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\markboth{Homework \#13, Phys623, Spring 1998, Prof.~Yakovenko}
{Homework \#13, Phys623, Spring 1998, Prof.~Yakovenko}
\begin{document}
\thispagestyle{empty}
\begin{center}
\Large\bf This is the last homework!
\end{center}
\noindent
\begin{minipage}[t]{3.5in}
{\bf Homework \#13} --- Phys623 --- Spring 1998 \\
{\bf Deadline: 5 p.m., Monday, May 11, 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*{Density Matrix (Chapter 20.2) and Spin-Dependent Scattering}
\end{center}
\bigskip
\begin{enumerate}
\item {\bf [3 points]} Read the material and do exercise 19.6.3 from
the book by Shankar (a copy of pages 560-561 is enclosed).
\item {\em Adapted from Qualifier, Spring 1988, II-3.}
Consider scattering of two particles with spin 1/2. Assume that
the interaction between the two particles has the form
\begin{equation}
W=U(r)+V(r)\,\hat{\bf s}_1\cdot\hat{\bf s}_2/\hbar^2,
\label{W}
\end{equation}
where $r$ is the distance between the particles, and $\hat{\bf
s}_1$ and $\hat{\bf s}_2$ are the spin operators of the
particles.
\begin{enumerate}
\item {\bf [3 points]} Show that
\begin{eqnarray}
\hat{P}_1&=&\frac34 + \frac{\hat{\bf s}_1\hat{\bf s}_2}{\hbar^2},\\
\hat{P}_0&=&\frac14 - \frac{\hat{\bf s}_1\hat{\bf s}_2}{\hbar^2}
\end{eqnarray}
are the projectors onto the states with the total spin
$S=1$ and $S=0$, respectively. \label{P}
\item {\bf [3 points]} Using results of \ref{P}, show that
potential (\ref{W}) can be represented in the form
\begin{equation}
W=W_1(r)\hat{P}_1 + W_0(r)\hat{P}_0,
\label{W12}
\end{equation}
and find expressions for $W_1(r)$ and $W_0(r)$ in terms
of $U(r)$ and $V(r)$.
\setcounter{continue}{\value{enumii}}
\end{enumerate}
Let $f_0(\theta)$ and $f_1(\theta)$ be the scattering amplitudes
that describe scattering of a spinless particle of the reduced
mass $\mu$ by the potentials $W_0(r)$ and $W_1(r)$
respectively. Let $\delta_l^{(0)}$ and $\delta_l^{(1)}$ be the
partial-waves phase shifts associated with $W_0(r)$ and $W_1(r)$
respectively.
Let us consider the scattering of the two particles in their
center-of-mass reference frame. The particle that approaches
from the left has the spin projection $|{+1/2}\rangle$, whereas
the particle that approaches from the right has the spin
projection $|{-1/2}\rangle$.
Let us assume that the two particles are {\bf distinguishable},
for example, the proton and the electron.
\begin{enumerate}
\setcounter{enumii}{\value{continue}}
\item {\bf [5 points]} Express, in terms of $f_0(\theta)$ and
$f_1(\theta)$, the differential cross section
$d\sigma/d\Omega$ of scattering at the angle $\theta$ with
simultaneous flip of the two spins. That is, the left
particle scatters at the angle $\theta$, and its spin
projection becomes $|{-1/2}\rangle$. The right particle
scatters in the opposite direction, and its spin
projection becomes $|{+1/2}\rangle$. \label{diff}
\item {\bf [5 points]} Express, in terms of the phases
$\delta_l^{(0)}$ and $\delta_l^{(1)}$, the total cross
section $\sigma$ of scattering with simultaneous flip of
the two spins, as described in \ref{diff}.
\setcounter{continue}{\value{enumii}}
\end{enumerate}
Let us assume now that the two particles are {\bf
indistinguishable}, and they are fermions, for example, the two
electrons.
\begin{enumerate}
\setcounter{enumii}{\value{continue}}
\item {\bf [5 points]} Express, in terms of $f_0(\theta)$ and
$f_1(\theta)$, the differential cross section
$d\sigma/d\Omega$ of a scattering process in which a
particle comes out at an angle $\theta$ with the spin
projection $|{-1/2}\rangle$. The angle $\theta$ is defined
in the figure.
\item {\bf [5 points]} Express, in terms of the phases
$\delta_l^{(0)}$ and $\delta_l^{(1)}$, the total cross
section of scattering $\sigma$.
\end{enumerate}
\setlength{\unitlength}{1cm}
\begin{picture}(10,4)(-6,-2)
\put(0,0){\circle{2}}
\put(-3,0){\vector(1,0){2}}
\put(-3.8,-0.1){$|\uparrow\rangle$}
\put(3,0){\vector(-1,0){2}}
\put(3.2,-0.1){$|\downarrow\rangle$}
\put(0.7,0.7){\vector(1,1){1.4}}
\put(0.8,1.8){$|\downarrow\rangle$}
\put(-0.7,-0.7){\vector(-1,-1){1.4}}
\put(-1.2,-2){$|\uparrow\rangle$}
\qbezier(2,0)(2,0.7)(1.4,1.4)
\put(2.2,0.7){$\theta$}
\end{picture}
\item {\bf [7 points]} A particle of spin 1/2 scatters on the
following potential:
\begin{equation}
\hat{U}=U_0(r)+U_1(r)\hat{\mbox{\boldmath$\sigma$}}\cdot\hat{\bf l},
\label{ls}
\end{equation}
where the Pauli matrices $\hat{\mbox{\boldmath$\sigma$}}$ act on
the spin indices of the particle, and $\hat{\bf l}=-i{\bf
r}\times\partial/\partial {\bf r}$ is proportional to the
orbital angular momentum operator relative to the scattering
center. The last term in Eq.\ (\ref{ls}) is due to the
spin-orbital interaction.
In the Born approximation, show that the scattering amplitude
has the form
\begin{eqnarray}
\hat{f}(\theta,\phi)&=&A(\theta)
+B(\theta)\,\mbox{\boldmath$\nu$}\hat{\mbox{\boldmath$\sigma$}},
\label{AB} \\
\mbox{\boldmath$\nu$}&=&{\bf k}_i\times{\bf k}_f
/|{\bf k}_i\times{\bf k}_f|, \label{nu}
\end{eqnarray}
where ${\bf k}_i$ and ${\bf k}_f$ are the initial and final wave
vectors of the scattered particle. They define a plane to which
both ${\bf k}_i$ and ${\bf k}_f$ belong, called the production
plane. The unit vector {\boldmath$\nu$} is perpendicular to the
production plane. The vector {\boldmath$\nu$} introduces a
dependence on the azimuthal angle $\phi$ in Eq.\ (\ref{AB}).
In the Born approximation, express the coefficients $A$ and $B$
in (\ref{AB}) in terms of the Fourier transforms
$\tilde{U}_0({\bf q})$ and $\tilde{U}_1({\bf q})$ of the
potentials in Eq.\ (\ref{ls}). \label{Born}
\item The scattering amplitude (\ref{AB}) is a $2\times2$ matrix with
respect to the spin, so we can write it as $f_{\alpha\beta}$,
where $\alpha$ and $\beta$ are the spin indices taking the
values ``up'' and ``down''. If the spin state of the incident
wave is characterized by a two-component spinor
$\psi_{\beta,i}$, the spin state of the scattered wave is
characterized by the two-component spinor
\begin{equation}
\psi_{\alpha,f}=\sum_\beta f_{\alpha\beta}\psi_{\beta,i}.
\label{psi:fi}
\end{equation}
This simply means that in the presence of the spin, formula
(18.9) from Schwabl should be written as
\begin{equation}
\psi_\alpha({\bf k})=e^{ikz}\psi_{\alpha,i}+\frac{e^{ikr}}{r}
\sum_\beta f_{\alpha\beta}(\theta,\phi)\psi_{\beta,i}.
\label{psi}
\end{equation}
Once we know the final spin state of the scattered particle
(\ref{psi:fi}), we can calculate the cross section of
scattering by summing over the spin indices of the final
state:
\begin{equation}
\frac{d\sigma}{d\Omega}=\sum_\alpha \psi_{\alpha,f}^*\psi_{\alpha,f}
=\sum_{\alpha,\beta,\gamma}\psi_{\beta,i}^*f_{\beta\alpha}^+
f_{\alpha\gamma}\psi_{\gamma,i}={\rm Tr}(\hat{f}^+\hat{f}\,\hat{\rho}_i),
\label{dsdo}
\end{equation}
where we introduced the {\em density matrix} $\hat{\rho}_i$ of
the incident wave (see Ch.\ 20.2):
\begin{equation}
\rho_{\alpha\beta,i}=\psi_{\alpha,i}\psi_{\beta,i}^*.
\label{rhoi}
\end{equation}
The spin polarization of the scattered wave is characterized by
the ratio of the spin density to the total density:
\begin{equation}
{\bf P}_f=\frac{\sum_{\alpha,\beta}
\psi_{\alpha,f}^*\mbox{\boldmath$\sigma$}_{\alpha\beta}\psi_{\beta,f}}
{\sum_\alpha \psi_{\alpha,f}^*\psi_{\alpha,f}}=\frac
{{\rm Tr}(\hat{f}^+\hat{\mbox{\boldmath$\sigma$}}\hat{f}\,\hat{\rho}_i)}
{{\rm Tr}(\hat{f}^+\hat{f}\,\hat{\rho}_i)}.
\label{Pf}
\end{equation}
To use formulas (\ref{dsdo}) and (\ref{Pf}), we need to know the
density matrix of the initial state. Formula (\ref{rhoi}) gives
the density matrix in the case of a {\em pure} state, which is
characterized by a single wave function $\psi_\alpha$. The most
general expression for the density matrix is:\
\begin{equation}
\rho_{\alpha\beta}=(\delta_{\alpha\beta}+
{\bf P}\mbox{\boldmath$\sigma$}_{\alpha\beta})/2,
\label{rho1/2}
\end{equation}
where {\bf P} is called the spin polarization vector and is
constrained to $0\leq P\leq 1$ (see Eq.\ (20.37)).
In Problem \ref{Born}, we derived general expression (\ref{AB})
for the scattering amplitude of a spin-1/2 particle. In the Born
approximation, $A$ is real, and $B$ is imaginary, but this is
not necessarily true in a more general case.
\begin{enumerate}
\item {\bf [5 points]} Using Eqs.\ (\ref{AB}) and (\ref{dsdo}),
show that the cross section of scattering is:
\begin{equation}
d\sigma/d\Omega=|A|^2+|B|^2+2{\rm Re}(AB^*)
\mbox{\boldmath$\nu$}{\bf P}_i,
\label{Pi}
\end{equation}
where ${\bf P}_i$ is the initial polarization of the beam
{\bf(see Hints)}.
Thus, in the presence of spin-orbital interaction and
initial beam polarization, the cross section of scattering
depends on the azimuthal angle $\phi$ due to the term
proportional to $\mbox{\boldmath$\nu$}{\bf P}_i$ in Eq.\
(\ref{Pi}). \label{polarized}
\item {\bf [5 points]} {\em Adapted from Qualifier, September
1995, II-4.} \label{unpolarized}
Suppose that initial beam is unpolarized: ${\bf
P}_i=0$. Using Eqs.\ (\ref{AB}) and (\ref{Pf}), show that
the scattered state acquires a spin polarization ${\bf
P}_f$ perpendicular to the production plane {\bf(see
Hints)}:
\begin{equation}
{\bf P}_f=\frac{2{\rm Re}(AB^*)}{|A|^2+|B|^2}\mbox{\boldmath$\nu$}.
\end{equation}
\end{enumerate}
\item {\bf [7 points]} In the Born approximation, find the amplitude
and the differential cross section of scattering of fast
neutrons by the Coulomb field. The neutrons are unpolarized
{\bf(see Hints)}. \label{neutrons}
\item {\em Adapted from Qualifier, Fall 1985, II-3, September 1995,
September 1993, II-4.}
Let us consider scattering of a spin-1/2 particle on a spinless
target from the symmetry point of view.
\begin{enumerate}
\item Assuming and using the rotational invariance and the
parity invariance of the system, prove the following
statements.
\begin{enumerate}
\item {\bf [3 points]} If the initial beam is not
spin-polarized, the final beam may be polarized only
along the vector {\boldmath$\nu$} perpendicular to
the production plane.
\item {\bf [3 points]} The differential cross sections of
scattering to a given angle $\theta$ of the beams
with the initial spin polarizations ${\bf P}_i$
parallel and antiparallel to the initial wave vector
${\bf k}_i$ are the same (and do not depend on the
azimuthal angle $\phi$).
\end{enumerate}
\item {\bf [3 points]} Assuming and using only the rotational
invariance of the system show that the most general form
of the scattering amplitude is
\begin{equation}
\hat{f}=A
+B\,[{\bf k}_i\times{\bf k}_f]\hat{\mbox{\boldmath$\sigma$}}
+C\,{\bf k}_i\hat{\mbox{\boldmath$\sigma$}}
+C'{\bf k}_f\hat{\mbox{\boldmath$\sigma$}},
\label{ABCC'}
\end{equation}
where the coefficients $A$, $B$, $C$, and $C'$
depend only on the angle $\theta$ between ${\bf
k}_i$ and ${\bf k}_f$.
\item {\bf [3 points]} Show that if the system is
parity-invariant, the last two terms in Eq.\ (\ref{ABCC'})
are forbidden, that is $C=C'=0$, so Eq.\ (\ref{ABCC'})
reduces to Eq.\ (\ref{AB}).
\item {\bf [5 points]} Assuming that $C=C'\neq0$ in Eq.\
(\ref{ABCC'}), calculate the asymmetry
$(\sigma_+-\sigma_-)/(\sigma_++\sigma_-)$, where
$\sigma_+$ and $\sigma_-$ are the differential cross
sections of scattering to a given angle $\theta$ of
completely spin-polarized beams with the initial
polarizations ${\bf P}_i$ parallel and antiparallel to the
initial wave vector ${\bf k}_i$.
{\em The asymmetry of scattering is used to experimentally
measure parity nonconservation in weak interactions.}
\end{enumerate}
\item {\bf [5 points]} Let us discuss a possible physical origin
of spin-unpolarized beams. {\em This Problem is about the
density matrix, not about scattering.}
\setlength{\unitlength}{1cm}
\begin{picture}(10,4)(-6,-2)
\put(0,0){\circle{2}}
\put(-0.15,-0.15){O}
\put(-1,0){\vector(-1,0){2}}
\put(-4.5,-0.1){B, $\downarrow$, $\uparrow$}
\put(1,0){\vector(1,0){2}}
\put(3.2,-0.1){A, $\uparrow$, $\downarrow$}
\end{picture}
A spinless particle O decays into two spin-1/2 particles,
A and B, that fly in the opposite directions. The
spin-zero wave function of the system after the decay is
\begin{equation}
|\psi(s_A,s_B)\rangle=\frac{1}{\sqrt{2}}
(|\uparrow\rangle_A|\downarrow\rangle_B
-|\downarrow\rangle_A|\uparrow\rangle_B),
\label{sAsB}
\end{equation}
where $s_A$ and $s_B$ are the spin variable of the
particles A and B taking the values $\uparrow$ and
$\downarrow$. Since the system is in a pure state, the
density matrix is given by Eq.\ (\ref{rhoi}):
\begin{equation}
\rho(s_A,s_B,s'_A,s'_B)=
|\psi(s_A,s_B)\rangle\langle\psi(s'_A,s'_B)|.
\label{rho}
\end{equation}
We use the beam of particles A for our experiments in
scattering, and we don't care what happen to particles
B. (They hit the wall of the reactor and get absorbed.)
So, for description of our experiments we need only the
\emph{reduced} density matrix $\rho_r$, which is obtained
by taking trace over the coordinates of the particle B and
depends only on the coordinates of the particle A:
\begin{equation}
\rho_r(s_A,s'_A)={\rm Tr}_{\scriptscriptstyle B}\rho
=\sum_{\displaystyle s_B}\rho(s_A,s_B,s'_A,s_B).
\label{rhor}
\end{equation}
Calculate $\rho_r$ using Eqs.\ (\ref{rhor}), (\ref{rho}),
and (\ref{sAsB}) and show that $\rho_r$ is the density
matrix of the spin-unpolarized state (Eq.\ (\ref{rho1/2})
with {\bf P}=0).
\end{enumerate}
%\newpage
\label{Hints}
\section*{\centerline{Hints}}
\begin{description}
\item[\ref{polarized}, \ref{unpolarized}] \quad $({\bf
a}\hat{\mbox{\boldmath$\sigma$}}) ({\bf
b}\hat{\mbox{\boldmath$\sigma$}})={\bf a}{\bf b}+ i\,[{\bf
a}\times{\bf b}]\,\hat{\mbox{\boldmath$\sigma$}}$ (Eq.\ (14.3.39)
from Shankar).
\item[\ref{neutrons}] Due to the Lorentz transformation of
electromagnetic field, a magnetic field ${\bf
B}=\mbox{\boldmath$\cal E$}\times{\bf v}/c$, where
$\mbox{\boldmath$\cal E$}=Ze{\bf r}/r^3$ is the Coulomb electric
field and ${\bf v}={\bf p}/m$ is the velocity of the neutron, is
present in the reference frame of the neutron. The magnetic field
interacts with the magnetic moment of the neutron,
$\hat{\mbox{\boldmath$\mu$}}=\mu_n\hat{\mbox{\boldmath$\sigma$}}$,
in the usual manner: $\hat{H}=-\hat{\mbox{\boldmath$\mu$}}{\bf
B}$. Similar consideration is done in the beginning of Sec.\
12.2. Consider the value of the neutron magnetic moment $\mu_n$
to be given and ignore possible complications, such as the Thomas
factor.
\end{description}
\end{document}