% begin preamble
\documentclass[11pt,titlepage]{article}
\usepackage[margin=1in,noheadfoot]{geometry}
\usepackage[parfill]{parskip}
\usepackage{graphicx}
\usepackage[colorlinks,urlcolor=blue]{hyperref}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{wrapfig}

\newcommand{\e}[1]{\times 10^{#1}}
\newcommand{\q}{\mathrm}
\newcommand{\where}[1][where]{\qquad\mbox{#1}\qquad}
\newcommand{\bs}[1]{\boldsymbol{#1}}
\newcommand{\bh}[1]{\hat{\boldsymbol{#1}}}
\newcommand{\bv}[1]{\vec{\boldsymbol{#1}}}
\newcommand{\bt}[1]{\tilde{\boldsymbol{#1}}}
\newcommand{\ad}{^{\dagger}}
\newcommand{\flow}[1]{\mathcal{E}_{#1}}
\newcommand{\flux}[1]{\Phi_{#1}}
\newcommand{\dslash}{d \hspace{-0.5ex}\rule[1.2ex]{0.6ex}{.1ex}}
\def\rcurs{{\mbox{$\resizebox{.16in}{.08in}{\includegraphics{ScriptR}}$}\!\!}}
\def\brcurs{{\mbox{$\resizebox{.16in}{.08in}{\includegraphics{BoldR}}$}\!\!}}
\def\hrcurs{{\mbox{$\hat \brcurs$}\!}}
%end preamble

\begin{document}
\begin{center}
  {\bf University of Kentucky, Physics 521 \\
    Homework \#11, Rev. A, due Monday, 2016-02-22 }
\end{center}

\vspace{.3 in}

{\bf 0.} Griffiths [2ed]
Ch.~4 \#26, %p177 *Clifford algebra
%      \#27, %p177 expectation value and uncertainty
      \#28, %p177 *most general spinor <S>
%      \#29, %p177 eigenstates of Sy
      \#30, %p178 *eigenstates of nhat
%      \#31, %p178 matrices of S=1
%      \#33, %p184 *Rabi flipping  see #1
      \#34, %p189 *apply S+/- to each state
%      \#36, %p189 S=1+2, probabilities
      \#53, %p195 *matrices for arbitrary S (cf #31)
      \#56, %p196 *Lz generator of rotation
      \#59, %p198 *minimal coupling -> H(A,phi)
      \#61% %p200 *gauge invariance dLambda
.

{\bf 1. Neutron in a magnetic field}--the neutron is spin-$\frac12$ with
magnetic moment $\mu_n=-1.91~\mu_N$ in units of the {\it nuclear magneton}
$\mu_N=\frac{e\hbar}{2m_p}$.  The corresponding operator is
$\bs\mu=\mu_n\bs\sigma$.  The gyromagnetic factor is $\gamma_n =
g_s\mu_N/\hbar$, where the Land\'e factor is $g_s=2 \mu_n$.  Due to spin
precision in the magnetic field, both the classical and quantum mechanical
equations of motion are greatly simplified in a rotating reference frame,
where $\pmatrix {\bh x' & \bh y'}=\pmatrix {\bh x & \bh y} {\scriptstyle
  \pmatrix{\cos\omega t & - \sin\omega t \cr \sin\omega t & \cos\omega
    t}}$. [Rabi, Ramsey, Schwinger, Rev. Mod. Phy. {\bf 26}, 167, 1954]

{\bf \quad a)} Solve the classical equation of motion [see our class notes]
\begin{equation} \label{bloch}
\frac{d\bs S}{dt} = \bs
\tau = \bs \mu_n\times \bs B = \gamma_n \bs S \times \bs B
\end{equation}
for {\it Larmor precession} of a neutron in a constant magnetic field $\bh z
B_0$, with spin $\bs S_0$ at $t=0$.  Since the operator $\bs\omega dt \times$
generates rotation, $\frac{d\bs S}{dt} = \frac{d\bs S'}{dt} + \bs\omega \times
\bs S$, where the second derivative is with respect to components in the
rotating frame.  Show that the effective field appearing in Eq.~\ref{bloch} is
$\bs B'=\bs B+\gamma_n\bs \omega$, which is zero in the frame rotating at the
Larmor frequency $\bs\omega_L = -\gamma_n \bs B$.  Thus the spin remains
constant $\bs S'=\bs S_0$ in this frame.  Reconcile this with \mbox{the
  solution in the static frame.}

{\bf \quad b)} Solve the same precession quantum mechanically [see our class
notes] starting from the initial state $\chi_0={\scriptstyle\pmatrix{a\cr
    b}}$.  Show that the expected value of spin $\langle \bs S\rangle (t)$
agrees with the classical result.  Using the results of Griffiths Ch.~4 \#56,
show that $\chi(t)=\exp(-i\bs\omega\cdot\bs S\, t)\chi'(t)$ for $\chi'$ in the
rotating frame.  Show that the same effective field $\bs B'=\bs B+\gamma_n\bs
\omega$ as above appears in the Hamiltonian in the rotating frame.  Thus the
state remains constant if frame rotates at the Larmor frequency.  Show that
this equivalent to the solution in the static frame.

{\bf \quad c)} Note that the spin state $m$ does not change in a constant
magnetic field $\bh z B_0$, which is thus called a {\it holding field}.  To
make a transition, we use an oscillatory (RF) field $B_1(\bh x\cos\omega t +
\bh y\sin\omega t)$.  Show that the total field is constant $\bs B'=\bh z
(B_0+\omega/\gamma_n)+\bh x B_1$ in the rotating frame.  Let $\theta$ be the
angle between $\bs B'$ and $\bh z$, and let the initial spin be $S_0=\bh z$.
Show classically that the $z$-component of the spin varies as $S_z(t) =
S_0(\cos^2\theta + \sin^2\theta \cos \gamma_nB_1t) =
S_0(1-\sin^2\theta\sin^2(\gamma_n B_1 t/2))$, which oscillates at the {\it
  Rabi flopping frequency} $\gamma_nB_1$.  Plot the amplitude of oscillations
as a function of the RF frequency $\omega$ and note the resonance at
$\omega=\omega_L$.  Thus this is called a resonant transition, and is the
basis of {\it nuclear magnetic resonance} (NMR) a technique used in MRI
machines.  Plot the 3-d trajectory of $\bs S$ in the lab frame over half a
Rabi cycle.

{\bf \quad d)} Solve the same problem quantum mechanically to show that the
probability $|a(t)|^2$ of measuring spin up at time $t$, is the same as the
classical result $S_z(t)/S_0$.

{\bf \quad e) [bonus]} An alternative to resonant transitions (with the RF
frequency tuned to the Larmour frequency for a specific time) are adiabatic
transitions, where the RF frequecy (or holding field $B_0$) is swept across
the Larmour frequency.  If the magnetic field direction varies much slower
than the precession frequency, the direction of $S$ will adiabatically follow
the direction of the field.  Plot the trajector of the total field vector $\bs
B'$ in the rotating frame as the frequency varies from $\omega\ll\omega_L$ to
$\omega\gg\omega_R$ to show that the spin transitions from spin up to spin
down.  What is the limit on how fast the fequency can be swept?

\end{document}
