\input phyzzx

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\noindent PHY--396~K.\quad
Problem set \#6.\quad
Due March 11, 1999.
\par\smallskip\hrule\kern1pt\hrule\vbox{}\medskip

\pointbegin
Consider a particle in a box with reflecting walls:
The particle may touch walls but may not penetrate beyond,
$0\le x(t)\le a$ at all times.
\spointbegin
Use path integral formalism to show that
$$
U_{\rm box}(t',x';t_0,x_0)\
= \sum_{n=-\infty}^{+\infty}\sum_\pm
U_{\rm free}(t',(2na\pm x');t_0,x_0)\
\eqn\Neumann
$$
Hint: Think mirrors.

\spoint
Extract the box's eigenstates and eigenenergies
from the evolution kernel~\Neumann\ (use Poisson resummation)
and show that they agree with the Schr\"odinger eigenstates
for the Neumann boundary conditions $\psi'(0)=\psi'(a)=0$.

\spoint
To implement the Dirichle boundary conditions $\psi(0)=\psi(a)=0$
the wave function should change its sign upon each reflection
of a wall.
Let us therefore consider
$$
U'_{\rm box}(t',x';t_0,x_0)\
= \mathop{\setbox0=\hbox{$\displaystyle\int$}\hbox to 1.5pt{\copy0\hss}\box0}
\limits_{x(t_0)=0}^{x(t')=x'}\!\!\!\!\!\!
{\cal D}[x(t)]\,\exp\left({im\over 2\hbar}\!\int\!\!dt\,{\dot x}^2\right)
\times (-1)^{\rm\#reflections} .
\eqn\Dirichle
$$
Express this kernel in terms of the free kernel, resum and show that it
indeed describes a quantum particle in a box with Dirichle boundary
conditions.

\bigskip
\point
Consider a harmonic oscillator subject to a time-dependent external force:
$$
L(x,\dot x,t)\ =\ \coeff{m}{2}\left( \dot x^2 -\omega^2 x^2\right)\
 +\ xf(t).
\eqn\Lagrange
$$
\spointbegin
Using path integral formalism, show that the evolution kernel of this
system is given by
$$
U(t',x';t_0,x_0))\ =\ A(t'-t_0)\times \exp(iS_{\rm cl}/\hbar),
\eqn\OSC
$$
where $S_{\rm cl}$ is the action of the classical trajectory that begins at
$x_0$ at time $t_0$, ends at $x'$ at time $t'$
and obeys the Newton's equation
$m \ddot x(t)=f(t)-m\omega^2 x(t)$ at all times $t_0\le t\le t'$
while $A(t'-t_0)$ is independent of $x_0$, $x'$ or the external force.

\spoint
To calculate the coefficient $A(t'-t_0)$, Fourier transform 
$$
x(t)\ -\ x_{\rm cl}(t)\ = \sum_{n=1}^\infty y_n\,
\sin{\pi n(t-t_0)\over t'-t_0}\,,
\eqno\eq
$$
do a similar transform for a free particle of the same mass
 and compare the two path integrals to show that
$$
{A_{\rm osc}(t'-t_0)\over A_{\rm free}(t'-t_0)}\
= \left[\prod_{n=1}^\infty
	{n^2\Omega^2-\omega^2\over n^2\Omega^2}
	\right]^{-1/2}
\eqno\eq
$$
where $\Omega=\pi/(t'-t_0)$ --- and therefore
$$
A_{\rm osc}(t'-t_0)\
=\ \sqrt{{m\over 2\pi i\hbar}\,{\omega\over\sin\omega(t'-t_0)}}\,.
\eqno\eq
$$

\spoint
Finally, dispense with the external force and consider the free harmonic
oscillator, $f\equiv0$.
Calculate the classical action for going from point 
$x_0$ at time $t_0$ to point $x'$ at time $t'$
and use the evolution kernel~\OSC\ to reproduce the oscillator's
energy spectrum.
\subpar
Hint: For a time-independent Hamiltonian,
$$
\tr(\hat U)\ =\sum_n\exp\left(-iE_n(t'-t_0)/\hbar\right).
\eqn\Hint
$$

\bye
