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\markboth{Homework \#10, Phys623, Spring 1998, Prof.~Yakovenko}
{Homework \#10, Phys623, Spring 1998, Prof.~Yakovenko}
\begin{document}
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\noindent
\begin{minipage}[t]{3.5in}
{\bf Homework \#10} --- Phys623 --- Spring 1998 \\
{\bf Deadline: 5 p.m., Monday, April 20, 1998.} \\
Return homework in class, by e-mail, or \\
put in the box on the door of Phys 2314.
\end{minipage}
\hfill
\begin{minipage}[t]{2.9in}
\raggedleft
Victor Yakovenko, Assistant Professor \\
Office: Physics 2314 \\
Phone: (301)--405--6151 \\
E-mail: yakovenk@physics.umd.edu
\end{minipage}
\medskip
\centerline{\bf Do not forget to write your name and the homework
number!}
\centerline{Equation numbers with the period, like (3.25), refer to the
equations of Schwabl.}
\centerline{Equation numbers without period, like (5), refer to
the equations of this homework.}
\begin{center}
\section*{Interaction with the Electromagnetic Field (Chapter 16.4)}
\end{center}
\bigskip
\begin{enumerate}
\item {\em Adapted from Qualifier, September 1993, II-2.}
Consider an empty rectangular box (the cavity) of the dimensions
$L_x$, $L_y$, and $L_z$ with perfectly conducting metallic
walls. The interior of the cavity is defined by the
inequalities $0\leq x\leq L_x$, $0\leq y\leq L_y$, and $0\leq
z\leq L_z$.
Electromagnetic field has various modes inside the cavity. One
of the modes is the so-called transverse electric (TE) mode. In
this mode, the electric field of an amplitude ${\cal E}_0$ is
directed along the $z$ axis and is modulated along the $x$ and
$y$ axes, so that it vanishes at the walls of the cavity
parallel to the $z$ axis:
\begin{equation}
\bfE={\bf e}_z{\cal E}_0 \sin\left(\frac{\pi x}{L_x}\right)
\sin\left(\frac{\pi y}{L_y}\right) 2\cos(\omega t),
\label{cTE}
\end{equation}
where $\omega=c\pi\sqrt{1/L_x^2+1/L_y^2}$. Eq.\ (\ref{cTE}) is
written for classical electrodynamics. As explained in Sec.\
16.4.2, in quantum electrodynamics the electric field becomes an
operator written in terms of creation and destruction operators:
\begin{equation}
\hat{\bfE}={\bf e}_z{\cal E}_0 \sin\left(\frac{\pi x}{L_x}\right)
\sin\left(\frac{\pi y}{L_y}\right)
(\hat{a}e^{-i\omega t}+\hat{a}^+e^{i\omega t}).
\label{TE}
\end{equation}
Unlike in classical electrodynamics, the amplitude ${\cal E}_0$
in Eq.\ (\ref{TE}) is not arbitrary, but corresponds, roughly
speaking, to the electric field of a single photon. The value of
${\cal E}_0$ can be determined using the procedure of Sec.\
16.4.2, but we will find it using a shortcut explained below.
\begin{enumerate}
\item {\bf [3 points]} In an electromagnetic wave, the energy
contained in the magnetic field is equal to the energy
contained in the electric field, so the average
electromagnetic energy is equal to
\begin{equation}
E=\frac{1}{4\pi}\int dx\,dy\,dz\,\langle\hat{\bfE}^2\rangle
\label{E^2}
\end{equation}
(see Eq.\ (16.48)). Substitute Eq.\ (\ref{TE}) into Eq.\
(\ref{E^2}) and calculate the average over the vacuum
state $|0\rangle$. The vacuum state is the oscillator
ground state and has the properties
\begin{equation}
\hat{a}|0\rangle=0,\qquad \hat{a}^+|0\rangle=|1\rangle.
\end{equation}
On the other hand, the energy $E$ should be equal to the
ground state energy of the oscillator: $E=\hbar\omega/2$.
Show that this condition requires that ${\cal
E}_0^2=8\pi\hbar\omega/L_xL_yL_z$. Persuade yourself that,
taking into account Eq.\ (16.47a), Eq.\ (\ref{TE}) agrees
with Eq.\ (16.49) up to insignificant redefinition
$\hat{a}\to i\hat{a}$.
\setcounter{continue}{\value{enumii}}
\end{enumerate}
Suppose that initially there are no photons in the cavity.
Suppose there is a hydrogen atom at the center of the cavity in
a highly excited state $|nlm\rangle$ with a very large $n$, $l$
equal to its maximal value $l=n-1$, and $m=0$. A static magnetic
field $B$ along the $z$ axis removes the degeneracy among
magnetic substates. The sizes of the box $L_x$ and $L_y$ are
selected in such a way that the frequency of the TE mode,
$\omega=c\pi\sqrt{1/L_x^2+1/L_y^2}$, is exactly equal to the
frequency of the atomic transition $n\rightarrow n'=n-1$ with
$m'=m$. The Hamiltonian of interaction between the
electromagnetic field and the electron of the hydrogen atom is
\begin{equation}
\hat{H}=-e{\bf r}\cdot\hat{\bfE},
\label{H}
\end{equation}
where {\bf r} is the coordinate of the electron.
\begin{enumerate}
\setcounter{enumii}{\value{continue}}
\item {\bf [3 points]} Show that the following two states of the
system $|\rm atom\otimes \mbox{radiation field}\rangle$
are degenerate (have equal energies) and are connected by
a matrix element of Hamiltonian (\ref{H}):
$|n,n-1,0\rangle\otimes|0\rangle$ and
$|n-1,n-2,0\rangle\otimes|1\rangle$.
\item {\bf [5 points]} Using the solution of Problem 6 from
Homework 8, show that the time evolution of the wave
function of this system is oscillatory and find an
expression for the oscillation frequency $\nu$ in terms of
the electric dipole matrix element $d$ and other
parameters of the problem.
\item {\bf [5 points]} Calculate $d$ taking into account that
the radial wave function is:
\begin{equation}
R_{n,n-1}(r)\propto r^{n-1}e^{-r/na_B},
\end{equation}
where $a_B$ is the Bohr radius.
\item {\bf [3 points]} Compute the numerical value of the
frequency $\nu$ for n=10 assuming that $L_x=L_y=L_z$.
\end{enumerate}
Useful information:
\[ E_n=-\frac{e^2}{2a_Bn^2}=-\frac{13.6\:{\rm eV}}{n^2} \]
\[ \int r^Ne^{-br}\,dr=\frac{N!}{b^{N+1}}\]
\[ \int d\Omega\, Y^*_{l'm'}\cos(\theta)\,Y_{lm}=
\left(\frac{l^2-m^2}{(2l+1)(2l-1)}\right)^{1/2}
\delta_{m,m'}\delta_{l-1,l'} \]
\item {\em Adapted from Qualifier, Fall 1994, January 1998, II-5}
In this problem we explore one of the principles of cooling
atoms by laser radiation (1997 Nobel Prize), the so-called
Doppler cooling (there is also the Sisyphus cooling).
\begin{enumerate}
\item {\bf [5 points]} Consider an atom at rest, which is illuminated
by a laser plane wave of the frequency $\omega$ tuned to the
transition frequency $\omega_0$ between two energy levels of the
atom: $\omega=\omega_0$.
When the atom absorbs a photon of the laser plane wave,
what momentum does the photon transfer to the atom? When
the atom emits a photon (as a spherical wave), does the
atom acquire any momentum? Using the answers to these
questions, calculate the force acting on the atom due to
absorption of the photons in terms of the absorption rate,
\begin{equation}
R(\omega)=B(\omega)I(\omega),
\label{RB}
\end{equation}
$B(\omega)$ is the Einstein coefficient, and $I(\omega)$
is the spectral density of the energy flux of the laser
(see Problem 5 of Homework 9). \label{force}
\item {\bf [5 points]} Now suppose that the atom moves with a
velocity {\bf v} and is illuminated by two laser beams
shining parallel and antiparallel to {\bf v}. Using the
conservation laws of energy and momentum or the Doppler
shift of frequencies, calculate the frequency of a photon
that the atom would need to absorb from the parallel, or
alternatively from the antiparallel, beam in order to make
a transition to the excited state.
Assuming that the momentum transfered by a photon is much
smaller than atom's momentum $mv$, calculate the
difference of the forces exerted on the atom by the two
beams. Express you answer in terms of derivatives the
$dI(\omega)/d\omega$ and $dB(\omega)/d\omega$. Argue that,
if the laser line is narrow, $\Delta\omega\ll\omega_0$,
where $\Delta\omega$ the width is the laser line, than
$dI(\omega)/d\omega$ dominates over $dB(\omega)/d\omega$.
Show that if the laser is tuned below the transition
frequency $\omega_0$, so that $dI(\omega)/d\omega<0$, the
net force exerted by the pair of beams is dissipative,
such that
\begin{equation}
{\bf F}=-\alpha{\bf v}.
\label{F}
\end{equation}
Find an expression for the friction coefficient $\alpha$.
\item {\bf [5 points]} Show that illumination by three pairs of
orthogonal beams decreases temperature $T$ of the atoms
and calculate the cooling rate $(dT/dt)/T$ assuming that
the velocity distribution remains approximately
Maxwellian. Estimate the lowest temperature that can be
achieved by this method. (When does the method stop to be
effective?)
\end{enumerate}
\begin{center}
\section*{The Central Potential II (Chapter 17)}
\end{center}
\item {\bf [10 points]} Schwabl's Problem 17.1.
In part (b), in order to solve the radial Schr\"{o}dinger
equation (6.11), make the substitution
\begin{equation}
u(r)=v(r)\exp(-r^2/2r_0^2),
\end{equation}
where $r_0$ (3.3) is the characteristic length for the
oscillator, and obtain an equation for $v(r)$. Try a solution
for $v(r)$ in terms of a power series:
\begin{equation}
v(r)=r^{l+1}\sum_{n=0}^\infty C_n r^n.
\label{series}
\end{equation}
Find a two-term recursion relation for the coefficients
$C_n$. Determine the energy levels by imposing a condition that
series (\ref{series}) terminates at a term $N$ (the radial
quantum number). Find degeneracies of the energy levels and
compare with those found in part (a).
\item {\bf [7 points]} Schwabl's Problem 17.2.
Don't do part (a), because we have done it already last
semester; do only parts (b) and (c). In part (c), take into
account that a marginally bound state has the energy $E=0$, and
the wave function of this state is given the equation that
precedes Eq.\ (6.13a).
\item {\bf [5 points]} Schwabl's Problem 17.3.
\end{enumerate}
\end{document}