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%% An attempt at LaTeXing Ma441 QM
%% Covers up till Friday 14th November 2008
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click get pdf!
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%% Blah blah blah. Chris Blair.
\documentclass{report}
\usepackage{fullpage}
\usepackage{amsmath,amsthm,amsfonts}
\usepackage{color}
\usepackage{hyperref}
\usepackage{tikz}
\usepackage{graphicx}
\numberwithin{equation}{chapter}
% new commands defined to make things esp. bras & kets easier
\newcommand{\cdc}{, \dots ,}
\newcommand{\pdp}{+ \dots +}
\newcommand{\bra}[1]{\langle #1 |}
\newcommand{\ket}[1]{ | #1 \rangle}
\newcommand{\bk}[2]{\left\langle #1 | #2 \right\rangle} % < 1 | 2 >
\newcommand{\bkk}[2]{\langle #1 | #2 \rangle} % the above command
specially for \vec{k}s as they didn't fit otherwise ...
\newcommand{\blk}[3]{\left\langle #1 \left| #2 \right| #3 \right\rangle} %
< 1 | 2 | 3 >
\newcommand{\md}[1]{\left| #1 \right|} % mod
\newcommand{\mds}[1]{\left| #1 \right|^2} % mod squared
\newcommand{\hil}{\mathcal{H}}
\newcommand{\hsp}{Hilbert space $\mathcal{H}$}
\newtheoremstyle{mythm}% name
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\theoremstyle{mythm}
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\title{Ma441 Quantum Mechanics}
\author{M.P. Fry}
\date{\scriptsize Transcription started by Chris Blair...}
\begin{document}
\maketitle
\tableofcontents
\chapter{Principles of Quantum Mechanics}
\section{Prologue}
Quantum mechanics is not just another physical theory; it is the framework
for all physical theories. Its structure is independent of what kind of
particles exist and their interactions. Its progress has no analogue in
the history of science and has reached a status of universal
applicability.
\section{Principles of Quantum Mechanics}
\begin{enumerate}
\item Classical dynamics including relativistic dynamics cannot answer
some obvious questions. Why is it when $10^{24}$ atoms are brought
together we don't have a violent explosion? Why does matter not shrink to
nothing under the Coulomb force between electrons and nuclei? Atoms have
finite size yet an electron bound to a proton accelerates and hence
classically must radiate energy at the rate $\frac{dE}{dt} = \frac{2}{3}
\frac{e^2 a^2}{c^2}$, where $a$ is its acceleration. This gives a
characteristic decay time of $O(10^{-11})$ seconds. So we conclude that
the stability of matter - indeed life - cannot be understood classically.
\item \textbf{Fact:} All processes so far known can be understood by the
following prescription: \textbf{To every process there corresponds an
amplitude $\psi$, where $\psi$ is a complex number. The probability $P$ of
the process is $|\psi|^2$ up to a normalisation constant.}
\item Quantum behaviour of matter (we shall use electrons as an example
but what we say is true for all particles)
\begin{figure}[h]
\centering
\begin{tikzpicture}
\draw (0,0)--(0,0.4);
\draw (0,0.8)--(0,1.4);
\draw (0,1.8)--(0,2.2);
\draw[thick] (5,-0.3)--(5,2.5);
\draw[color=red,->, densely dashed] (-2.35,0.6)--(-0.2,0.6);
\draw[color=red,->, densely dashed] (-2.35,1.1)--(-0.2,1.1);
\draw[color=red,->, densely dashed] (-2.35,1.6)--(-0.2,1.6);
\draw (-4,1.1) node[text width=4cm] {Electrons with definite momentum};
\draw (0.4,0.6) node[font=\scriptsize] {Slit 2};
\draw (0.4,1.6) node[font=\scriptsize] {Slit 1};
\draw (5.6,1.1) node[font=\scriptsize] {Screen};
\end{tikzpicture}
\label{fig:ds}
\caption{Double slit experiment}
\end{figure}
An electron hits the screen. We cannot ask which slit, 1 or 2, it passed
unless we insert a device to observe the electron's path, e.g. a light
source. But this is a different experiment because photons from light
scatter from the electrons. Hence remove the light source.
Let $\psi_1$ (resp. $\psi_2$) be the amplitude for the electron to go
through slit 1 (resp. 2). When slit 2 is closed, probability for electron
to hit the screen is $P_1 = | \psi_1 |^2$; similarly with slit 1 closed we
have $P_2 = | \psi_2 |^2$. When both slits are open, Nature's rule is
\begin{align*}
\psi & = \psi_1 + \psi_2 \\ & = \mbox{probability amplitude for electron
to hit the screen with both slits open}
\end{align*}
This is a special case of the principle of superposition in quantum
mechanics. The probability for the electron to hit the screen is
\[
P_{12} = |\psi|^2 = |\psi_1 + \psi_2|^2 \neq P_1 + P_2
\]
due to interference of processes 1 and 2.
%\begin{center}
%\begin{tikzpicture}
%\draw (0,0)--(0,0.4);
%\draw (0,0.8)--(0,1.4);
%\draw (0,1.8)--(0,2.2);
%\draw[thick] (5,-0.3)--(5,2.5);
%\draw[color=red,->, densely dashed] (-2.35,0.6)--(-0.2,0.6);
%\draw[color=red,->, densely dashed] (-2.35,1.1)--(-0.2,1.1);
%\draw[color=red,->, densely dashed] (-2.35,1.6)--(-0.2,1.6);
%\draw (-3.5,1.1) node {Electrons};
%\draw (0.4,0.6) node[font=\scriptsize] {Slit 2};
%\draw (0.4,1.6) node[font=\scriptsize] {Slit 1};
%\end{tikzpicture}
%\end{center}
So electrons (particles) can behave like waves; they can interfere. The
associated wavelength $\lambda$ is
\[
\lambda= \frac{h}{p}
\]
known as the de Broglie wavelength. Here $p$ is momentum and $h$ is
Plank's constant, $h = 6.626069 \times 10^{-34}$Js, in SI units. We will
also use $\hbar = \frac{h}{2\pi} = 6.582119 \times 10^{-16}$eVs.
Thus in quantum mechanics there is no such concept as the path of a
particle.
Notes:
\begin{enumerate}
\item Suppose we fired bullets at the slits, then clearly $P = P_1 +
P_2$. Why? Because
\[
\lambda_{bullet} = \frac{h}{mv} \approx \frac{6.626 \times
10^{-34}\mbox{Js}}{10^{-3}\mbox{kg} \times 300 \mbox{ms}^{-1}} \approx
10^{-33} \mbox{m} << \mbox{bullet size}
\]
so there is no interference.
\item $\lambda \propto \frac{1}{m}$ and $\lambda \propto
\frac{1}{\sqrt{T}}$ since $\langle \frac{1}{2}mv^2 \rangle = \frac{3}{2} k
T$
\end{enumerate}
\item \textbf{Fact:} All matter must be ultimately described in terms of
waves. Interferences results when there are 2 or more indistinguishable,
coexisting alternatives leading to the same final outcome.
\item \textbf{Summary:}
\begin{itemize}
\item The probability of a process (event) $P$ is related to the
probability amplitude $\psi$ for the process by $P=|\psi|^2$, where $\psi$
is a complex number.
\item When a process (event) can occur in several indistinguishable
alternative ways, $1,2 \cdc n$ then $\psi = \psi_1 + \psi_2 \pdp \psi_n$ -
we sum over the alternatives.
\item If the experiment can determine which alternative is chosen then
there is no superposition and $P = |\psi_1|^2 + |\psi_2|^2 \pdp
|\psi_n|^2$.
\end{itemize}
\end{enumerate}
\section{Examples}
\subsection{The double slit experiment again}
a) Slit 2 closed
\begin{figure}[h]
\centering
\begin{tikzpicture}
\draw (0,0)--(0,0.4);
\draw[blue] (0,0.4)--(0,0.8);
\draw (0,0.8)--(0,1.4);
\draw (0,1.8)--(0,2.2);
\draw[thick] (5,-0.3)--(5,2.5);
\draw[color=red,->] (-3.5,1.1)--(0,1.6)--(5,0.7);
\draw (-3.7,1.1) node {$S$};
\draw (5.5,0.7) node {$x$};
\draw (1,0.6) node[font=\scriptsize] {Slit 2 (closed)};
\draw (0.5,1.8) node[font=\scriptsize] {Slit 1};
\end{tikzpicture}
\label{fig:ds2c}
\caption{Double slit experiment - slit 2 closed}
\end{figure}
``The amplitude that the particle from $S$ will arrive at $x$'' can be
written in Dirac notation as \begin{align*}
\psi_1(x) & = \bk{\mbox{particle arrives at } x}{\mbox{particle leaves }
S} \\ & = \bk{x}{S}
\end{align*}
Here $\langle \, \rangle = $ ``the amplitude that'' and $\bk{x}{S}$ is a
complex number. The first entry is the final condition (state), and the
second is the initial condition (state).
Because only slit 1 is open we have just one alternative:
\[
\bk{x}{S} = \bk{x}{\mbox{slit} \, 1} \bk{\mbox{slit} \, 1}{S}
\]
and
\[
P_1(x) = | \bk{x}{1} |^2 | \bk{1}{S} |^2
\]
or in words, the probability of arrival at $x$ equals (probability of
passing slit 1 to $x$) multiplied by (probability of arriving at slit 1
from $S$).
b) Slit 1 closed
In this case,
\[
\bk{x}{S} = \bk{x}{2} |^2 | \bk{2}{S}
\]
and
\[
P_2(x) = | \bk{x}{2} |^2 | \bk{2}{S} |^2
\]
c) Slits 1 and 2 open
We now have that
\begin{align*}
\bk{x}{S} & = \mbox{sum over alternatives} \cr
& = \sum_{i=1,2} \bk{x}{\mbox{slit} \, i} \bk{\mbox{slit} \, i}{S} \cr
& = \bk{x}{1} \bk{1}{S} + \bk{x}{2} \bk{2}{S}
\end{align*}
So
\[
P(x) = \left| \bk{x}{1} \bk{1}{S} + \bk{x}{2} \bk{2}{S} \right|^2 \neq
P_1(x) + P_2(x)
\]
\subsection{Coulomb scattering of O and He nuclei in their centre of mass
frame}
\begin{figure}[h]
\centering
\begin{tikzpicture}
\begin{scope}[scale=0.55]
\draw (-5,0)--(5,0);
\draw (-2,-3)--(2,3);
\draw (0.8,0) arc (0:57:0.8);
\draw (0.9,0.6) node {$\theta$};
\draw[blue] (-5,0.5) node[font=\scriptsize] {$\alpha$-particle};
\draw[green] (4,-0.5) node[font=\scriptsize] {$O$ nucleus};
\draw[blue,->] (-5,0.15) .. controls (0,0.15) .. (1.85,3);
\draw[green,->] (5,-0.15) .. controls (0,-0.15) .. (-1.85,-3);
\shadedraw (2.65,4) circle (0.4cm);
\draw (2.65,4.7) node[font=\scriptsize] {Detector};
\begin{scope}[xshift=12cm]
\draw (-5,0)--(5,0);
\draw (-2,-3)--(2,3);
\draw (0.8,0) arc (0:57:0.8);
\draw (0.9,0.6) node {$\theta$};
\draw[blue] (-4.5,-0.5) node[font=\scriptsize] {$\alpha$-particle};
\draw[green] (5,0.5) node[font=\scriptsize] {$O$ nucleus};
\draw[blue,->] (-5,-0.15) .. controls (-0.1,-0.15) .. (-2.15,-3);
\draw[green,->] (5,0.15) .. controls (0.1,0.15) .. (2.15,3);
\shadedraw (2.65,4) circle (0.4cm);
\draw (2.65,4.7) node[font=\scriptsize] {Detector};
\end{scope}
\end{scope}
\end{tikzpicture}
\label{fig:ohescat}
\caption{Scattering and back-scattering of $\alpha$-particle against
oxygen nucleus}
\end{figure}
Consider the scattering of $\alpha$-particles and oxygen nuclei in their
centre of mass frame. There are two possibilities for the scattering, as
shown above. Let $f(\theta) = $ amplitude to scatter $\alpha$-particle
through $\theta$, implying that $f(\pi - \theta) = $ amplitude to scatter
$O$ nucleus through $\theta$. The probability of \textbf{a} particle at
the detector is $|f(\theta)|^2 + |f(\pi - \theta)|^2$. This is because the
two alternatives are distinguishable in principle - by appropriate analysis
you can determine if it was the oxygen or the helium which you observed,
even years later. Hence we have the following:
\begin{center}
\textbf{Rule:} Never add amplitudes for distinguishable final states.
\end{center}
Now suppose we scatter $\alpha$-particles on $\alpha$-particles. This
means we can no longer determine if it was the incident or the target
particle that scattered into the detector. There are two indistinguishable
alternatives, so that
\[
\mbox{probability of} \, \alpha \mbox{-particle at detector} \, = |
f(\theta) + f(\pi - \theta) |^2
\]
which means there is twice as much scattering at $90^{\circ}$ if the
particles are indistinguishable. This is observed.
\subsection{Bragg reflection}
Consider a low energy ( $\approx \frac{1}{10}$ eV) neutron with
spin-$\frac{1}{2}$ scattering from a crystal of identical
spin-$\frac{1}{2}$ nuclei. Assume all the nuclei have spin up and that the
incident neutron has spin down. The neutron can scatter without flipping
its spin (amplitude $a$), or by flipping its spin (amplitude $b$),
provided the nucleus' spin also flips to conserve angular momentum.
a) What is the probability for detecting a neutron at the counter $C$ with
spin down?
b) What is the probability for detecting a neutron at the counter $C$ with
spin up?
c) What is the probability for detecting \textbf{a} neutron at the counter
$C$?
\begin{figure}[h]
\centering
\begin{tikzpicture}
\draw[->] (-1,2.6) -- (-0.2,2.6);
\filldraw (-1,2.6) circle (0.17cm);
\draw[thick,->] (-1,2.6)--(-1,2.3);
\foreach \x in {1,1.8,2.6}
\foreach \y in {1,1.8,2.6}
{
\draw (\x,\y) circle (0.2cm);
\draw[thick,->] (\x,\y) -- ++(0,0.35);
}
\begin{scope}[xshift=4cm]
\foreach \x in {1,2.6}
\foreach \y in {1,1.8,2.6}
{
\draw (\x,\y) circle (0.2cm);
\draw[thick,->] (\x,\y) -- ++(0,0.35);
}
\foreach \y in {1,2.6}
{
\draw (1.8,\y) circle (0.2cm);
\draw[thick,->] (1.8,\y) -- ++(0,0.35);
}
\draw (1.8,1.8) circle (0.2cm);
\draw[red,thick,->] (1.8,1.8) -- ++(0,-0.35);
\draw (3,1.8) -- (5,1.8);
\draw[->] (3,1.8) -- (5.5,1);
\shadedraw (6,0.85) circle (0.2cm);
\draw (7.1,0.85) node[font=\scriptsize] {Counter $C$};
\filldraw (4.6,1.3) circle (0.17cm);
\draw[thick,->] (4.6,1.3)-- ++ (0.11,0.29);
\draw (3.8,1.8) arc (360:343:0.8);
\draw (4,1.64) node[font=\scriptsize] {$\theta$};
\end{scope}
\end{tikzpicture}
\label{fig:bragg}
\caption{Bragg reflection - showing the case where the neutron scatters
and flips spin}
\end{figure}
\textbf{Solution:}
Remember the rule: never add amplitudes for distinguishable final states.
a) Label nuclei in crystal by $i=1,2 \cdc N$. Then amplitude for a neutron
to be scattered into $C$ by nucleus $i$ without a spin flip is
\[
\bk{\mbox{neutron at } \, C}{\mbox{neutron from }\, S} = \bk{c}{i} a
\bk{i}{s}
\]
There is no way of determining from which nucleus the neutron scattered.
Hence, total probability amplitude for neutron to arrive at $C$ with spin
down is
\[
\bk{C}{S} = \sum_{i=1}^N \bk{c}{i} a \bk{i}{s}
\]
and so
\[
P_{\mbox{no-flip}} = \mds{\sum_{i=1}^N \bk{c}{i} a \bk{i}{s}}
\]
- coherent scattering.
b) Now consider the case where the neutron arrives at the counter with
spin up. The nucleus from which scattering took place can be identified,
let us denote it as nucleus $k$. We then have
\[
P_{\mbox{spin-flip}} =\sum_{k=1}^N \mds{\bk{c}{k} b \bk{k}{s}}
\]
- incoherent scattering. The amplitudes do not add because the final
states are distinct.
c) The probability for detecting \textbf{a} nucleus at $C$ is
\[
P = P_{\mbox{no-flip}} + P_{\mbox{spin-flip}} = \mds{\sum_{i=1}^N
\bk{c}{i} a \bk{i}{s}} + \sum_{k=1}^N \mds{\bk{c}{k} b \bk{k}{s}}
\]
\textbf{Notes:} i) Consider the coherent scattering term,
$\mds{\sum_{i=1}^N \bk{c}{i} a \bk{i}{s}}$. This contains cross terms
(interference terms). In the case of randomly positioned nuclei these tend
to cancel, giving scattering proportional to $N$. If the nuclei are on a
regular lattice, there are directions known as Bragg angles given by
\[
2 d \sin \theta = n \lambda \qquad d = \mbox{lattice spacing} \quad n =
1,2, \dots
\]
where all scattered waves are in phase. Then the coherent term gives a
scattered beam strength proportional to $N^2$.
ii) It is not necessary that the crystal be polarised, i.e. spins can be
up or down. In this case,
\[
P_{\mbox{spin-flip}} =\sum_{k=1}^N \mds{\bk{c}{k} b \bk{k}{s}}
\]
but we sum only over nuclei with spin initially up. The other probability
term $P_{\mbox{no-flip}}$ is the same as before.
iii) Multiple scattering may also occur. In principle this should be
included in the probability terms. But such events are proportional to
powers of $a$ and $b$ and these are small.
\subsection{Other examples}
i) A particle is prepared in a state $\psi$ and arrives at $x$, $\psi(x) =
\bk{x}{\psi}$.
ii) Hydrogen molecular ion, $\mbox{H}_2^+$.
\begin{figure}[h]
\centering
\begin{tikzpicture}
\draw (0,0) circle (0.3cm);
\draw (3,0) circle (0.3cm);
\draw (0.2,0.7) circle (0.15cm);
\draw (0,0) node {$+$};
\draw (3,0) node {$+$};
\draw (0.2,0.7) node[font=\scriptsize] {$-$};
\draw (1.4,-0.7) node {$\ket{1}$};
\draw (8,0) circle (0.3cm);
\draw (11,0) circle (0.3cm);
\draw (11.2,0.7) circle (0.15cm);
\draw (8,0) node {$+$};
\draw (11,0) node {$+$};
\draw (11.2,0.7) node[font=\scriptsize] {$-$};
\draw (9.4,-0.7) node {$\ket{2}$};
\end{tikzpicture}
\label{fig:hion}
\caption{Hydrogen molecular ion}
\end{figure}
The ion is in some state $\ket{\psi}$. The amplitude it is in state 1 is
$\bk{1}{\psi}$.
iii) A system is in some general state $\ket{\psi}$ and we wish to find
the probability it is in a particular state $\ket{\varphi}$. Then we have
probability amplitude $\bk{\varphi}{\psi}$, and the probability system is
in the state $\ket{\varphi}$ is $\mds{\bk{\varphi}{\psi}}$.
\section{Mathematical formalism}
The above notation may remind you of the inner product in mathematics -
partially because that's precisely what it is. In this next section we
will set out the basic mathematical notation and formalism used in quantum
mechanics.
\subsection{State space}
Consider a vector $\vec{A}$ in an $n$-dimensional Euclidean space. Let
$\lbrace \vec{a}_i \rbrace_{i=1}^n$ be an orthonormal basis of vectors,
that is $\vec{a}_i^* \vec{a}_j = \delta_{ij}$ where $*$ denotes the
complex conjugate. Then $\vec{A}$ can be expanded in the basis
\[
\vec{A} = \sum_{i=1}^n \vec{a}_i^* \cdot \vec{A} \vec{a}_i = \sum_{i=1}^n
\underbrace{\left( \vec{a}_i, \vec{A} \right)}_{\scriptsize \mbox{inner
product}} \vec{a}_i
\]
Likewise the state $\ket{\psi}$ of a system can exist in some space, and
we can express it as a set of basis states. Consider again the molecular
hydrogen ion mentioned above. Clearly a set of basis states are $\ket{1}$
and $\ket{2}$.
In analogy with a vector, any state of $\mbox{H}_2^+$ can be expanded in a
set of basis states:
\[
\ket{\psi} = \bk{1}{\psi} \ket{1} + \bk{2}{\psi} \ket{2}
\]
where the probability amplitude for $\mbox{H}_2^+$ to be in state
$\ket{1}$ is
\[
\bk{1}{\psi} = \left( \ket{1}, \ket{\psi}\right)
\]
which is the inner product of $\ket{1}$ and $\ket{\psi}$, or the overlap
of $\ket{1}$ and $\ket{\psi}$, and is a complex number.
\subsection{Ket space, bra space, inner product}
In quantum mechanics a physical state, e.g. an electron with definite spin
orientation, is represented by a state vector in a complex vector space.
Following Dirac we call such a vector a \textbf{ket} and denote it by
$\ket{\cdot}$. The state vector is postulated to contain complete
information about the the physical state.
A basic principle of quantum mechanics is the of principle of
superposition: if $\ket{\psi_1}$ and $\ket{\psi_2}$ are possible states of
a system then $C_1 \ket{\psi_1} + C_2 \ket{\psi_2}$, where $C_1, C_2 \in
\mathbb{C}$, is also a possible state. Furthermore, the kets $\ket{\psi}$
and $c \ket{\psi}$, for $c$ a constant, represent the same physical state.
The vector space of all kets is called \textbf{ket space}. We now
introduce \textbf{bra space} as a vector space dual to the ket space. We
postulate that to every ket $\ket{\psi}$ there exists a bra $\bra{\psi}$
in this dual space. The bra space is spanned by the bra basis $\lbrace
\bra{i} \rbrace$ which corresponds to the ket basis $\lbrace \ket{i}
\rbrace$. The correspondence is one-to-one. Roughly, bra space is some
kind of ``mirror image'' of ket space.
The bra vector dual to $c \bra{\psi}$ is postulated to be $c^*
\bra{\psi}$. We denote the inner product of a bra and a ket by
\[
\bk{\psi}{\varphi} = \left(\bra{\psi} \right) \cdot \left( \ket{\varphi}
\right) \quad \mbox{(symbolically)}
\]
which gives a complex number.
We postulate the following properties:
\vspace{-0.75em}
\begin{itemize}
\item $\bk{\psi}{\varphi} = \bk{\varphi}{\psi}^*$
\item $\bk{\psi}{\psi} \geq 0$ with equality only if $\ket{\psi}=0$
\item $\bk{\psi}{c \varphi} = c \bk{\psi}{\varphi}$ and $\bk{c
\psi}{\varphi} = c^* \bk{ \psi}{\varphi}$ for $c \in \mathbb{C}$.
\item $\left( \bra{\psi_1} + \bra{\psi_2} \right) \ket{\varphi} =
\bk{\psi_1}{\varphi} + \bk{\psi_2}{\varphi}$
\end{itemize}
Note: suppose that $\ket{\psi} = C_1 \ket{\psi_1} + C_2 \ket{\psi_2}$ and
that $\bk{\psi_i}{\psi_j} = \delta_{ij}$, then
\[
\bk{\psi_1}{\psi} = C_1 \qquad \bk{\psi_2}{\psi} = C_2
\]
and the probability $\ket{\psi}$ is in state $\ket{\psi_i}$ is
$\mds{C_i}$. Then
\[
\bk{\psi}{\psi} = \mds{C_1} + \mds{C_2} = 1
\]
so $\bk{\psi}{\psi} \leq 0$ for non-zero $\ket{psi}$ is inconsistent with
a probability interpretation of quantum mechanics.
Two kets $\ket{\psi}$, $\ket{\varphi}$ are said to be orthogonal if
$\bk{\varphi}{\psi} = 0$.
The norm of a ket $\ket{\psi}$ is $\bk{\psi}{\psi}$ and so we can form a
normalised ket $\ket{\overline{\psi}}$ by
\[
\ket{\overline{\psi}} = \frac{1}{\sqrt{\bk{\psi}{\psi}}} \ket{\psi}
\]
The dimensionality of the vector space depends on what is being described.
For example, an electron's spin state can be ``spin up'' $\ket{\!\uparrow}$
or ``spin down'' $\ket{\!\downarrow}$, and so the spin space of an electron
is two-dimensional. A general spin state $\ket{\psi}$ is represented by
\[
\psi = \bk{\uparrow\!}{\psi}\ket{\!\uparrow} +
\bk{\downarrow\!}{\psi}\ket{\!\downarrow}
\]
where $\bk{\uparrow\!}{\!\uparrow} = \bk{\downarrow\!}{\!\downarrow} = 1$,
$\bk{\uparrow\!}{\!\downarrow}=0$ and $\mds{\bk{\uparrow\!}{\psi}}$ is the
probability the electron has ``spin up'', etc.
Now consider a spinless free particle. Its momentum $p$ can be anything,
$-\infty < p < \infty$. So its momentum state space is nondenumerably
infinite and a general state can be represented as $\ket{\psi} =
\int_{-\infty}^{\infty} dp \bk{p}{\psi} \ket{p}$.
A complex vector space with the following properties is called a
\textbf{Hilbert space} and is denoted by $\hil$:
\vspace{-0.75em}
\begin{itemize}
\item $\hil$ is complete: there exists an orthonormal set of vectors
$\ket{i}$ in $\hil$ such that for all $\ket{\psi} \in \hil$ we can write
$\ket{\psi} = \sum_i \bk{i}{\psi} \ket{i}$. The index $i$ may be
countable, in which case we say $\hil$ is separable, or uncountable, in
which case we say $\hil$ is inseparable.
\item $\hil$ is an inner product space whose properties we have listed.
\end{itemize}
\vspace{-0.75em}
A Hilbert space is a wonderful thing, and you can get carried away with
its beauty. We should recall the wisdom of Asher Peres: ``Quantum
phenomena do not occur in Hilbert space; they occur in a laboratory.''
\subsection{Observables}
Let $A$ be an operator representing some physical quantity, such as energy
or angular momentum, that characterises the state of a quantum system.
Require $A$ to be a linear operator acting on $\hil$, that is $A: \hil
\rightarrow \hil$ and $A(a_1 \ket{\varphi} + a_2 \ket{\psi}) = a_1 A
\ket{\varphi} + a_2 A \ket{\psi}$.
Suppose $A$ is such an operator on $\hil$. Let $\ket{\varphi} \in \hil$
and $\ket{\varphi^{\prime}} = A \ket{\varphi} \in \hil$. Then the dual of
$\ket{\varphi^{\prime}}$ is $\bra{\varphi^{\prime}} = \bra{\varphi}
A^{\dagger}$, where $\bra{\varphi}$ is the dual of $\ket{\varphi}$, and
$A^{\dagger}$ is called the adjoint of $A$.
The values an observable can take are called its eigenvalues. A state
$\ket{\psi} \in \hil$ for which $A$ has the eigenvalue $a$ is called an
eigenstate of $A$. Eigenstates can be of two types: non-degenerate, if
there is only one $\ket{\psi}$ for each $a$, or degenerate, if there is
more than one $\ket{\psi}$ for each $a$. (We exclude the trivial case of
states $\ket{\psi^{\prime}} = c \ket{\psi}$, $c \in \mathbb{C}$.)
Consider a general linear operator $A$ acting on $\ket{\psi} \in \hil$,
with $\ket{\psi^{\prime}} = A \ket{\psi}$. Form the inner product of
$\ket{\psi^{\prime}}$ with $\ket{\varphi} \in \hil$, and take the complex
conjugate:
\[
\blk{\varphi}{A}{\psi}^* = \bk{\varphi}{\psi^{\prime}}^* =
\bk{\psi^{\prime}}{\varphi}
\]
but the dual of $\ket{\psi^{\prime}}$ is $\bra{\psi^{\prime}} = \bra{\psi}
A^{\dagger}$, so
\[
\blk{\varphi}{A}{\psi}^* = \blk{\psi}{A^{\dagger}}{\varphi}
\]
If $\blk{\varphi}{A}{\psi}^* = \blk{\psi}{A^{\dagger}}{\varphi} =
\blk{\psi}{A}{\varphi}$ then $A$ is said to be \textbf{Hermitian}. Note
that these definitions parallel the matrix definitions, $a_{ij}^{\dagger}
= a_{ji}^*$ and $a_{ij}^{\dagger} = a_{ij} \Rightarrow$ Hermitian. Note
also that a Hermitian operator is not a self-adjoint operator contrary to
what many physicists say, though this is true for finite dimensional
matrices.
We may write the inner product as mathematicians do, $\bk{\varphi}{\psi} =
(\varphi, \psi)$, so that $\blk{\varphi}{A}{\psi}^* =
\blk{\psi}{A}{\varphi}$ can be written as $(\varphi, A \psi)^* = (\psi,
A^{\dagger} \varphi)$, or $(A \psi, \varphi) = (\psi, A^{\dagger}
\varphi)$, or, taking the complex conjugage of both sides, $(A^{\dagger}
\varphi, \psi) = (\varphi, A \psi)$.
If $A,B$ are Hermitian, then $\blk{\varphi}{AB}{\psi}^* =
\blk{\psi}{BA}{\varphi}$.
\begin{thm}
The eigenvalues $a_1, a_2 \dots$ of a Hermitian operator are real.
\end{thm}
\begin{proof} We have
\[
A \ket{\psi} = a \ket{\psi} \Rightarrow \blk{\psi}{A}{\psi} = a
\bk{\psi}{\psi}
\]
So $a$ is real if $\blk{\psi}{A}{\psi}$ is real, but
$\blk{\psi}{A}{\psi}^* = \blk{\psi}{A}{\psi}$ by the assumption $A$ is
Hermitian.
\end{proof}
Since the eigenvalues of an observable $A$ are the result of measurement
they must be real, hence the observable $A$ cannot be an arbitrary linear
operator; it must be Hermitian.
\begin{thm}
If $A$ is Hermitian and $a_m \neq a_n$ for $a_m$, $a_n$ eigenvalues of
$A$, then $\bk{\psi_m}{\psi_n} = 0$, i.e. eigenstates of $A$ are
orthogonal.
\end{thm}
\begin{proof}
We have
\[
A \ket{\psi_m} = a_m \ket{\psi_m} \quad (1) \qquad A \ket{\psi_n} = a_n
\ket{\psi_n} \quad (2)
\]
From (1), $\blk{\psi_n}{A}{\psi_m} = a_m \bk{\psi_n}{\psi_m} (3)$. Take
the complex conjugate to obtain
\[
\blk{\psi_n}{A}{\psi_m}^* = a_m^* \bk{\psi_n}{\psi_m}^* \quad (4)
\]
Now $A$ is Hermitian so $a_m$, $a_n$ are real. So,
\[
\blk{\psi_m}{A}{\psi_n} = a_m \bk{\psi_m}{\psi_n} \quad (5)
\]
and from (2),
\[
\blk{\psi_m}{A}{\psi_n} = a_n \bk{\psi_m}{\psi_n} \quad (6)
\]
Taking (6) from (5) gives
\[
(a_m - a_n) \bk{\psi_m}{\psi_n} = 0
\]
hence for distinct eigenvalues we must have $\bk{\psi_m}{\psi_n} = 0$.
\end{proof}
Now suppose degeneracy, e.g. double degeneracy: $A \ket{\psi_1} = a
\ket{\psi_1}$, $A \ket{\psi_2} = a \ket{\psi_2}$ (where we are assuming
that $\ket{\psi_1} \neq c \ket{\psi_2}$. It is not necessarily true that
the eigenstates $\ket{\psi_1}$ and $\ket{\psi_2}$ are orthogonal. However
they can be made orthogonal as follows: let
\[
\ket{\psi_1^{\mbox{ \scriptsize new}}} = \ket{\psi_1} \qquad
\ket{\psi_2^{\mbox{ \scriptsize new}}} = c_1 \ket{\psi_1} + c_2
\ket{\psi_2}
\]
then $\bk{\psi_1^{\mbox{ \scriptsize new}}}{\psi_2^{\mbox{ \scriptsize
new}}} = 0$ if
\[
\frac{c_1}{c_2} = - \frac{\bk{\psi_1}{\psi_2}}{\bk{\psi_1}{\psi_1}}
\]
and
\[
A \ket{\psi_{1,2}^{\mbox{ \scriptsize new}}} = a \ket{\psi_{1,2}^{\mbox{
\scriptsize new}}}
\]
This procedure can be generalised to any level of degeneracy.
Suppose all linearly independent eigenstates of a Hermitian operator are
normalised to unity and orthogonalised if there is degeneracy. Then we
have
\begin{thm}[Spectral theorem]
Let $A$ be a Hermitian operator acting on a finite dimensional \hsp and
suppose the eigenvectors $\ket{\psi_n}$ of $A$ satisfy the orthonormality
condition $\bk{\psi_n}{\psi_m} = \delta_{mn}$, with eigenvalues $\lbrace
a_n \rbrace$. Then for every state $\ket{\varphi} \in \hil$,
\[
\ket{\varphi} = \sum_{n=1}^{N} \ket{\psi_n} \bk{\psi_n}{\varphi} \quad (*)
\]
\end{thm}
The set $\lbrace a_n \rbrace$ is called the \textbf{spectrum} of $A$. The
expansion (*) is called the spectral decomposition of $\ket{\varphi}$, or
the eigenvector expansions of $\ket{\varphi}$. The eigenvectors $\lbrace
\ket{\psi_n} \rbrace$ span $\hil$; they are a basis set, or a complete
orthonormal set.
\textbf{Note:} In the infinite dimensional case, the theorem should read
``Let $A$ be a self-adjoint compact operator on $\hil$. Then there exists
a complete orthonormal basis $\lbrace \ket{\psi_n} \rbrace$ for $\hil$
such that $A \ket{\psi_n} = a_n \ket{\psi_n}$, and for any $\ket{\varphi}
\in \hil$,
\[
\ket{\varphi} = \sum_{n=1}^{\infty} \ket{\psi_n} \bk{\psi_n}{\varphi}
\quad (*)
\]
The two spectral decompositions above allow us to state the completeness
of the eigenvectors $\ket{\psi_n}$ of $A$ in the form
\[
\sum_{n=1}^{N \, \mbox{\scriptsize or} \, \infty } \ket{\psi_n}
\bra{\psi_n} = \mathbb{I}
\]
\textbf{Examples:}
\begin{enumerate}
\item As we shall see, for an electron a complete set of spin states is
\[
\ket{\!\uparrow} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad
\ket{\!\downarrow} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}
\]
corresponding to spin-up and spin-down along the $z$-axis, and an
arbitrary spin state $\ket{\psi}$ is
\[
\psi = \bk{\uparrow\!}{\psi}\begin{pmatrix} 1 \\ 0 \end{pmatrix} +
\bk{\downarrow\!}{\psi}\begin{pmatrix} 0 \\ 1 \end{pmatrix}
\]
Note that we have
\[
\ket{\! \uparrow}\bra{\uparrow \!} + \ket{ \! \downarrow}\bra{\downarrow
\!} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0
\end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1
\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\]
\item (Complete set of functions) In two dimensions, let
\[
L_z \Phi_m(\varphi) = m \hbar \Phi_m (\varphi)
\]
where $L_z = \frac{\hbar}{i} \frac{\partial}{\partial \varphi}$ is the $z$
component of the orbital angular momentum operator. We then have that
\[
\Phi_m(\varphi) = \frac{1}{\sqrt{2 \pi}} e^{i m \varphi}
\]
and
\[
\sum_{m=-\infty}^{\infty} \Phi_m(\varphi) \Phi_m^* (\varphi) = \frac{1}{2
\pi} \sum_{m=-\infty}^{\infty} e^{im(\varphi - \varphi^{\prime})} = \delta
( \varphi - \varphi^{\prime} )
\]
If $f(\varphi)$ a smooth function on $0 \leq \varphi < 2\pi$, then
\[
f(\varphi) = \sum_{m=-\infty}^{\infty} ( \Phi_m, f) \Phi_m (\varphi)
\]
where the inner product is
\[
(\Phi_m,f) = \int_0^{2 \pi} d \varphi \, \Phi_m^* (\varphi) f(\varphi)
\]
- note that this is Fourier analysis.
\end{enumerate}
Sometimes the spectrum of $A$ has a continuum part. The spectral theorem
generalises to
\[
\ket{\varphi} = \sum_{m} \ket{\psi_n} \bk{\psi_n}{\varphi} + \int d\lambda
\, \ket{\psi_n} \bk{\psi_n}{\varphi}
\]
and
\[
\sum_{m} \ket{\psi_n} \bra{\psi_n} + \int d\lambda \, \ket{\psi_n}
\bra{\psi_n} = \mathbb{I}
\]
\subsection{Unitary Operators}
A linear operator $U$ is said to be \textbf{unitary} if $U^{-1} =
U^{\dagger}$, or $UU^{\dagger} = U^{\dagger}U = \mathbb{I}$ - this last
statement must be satisfied in the infinite dimensional case as there one
in general has that $AB = \mathbb{I}$ does not imply $BA = \mathbb{I}$.
\begin{thm}
Any Hermitian matrix can be diagonalised by a unitary transformation.
\end{thm}
\begin{proof}
See any book on matrices.
\end{proof}
Now let $A,B$ be Hermitian operators acting on a \hsp. Then the
eigenstates $\lbrace \ket{\psi_k} \rbrace_{k=1}^n$ and $\lbrace
\ket{\varphi_k} \rbrace_{k=1}^n$ of $A$ and $B$ are complete orthonormal
sets spanning the Hilbert space (by the spectral theorem).
\textbf{Problem:} Find the operator $U$ that converts $\ket{\varphi_k}$ to
$\ket{\psi_k}$,
\[
U \ket{\varphi_k} = \ket{\psi_k} \qquad k=1,2, \dots
\]
\textbf{Solution:} Multiply by $\bra{\varphi_k}$ from the right,
\[
U \ket{\varphi_k}\bra{\varphi_k} = \ket{\psi_k}\bra{\varphi_k}
\]
and sum over $k$. Using the completeness of the set $\lbrace
\ket{\varphi_k} \rbrace$
\[
\sum_k \ket{\varphi_k}\bra{\varphi_k} = \mathbb{I}
\]
we find that
\[
U = \sum_k \underbrace{\ket{\psi_k}}_{\mbox{ \scriptsize new}}
\underbrace{\bra{\varphi_k}}_{\mbox{ \scriptsize old}}
\]
Note that we may write $U$ as a matrix. From definition,
\[
U \ket{\varphi_k} = \ket{\psi_k}
\]
We use completeness of the set $\lbrace \ket{\varphi_k} \rbrace$ to get
\[
\sum_l \ket{\varphi_l} \blk{\varphi_l}{U}{\varphi_k} = \ket{\psi_k}
\]
or
\[
\sum_l U_{lk} \ket{\varphi_l} = \ket{\psi_k}
\]
where
\[
U_{lk} = \blk{\varphi_l}{U}{\varphi_k}
\]
\begin{thm}
This $U$ is unitary.
\end{thm}
\begin{proof}
We have that
\[
U^{\dagger} = \sum_k \ket{\varphi_k}\bra{\psi_k}
\]
then
\[
U^{\dagger}U = \sum_{k,l} \ket{\varphi_l}
\underbrace{\bk{\psi_k}{\psi_l}}_{=\delta_{kl}} \bra{\varphi_l} = \sum_k
\ket{\varphi_k}\bra{\varphi_k} = \mathbb{I}
\]
and similarly for $UU^{\dagger}$, hence $U^{\dagger} = U^{-1}$.
\end{proof}
Hence we see that a unitary transformations transforms one orthonormal
basis into another.
\textbf{Example:} An alternative set of spin states for an electron is
\[
\ket{\!\uparrow}_x = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1
\end{pmatrix} \quad \frac{1}{\sqrt{2}} \ket{\!\downarrow}_x =
\begin{pmatrix} 1 \\ -1 \end{pmatrix}
\]
corresponding to spin-up and spin-down along the $x$-axis. A unitary
operators that transforms from the $x$-basis to the $z$-basis
\[
\ket{\!\uparrow}_z = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad
\ket{\!\downarrow}_z = \begin{pmatrix} 0 \\ 1 \end{pmatrix}
\]
is
\begin{align*}
U & = \ket{\!\uparrow}_z \bra{\uparrow \!}_x + \ket{\!\downarrow}_z
\bra{\downarrow \! }_x \\
& = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \end{pmatrix}
\begin{pmatrix} 1 & 1 \end{pmatrix} + \frac{1}{\sqrt{2}} \begin{pmatrix} 0
\\ 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \end{pmatrix} \\
& = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} +
\frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 0 \\ 1 & -1 \end{pmatrix} \\
& = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}
\end{align*}
\begin{thm}
A unitary transformation preserves probability amplitudes.
\end{thm}
\begin{proof}
Let $\ket{a}$, $\ket{b}$ be in $\hil$, and $\ket{a^{\prime}} = U \ket{a}$,
$\ket{b^{\prime}} = U \ket{b}$. Then $\bra{b^{\prime}} = \bra{b}
U^{\dagger}$ and
\[
\bk{b^{\prime}}{a^{\prime}} = \blk{b}{U^{\dagger}U}{a} = \bk{a}{b}
\qedhere
\]\end{proof}
Now consider the matrix elements of an operator $A$ in the $\varphi$
basis. Then
\[
\blk{\varphi_k}{A}{\varphi_l} =
\blk{\varphi_k}{U^{\dagger}UAU^{\dagger}U}{\varphi_l} =
\blk{\psi_k}{UAU^{\dagger}}{\psi_l}
\]
as $U \ket{\varphi_k} = \ket{\psi_k}$. Hence the operator $A$ is given by
$A$ in the $\varphi$ basis and by $UAU^{\dagger}$ in the $\psi$ basis. We
say that $A$ and $UAU^{\dagger}$ are unitary equivalent.
\textbf{Example:} We found $U$ that transforms the $x$-basis to the
$z$-basis for a spin $\frac{1}{2}$ particle. For the $x$-basis we had
\[
\ket{\! \uparrow}_x = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1
\end{pmatrix} \qquad U \ket{\! \uparrow}_x = \ket{\! \uparrow}_z
\]
with
\[
U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}
\]
Now introduce the operator $S_x$ such that
\[
S_x \ket{\! \uparrow}_x = \frac{\hbar}{2} \ket{\! \uparrow}_x
\]
This operator corresponds to the $x$-component of spin angular momentum
and is given in matrix form by
\[
S_x = \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\]
To find $S_z$, we use
\[ S_x \ket{\! \uparrow}_x = \frac{\hbar}{2} \ket{\! \uparrow}_x
\Rightarrow U S_x U^{\dagger} U \ket{\! \uparrow}_x = \frac{\hbar}{2} U
\ket{\! \uparrow}_x \Rightarrow U S_x U^{\dagger}\ket{\! \uparrow}_z =
\frac{\hbar}{2} U \ket{\! \uparrow}_z
\]
so that
\[
S_x \rightarrow US_xU^{\dagger} = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\
0 & -1 \end{pmatrix} = S_z
\]
\begin{thm}
Unitary equivalent observables have the same spectrum, i.e. unitary
transformations do not alter the spectrum of an observable.
\end{thm}
\begin{proof}
Given $A \ket{\varphi_n} = a_n \ket{\varphi_n}$ and $U \ket{\varphi_n} =
\ket{\psi_n}$, then
\[
A U^{\dagger}U \ket{\varphi_n} = a_n \ket{\varphi_n}
\]
and
\[
UAU^{\dagger}U \ket{\varphi_n} = a_n U \ket{\varphi_n}
\]
or
\[
UAU^{\dagger} \ket{\psi_n} = a_n \ket{\psi_n} \qedhere
\]
\end{proof}
\chapter{Quantum Dynamics and Measurement}
\section{Quantum Dynamics}
The eigenstates $\lbrace \ket{a_n} \rbrace$ of an observable $A$ form a
basis in a \hsp. A general state $\ket{\psi(t)} \in \hil$ at time $t$ may
therefore be expanded in $A$'s eigenstates
\[
\ket{\psi(t)} = \sum_n \bk{a_n}{\psi(t)} \ket{a_n}
\]
where $\bk{a_n}{\psi(t)}$ is the probability amplitude $\psi$ is in state
$\ket{a_n}$ at time $t$, and so
\[
\bk{\psi(t)}{\psi(t)} = \sum_n \mds{\bk{a_n}{\psi(t)}} = 1
\]
Now demand that
\[
\bk{\psi(t)}{\psi(t)} = \bk{\psi(t^{\prime})}{\psi(t^{\prime})}
\]
i.e. demand probability conservation. This is ensured if
$\ket{\psi(t+dt)}$ is obtained from $\ket{\psi(t)}$ by a unitary
transformation:
\[
\ket{\psi(t+dt)}= U(dt) \ket{\psi(t)} \qquad U^{\dagger}(dt) = U^{-1} (dt)
\]
For an infinitesimal time displacement write
\[
U(dt) = 1 - \frac{i}{\hbar} H dt
\]
where the operator $H$ is self-adjoint ($H^{\dagger}=H$), so that
$U^{\dagger} = U(dt) = 1 + \frac{i}{\hbar} H dt$ and $U^{\dagger}U = 1 +
)(dt)^2$. By construction $H$ has dimension of energy. Hence we have
\[
\ket{\psi(t+dt)} = \left(1 - \frac{i}{\hbar} H dt\right) \ket{\psi(t)}
\]
and using
\[
\ket{\psi(t+dt)} - \ket{\psi(t)} = \frac{\partial}{\partial t}
\ket{\psi(t)} dt + O(dt)^2
\]
we see that the time development of the state is determined by
\textbf{Schrodinger's equation}
\[
i \hbar \frac{\partial}{\partial t} \ket{\psi(t)} = H \ket{\psi(t)}
\]
We now know that $H$ is the system's Hamiltonian operator. For a single
particle interacting with a potential $V$,
\[
H = \frac{1}{2m} \vec{p}\cdot\vec{p} + V(\vec{x},t)
\]
where $[x_i,p_j] = i \hbar \delta_{ij}$. For an isolated system, $H$ is
time-independent.
\section{Quantum Measurements}
A system is assumed to be prepared in some initial state $\ket{\psi_0}$.
It evolves in time according to Schrodinger's equation, $\ket{\psi_0}
\stackrel{H}{\rightarrow} \ket{\psi}$. We wish to measure an observable
$A$. By the spectral theorem, $\ket{\psi}$ can be expanded in terms of the
observable's eigenvectors, $\ket{\psi_n}$.
\[
\ket{\psi} = \sum_n \bk{\psi_n}{\psi} \ket{\psi_n} \qquad A \ket{\psi_n}
= a_n \ket{\psi_n}
\]
The only possible result of a measurement of $A$ is one of its
eigenvectors. A measurement always causes a system to jump into an
eigenstate of the observable $A$ - this is a postulate of Dirac and von
Neumann. The possibility for jumping into $\ket{\psi_n}$ is
$\mds{\bk{\psi_n}{\psi}}$. We cannot predict which state $\ket{\psi_n}$ a
system will be thrown into by the measurement.
The expectation value of an observable $A$ is
\[
\langle A \rangle \equiv \blk{\psi}{A}{\psi} = \sum_n
\mds{\bk{\psi_n}{\psi}} a_n
\]
where $\bk{\psi}{\psi} = \sum_n \mds{\bk{\psi_n}{\psi}} = 1$.
The experimental significance of $\langle A \rangle$ is claimed to be the
following: consider $N$ identical systems prepared in identical fashion
and let $a_i$ be the value of $A$ measured on the $i^{th}$ member of the
ensemble. Then $\langle A \rangle = \lim_{N\rightarrow \infty} \frac{1}{N}
\sum_{i=1}^N a_i$. Thus $\langle A \rangle$ has a statistical significance
involving many systems or many measurements of one system. The measurement
postulate of Dirac and von Neumann holds for a single solitary event.
\textbf{Examples:}
\begin{enumerate}
\item Suppose $\ket{\psi} = \frac{1}{\sqrt{2}} ( \ket{1} + \ket{2})$. Then
there is a 50-50 chance that a measurement of $\ket{\psi}$ will produce
$\ket{1}$, but one or a few measurements will not establish this. The
probability that $\ket{1}$ will be found $r$ times in $n$ independent
trials is
\[
B(r) = \frac{n!}{r!(n-r)!} p^r q^{n-r}
\]
where $p$ is the probability of measurement $\ket{1}$ and $q$ is the
probability of measurement $\ket{2}$. Here $p=q=\frac{1}{2}$.
For $r$, $n$, $n-r$ large we can use Stirling's formula
\[
B(r) = \frac{1}{\sqrt{2 \pi n p q}} \exp \left( - \frac{(r-np)^2}{2npq}
\right)
\]
This is in accord with the law of large numbers: let $x$ be the number of
successes in $n$ independent trials with constant probability. If
$\epsilon$ is any positive number, then the probability for the inequality
\[
\md{\frac{x}{n} - p} < \epsilon
\]
tends to $1$ as $n \rightarrow \infty$. This implies the relativity
frequency of an event is almost sure to be close to probability for event
when the number of trials is large.
\item A spin zero particle decay into two spin half particles e.g. $\pi^0
\rightarrow e^+ + e^-$ in $(1.8 \pm 0.7) \times 10^{-3}$ \% of decays.
Their state is
\[
\ket{\psi} = \frac{1}{\sqrt{2}} \left(\ket{\!\uparrow}_1
\ket{\!\downarrow}_2 - \ket{\!\downarrow}_1\ket{\!\uparrow}_2 \right)
\]
called an entangled state as it cannot be factorised to $\ket{\psi} =
\ket{\mbox{particle }\,1} \ket{\mbox{particle }\,2}$. This predicts
correlation between spin measurements made on the two entangled state
particles. The two particles always have opposite spin projections when
measured along some quantisation axis. Thus, particle one has a 50-50
chance of having spin-up along, say, the $z$-axis. Suppose observer one
finds spin-up, then particle two must have spin-down along the $z$-axis.
\textbf{Comments:} i) This is not like pulling a white and a black ball
out of a hat. I may decided which axis to measure the spin along long
after the decay. Thus
\[
\ket{\psi} = \bk{\uparrow_z\!}{\psi} \ket{\!\uparrow}_z +
\bk{\downarrow_z\!}{\psi} \ket{\!\downarrow}_z
\]
or
\[
\ket{\psi} = \bk{\uparrow_x\!}{\psi} \ket{\!\uparrow}_x +
\bk{\downarrow_x\!}{\psi} \ket{\!\downarrow}_x
\]
or
\[
\ket{\psi} = \bk{\uparrow_{\hat{n}} \!}{\psi} \ket{\!\uparrow}_{\hat{n}}
+ \bk{\downarrow_{\hat{n}} \!}{\psi} \ket{\!\downarrow}_{\hat{n}}
\]
where $\hat{n}$ is any axis.
ii) If quantum mechanics is a complete theory then Nature allows non-local
effects. Experiments support this.
iii) Another entangled state
PICTURE OF SLITS AND A DECAY
Here $\Omega$ decays into two particles. One can't tell if pair passes
through holes $AA^{\prime}$ or $BB^{\prime}$. The two particle state is
$\ket{\psi} = \frac{1}{\sqrt{2}} \left(\ket{A}_1 \ket{A^{\prime}}_2 +
\ket{B}_1 \ket{B^{\prime}}_2 \right)$.
iv) Schrodinger's comment: Entanglement is ``not one but rather the
characteristic trait of quantum mechanics.''
\item A beam of silver atoms from a Stern-Gerlach apparatus. Ag(Z=47) has
all electron subshells filled except for the 5s electron. We are referring
to this electron in what follows. Let us imagine passing the beam of atoms
through a magnetic field oriented in the $z$-direction.
PICTURE OF STUFF INCLUDING A MAGNETIC FIELD
The emitted Ag has spin $\frac{\hbar}{2}$ up or down. Its magnetic dipole
moment due to 5s electron is $\vec{\mu} = (2) \frac{e}{2mc} \vec{S}$ where
$\vec{S}$ is the electron's spin angular momentum. Its interaction
Hamiltonian in the presence of the magnetic field $\vec{B}$ is $-
\vec{\mu}\cdot \vec{B} = -\frac{e}{mc} \vec{B} \cdot \vec{S}$. Since $e<0$
for an electron, spin up electrons want to avoid strong magnetic fields.
Consider the spin-up ensemble. if we measure the $z$-component of their
spin, $S_z$, we know what the result will be.
PICTURE OF DETECTORS AND STUFF
But suppose we impose an $x$-filter (i.e. another magnetic filed oriented
in $x$-direction) on the $\ket{\!\uparrow}_z$ atoms. We get two beams as a
result.
Spin-up and spin-down in the $x$-direction is a basis set, so we can write
\[
\ket{\!\uparrow}_z = \bk{\uparrow\!}{\!\uparrow}_z \ket{\!\uparrow}_x +
\bk{\downarrow\!}{\!\uparrow}_z \ket{\!\downarrow}_x
\]
where the two probability amplitudes equal $\frac{1}{\sqrt{2}}$. When
$\ket{\!\uparrow}_z$ enters the $x$-filter, we do not know the outcome a
priori. The probability of that result is
\[
\ket{\!\uparrow}_x = \frac{1}{2} \qquad \ket{\!\downarrow}_x = \frac{1}{2}
\]
Note that $\ket{\!\uparrow}_x$ can also be expanded in the $z$-basis.
\end{enumerate}
Let us now consider the time-dependence of the expectation value:
\[
\frac{\partial}{\partial t} \blk{\psi}{A}{\psi} = \left(
\frac{\partial}{\partial t} \bra{\psi}\right) A\ket{\psi} +
\blk{\psi}{\frac{\partial A}{\partial t}}{\psi} + \bra{\psi}A \left(
\frac{\partial}{\partial t} \ket{\psi} \right)
\]
Using the Schrodinger equation,
\[
i \hbar \frac{\partial}{\partial t} \ket{\psi} = H \ket{\psi} \Rightarrow
- i \hbar \frac{\partial}{\partial t} \bra{\psi} = \bra{\psi} H
\]
we get
\[
\frac{\partial}{\partial t} \blk{\psi}{A}{\psi} = \frac{i}{\hbar}
\blk{\psi}{HA}{\psi} - \frac{i}{\hbar} \blk{\psi}{AH}{\psi} +
\blk{\psi}{\frac{\partial A}{\partial t}}{\psi}
\]
or
\[
\frac{\partial}{\partial t} \blk{\psi}{A}{\psi} = \frac{1}{i \hbar}
\blk{\psi}{[A,H]}{\psi} + \blk{\psi}{\frac{\partial A}{\partial t}}{\psi}
\]
so $\langle A \rangle$ is constant if $\frac{\partial}{\partial t}A=0$ and
$[A,H] = AH-HA = 0$.
Consider now a system with observables $A$, $B$ such $[A,B] \neq 0$. The
states of the system can be eigenstates of $A$ \textbf{or} $B$.
\[
A \ket{a} = a \ket{a} \quad \mbox{or} \quad B \ket{b} = b \ket{b}
\]
The states $\ket{a}$, $\ket{b}$ are connected by a unitary transformation,
$\ket{b} = U \ket{a}$. Suppose we insisted on simultaneous states of $A$,
$B$.
\[
AB \ket{a,b} = ab \ket{a,b} \qquad BA \ket{a,b} = ba \ket{a,b}
\]
which implies
\[
(AB-BA) \ket{a,b} = 0
\]
this is not consistent with $[A,B] \neq 0$ for all states. This is a
heuristic result.
Note that elementary quantum precepts insist that we cannot assign
definite values to non-commuting observables. It was not until 1967 that
this was derived directly from elementary quantum precepts.
\section{More dynamics}
\subsection{Stationary states}
A state of an isolated system with a definite energy is called a
stationary state; it is an eigenvector of $H$. The Schrodinger equation
becomes
\[
i \hbar \frac{\partial}{\partial t} \ket{E_H,t} = H \ket{E_H, t} = E_H
\ket{E_H,t}
\]
\[
\Rightarrow \ket{E_H,t} = \exp \left(-\frac{E_H t}{\hbar}\right)
\ket{E_H,0} = \exp \left(-\frac{E_H t}{\hbar}\right) \ket{E_H}
\]
Recall that $c \ket{\psi}$ and $\ket{\psi}$ both represent the same state,
hence this system does not change with time.
Consider the expansion of an arbitrary time-dependent state
$\ket{\psi(t)}$. Expand $\ket{\psi(t)}$ in terms of the eigenstates
$\ket{E_n}$ of $H$:
\[
\ket{\psi(t)} = \sum_n c_n(t) \ket{E_n}
\]
where the expansion coefficients are to be determined. Then,
\begin{align*}
i \hbar \frac{\partial}{\partial t} \ket{\psi(t)} & = \sum_n i \hbar
\dot{c}_n \ket{E_n} \\
& = \sum_n c_n H \ket{E_n} \\
& = \sum_n c_n \ket{E_n}
\end{align*}
We now project out the $n^{th}$ term by taking the scalar product with
$\bra{E_m}$, to get
\[
\sum_n i \hbar \dot{c}_n \underbrace{\bk{E_m}{E_n}}_{\delta_{mn}} = \sum_n
c_n E_n \underbrace{\bk{E_m}{E_n}}_{\delta_{mn}}
\]
\[
\Rightarrow i \hbar \dot{c}_m = E_m c_m \Rightarrow c_m(t) = \exp \left( -
\frac{i E_m t}{\hbar} \right) c_m(0)
\]
and thus
\[
\ket{\psi(t)} = \sum_n c_n(0)\exp \left( - \frac{i E_m t}{\hbar} \right)
\ket{E_n}
\]
The $c_n(0)$'s are fixed by the initial state vector
\[
\ket{\psi(0)} = \sum_n c_n(0) \ket{E_n} \Rightarrow c_n(0) =
\bk{E_n}{\psi(0)}
\]
We might have expanded $\ket{\psi}$ in some other basis that spans $\hil$,
say $\lbrace \ket{n} \rbrace$, and in which $\hil$ is not diagonal. Then
\[
i \hbar \frac{\partial}{\partial t} \bk{m}{\psi(t)} = \sum_n \blk{m}{H}{n}
\bk{n}{\psi(t)}
\]
or
\[
i \hbar \dot{c}_m(t) = \sum_n H_{mn} c_n(t)
\]
where $c_n(t) = \bk{n}{\psi(t)}$. A unitary transformation connects the
two bases,
\[
U \ket{n} = \ket{E_n}
\]
Note that $H_{mn} = \blk{m}{H}{n}$ satisfies $H_{mn}^* = \blk{n}{H^*}{m} =
\blk{n}{H}{m}$ hence $H_{mn}$ is an Hermitian matrix.
\subsection{The time-evolution operator}
We will now take up our earlier remark that time evolution is a unitary
transformation on the states. Recall the Schrodinger equation for a ket
state, $i \hbar \frac{\partial}{\partial t} \ket{a,t} = H \ket{a,t}$,
where $\lim_{t\rightarrow t_0} \ket{a,t} = \ket{a}$. We are interested in
the time evolution of the ket $\ket{a} \stackrel{\mbox{\tiny
time}}{\rightarrow} \ket{a,t}$. Introduce the time evolution operator
$U(t,t_0)$ such that
\[
U(t,t_0) \ket{a} = \ket{a,t}
\]
with $U(t,t_0) = \mathbb{I}$. Substituting this into the Schrodinger
equation we get
\[
i \hbar \frac{\partial U(t,t_0)}{\partial t} = H U (t,t_0)
\]
\textbf{Properties of the time evolution operator:} (we assume $H$ is
Hermitian) i) $U$ is unitary. To see this, consider
\begin{align*}
\frac{\partial}{\partial t} \bk{a,t}{a,t} & = \frac{\partial}{\partial t}
\Big( \bra{a,t} \Big) \ket{a,t} + \bra{a,t} \frac{\partial}{\partial t}
\Big( \ket{a,t} \Big) \\
& = \frac{i}{\hbar} \blk{a,t}{H}{a,t} -
\frac{i}{\hbar}\blk{a,t}{H}{a,t}\\
& = 0
\end{align*}
thus $\bk{a,t}{a,t} = \blk{a,t_0}{U^{\dagger} (t,t_0) U(t,t_0)}{a,t_0} =
\bk{a,t_0}{a,t_0}$. Since the initial state $\ket{a,t_0}$ is arbitrary we
then have $U^{\dagger}(t,t_0)U(t,t_0) = \mathbb{I}$.
Note no assumptions were made about the time dependence of $H$. In the
infinite-dimensional case, $UU^{\dagger} \neq 1$ implies that if one
begins in a basis set $\lbrace \ket{a_n,t_0} \rbrace_{n=1}^{\infty}$ at
$t=t_0$, the set is no longer a basis set for $t>t_0$,
$\sum_{n=1}^{\infty} \ket{a_n,t}\bra{a_n,t} \neq \mathbb{I}$. However if
$H$ is self-adjoint, this cannot happen.
ii) $U$ has the group property, $U(t_2,t_0) = U(t_2,t_1) U(t_1,t_0)$. To
see this, write
\[
\ket{a,t^{\prime}} = U(t^{\prime},t) \ket{a,t} = U(t^{\prime},t) U (t,t_0)
\ket{a,t_0} = U (t^{\prime},t_0) \ket{a,t_0}
\]
hence $U(t^{\prime},t_0) = U(t^{\prime},t) U (t,t_0)$.
\textbf{Construction of the time-evolution operator:} There are three
cases:
\vspace{-0.75em}
\begin{itemize}
\item $H$ is time independent, implying an isolated system. Then
\[
U(t,t_0) = \exp \left(-i\frac{H(t-t_0)}{\hbar}\right)
\]
Note the $(t-t_0)$ time-dependence, characteristic of an isolated system.
\item $H$ is time dependent, but $[H(t), H(t^{\prime})]=0$, then
\[
U(t,t_0) = \exp \left(-\frac{i}{\hbar} \int_{t_0}^t H(t-t_0)\right)
\]
\item $H$ is time dependent, and $[H(t), H(t^{\prime})]\neq0$. (For
example, $H = - \vec{\mu}\cdot\vec{B}(t) = - \frac{eg}{2mc}
\vec{S}\cdot\vec{B}(t)$, where $[S_x,S_y] = i \hbar S_z$ and cyclic
permutations - if $\vec{B}$ changes direction with time then in general
$[H(t), H(t^{\prime})]\neq0$.) In this case we obtain
\[
U(t,t_0) = 1 - \frac{i}{\hbar} \int_{t_0}^t d t^{\prime} H(t^{\prime})
U(t^{\prime},t_0)
\]
Iterating, we get
\[
U(t,t_0) = 1 - \frac{i}{\hbar} \int_{t_0}^t d t^{\prime} H(t^{\prime}) +
\left(-\frac{i}{\hbar}\right)^2 \int_{t_0}^t d t^{\prime}
\int_{t_0}^{t^{\prime}} d t^{\prime\prime} H(t^{\prime}) H(t^{\prime
\prime}) + \dots
\]
a Dyson series. This can be simplified to
\[
U(t,t_0) = T \left( \exp \left[ - \frac{i}{\hbar} \int_{t_0}^t d
t^{\prime} H(t^{\prime}) \right] \right)
\]
where $T$ is the time ordering operator:
\[
T\Big( H(t_1), \dots , H(t_n) \Big) = H(t_i) \dots H(t_j) \dots H(t_k)
\qquad t_i > \dots > t_j > \dots > t_k
\]
\end{itemize}
\section{Schrodinger and Heisenberg descriptions}
There are several approaches to quantum dynamics. The main two that we
will consider are as follows:
\vspace{-0.75em}
\begin{itemize}
\item \textbf{The Schrodinger approach:} bras and kets evolve in time,
observables remain fixed, thus $U(t,t_0) \ket{\alpha} = \ket{\alpha,t}$.
\item \textbf{The Heisenberg approach:} bras and kets remain fixed,
observables evolve in time.
\end{itemize}
Thus consider the matrix elements of an observable $X$:
\[
\blk{\alpha}{X}{\beta} \stackrel{\mbox{\scriptsize time}}{\rightarrow}
\blk{\alpha,t}{X}{\beta,t} = \blk{\alpha}{U^{\dagger}(t,t_0) X
U(t,t_0)}{\beta}
\]
so time-evolution can be viewed equally well as
\[
X \rightarrow U^{\dagger}(t,t_0) X U(t,t_0) = X(t)
\]
with $X(t_0) = X$, and $\ket{\alpha}, \ket{\beta}$ fixed. We can write
this equation as
\[
H_{H}(t) = U^{\dagger}(t,t_0) X_S U(t,t_0)
\]
Note that commutation relations are preserved, if $[X_S,P_S] = i \hbar$
then $ U^{\dagger}(t,t_0)[X_S,P_S]U(t,t_0) = [X_H(t), P_H(t)] = i \hbar$.
\subsection{Heisenberg equation of motion}
Consider the simplest case where $H$ is constant and self-adjoint, and let
$X_S$ have no explicit time-dependence. Then
\[
U(t,t_0) = \exp \left(-i\frac{H(t-t_0)}{\hbar}\right)
\]
and
\begin{align*}
\frac{d}{dt}X_H(t) & = \dot{U}^{\dagger} X_S U + U^{\dagger} X_S \dot{U}
\\
& = - \frac{1}{i \hbar} U^{\dagger}HUU^{\dagger}X_SU + \frac{1}{i \hbar}
U^{\dagger}X_S UU^{\dagger}X_SU\\
& = - \frac{1}{i \hbar} H X_H + \frac{1}{i \hbar} X_H H \\
\end{align*}
and so we get the Heisenberg equation of motion
\[
i \hbar \frac{d}{dt} X_H(t) = [X_H, H]
\]
\textbf{Exercise:} Suppose $H$ is explicitly time-dependent. From
$X_H(t,t_0) = U^{\dagger}(t,t_0) X U(t,t_0)$ show that
\[
i \hbar \frac{d}{dt} X_H(t) = [X_H(t,t_0), H_H(t,t_0)]
\]
where $H_H(t,t_0) = U^{\dagger}(t,t_0) H_S(t) U(t,t_0)$.
\textbf{Example (harmonic oscillator):} A harmonic oscillator with unit
mass and frequency $\omega$ has Hamiltonian
\[
H = \frac{1}{2} p^2 + \frac{1}{2} \omega^2 x^2 \qquad [x,p] = i \hbar
\]
We assume the oscillator is isolated $\Rightarrow H(t) = H(0) = H$.
Setting $t_0 = 0$ we define the Heisenberg operators
\[
p(t) = U^{\dagger}p \,U(t) = \exp \left(\frac{i H t}{\hbar}\right) p \exp
\left(\frac{- i H t}{\hbar}\right)
\]
\[
x(t) = U^{\dagger}x \,U(t) = \exp \left(\frac{i H t}{\hbar}\right) x \exp
\left(\frac{- i H t}{\hbar}\right)
\]
and then clearly $\exp \left(\frac{i H t}{\hbar}\right) [x,p] \exp
\left(\frac{- i H t}{\hbar}\right) = [x(t),p(t)] = i \hbar$. We now wish
to find $x(t)$ and $p(t)$ explicitly.
First we write the Heisenberg equations of motion for the operators
\[
i \hbar \dot{p}(t) = [p(t),H] = - i \hbar \omega^2 x(t)
\]
\[
i \hbar \dot{x}(t) = [x(t),H] = i \hbar p(t)
\]
using the commutation relation and our Hamiltonian. Now let
\[
a(t) = \frac{p(t) - i \omega x(t)}{\sqrt{2 \hbar \omega}}
\]
then
\[
\dot{a}(t) = \frac{\dot{p}(t) - i \omega \dot{x}(t)}{\sqrt{2 \hbar
\omega}} = \frac{- \omega^2 x(t) - i \omega p(t) }{\sqrt{2 \hbar \omega}}
= -i \omega a(t)
\]
\[
\Rightarrow a(t) = a(0) \exp \left(-i \omega t\right) = \frac{p - i \omega
x}{\sqrt{2 \hbar \omega}} \exp \left(-i \omega t\right)
\]
where $p=p(0)$, $x=x(0)$. We then find that
\[
p(t) = \sqrt{\frac{\hbar \omega}{2}} \Big(a(t) + a^{\dagger}(t)\Big) = p
\cos \omega t - \omega x \sin \omega t
\]
\[
x(t) = i \sqrt{\frac{\hbar}{2 \omega}} \Big(a(t) - a^{\dagger}(t)\Big) = x
\cos \omega t - \frac{p}{\omega} \sin \omega t
\]
where $[x,p]=i \hbar$.
\emph{Notes:} i) Consider the classical equation $\ddot{x} + \omega^2 x =
0$, with initial conditions $x(0) = x$, $\dot{x}(0) = \dot{p}(0) = p$, and
solution $x(t) = x \cos \omega t - \frac{p}{\omega} \sin \omega t$. We see
that the quantum result is obtained by replacing $x$, $p$ with operators
such that $[x,p]=i \hbar$.
ii) Note that
\[
[x(t), x(t^{\prime})] = \frac{i \hbar}{\omega} \sin [ \omega(t-t^{\prime})
]\qquad [x(t), p(t^{\prime})] = i \hbar \cos [ \omega(t-t^{\prime}) ]
\qquad [p(t), p(t^{\prime})] = i \hbar \omega \sin [ \omega(t-t^{\prime})
]
\]
\chapter{Wave Mechanics}
\section{Coordinate and momentum representations}
Introduce the notation $\bk{\vec{r}}{\psi} = \psi(\vec{r}) = $ wave
function. Here $\bk{\vec{r}}{\psi} \equiv \bk{x,y,z}{\psi}$. Express
$\psi$ in terms of its Fourier transform:
\[
\psi(\vec{r}) = \frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3k e^{i
\vec{k}\cdot\vec{r}} \hat{\psi}(\vec{k})
\]
\[
\Rightarrow \hat{\psi}(\vec{k}) = \frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3r
e^{-i \vec{k}\cdot\vec{r}} \psi(\vec{r})
\]
Define the symbols
\[
\bkk{\vec{k}}{\psi} = \hat{\psi}(\vec{k})
\]
\[
\bkk{\vec{r}}{\vec{k}} = \frac{1}{(2\pi)^{\frac{3}{2}}} e^{i
\vec{k}\cdot\vec{r}}
\]
\[
\bkk{\vec{k}}{\vec{r}} = \frac{1}{(2\pi)^{\frac{3}{2}}} e^{-i
\vec{k}\cdot\vec{r}} \Rightarrow \bkk{\vec{r}}{\vec{k}}^* =
\bkk{\vec{k}}{\vec{r}})
\]
The Fourier transforms can then be written as
\[
\bk{\vec{r}}{\psi} = \int d^3k \, \bkk{\vec{r}}{\vec{k}}
\bkk{\vec{k}}{\psi}
\]
\[
\bkk{\vec{k}}{\psi} = \int d^3r \, \bkk{\vec{k}}{\vec{r}}
\bk{\vec{r}}{\psi}
\]
This suggests the completeness relations
\[
\int d^3k \, \ket{\vec{k}}\bra{\vec{k}} = 1
\]
\[
\int d^3r \, \ket{\vec{r}}\bra{\vec{r}} = 1
\]
Note that
\[
\int d^3k \, \bkk{\vec{r}}{\vec{k}} \bkk{\vec{k}}{\vec{r^{\prime}}} =
\frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3k e^{i
\vec{k}\cdot(\vec{r}-\vec{r}^{\prime})} = \delta(\vec{r} -
\vec{r}^{\prime})
\]
\[
\int d^3r \, \bkk{\vec{k}}{\vec{r}} \bkk{\vec{r}}{\vec{k}^{\prime}} =
\frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3k e^{-i
\vec{k}\cdot(\vec{k}-\vec{k}^{\prime})} = \delta(\vec{k} -
\vec{k}^{\prime})
\]
so we have the orthogonality relations
\[
\bk{\vec{r}}{\vec{r}^{\prime}} = \delta (\vec{r} - \vec{r}^{\prime})
\qquad \bkk{\vec{k}}{\vec{k}^{\prime}} = \delta (\vec{k} -
\vec{k}^{\prime})
\]
The completeness relations, above, imply that
\[
\ket{\psi} = \int d^3k \, \ket{\vec{k}}\bkk{\vec{k}}{\psi} \quad
\mbox{and} \quad \ket{\psi} = \int d^3r \, \ket{\vec{r}}\bk{\vec{r}}{\psi}
\]
This suggests the following interpretation: $\ket{\vec{k}}$ and
$\ket{\vec{r}}$ are basis vectors in which the state $\ket{\psi}$ can be
represented. This suggests that $\bk{\vec{r}}{\psi}$ and
$\bkk{\vec{k}}{\psi}$ are scalar products of the state vector $\ket{\psi}$
and the $\vec{r}$ and $\vec{k}$ bases:
\[
\bk{\vec{r}}{\psi} = \psi(\vec{r}) = \mbox{coordinate representation of
the state } \ket{\psi}
\]
\[
\bkk{\vec{k}}{\psi} = \psi(\vec{k}) = \mbox{momentum representation of
the state } \ket{\psi}
\]
\textbf{Note:} the vector $\ket{\psi}$ is an element of a Hilbert space
$\hil$. The generalised eigenvectors $\ket{\vec{r}}$ ($\vec{r}$ can take
continuous values) are not vectors in $\hil$ (they have infinite norms).
Rather, they are continuous linear functionals on $\hil$, i.e. mappings
from $\hil$ to the complex numbers.
It is convenient to interpret $\ket{\vec{r}}$ and $\ket{\vec{k}}$ as
eigenstates of position and $\frac{\mbox{\scriptsize momentum}}{\hbar}$:
\[
\vec{r}_{op} \ket{\vec{r}} = \vec{r} \ket{\vec{r}} \qquad \vec{k}_{op}
\ket{\vec{k}} = \vec{k} \ket{\vec{k}} \quad \vec{p}_{op} \equiv \hbar
\vec{k}_{op}
\]
Here $\vec{r}_{op} \equiv x_{op} \hat{i} + y_{op} \hat{j} + z_{op}
\hat{k}$, $x_{op} \ket{\vec{r}} = x \ket{\vec{r}}$ etc.
\textbf{Example:} In the $\vec{r}$-representation, we have
\begin{align*}
\blk{\vec{r}}{\vec{P}_{op}}{\vec{r}^{\prime}} & = \hbar
\blk{\vec{r}}{\vec{k}_{op}}{\vec{r}^{\prime}} \\
& = \hbar \int d^3 k \, \blk{\vec{r}}{\vec{k}_{op}}{\vec{k}}
\bkk{\vec{k}}{\vec{r}^{\prime}} \\
& = \hbar \int d^3 k \, \vec{k} \bkk{\vec{r}}{\vec{k}}
\bkk{\vec{k}}{\vec{r}^{\prime}} \\
& = \hbar \frac{1}{(2\pi)^{\frac{3}{2}}} \int d^3 k \, \vec{k}
e^{i\vec{k} \cdot ( \vec{r} - \vec{r}^{\prime})}\\
& = \frac{\hbar}{i} \frac{1}{(2\pi)^{\frac{3}{2}}} \vec{\nabla}_r \int
d^3 k \, e^{i\vec{k} \cdot ( \vec{r} - \vec{r}^{\prime})}\\
& = \frac{\hbar}{i} \vec{\nabla}_r \delta ( \vec{r} - \vec{r}^{\prime})\\
\end{align*}
hence the coordinate represention of $\vec{p}_{op}$ is the gradient
operator
\[
\frac{\hbar}{i} \vec{\nabla}_r = \frac{\hbar}{i} \left( \hat{i}
\frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} +
\hat{k} \frac{\partial}{\partial z} \right)
\]
Thus we have
\[
[x_{op,i}, p_{op,j}] = i \hbar \delta_{ij}
\]
as
\[
[x_i,p_j] = \left[x_i, \frac{\hbar}{i} \frac{\partial}{\partial
x_j}\right] = i \hbar \delta_{ij}
\]
Likewise in $\vec{k}$-space
\[
\blk{\vec{k}}{\vec{p}_{op}}{\vec{k}^{\prime}} = \hbar \vec{k}^{\prime}
\delta ( \vec{k} - \vec{k}^{\prime} ) = \hbar \vec{k} \, \delta ( \vec{k}
- \vec{k}^{\prime} )
\]
and so the momentum representation of $\vec{p}_{op}$ is $\hbar \vec{k}
\equiv \vec{p} = $ momentum.
\textbf{Exercise:} Show that the momentum space representation of
$\vec{r}_{op}$ is
\[
i \hbar \vec{\nabla}_p = i \hbar \left(\hat{i} \frac{\partial}{\partial
p_x} + \hat{j} \frac{\partial}{\partial p_y} + \hat{k}
\frac{\partial}{\partial p_z} \right)
\]
Note also that the de Broglie wavelength is $\lambda = \frac{h}{p}$, so
from $\md{\vec{p}} = \hbar |\vec{k}| = \frac{h}{2 \pi} |\vec{k}|$ we get
$|\vec{k}| = \frac{2\pi}{\lambda} = $ wave number.
\section{Schrodinger's wave equation}
Consider the Schrodinger equation, $i \hbar \frac{\partial}{\partial t}
\ket{\psi(t)} = H \ket{\psi(t)}$. For a single particle interacting with a
potential $V$, we have $H = \frac{1}{2m} p_{op}^2 + V(\vec{r}_{op},t)$.
Going to the $\vec{r}$ basis, we have
\[
i \hbar \frac{\partial}{\partial t} \bk{\vec{r}}{\psi(t)} =
\blk{\vec{r}}{H}{\psi(t)} = \int d^3r^{\prime}
\blk{\vec{r}}{H}{r^{\prime}}\bk{r^{\prime}}{\psi(t)}
\]
Let us now calculate the matrix element
\[
\blk{\vec{r}}{H}{r^{\prime}} = \frac{1}{2m}
\blk{\vec{r}}{p_{op}^2}{r^{\prime}} +
\blk{\vec{r}}{V(\vec{r}_{op},t)}{r^{\prime}}
\]
We have $\blk{\vec{r}}{V(\vec{r}_{op},t)}{r^{\prime}} = V(\vec{r},t)
\delta(\vec{r} - \vec{r}^{\prime})$, while
\begin{align*}
\blk{\vec{r}}{p_{op}^2}{r^{\prime}} & = \int d^3r^{\prime \prime}
\blk{\vec{r}}{p_{op}}{r^{\prime\prime}}\blk{\vec{r}^{\prime
\prime}}{p_{op}}{r^{\prime}}\\
& = \left(\frac{\hbar}{i}\right)^2 \int d^3r^{\prime \prime}
\vec{\nabla}_r \delta(\vec{r} - \vec{r}^{\prime\prime})
\vec{\nabla}_{r^{\prime \prime}} \delta(\vec{r}^{\prime\prime} -
\vec{r}^{\prime})\\
& = \hbar^2
\end{align*}
\end{document}
