big (~18M).  Some of my mathematica generated 3d images appear to result in very large pdfs.

Changelog for this update (relative to the the first, and second, and third Changelogs) :

December 02, 2013 Lecture 23, Superconductivity

December 01, 2013 Lecture 22, Intro to semiconductor physics

December 01, 2013 Lecture 21, Electron-phonon scattering

November 26, 2013 Problem Set 9, Electron band structure, density of states, and effective mass

November 22, 2013 Lecture 20, Electric current (cont.)

November 20, 2013 Problem Set 8, Tight Binding.

November 18, 2013 Lecture 19, Electrical transport (cont.)

## PHY450H1S. Relativistic Electrodynamics Tutorial 4 (TA: Simon Freedman). Waveguides: confined EM waves.

Posted by peeterjoot on March 14, 2011

# Motivation

While this isn’t part of the course, the topic of waveguides is one of so many applications that it is worth a mention, and that will be done in this tutorial.

We will setup our system with a waveguide (conducting surface that confines the radiation) oriented in the $\hat{\mathbf{z}}$ direction. The shape can be arbitrary

PICTURE: cross section of wacky shape.

## At the surface of a conductor.

At the surface of the conductor (I presume this means the interior surface where there is no charge or current enclosed) we have

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} &= - \frac{1}{{c}} \frac{\partial {\mathbf{B}}}{\partial {t}} \\ \boldsymbol{\nabla} \times \mathbf{B} &= \frac{1}{{c}} \frac{\partial {\mathbf{E}}}{\partial {t}} \\ \boldsymbol{\nabla} \cdot \mathbf{B} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} &= 0\end{aligned} \hspace{\stretch{1}}(1.1)

If we are talking about the exterior surface, do we need to make any other assumptions (perfect conductors, or constant potentials)?

## Wave equations.

For electric and magnetic fields in vacuum, we can show easily that these, like the potentials, separately satisfy the wave equation

Taking curls of the Maxwell curl equations above we have

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{E}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{E}}}{\partial {{t}}^2} \\ \boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{B}) &= - \frac{1}{{c^2}} \frac{\partial^2 {\mathbf{B}}}{\partial {{t}}^2},\end{aligned} \hspace{\stretch{1}}(1.5)

but we have for vector $\mathbf{M}$

\begin{aligned}\boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{M})=\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{M}) - \Delta \mathbf{M},\end{aligned} \hspace{\stretch{1}}(1.7)

which gives us a pair of wave equations

\begin{aligned}\square \mathbf{E} &= 0 \\ \square \mathbf{B} &= 0.\end{aligned} \hspace{\stretch{1}}(1.8)

We still have the original constraints of Maxwell’s equations to deal with, but we are free now to pick the complex exponentials as fundamental solutions, as our starting point

\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{i k^a x_a} = \mathbf{E}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{i k^a x_a} = \mathbf{B}_0 e^{ i (k^0 x_0 - \mathbf{k} \cdot \mathbf{x}) },\end{aligned} \hspace{\stretch{1}}(1.10)

With $k_0 = \omega/c$ and $x_0 = c t$ this is

\begin{aligned}\mathbf{E} &= \mathbf{E}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) } \\ \mathbf{B} &= \mathbf{B}_0 e^{ i (\omega t - \mathbf{k} \cdot \mathbf{x}) }.\end{aligned} \hspace{\stretch{1}}(1.12)

For the vacuum case, with monochromatic light, we treated the amplitudes as constants. Let’s see what happens if we relax this assumption, and allow for spatial dependence (but no time dependence) of $\mathbf{E}_0$ and $\mathbf{B}_0$. For the LHS of the electric field curl equation we have

\begin{aligned}0 &= \boldsymbol{\nabla} \times \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 - \mathbf{E}_0 \times \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + \mathbf{E}_0 \times \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \times \mathbf{E}_0 + i \mathbf{E}_0 \times \mathbf{k} ) e^{i k_a x^a}.\end{aligned}

Similarly for the divergence we have

\begin{aligned}0 &= \boldsymbol{\nabla} \cdot \mathbf{E}_0 e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \boldsymbol{\nabla}) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 + \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k_a \partial_\alpha x^a) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - \mathbf{E}_0 \cdot \mathbf{e}^\alpha i k^a {\delta_\alpha}^a ) e^{i k_a x^a} \\ &= (\boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 ) e^{i k_a x^a}.\end{aligned}

This provides constraints on the amplitudes

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 - i \mathbf{k} \times \mathbf{E}_0 &= -i \frac{\omega}{c} \mathbf{B}_0 \\ \boldsymbol{\nabla} \times \mathbf{B}_0 - i \mathbf{k} \times \mathbf{B}_0 &= i \frac{\omega}{c} \mathbf{E}_0 \\ \boldsymbol{\nabla} \cdot \mathbf{E}_0 - i \mathbf{k} \cdot \mathbf{E}_0 &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{B}_0 - i \mathbf{k} \cdot \mathbf{B}_0 &= 0\end{aligned} \hspace{\stretch{1}}(1.14)

Applying the wave equation operator to our phasor we get

\begin{aligned}0 &=\left(\frac{1}{{c^2}} \partial_{tt} - \boldsymbol{\nabla}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})} \\ &=\left(-\frac{\omega^2}{c^2} - \boldsymbol{\nabla}^2 + \mathbf{k}^2 \right) \mathbf{E}_0 e^{i (\omega t - \mathbf{k} \cdot \mathbf{x})}\end{aligned}

So the momentum space equivalents of the wave equations are

\begin{aligned}\left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \boldsymbol{\nabla}^2 +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(1.18)

Observe that if $c^2 \mathbf{k}^2 = \omega^2$, then these amplitudes are harmonic functions (solutions to the Laplacian equation). However, it doesn’t appear that we require such a light like relation for the four vector $k^a = (\omega/c, \mathbf{k})$.

# Back to the tutorial notes.

In class we went straight to an assumed solution of the form

\begin{aligned}\mathbf{E} &= \mathbf{E}_0(x, y) e^{ i(\omega t - k z) } \\ \mathbf{B} &= \mathbf{B}_0(x, y) e^{ i(\omega t - k z) },\end{aligned} \hspace{\stretch{1}}(2.20)

where $\mathbf{k} = k \hat{\mathbf{z}}$. Our Laplacian was also written as the sum of components in the propagation and perpendicular directions

\begin{aligned}\boldsymbol{\nabla}^2 = \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} + \frac{\partial^2 {{}}}{\partial {{z}}^2}.\end{aligned} \hspace{\stretch{1}}(2.22)

With no $z$ dependence in the amplitudes we have

\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{E}_0 &= 0 \\ \left( \frac{\partial^2 {{}}}{\partial {{x_\perp}}^2} +\frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \mathbf{B}_0 &= 0.\end{aligned} \hspace{\stretch{1}}(2.23)

# Separation into components.

It was left as an exercise to separate out our Maxwell equations, so that our field components $\mathbf{E}_0 = \mathbf{E}_\perp + \mathbf{E}_z$ and $\mathbf{B}_0 = \mathbf{B}_\perp + \mathbf{B}_z$ in the propagation direction, and components in the perpendicular direction are separated

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E}_0 &=(\boldsymbol{\nabla}_\perp + \hat{\mathbf{z}}\partial_z) \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_0 \\ &=\boldsymbol{\nabla}_\perp \times (\mathbf{E}_\perp + \mathbf{E}_z) \\ &=\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=( \hat{\mathbf{x}} \partial_x +\hat{\mathbf{y}} \partial_y ) \times ( \hat{\mathbf{x}} E_x +\hat{\mathbf{y}} E_y ) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z \\ &=\hat{\mathbf{z}} (\partial_x E_y - \partial_z E_z) +\boldsymbol{\nabla}_\perp \times \mathbf{E}_z.\end{aligned}

We can do something similar for $\mathbf{B}_0$. This allows for a split of 1.14 into $\hat{\mathbf{z}}$ and perpendicular components

\begin{aligned}\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_z \\ \boldsymbol{\nabla}_\perp \times \mathbf{E}_z - i \mathbf{k} \times \mathbf{E}_\perp &= -i \frac{\omega}{c} \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \times \mathbf{B}_z - i \mathbf{k} \times \mathbf{B}_\perp &= i \frac{\omega}{c} \mathbf{E}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= i k E_z - \partial_z E_z \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i k B_z - \partial_z B_z.\end{aligned} \hspace{\stretch{1}}(3.25)

So we see that once we have a solution for $\mathbf{E}_z$ and $\mathbf{B}_z$ (by solving the wave equation above for those components), the components for the fields in terms of those components can be found. Alternately, if one solves for the perpendicular components of the fields, these propagation components are available immediately with only differentiation.

In the case where the perpendicular components are taken as given

\begin{aligned}\mathbf{B}_z &= i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp \\ \mathbf{E}_z &= -i \frac{ c }{\omega} \boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.31)

we can express the remaining ones strictly in terms of the perpendicular fields

\begin{aligned}\frac{\omega}{c} \mathbf{B}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) + \mathbf{k} \times \mathbf{E}_\perp \\ \frac{\omega}{c} \mathbf{E}_\perp &= \frac{c}{\omega} \boldsymbol{\nabla}_\perp \times (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) - \mathbf{k} \times \mathbf{B}_\perp \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{E}_\perp &= -i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) \\ \boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp &= i \frac{c}{\omega} (i k - \partial_z) \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp).\end{aligned} \hspace{\stretch{1}}(3.33)

Is it at all helpful to expand the double cross products?

\begin{aligned}\frac{\omega^2}{c^2} \mathbf{B}_\perp &= \boldsymbol{\nabla}_\perp (\boldsymbol{\nabla}_\perp \cdot \mathbf{B}_\perp) -{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ &= i \frac{c}{\omega}(i k - \partial_z)\boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp)-{\boldsymbol{\nabla}_\perp}^2 \mathbf{B}_\perp + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \end{aligned}

This gives us

\begin{aligned}\left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{B}_\perp &= - \frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{E}_\perp) + \frac{\omega}{c} \mathbf{k} \times \mathbf{E}_\perp \\ \left( {\boldsymbol{\nabla}_\perp}^2 + \frac{\omega^2}{c^2} \right) \mathbf{E}_\perp &= -\frac{c}{\omega} (k + i\partial_z) \boldsymbol{\nabla}_\perp \hat{\mathbf{z}} \cdot (\boldsymbol{\nabla}_\perp \times \mathbf{B}_\perp) - \frac{\omega}{c} \mathbf{k} \times \mathbf{B}_\perp,\end{aligned} \hspace{\stretch{1}}(3.37)

but that doesn’t seem particularly useful for completely solving the system? It appears fairly messy to try to solve for $\mathbf{E}_\perp$ and $\mathbf{B}_\perp$ given the propagation direction fields. I wonder if there is a simplification available that I am missing?

# Solving the momentum space wave equations.

Back to the class notes. We proceeded to solve for $\mathbf{E}_z$ and $\mathbf{B}_z$ from the wave equations by separation of variables. We wish to solve equations of the form

\begin{aligned}\left( \frac{\partial^2 {{}}}{\partial {{x}}^2} + \frac{\partial^2 {{}}}{\partial {{y}}^2} + \frac{\omega^2}{c^2} - \mathbf{k}^2 \right) \phi(x,y) = 0\end{aligned} \hspace{\stretch{1}}(4.39)

Write $\phi(x,y) = X(x) Y(y)$, so that we have

\begin{aligned}\frac{X''}{X} + \frac{Y''}{Y} = \mathbf{k}^2 - \frac{\omega^2}{c^2}\end{aligned} \hspace{\stretch{1}}(4.40)

One solution is sinusoidal

\begin{aligned}\frac{X''}{X} &= -k_1^2 \\ \frac{Y''}{Y} &= -k_2^2 \\ -k_1^2 - k_2^2&= \mathbf{k}^2 - \frac{\omega^2}{c^2}.\end{aligned} \hspace{\stretch{1}}(4.41)

The example in the tutorial now switched to a rectangular waveguide, still oriented with the propagation direction down the z-axis, but with lengths $a$ and $b$ along the $x$ and $y$ axis respectively.

Writing $k_1 = 2\pi m/a$, and $k_2 = 2 \pi n/ b$, we have

\begin{aligned}\phi(x, y) = \sum_{m n} a_{m n} \exp\left( \frac{2 \pi i m}{a} x \right)\exp\left( \frac{2 \pi i n}{b} y \right)\end{aligned} \hspace{\stretch{1}}(4.44)

We were also provided with some definitions

\begin{definition}TE (Transverse Electric)

$\mathbf{E}_3 = 0$.
\end{definition}
\begin{definition}
TM (Transverse Magnetic)

$\mathbf{B}_3 = 0$.
\end{definition}
\begin{definition}
TM (Transverse Electromagnetic)

$\mathbf{E}_3 = \mathbf{B}_3 = 0$.
\end{definition}

\begin{claim}TEM do not existing in a hollow waveguide.
\end{claim}

Why: I had in my notes

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{E} = 0 & \implies \frac{\partial {E_2}}{\partial {x^1}} -\frac{\partial {E_1}}{\partial {x^2}} = 0 \\ \boldsymbol{\nabla} \cdot \mathbf{E} = 0 & \implies \frac{\partial {E_1}}{\partial {x^1}} +\frac{\partial {E_2}}{\partial {x^2}} = 0\end{aligned}

and then

\begin{aligned}\boldsymbol{\nabla}^2 \phi &= 0 \\ \phi &= \text{const}\end{aligned}

In retrospect I fail to see how these are connected? What happened to the $\partial_t \mathbf{B}$ term in the curl equation above?

It was argued that we have $\mathbf{E}_\parallel = \mathbf{B}_\perp = 0$ on the boundary.

So for the TE case, where $\mathbf{E}_3 = 0$, we have from the separation of variables argument

\begin{aligned}\hat{\mathbf{z}} \cdot \mathbf{B}_0(x, y) =\sum_{m n} a_{m n} \cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.45)

No sines because

\begin{aligned}B_1 \propto \frac{\partial {B_3}}{\partial {x_a}} \rightarrow \cos(k_1 x^1).\end{aligned} \hspace{\stretch{1}}(4.46)

The quantity

\begin{aligned}a_{m n}\cos\left( \frac{2 \pi i m}{a} x \right)\cos\left( \frac{2 \pi i n}{b} y \right).\end{aligned} \hspace{\stretch{1}}(4.47)

is called the $TE_{m n}$ mode. Note that since $B = \text{const}$ an ampere loop requires $\mathbf{B} = 0$ since there is no current.

Writing

\begin{aligned}k &= \frac{\omega}{c} \sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 } \\ \omega_{m n} &= 2 \pi c \sqrt{ \left(\frac{m}{a} \right)^2 + \left(\frac{n}{b} \right)^2 }\end{aligned} \hspace{\stretch{1}}(4.48)

Since $\omega < \omega_{m n}$ we have $k$ purely imaginary, and the term

\begin{aligned}e^{-i k z} = e^{- {\left\lvert{k}\right\rvert} z}\end{aligned} \hspace{\stretch{1}}(4.50)

represents the die off.

$\omega_{10}$ is the smallest.

Note that the convention is that the $m$ in $TE_{m n}$ is the bigger of the two indexes, so $\omega > \omega_{10}$.

The phase velocity

\begin{aligned}V_\phi = \frac{\omega}{k} = \frac{c}{\sqrt{ 1 - \left(\frac{\omega_{m n}}{\omega}\right)^2 }} \ge c\end{aligned} \hspace{\stretch{1}}(4.51)

However, energy is transmitted with the group velocity, the ratio of the Poynting vector and energy density

\begin{aligned}\frac{\left\langle{\mathbf{S}}\right\rangle}{\left\langle{{U}}\right\rangle} = V_g = \frac{\partial {\omega}}{\partial {k}} = 1/\frac{\partial {k}}{\partial {\omega}}\end{aligned} \hspace{\stretch{1}}(4.52)

(This can be shown).

Since

\begin{aligned}\left(\frac{\partial {k}}{\partial {\omega}}\right)^{-1} = \left(\frac{\partial {}}{\partial {\omega}}\sqrt{ (\omega/c)^2 - (\omega_{m n}/c)^2 }\right)^{-1} = c \sqrt{ 1 - (\omega_{m n}/\omega)^2 } \le c\end{aligned} \hspace{\stretch{1}}(4.53)

We see that the energy is transmitted at less than the speed of light as expected.

# Final remarks.

I’d started converting my handwritten scrawl for this tutorial into an attempt at working through these ideas with enough detail that they self contained, but gave up part way. This appears to me to be too big of a sub-discipline to give it justice in one hours class. As is, it is enough to at least get an concept of some of the ideas involved. I think were I to learn this for real, I’d need a good text as a reference (or the time to attempt to blunder through the ideas in much much more detail).

## My submission for PHY356 (Quantum Mechanics I) Problem Set 3.

Posted by peeterjoot on November 30, 2010

# Problem 1.

## Statement

A particle of mass $m$ is free to move along the x-direction such that $V(X)=0$. The state of the system is represented by the wavefunction Eq. (4.74)

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k x} e^{- i \omega t} f(k)\end{aligned} \hspace{\stretch{1}}(1.1)

with $f(k)$ given by Eq. (4.59).

\begin{aligned}f(k) &= N e^{-\alpha k^2}\end{aligned} \hspace{\stretch{1}}(1.2)

Note that I’ve inserted a $1/\sqrt{2\pi}$ factor above that isn’t in the text, because otherwise $\psi(x,t)$ will not be unit normalized (assuming $f(k)$ is normalized in wavenumber space).

\begin{itemize}
\item
(a) What is the group velocity associated with this state?
\item
(b) What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(c) What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(d) Explain the physical meaning of the above results.
\end{itemize}

## Solution

### (a). group velocity.

To calculate the group velocity we need to know the dependence of $\omega$ on $k$.

Let’s step back and consider the time evolution action on $\psi(x,0)$. For the free particle case we have

\begin{aligned}H = \frac{\mathbf{p}^2}{2m} = -\frac{\hbar^2}{2m} \partial_{xx}.\end{aligned} \hspace{\stretch{1}}(1.3)

Writing $N' = N/\sqrt{2\pi}$ we have

\begin{aligned}-\frac{i t}{\hbar} H \psi(x,0) &= \frac{i t \hbar }{2m} N' \int_{-\infty}^\infty dk (i k)^2 e^{i k x - \alpha k^2} \\ &= N' \int_{-\infty}^\infty dk \frac{-i t \hbar k^2}{2m} e^{i k x - \alpha k^2}\end{aligned}

Each successive application of $-iHt/\hbar$ will introduce another power of $-it\hbar k^2/2 m$, so once we sum all the terms of the exponential series $U(t) = e^{-iHt/\hbar}$ we have

\begin{aligned}\psi(x,t) =N' \int_{-\infty}^\infty dk \exp\left( \frac{-i t \hbar k^2}{2m} + i k x - \alpha k^2 \right).\end{aligned} \hspace{\stretch{1}}(1.4)

Comparing with 1.1 we find

\begin{aligned}\omega(k) = \frac{\hbar k^2}{2m}.\end{aligned} \hspace{\stretch{1}}(1.5)

This completes this section of the problem since we are now able to calculate the group velocity

\begin{aligned}v_g = \frac{\partial {\omega(k)}}{\partial {k}} = \frac{\hbar k}{m}.\end{aligned} \hspace{\stretch{1}}(1.6)

## (b). What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

In order to evaluate the probability, it looks desirable to evaluate the wave function integral 1.4.
Writing $2 \beta = i/(\alpha + i t \hbar/2m )$, the exponent of that integral is

\begin{aligned}-k^2 \left( \alpha + \frac{i t \hbar }{2m} \right) + i k x&=-\left( \alpha + \frac{i t \hbar }{2m} \right) \left( k^2 - \frac{i k x }{\alpha + \frac{i t \hbar }{2m} } \right) \\ &=-\frac{i}{2\beta} \left( (k - x \beta )^2 - x^2 \beta^2 \right)\end{aligned}

The $x^2$ portion of the exponential

\begin{aligned}\frac{i x^2 \beta^2}{2\beta} = \frac{i x^2 \beta}{2} = - \frac{x^2 }{4 (\alpha + i t \hbar /2m)}\end{aligned}

then comes out of the integral. We can also make a change of variables $q = k - x \beta$ to evaluate the remainder of the Gaussian and are left with

\begin{aligned}\psi(x,t) =N' \sqrt{ \frac{\pi}{\alpha + i t \hbar/2m} } \exp\left( - \frac{x^2 }{4 (\alpha + i t \hbar /2m)} \right).\end{aligned} \hspace{\stretch{1}}(1.7)

Observe that from 1.2 we can compute $N = (2 \alpha/\pi)^{1/4}$, which could be substituted back into 1.7 if desired.

Our probability density is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 &=\frac{1}{{2 \pi}} N^2 {\left\lvert{ \frac{\pi}{\alpha + i t \hbar/2m} }\right\rvert} \exp\left( - \frac{x^2}{4} \left( \frac{1}{{(\alpha + i t \hbar /2m)}} + \frac{1}{{(\alpha - i t \hbar /2m)}} \right) \right) \\ &=\frac{1}{{2 \pi}} \sqrt{\frac{2 \alpha}{\pi} } \frac{\pi}{\sqrt{\alpha^2 + (t \hbar/2m)^2 }} \exp\left( - \frac{x^2}{4} \frac{1}{{\alpha^2 + (t \hbar/2m)^2 }} \left( \alpha - i t \hbar /2m + \alpha + i t \hbar /2m \right)\right) \\ &=\end{aligned}

With a final regrouping of terms, this is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 =\sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right).\end{aligned} \hspace{\stretch{1}}(1.8)

As a sanity check we observe that this integrates to unity for all $t$ as desired. The probability that we find the particle at position $x > x_0$ is then

\begin{aligned}P_{x>x_0}(t) = \sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\int_{x=x_0}^\infty dx \exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right)\end{aligned} \hspace{\stretch{1}}(1.9)

The only simplification we can make is to rewrite this in terms of the complementary error function

\begin{aligned}\text{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^\infty e^{-t^2} dt.\end{aligned} \hspace{\stretch{1}}(1.10)

Writing

\begin{aligned}\beta(t) = \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 },\end{aligned} \hspace{\stretch{1}}(1.11)

we have

\begin{aligned}P_{x>x_0}(t_0) = \frac{1}{{2}} \text{erfc} \left( \sqrt{\beta(t_0)/2} x_0 \right)\end{aligned} \hspace{\stretch{1}}(1.12)

Sanity checking this result, we note that since $\text{erfc}(0) = 1$ the probability for finding the particle in the $x>0$ range is $1/2$ as expected.

## (c). What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

This unit length probability is thus

\begin{aligned}P_{x>x_0+1/2}(t_0) - P_{x>x_0-1/2}(t_0) &=\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0+\frac{1}{{2}} \right) \right) -\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0-\frac{1}{{2}} \right) \right) \end{aligned} \hspace{\stretch{1}}(1.13)

## (d). Explain the physical meaning of the above results.

To get an idea what the group velocity means, observe that we can write our wavefunction 1.1 as

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k (x - v_g t)} f(k)\end{aligned} \hspace{\stretch{1}}(1.14)

We see that the phase coefficient of the Gaussian $f(k)$ “moves” at the rate of the group velocity $v_g$. Also recall that in the text it is noted that the time dependent term 1.11 can be expressed in terms of position and momentum uncertainties $(\Delta x)^2$, and $(\Delta p)^2 = \hbar^2 (\Delta k)^2$. That is

\begin{aligned}\frac{1}{{\beta(t)}} = (\Delta x)^2 + \frac{(\Delta p)^2}{m^2} t^2 \equiv (\Delta x(t))^2\end{aligned} \hspace{\stretch{1}}(1.15)

This makes it evident that the probability density flattens and spreads over time with the rate equal to the uncertainty of the group velocity $\Delta p/m = \Delta v_g$ (since $v_g = \hbar k/m$). It is interesting that something as simple as this phase change results in a physically measurable phenomena. We see that a direct result of this linear with time phase change, we are less able to find the particle localized around it’s original time $x = 0$ position as more time elapses.

# Problem 2.

## Statement

A particle with intrinsic angular momentum or spin $s=1/2$ is prepared in the spin-up with respect to the z-direction state ${\lvert {f} \rangle}={\lvert {z+} \rangle}$. Determine

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.16)

and

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.17)

and explain what these relations say about the system.

## Solution: Uncertainty of $S_z$ with respect to ${\lvert {z+} \rangle}$

Noting that $S_z {\lvert {f} \rangle} = S_z {\lvert {z+} \rangle} = \hbar/2 {\lvert {z+} \rangle}$ we have

\begin{aligned}{\langle {f} \rvert} S_z {\lvert {f} \rangle} = \frac{\hbar}{2} \end{aligned} \hspace{\stretch{1}}(2.18)

The average outcome for many measurements of the physical quantity associated with the operator $S_z$ when the system has been prepared in the state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$ is $\hbar/2$.

\begin{aligned}\Bigl(S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \Bigr) {\lvert {f} \rangle}&= \frac{\hbar}{2} {\lvert {f} \rangle} -\frac{\hbar}{2} {\lvert {f} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(2.19)

We could also compute this from the matrix representations, but it is slightly more work.

Operating once more with $S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ on the zero ket vector still gives us zero, so we have zero in the root for 2.16

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = 0\end{aligned} \hspace{\stretch{1}}(2.20)

What does 2.20 say about the state of the system? Given many measurements of the physical quantity associated with the operator $V = (S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1})^2$, where the initial state of the system is always ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, then the average of the measurements of the physical quantity associated with $V$ is zero. We can think of the operator $V^{1/2} = S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ as a representation of the observable, “how different is the measured result from the average ${\langle {f} \rvert} S_z {\lvert {f} \rangle}$”.

So, given a system prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, and performance of repeated measurements capable of only examining spin-up, we find that the system is never any different than its initial spin-up state. We have no uncertainty that we will measure any difference from spin-up on average, when the system is prepared in the spin-up state.

## Solution: Uncertainty of $S_x$ with respect to ${\lvert {z+} \rangle}$

For this second part of the problem, we note that we can write

\begin{aligned}{\lvert {f} \rangle} = {\lvert {z+} \rangle} = \frac{1}{{\sqrt{2}}} ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ).\end{aligned} \hspace{\stretch{1}}(2.21)

So the expectation value of $S_x$ with respect to this state is

\begin{aligned}{\langle {f} \rvert} S_x {\lvert {f} \rangle}&=\frac{1}{{2}}( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) S_x ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) \\ &=\hbar ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( {\lvert {x+} \rangle} - {\lvert {x-} \rangle} ) \\ &=\hbar ( 1 + 0 + 0 -1 ) \\ &= 0\end{aligned}

After repeated preparation of the system in state ${\lvert {f} \rangle}$, the average measurement of the physical quantity associated with operator $S_x$ is zero. In terms of the eigenstates for that operator ${\lvert {x+} \rangle}$ and ${\lvert {x-} \rangle}$ we have equal probability of measuring either given this particular initial system state.

For the variance calculation, this reduces our problem to the calculation of ${\langle {f} \rvert} S_x^2 {\lvert {f} \rangle}$, which is

\begin{aligned}{\langle {f} \rvert} S_x^2 {\lvert {f} \rangle} &=\frac{1}{{2}} \left( \frac{\hbar}{2} \right)^2 ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( (+1)^2 {\lvert {x+} \rangle} + (-1)^2 {\lvert {x-} \rangle} ) \\ &=\left( \frac{\hbar}{2} \right)^2,\end{aligned}

so for 2.22 we have

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = \frac{\hbar}{2}\end{aligned} \hspace{\stretch{1}}(2.22)

The average of the absolute magnitude of the physical quantity associated with operator $S_x$ is found to be $\hbar/2$ when repeated measurements are performed given a system initially prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$. We saw that the average value for the measurement of that physical quantity itself was zero, showing that we have equal probabilities of measuring either $\pm \hbar/2$ for this experiment. A measurement that would show the system was in the x-direction spin-up or spin-down states would find that these states are equi-probable.

I lost one mark on the group velocity response. Instead of 3.23 he wanted

\begin{aligned}v_g = {\left. \frac{\partial {\omega(k)}}{\partial {k}} \right\vert}_{k = k_0}= \frac{\hbar k_0}{m} = 0\end{aligned} \hspace{\stretch{1}}(3.23)

since $f(k)$ peaks at $k=0$.

I’ll have to go back and think about that a bit, because I’m unsure of the last bits of the reasoning there.

I also lost 0.5 and 0.25 (twice) because I didn’t explicitly state that the probability that the particle is at $x_0$, a specific single point, is zero. I thought that was obvious and didn’t have to be stated, but it appears expressing this explicitly is what he was looking for.

Curiously, one thing that I didn’t loose marks on was, the wrong answer for the probability per unit length. What he was actually asking for was the following

\begin{aligned}\lim_{\epsilon \rightarrow 0} \frac{1}{{\epsilon}} \int_{x_0 - \epsilon/2}^{x_0 + \epsilon/2} {\left\lvert{ \Psi(x_0, t_0) }\right\rvert}^2 dx = {\left\lvert{\Psi(x_0, t_0)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.24)

That’s a whole lot more sensible seeming quantity to calculate than what I did, but I don’t think that I can be faulted too much since the phrase was never used in the text nor in the lectures.