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\heading{E473: Particle on a spherical shell, dynamics} 
\auname{Ohad Akiva}


{\bf The problem:}

\Dn

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

(1) What is the energy of each of the base states. Define the ground state as zero-energy.
\\
\\
The particle is prepared as ''concentrated'' as possible at the ”north pole” of the shell.
\\
(2) Write the wave function ${\psi(\theta,\varphi)}$ of the particle.
\\
\\
The state you wrote is not stationary.
The particle oscillates between the ''north pole'' and the ''south pole''.
\\
(3) what is the 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 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 particale will be in the initial state.
\\


\Dn\Dn


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{\bf The solution:}

\Dn

(1) The Hamiltonian of the particle with mass M restricted to move on a sphere with radius R 
\\is given by:
\[ \hat{H_0}= \frac{\hat{P^2}}{2M}= \frac{\hat{L^2}}{2MR^2} \]

The solutions for this hamiltonian are the ${Y_{l,m}}$ functions.
\\
We know that the particle can be in the ground state or the first state so l=0,1 ,and
the possible energies are:
\\
\\
${E_{l,m}}$=${\frac{l(l+1)}{2MR^2}}$
\\
\\
And the states:
\\
\\
${\mid}$0${\rangle}$=${Y_{0,0}}$ , ${\mid}$${\uparrow}$${\rangle}$=${Y_{1,1}}$ , ${\mid}$${\downarrow}$${\rangle}$=${Y_{1,-1}}$ , ${\mid}$${\updownarrow}$${\rangle}$=${Y_{1,0}}$
\\\\\\\\
so the energy of each of the base states:
\\
\\
${E_0}$=${\hat{H_0}}$${\mid}$0${\rangle}$=0, 
${E_1}$=${\hat{H_0}}$${\mid}$${\uparrow}$${\rangle}$=${\hat{H_0}}$${\mid}$${\downarrow}$${\rangle}$=${\hat{H_0}}$${\mid}$${\updownarrow}$${\rangle}$=${\frac{1}{MR^2}}$
%%%%%%%%%%%%%%%
\Dn
\\
\\
(2)
The particle is prepared as ”concentrated” as possible at the ''north pole'' of the shell.\\
We will look for a function that gets a maximum value on the ''north pole'' (${\theta}$=0).\\
A good guess for that function is cos(${\theta}$) but we need to remember that at the ''south pole'' (${\theta}$=${\frac{\pi}{2}}$)\\ 
cos(${\frac{\pi}{2}}$)=-1 so a better guess is 1+cos(${\theta}$) so now we get 2 at the ''north pole'' and 0 at the ''south pole''.\\
We will write this function using the ${Y_{l,m}}$ base.\\
cos(${\theta}$)=${\sqrt{\frac{4\pi}{3}}}$${Y_{1,0}}$ , 1=${\sqrt{4\pi}}$${Y_{0,0}}$\\ \\
And the Normalized wave function :
\\
${\psi(\theta,\varphi)}$=${\frac{1}{2}}$(${\sqrt{3}}$${Y_{0,0}}$+${Y_{1,0}}$)
\\





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\Dn

(3)
We will write ${\psi(t)}$ by using the evolution operator.\\
${\psi(\theta,\varphi,t)}$=${\frac{1}{2}}$(${\sqrt{3}}$${Y_{0,0}}$${e^{-i{E_0}t}}$+${Y_{1,0}}$${e^{-i{E_1}t}}$)=${\frac{1}{2}}$(${\sqrt{3}}$${Y_{0,0}}$+${Y_{1,0}}$${e^{\frac{-it}{MR^2}}}$)
\\ \\
Now there are two ways to find the period of the oscillations.\\
The easy way is due to the fact we have only two energy levels in this question, so we will use the relation:\\
${\omega}$=${E_1}$-${E_0}$=${\frac{1}{MR^2}}$\\ \\
and\\
${T_0}$=${\frac{2\pi}{\omega}}$=${2\pi MR^2}$
\\ \\
Second way is to calculate the probability of ${\psi(t)}$ to return to ${\psi(t=0)}$.\\
\[ | \langle \psi(0) \left | \psi(t) \right \rangle|^2 = |\frac{1}{2}(\sqrt{3}Y_{0,0}+Y_{1,0})\frac{1}{2}(\sqrt{3}Y_{0,0}+Y_{1,0}e^{\frac{-it}{MR^2}})|^2 \]
\[ = |\frac{1}{4}(3+e^{\frac{-it}{MR^2}})|^2 =\frac{1}{16}(3+e^{\frac{-it}{MR^2}})(3+e^{\frac{it}{MR^2}}) \]
\[ =\frac{1}{16}(10+3(e^{\frac{it}{MR^2}}+e^{\frac{-it}{MR^2}}))=\frac{1}{16}(10+6cos(\frac{t}{MR^2})) \]
And again we can note that:\\
${\omega}$=${\frac{1}{MR^2}}$\\ \\
And get\\
${T_0}$=${\frac{2\pi}{\omega}}$=${2\pi MR^2}$ \\

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

(4)
If we add a constant electric field, ${\epsilon}$ ,in the z direction, we need \\
to add the potential it creates to the Hamiltonian.\\
\[ V(z) \ = -q\epsilon z \] \\
We will use z=Rcos(${\theta}$) and get:\\
\[ V(\theta) \ = -q\epsilon Rcos(\theta) \] \\
And to find V(${\theta}$) components we need to calculate:\\
\[  \langle \ Y_{l`,m`} \left | V(\theta) | \ Y_{l,m} \right \rangle \]
\\
Again there are two possibilities to calculate the V components.\\
First way is to solve all possible integrals for l=1,0 and m=-1,0,1 (if l=0 then m is only 0).\\
In that way we find that only two integrals are different from zero.\\
\[  \langle \ Y_{1,0} \left | V(\theta) | \ Y_{0,0} \right \rangle = \langle \ Y_{0,0} \left | V(\theta) | \ Y_{1,0} \right \rangle = - \frac{q \epsilon R}{\sqrt{3}} \] \\
The seconed way is to define:\\
\[ \hat{V(\theta)} \ = -q\epsilon Rcos(\theta) =  - \sqrt {\frac{4\pi}{3}}q\epsilon RY_{1,0}  \] \\
and to see that:\\
\[ \hat{V(\theta)} \ | \ Y_{0,0} \rangle = - \frac{q\epsilon R}{\sqrt {3}} | \ Y_{1,0} \rangle   \]\\
And again we get:\\
\[  \langle \ Y_{0,0} \left | \hat{V(\theta)} | \ Y_{1,0} \right \rangle = \langle \ Y_{1,0} \left | \hat{V(\theta)} | \ Y_{0,0} \right \rangle = - \frac{q \epsilon R}{\sqrt{3}}\langle \ Y_{1,0} \left | \ Y_{1,0} \right \rangle   = - \frac{q \epsilon R}{\sqrt{3}} \] \\
 
%%%%%%%%%%%%%
\Dn

(5)
Now after we found the V(${\theta}$) components we can write the Hamiltonian matrix.

\[
 H =
 \begin{pmatrix}
  0 & - \frac{q \epsilon R}{\sqrt{3}} \\
  - \frac{q \epsilon R}{\sqrt{3}} & \frac{1}{MR^{2}} \\
 \end{pmatrix}
\]

And with appropriate gauge we can write the Hamiltonian as:

\[
 H =
 \begin{pmatrix}
  -\frac{1}{2MR^{2}} & - \frac{q \epsilon R}{\sqrt{3}} \\
  - \frac{q \epsilon R}{\sqrt{3}} & \frac{1}{2MR^{2}} \\
 \end{pmatrix}
\]

By writing the Hamiltonian with Pauli matrices:
\[ H= -\frac{1}{2MR^{2}}\sigma_{z} - \frac{q \epsilon R}{\sqrt{3}}\sigma_{x} =\overrightarrow{\Omega} \cdot \overrightarrow{S}  \]   

Where \[ \overrightarrow{\Omega}=(- \frac{2q \epsilon R}{\sqrt{3}},0,-\frac{1}{MR^{2}})\]
 And \[ \overrightarrow{S}=(\frac{\sigma_{x}}{2},\frac{\sigma_{y}}{2},\frac{\sigma_{z}}{2}) . \]
Then it is clear, just by looking at the Hamiltonian that the frequency of oscillations
is:
\[ \omega = |\overrightarrow{\Omega}|= \sqrt{\frac{1}{M^{2}R^{4}}+\frac{4q^{2} \epsilon^{2} R^{2}}{3}} \] 
And ${T}$ is: \\
${T}$=${\frac{2\pi}{\omega}}$\\
 
(6)
If we add a constant magnetic field, ${\overrightarrow{B}}$ ,in the x direction, we need \\
to add the Zeeman term, remeber this is a spinless particle.
\[ H_{Zeeman} = -\frac{e}{2mc} \overrightarrow{B}\cdot \overrightarrow{L}   \]

So in our case the perturbation component is:
\[ V = -\frac{e}{2mc}B_{x}\cdot L_{x}   \]

(7)
We will write the time dependent wave function after adding the magnetic field. \\
We notice that ${Y_{0,0}}$ is not affected from the magnetic field due to its Spherical symmetry.\\
\[ \psi(t)=\frac{\sqrt{3}}{2}Y_{0,0}+\frac{1}{2}e^{-i\epsilon t}e^{-i\Omega tL_{x}}Y_{1,0}  \]
\[  \Omega=\frac{eB_{x}}{2Mc}   \]
The element ${e^{-i\Omega tL_{x}}}$ is a turn around the x axis with ${\Omega}$t degrees.\\
At time ${\frac{T_0}{2}}$ without the magnetic field the wave function is:
\[ \psi(\frac{T_0}{2})=\frac{\sqrt{3}}{2}Y_{0,0}-\frac{1}{2}Y_{1,0}  \]
So we want that ${\Omega \frac{T_0}{2}}$ will be ${\pi}$ (a turn of 180 degrees around the x axis) so we will get:
\[ \psi(\frac{T_0}{2})=\frac{\sqrt{3}}{2}Y_{0,0}+\frac{1}{2}Y_{1,0}  \]
\[ \Omega \frac{T_0}{2} = \pi \]
\[ \Omega = \frac{2\pi}{T_0} = \frac{eB_{x}}{2Mc}  \]
\[ B_{x}= \frac{4Mc\pi}{eT_0} \]





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