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# Phase and Group Velocity

## Application to the SchrÃ¶dinger Equation

The most general (1-dimensional) wave function is a function of the form \(f(kx - \omega t)\), where \(f:\RR\rightarrow\CC\). Any particular value of \(f\), say \(f(c)\), moves to the right at a speed \(\omega/k\), because this is how \(x\) and \(t\) evolve on the line \(kx - \omega t = c\).

In particular a *plane wave*, \(e^{i (kx - \omega t)}\), propagates to the right with speed
\(\omega/k\). In this case, \(k\) is the *wave number*, the number of cycles in a spatial interval
of length \(2\pi\), and \(\omega\) is the (normalized) frequency, the number of cycles in
a temporal interval of length \(2\pi\). From this we have the wavelength \(\lambda = 2\pi/k\)
and the frequency \(\nu = \omega / 2\pi\) and recover the formula
*velocity = frequency \(\times\) wavelength*.

Physically, the refractive index of a medium will vary with the wavelength; this is the
phenomenon which enables a prism to split a beam of white light. Mathematically we
typically make \(\omega\) depend on \(k\) (called the *dispersion relation__), and we can
assemble a _wave packet* of the form:
\[
\Psi(x, t) = \int_{-\infty}^\infty c(k) e^{i (kx - \omega(k) t)} dk
\]

Now suppose that \(c(k)\) is narrowly peaked about some value \(k_0\). If the peak is narrow enough
then \(\omega(k)\) is well-approximated by the first two terms of its Taylor series for
all values of \(k\) for which \(c(k)\) is non-zero. In this case
\[
\begin{align}
\Psi(x, t)
&= \int_{-\infty}^\infty c(k) e^{i (kx - (\omega(k_0) + \omega'(k_0)(k-k_0) )t)} dk \\
&=
e^{i (k_0x - \omega(k_0)t}
\int_{-\infty}^\infty c(k) e^{i ((k-k_0)x - \omega'(k_0)(k-k_0)t)} \\
&=
e^{i (k_0x - \omega(k_0)t}
\int_{-\infty}^\infty c(k) e^{i (x - \omega'(k_0)t)(k-k_0)} \\
&=
e^{i (k_0x - \omega(k_0)t} F(x - \omega'(k_0)t)
\end{align}
\]
The first term of these is a plane wave which propagates with velocity
\(\omega(k_0) / k_0\), and this is the phase velocity. This is modulated by
the seond term, which propagates with velocity \(\omega'(k_0) / 1 = \omega'(k_0)\).
This is the *group velocity*.

For a *free particle* (i.e. \(V(x) = 0\) for all \(x\)) the
time-independent SchrÃ¶dinger Equation
has solutions of the form
\[
\Psi(x,t) = e^{-iEt/\hbar} \psi(x)
\]
where \(\psi\) is a solution of
\[
-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi
\]
and so
\[
\frac{d^2\psi}{dx^2} = -\frac{2mE}{\hbar^2}\psi = - k^2 \psi
\]
where
\[
k^2 = 2mE/\hbar^2.
\]

Thus \(\psi(x) = e^{ikx}\) (traveling in the positive direction; there's also a solution traveling in the negative direction of course), and \[ \Psi(x, t) = e^{i(kx - Et/\hbar)} \] It follows that \[ \omega(k) = \frac{E}{\hbar} = \frac{\hbar k^2}{2m} \]

Thus the *phase* velocity is
\[
\frac{\omega(k)}{k} = \frac{\hbar k}{2m}
\]
and the *group* velocity is
\[
\omega'(k) = \frac{\hbar k}{m}
\]

Now the classical (macro) velocity \(v\) satisfies \(E = \frac{1}{2}mv^2\), and so \[ v^2 = \frac{2E}{m} = \frac{\hbar^2 k^2}{m^2} \] Thus we see the group velocity corresponds to the classical velocity of the particle/wave group, while the phase velocity is off by a factor of \(\sqrt{2}\).