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\begin{center}
  {\bf University of Kentucky, Physics 521 \\
    Homework \#14, Rev. A, due Wednesday, 2018-01-31 }
\end{center}

\vspace{.3 in}

{\bf 0.} Griffiths [2ed]
Ch.~4 \#28, %p177 *most general spinor <S>
%      \#33, %p184 *Rabi flipping  see #1
      \#58, %p198 equality uncertainty SxSy
      \#59, %p198 *minimal coupling -> H(A,phi)
      \#61. %p200 *gauge invariance dLambda

{\bf 1. Neutron in a magnetic field}--the neutron is a spin-$\frac12$ particle
with a magnetic moment (for the highest $m$ state) of $\mu_n=-1.91~\mu_N$ in
units of the {\it nuclear magneton} $\mu_N=\frac{e\hbar}{2m_p}$.  The
corresponding operator is $\bs\mu=\mu_n\bs\sigma=\gamma_{n}\bs S$, with
{\it gyromagnetic ratio} $\gamma_n = g_n\mu_N/\hbar$, and dimensionless {\it g-factor}
$g_n=2 \mu_n/\mu_{N}=-3.826$.  Along the $z$-axis, $\mu_{z}=g_{s}\mu_{N}m$.
Due to spin precession 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 initial spin $\bs S_0$ at $t=0$.  Using the following steps, solve
for the same motion in the rotating frame: 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 $\frac{d\bs S'}{dt}$ is with respect to
components in the rotating frame.  Substite $\frac{d\bs S}{dt}$ into
Eq.~\ref{bloch} and show that the form remains the same except for the
replacement of $\bs B$ with the effective field $\bs B'=\bs B+\bs
\omega/\gamma_n$.  Note this field is zero if the frame is rotating at the
Larmor frequency $\bs\omega_L = -\gamma_n \bs B$, and thus the spin remains
constant $\bs S'=\bs S_0$.  Reconcile this picture with 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 $\chi(t)=\exp(-i\bs\omega\cdot\bs S\, t/\hbar)\chi'(t)$ for $\chi'$ in the
rotating frame.  Show that the same effective field $\bs B'=\bs B+\bs
\omega/\gamma_n$ as above appears in the Hamiltonian in the rotating frame if $\bs B$
is parallel to $\bs \omega$.  Thus the state remains constant in the frame
that rotates at the Larmor frequency.  Show that this is 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 $\bh z B_0$ (called a {\it holding field} because it preserves
the spin state).  To make a transition, we must use an oscillatory (RF) field
$B_1(\bh x\cos\omega t + \bh y\sin\omega t)$.  In the rotating frame with
angular velocity $\bh z\omega$, show that the total field is
$\bs B'=\bh z' (B_0+\omega/\gamma_n)+\bh x' B_1$, which is constant.  Let
$\theta$ be the angle between $\bs B'$ and $\bh z$, and let the initial spin
be $\bs S_0=\bh z$.  Show classically that the $z$-component of the spin
varies as
$S_z(t) = S_0(\cos^2\theta + \sin^2\theta \cos \gamma_nB_1t) =
S_0(1-2\sin^2\theta\sin^2(\gamma_n B_1 t/2))$, 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 frequency (or holding field $B_0$) is swept across
% the Larmor 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 trajectory 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 frequency can be swept?
% 
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