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\heading{E4730: Particle on a spherical shell, dynamics} 
\auname{Almog Hagag and Anne Hausmann}


{\bf The problem:}

\Dn

A particle with mass $M$ and charge e is placed on a spherical
shell with radius $R$. Note that the standard observation variables
that describe the state of the particle are $\left(\theta,\varphi,L_{x},L_{y},L_{z}\right)$.
In this question we assume that the particle can be excited from the ground
state to the first energy level, but not beyond that, so the states space
is four dimensional. Mark the states of the base with $\left|0\right\rangle ,\left|\uparrow\right\rangle ,\left|\updownarrow\right\rangle$ and $\left|\downarrow\right\rangle$.

\Dn

(1) What is the energy of each of the base states. Define
the ground state as zero-energy.

The particle is prepared as \textquotedblright{}concentrated\textquotedblright{}
as possible at the north pole of the shell.

(2) Write the wave function $\psi\left(\theta,\varphi\right)$ of
the particle.

The state you wrote is not stationary.

The particle oscillates between the \textquotedblright{}north
pole\textquotedblright{} and the \textquotedblright{}south pole\textquotedblright{}.

(3) What is the time period, $T_{0}$, of the oscillations?

First case: We add a constant electric field, $\epsilon$, in the z direction.

(4) Write Formally the perturbation element using the observation
variables.

(5) What is the time period, T, of the oscillations?

Second case: We add a constant magnetic field, B, in the x
direction.

(6) Write Formally the perturbation element using the observation
variables.

(7) Find the value of B so that in Time $T_{0}/2$ the particle
will be in the initial state.


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
The Hamiltonian of a particle with mass M restricted to
move on a spherical shell with radius R is given by:

\[
\hat{H}=\frac{\hat{L^{2}}}{2MR^{2}}
\]

The solutions for this Hamiltonian are the spherical harmonic
functions. We know that the particle can be in the ground state or
the first excited state, and that's why l=0,1. The eigenstates are:

\[
\left|0\right\rangle =Y^{0,0},\left|\updownarrow\right\rangle =Y^{1,0},\left|\uparrow\right\rangle =Y^{1,1},\left|\downarrow\right\rangle =Y^{1,-1}
\]

and the energies are:

\[
E_{l,m}=\frac{l\left(l+1\right)}{2MR^{2}}
\]

In our case we get:

\[
\hat{H}\left|0\right\rangle =0=E_{0,0},\hat{H}\left|\updownarrow\right\rangle =\hat{H}\left|\uparrow\right\rangle =\hat{H}\left|\downarrow\right\rangle =\frac{1}{MR^{2}}=E_{1,0}=E_{1,1}=E_{1,-1}
\]

%%%%%%%%%%%%%%%
\Dn

(2)
The particle is prepared as concentrated as possible at
the ``north pole'' of the shell. That's why we should look for a
function that is a superposition of the eigenfunctions and gets its
maximum absolute value at the ``north pole''. The only function
that satisfies this condition is: $\psi=1+\cos\theta$. Choosing this
function avoids the case that we would get if we choose $\psi=\cos\theta$,
which would give us a maximum in both, the ``north pole'' and the
``south pole''. Let's express the wave function in terms of the
eigenfunctions:

\[
\psi=\sqrt{3}Y^{0,0}+Y^{1,0}
\]

The normalized function is given by:

\[
\psi=\frac{1}{2}\left(\sqrt{3}Y^{0,0}+Y^{1,0}\right)
\]


%%%%%%%%%%%%%%
\Dn

(3)
Let's express $\psi\left(t\right)$ using the evolution
operator, taking in account that $E_{0,0}=0,E_{1,0}=\frac{1}{MR^{2}}$:

\[
\psi\left(t\right)=\frac{1}{2}\left(\sqrt{3}Y^{0,0}+Y^{1,0}e^{-i\frac{1}{MR^{2}}t}\right)
\]

In order to find the period time of the oscillations, we will
use the relation $\omega=E_{1,0}-E_{0,0}=\frac{1}{MR^{2}}$. Using
the fact that $T=\frac{2\pi}{\omega}$, we get:

\[
T_{0}=2\pi MR^{2}
\]


%%%%%%%%%%%%%
\Dn

(4)
Adding an electric field in the z direction creates a
potential $V=-e\epsilon R\cos\theta$ . We want to find this perturbation
matrix elements. Keeping in mind that $V=-e\epsilon R\sqrt{\frac{4\pi}{3}}Y^{1,0}$,
we can realize that this perturbation couples only the eigenfunctions
$Y^{0,0},Y^{1,0}$, and all other couplings vanish. From calculation
of this coupling we get:

\[
\left\langle 0\right|V\left|\updownarrow\right\rangle =\left\langle \updownarrow\right|V\left|0\right\rangle =\int -Y^{0,0}e\epsilon R\sqrt{\frac{4\pi}{3}}Y^{1,0}Y^{1,0}d\Omega=-\frac{e\epsilon R}{\sqrt{3}}
\]


%%%%%%%%%%%%%
\Dn

(5)
According to what we found, we can write the Hamiltonian
matrix as follows:

\[
\hat{H}=\left(\begin{array}{cccc}
0 &- \frac{e\epsilon R}{\sqrt{3}} & 0 & 0\\
-\frac{e\epsilon R}{\sqrt{3}} & \frac{1}{MR^{2}} & 0 & 0\\
0 & 0 & \frac{1}{MR^{2}} & 0\\
0 & 0 & 0 & \frac{1}{MR^{2}}
\end{array}\right)
\]

Examining only the subspace of the states $\left|0\right\rangle ,\left|\updownarrow\right\rangle $we
get:

\[
\hat{H}=\left(\begin{array}{cc}
0 & -\frac{e\epsilon R}{\sqrt{3}}\\
-\frac{e\epsilon R}{\sqrt{3}} & \frac{1}{MR^{2}}
\end{array}\right)
\]

In order to express this Hamiltonian using Pauli matrices, we would like to gauge the Hamiltonian as follows:

\[
\hat{H}=\left(\begin{array}{cc}
-\frac{1}{2MR^{2}} &- \frac{e\epsilon R}{\sqrt{3}}\\
-\frac{e\epsilon R}{\sqrt{3}} & \frac{1}{2MR^{2}}
\end{array}\right)
\]

which yields:

\[
\hat{H}=-\frac{e\epsilon R}{\sqrt{3}}\sigma_{x}-\frac{1}{2MR^{2}}\sigma_{z}
\]

On the other hand, we know that $\hat{H}=\vec{\Omega}\cdot\vec{S}$
and $\vec{S}=\frac{1}{2}\left(\sigma_{x},\sigma_{y},\sigma_{z}\right)$,
which yields:

\[
\vec{\Omega}=\left(-\frac{2e\epsilon R}{\sqrt{3}},0,-\frac{1}{MR^{2}}\right)
\]

In effect, the frequency of the oscillations is the norm of this
vector:

\[
\omega=\left|\vec{\Omega}\right|=\sqrt{\frac{4e^{2}\epsilon^{2}R^{2}}{3}+\frac{1}{M^{2}R^{4}}}
\]

We recall that $T=\frac{2\pi}{\omega}$, and that's why

\[
T=\frac{2\pi}{\sqrt{\frac{4e^{2}\epsilon^{2}R^{2}}{3}+\frac{1}{M^{2}R^{4}}}}
\]

%%%%%%%%%%%%%
\Dn

(6)
Adding a magnetic field, while taking in account that
this is a spinless particle, requires adding of the Zeeman term: 

\[
\hat{H}_{Zeeman}=-\frac{e}{2M}\vec{B}\cdot\vec{L}
\]

Note that we neglect the diamagnetic term, because this is
a second order perturbation.\\

Because the magnetic field is in the x direction, the perturbation
term is:

\[
\hat{V}=-\frac{e}{2M}B\hat{L}_{x}
\]

%%%%%%%%%%%%%
\Dn

(7)
Now, the Hamiltonian is given by:

\[
\hat{H}=\frac{\hat{L^{2}}}{2MR^{2}}-\frac{e}{2M}B\hat{L}_{x}
\]

The term of the state $\left|0\right\rangle $ in the wave
function is not affected by the operator $\hat{L}_{x}$, because it
has a spherical symmetry. Therefore:

\[
\psi\left(t\right)=e^{-i\frac{\hat{L^{2}}}{2MR^{2}}t}e^{i\frac{e}{2M}B\hat{L}_{x}t}\frac{1}{2}\left(\sqrt{3}\left|0\right\rangle +\left|\updownarrow\right\rangle \right)=\frac{1}{2}\left(\sqrt{3}\left|0\right\rangle +e^{-i\frac{1}{MR^{2}}t}e^{i\frac{e}{2M}B\hat{L}_{x}t}\left|\updownarrow\right\rangle \right)
\]

We know that a rotation operator is given by $R=e^{-i\vec{L}\cdot\vec{\phi}}$and
that $\phi=\Omega t$. That's why the term $e^{i\frac{e}{2M}B\hat{L}_{x}t}$
represents a rotation operator around the (-x) axis with $\Omega=\frac{eB}{2M}$.

It's easy to see that at $t=\frac{T_{0}}{2}$ the wave function
is:

\[
\psi\left(t=\frac{T_{0}}{2}\right)=\frac{1}{2}\left(\sqrt{3}\left|0\right\rangle -e^{i\frac{e}{2M}B\hat{L}_{x}t}\left|\updownarrow\right\rangle \right)
\]

In order that at this time the wave function will return to
it's initial state, we need that the rotation operator will rotate
the state $\left|\updownarrow\right\rangle $ by 180 degrees around
the (-x) axis.

Therefore:

\[\Omega\frac{T_{0}}{2}=\pi\]

After substituting we get:

\[\frac{eB}{2M}\frac{2\pi MR^{2}}{2}=\frac{\pi eR^{2}B}{2}=\pi\]

Finally we get:

\[B=\frac{2}{eR^{2}}\]

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