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\begin{document}

\section{Group Velocity}

In quantum mechanics we deal with plane waves that are expressed as
complex exponentials rather than sines and cosines.  We first explain
the mathematics of complex exponentials and Fourier transforms.  We
then go on to the idea of group velocity, which is the velocity at
which a wave packet moves.  We end with the definition of a special
function of great importance in quantum mechanics called the Dirac
delta function.

\subsection{Mathematical Interlude}

\subsubsection{Euler's Equation}

An important mathematical result is that
\begin{equation}
     \exp (i \phi ) = \cos ( \phi ) + i \sin (\phi ) .
\label{euler}
\end{equation}
This is called Euler's equation and may be demonstrated by comparing
the Taylor expansions of both sides term by term.  The Taylor
expansion of the complex exponential is
\begin{equation}
     \exp (i \phi ) = 1 + (i \phi ) + (i \phi )^2 /2 + (i \phi )^3 /6
	\ldots = 1 - \phi^2 /2 + i( \phi - \phi^3 /6 ) \ldots
\end{equation}
which upon comparison with
\begin{equation}
     \cos ( \phi ) = 1 - \phi^2 /2 \ldots
\end{equation}
and
\begin{equation}
     \sin ( \phi ) = \phi - \phi^3 /6 \ldots
\end{equation}
shows that equation \ref{euler} is true.  It is easy to show from
equation \ref{euler} that
\begin{equation}
     \cos \phi = \frac{ \exp (i \phi ) + \exp (-i \phi )}{2}
\end{equation}
and
\begin{equation}
     \sin \phi = \frac{ \exp (i \phi ) - \exp (-i \phi )}{2i} .
\end{equation}

\subsubsection{Fourier Transform}

In this section we extend the ordinary Fourier series representation
of a function in two ways, first by converting it into a series of
complex exponentials, second by making it cover an infinite domain.

The ordinary Fourier series representation of a function $f(x)$ which
is defined over $-L/2 \le x \le L/2$ is
\begin{equation}
	f(x) = A_0 + \sum_{m = 1}^{\infty} [
	A_m \cos (2 \pi mx/L) + B_m \sin (2 \pi mx/L)] ,
\label{fseries}
\end{equation}
where the $A_i$ and $B_i$ are constants which depend on the form of
the function.  This representation turns out to be equivalent to
\begin{equation}
	f(x) = \sum_{m = -\infty}^{\infty} C_m \exp (2 \pi i mx/L) ,
\label{cseries}
\end{equation}
where $i = (-1)^{1/2}$ and the $C_i$ form a possibly complex set of
constants.

Equation (\ref{cseries}) may be written
\begin{equation}
	f(x) = C_0 + \sum_{m = 1}^{\infty} [
	C_m \exp (2 \pi i mx/L) + C_{-m} \exp (-2 \pi i mx/L)] .
\end{equation}
Using Euler's equation, this may be rewritten
\begin{eqnarray}
	f(x) = C_0 & + & \sum_{m = 1}^{\infty} [
	(C_m + C_{-m} ) \cos (2 \pi mx/L) \nonumber \\
	& & + i(C_m - C_{-m} ) \sin (2 \pi mx/L) ] .
\end{eqnarray}
This is equivalent to the ordinary Fourier series if we identify $A_0
= C_0$, $A_m = C_m + C_{-m}$, and $B_m = i(C_m - C_{-m} )$.

The inverse transformation may be obtained by multiplying equation
(\ref{cseries}) by $\exp (-2 \pi i n x/L)$ and integrating over the
domain $[-L/2,L/2]$.  All terms in the sum are zero after integration
except $m = n$, which means that we can solve the result for $C_n$:
\begin{equation}
	C_n = \frac{1}{L} \int_{-L/2}^{L/2} f(x) \exp (-2 \pi i nx/L) dx .
\label{ccoefs}
\end{equation}

We now redefine $C_m$ in preparation for extension to an infinite
domain.  We first define $k_n = 2 \pi n/L$ and $\Delta k = 2 \pi/L$.
We then write $\tilde{f} (k_m ) = (2 \pi )^{1/2} C_m / \Delta k$, so
that equations (\ref{cseries}) and (\ref{ccoefs}) become
\begin{equation}
	f(x) = \frac{1}{(2 \pi )^{1/2}} \sum_{m = -\infty}^{\infty}
		\tilde{f} (k_m ) \exp (ik_m x) \Delta k
\label{cseries2}
\end{equation}
and
\begin{equation}
	\tilde{f} (k_n ) = \frac{1}{(2 \pi )^{1/2}} \int_{-L/2}^{L/2}
		f(x) \exp (-i k_n x) dx .
\label{ccoefs2}
\end{equation}

Finally, we let $L \rightarrow \infty$ so that the difference $\Delta
k$ becomes the differential $dk$ and the sum in equation
(\ref{cseries2}) becomes an integral:
\begin{equation}
	f(x) = \frac{1}{(2 \pi )^{1/2}}
	\int_{-\infty}^{\infty} \tilde{f} (k) \exp (ikx) dk
\label{inverse}
\end{equation}
and
\begin{equation}
	\tilde{f} (k) = \frac{1}{(2 \pi )^{1/2}} \int_{-\infty}^{\infty}
		f(x) \exp (-ikx) dx .
\label{forward}
\end{equation}
These equations are respectively the \emph{inverse and forward Fourier
transforms}.  They show that any waveform can be expressed as a
superposition of complex exponential plane waves.  The parameter $k$
is just the wavenumber, i. e., $2 \pi$ over the wavelength, associated
with each wave.

\subsection{Plane Waves and Wave Packets}

In quantum mechanics we must represent a plane wave by a complex
exponential function rather than by a sine or a cosine.  Thus,
\begin{equation}
     \psi = \exp [i(kx - \omega t)]
\end{equation}
represents the probability amplitude for a particle of definite
wavenumber $k$ and angular frequency $\omega$.  The phase speed of
this wave is $v_{\phi} = \omega / k$.  Notice that the probability of
finding this particle at any point in space is $P = |\psi |^2 = 1$ ---
i. e., there is equal probability of finding the particle anywhere in
the universe.

Quantum mechanical waves obey the {\em superposition principle},
i. e., if $\psi_A$ and $\psi_B$ are solutions to the quantum
mechanical governing equations, then so is the arbitrary linear
combination $A \psi_A + B \psi_B$, where $A$ and $B$ are constants.

Let us see what happens when two waves with slightly different
wavenumber and angular frequency are added  together.  Suppose
$\omega_{1,2} = \omega \pm \Delta \omega$ and
$k_{1,2} = k \pm \Delta k$.  The sum of these two waves becomes
\begin{eqnarray}
     \psi & = & \exp [i(kx + \Delta k x - \omega t - \Delta \omega t)] +
		\exp [i(kx - \Delta k x - \omega t + \Delta \omega t)]
		\nonumber \\
          & = & \exp [i(kx - \omega t)] \left\{
			\exp [i( \Delta k x - \Delta \omega t)] +
			\exp [-i( \Delta k x - \Delta \omega t)]
		\right\} \nonumber \\
          & = & 2 \exp [i(kx - \omega t)]
		 \cos ( \Delta k x - \Delta \omega t) ,
\end{eqnarray}
and the probability distribution is
\begin{equation}
     P = | \psi^2 | = 4 \cos^2 ( \Delta k x - \Delta \omega t ) .
\label{packet}
\end{equation}
Constructive and destructive interference between these two waves
results in a pattern of peaks and valleys in the probability
distribution.  The peaks are called {\em wave packets}, and they move
with a speed called the {\em group velocity},
\begin{equation}
     u = \frac{\Delta \omega}{\Delta k} \rightarrow \frac{d \omega}{dk} ,
\end{equation}
where in the rightmost expression we have taken the limit as
$\Delta k$ becomes very small.

\begin{figure}
\psfig{figure=beats.eps}
\caption{Example of adding two sine waves, the upper with $k = 0.9$
and the lower with $k = 1.1$.  The middle plot shows the sum of the
two.}
\label{beats}
\end{figure}

Figure \ref{beats} shows the results of adding two sine waves together
of slightly differing wavenumber.  This isn't precisely the same as
adding two complex exponentials, but is close enough for the purposes
of illustration.  Notice that the amplitude of the sum varies
systematically with the phase relationship between the two waves,
resulting in a series of wave packets, as described above.

In this example we have added only two plane waves together.  If many
waves of different wavelength and frequency are added together with
the right amplitude and phase, one can obtain an isolated wave packet
rather than the sequence of wave packets that we see here.  Figure
\ref{waves9} shows an example in which 9 sine waves have been
added together with wavenumbers ranging from $0.8$ to $1.2$

\begin{figure}
\psfig{figure=packet.eps}
\caption{Example of adding nine sine waves with wavenumbers ranging
from $0.8$ to $1.2$.}
\label{waves9}
\end{figure}

Physically, the wave packet represents a particle whose location has
been restricted to some finite region, and the group velocity is the
velocity with which this region moves.  The group velocity depends on
the functional relationship between the angular frequency and the
wavenumber.  For historical reasons this relationship is called the
{\em dispersion relation}.  Note that the phase and group velocities
are only equal when the relationship between the frequency and
wavenumber takes the form $\omega = v_{\phi} k$.

\subsection{Integral Representation of a Wave Packet and Dirac Delta Function}

We can represent a wave packet analytically as a superposition of an
infinite number of plane waves by replacing the sum by an integral:
\begin{equation}
	\psi (x,t) = \frac{1}{(2 \pi )^{1/2}} \int_{-\infty}^{\infty}
		\phi (k) \exp [i(kx - \omega t)] dk .
\label{packetint}
\end{equation}

If $\phi (k)$ doesn't oscillate rapidly, but has a peak at $k = k_0$
and approaches zero for larger and smaller values of $k$, we can
estimate the group velocity in another way.  Under these conditions
the complex exponential in the integrand of equation (\ref{packetint})
oscillates rapidly for most values of $k$.  This oscillation causes
the contribution of these regions to the integral to be small, because
the positive and negative phases of the integrand tend to cancel.  The
only place where this isn't true is where the oscillation of the
integrand isn't very rapid.  This occurs near the point where
\begin{equation}
	\frac{\partial}{\partial k} (kx - \omega t) = 0 .
\end{equation}
Expanding and rearranging this yields the condition
\begin{equation}
	\left. \frac{\partial \omega}{\partial k} \right|_{k = k_0} =
		\frac{x}{t} .
\end{equation}
The integral is the largest when this minimum in the oscillation of
the complex exponential occurs near the peak in $\phi (k)$, which is
why we have imposed the condition $k = k_0$.

The above result tells us that the wave packet $\psi (x,t)$ has the
largest amplitude where $x = ( \partial \omega / \partial k)_{k = k_0}
t$, which simply indicates that the wave packet moves at velocity $(
\partial \omega / \partial k)_{k = k_0}$.  However, this is precisely
the previously defined group velocity.

At time $t = 0$ equation (\ref{packetint}) reduces to an ordinary
inverse Fourier transform
\begin{equation}
	\psi (x,0) = \frac{1}{(2 \pi )^{1/2}}
	\int_{-\infty}^{\infty} \phi (k) \exp (ikx) dk ,
\end{equation}
which has an associated forward transform
\begin{equation}
	\phi (k) = \frac{1}{(2 \pi )^{1/2}} \int_{-\infty}^{\infty}
		\psi (x,0) \exp (-ikx) dx .
\end{equation}
Substituting this into equation (\ref{packetint}) and replacing $x$
by $x'$ to avoid confusion, we obtain
\begin{equation}
	\psi (x,t) = \int_{-\infty}^{\infty} \left( \frac{1}{2 \pi}
		\int_{-\infty}^{\infty} \psi (x' ,0) \exp (-ikx' ) dx'
		\right) \exp [i(kx - \omega t)] dk .
\end{equation}

By interchanging the order of integration, we come up with an equation
for $\psi (x,t)$ in terms of $\psi (x,0)$ and a function $K(x,t)$
which is independent of $\psi$:
\begin{equation}
	\psi (x,t) = \int_{-\infty}^{\infty} K(x - x' ,t)
		\psi (x' ,0) dx' ,
\label{propagation}
\end{equation}
where
\begin{equation}
	K(x,t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}
		\exp [ikx - i \omega (k) t] dk .
\end{equation}
This function, called the \emph{propagator}, depends only on the
dispersion relation for the waves, $\omega (k)$, and contains all
information needed to compute the evolution of a wave packet from its
initial state at time $t = 0$ to an arbitrary time in the future.

\begin{figure}
\begin{center}
\psfig{figure=dirac.eps}
\end{center}
\caption{Plot of $\epsilon / [ \pi (x^2 + \epsilon^2 )]$ for various
values of $\epsilon$.  The smallest value of $\epsilon$ yields the
highest peak.}
\label{dirac}
\end{figure}

For $t = 0$, $K$ has a special behavior.  At zero time we formally
have
\begin{equation}
	K(x,0) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}
		\exp (ikx) .
\label{kernel}
\end{equation}
This integral is undefined, which means that the interchange of the
order of integration we made to obtain the propagator isn't
mathematically justified.  However, a sleazy mathematical trick allows
us to obtain a definite value of $K$ under these circumstances.  If we
replace $ikx$ by $ikx - \epsilon |k|$, where $\epsilon$ is a small
constant, then the integral in equation (\ref{kernel}) becomes well
defined and takes on the value $\epsilon / [ \pi (x^2 + \epsilon^2
)]$.  This is plotted as a function of $x$ for various values of
$\epsilon$ in figure \ref{dirac}.  Notice that as $\epsilon
\rightarrow 0$, this function approaches zero everywhere except at $x
= 0$, where it approaches infinity.  Furthermore, notice that the $x$
integral of the function is unity for all values of $\epsilon$.  In
the limit as $\epsilon \rightarrow 0$ this function has a special name
--- it is called the \emph{Dirac delta function}, written $\delta
(x)$.  Thus, for $t = 0$ we have $K(x,0) = \delta (x)$.

Let us return to equation (\ref{propagation}) and set $t = 0$.  We
have
\begin{equation}
	\psi (x,0) = \int_{-\infty}^{\infty} \delta (x - x' )
		\psi (x' ,0) dx' .
\end{equation}
Thus, convoluting a Dirac delta function with any other function
returns the original function back again.  How does this work?  Since
the delta function $\delta (x - x' )$ is zero everywhere except at $x'
= x$, we can rewrite the above equation by substituting $x$ for $x'$
in $\psi (x' ,0)$.  This function becomes a constant as far as the
integration goes, and it may therefore be extracted from the integral.
Thus,
\begin{equation}
	\psi (x,0) = \psi (x,0) \int_{-\infty}^{\infty}
	\delta (x - x' ) dx' = \psi (x,0) ,
\end{equation}
where the last step is justified because the integral of the Dirac
delta function is unity.

Though the above analysis is slipshod, there exists a mathematically
sound theory of functions such as the Dirac delta function, and all of
the above results can be rigorously justified as long as $\psi (x'
,0)$ is reasonably smooth.

\end{document}
