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\heading{E4710: Motion on a spherical shell} 
\auname{Kurtser Eitan}


{\bf The problem:}

\Dn
Spherical shell with radius $R$ and a particle with mass $M$ and charge $q$.
In this question you have to assume that the particle can be existed from the ground state ($l=0$)
 to the first energy level ($l=1$), but not beyond that, which means that the state space 
is four dimensional. 

A particle is prepared  as "concentrated" as possible in the "north pole" of the shell.


\Dn

(1) Write the wave function $\Psi(\theta,\phi)$.

(2) Find the periodic time of the oscillations.

The polarization of the system is defined as $P(t)=\bra\Psi(t)|\hat{z}|\Psi(t)\ket $

(3) Find the polarization as function of time $P(t) $.

Given: $Y_{00}=\frac{1}{\sqrt{4\pi}};\ Y_{10}(\theta,\phi)=\sqrt{\frac{3}{4\pi}}\cos\theta;\ Y_{1,\pm 1}\propto \sin\theta$ 


\Dn\Dn


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{\bf The solution:}

\Dn

(1) 
We want that probability $|\Psi|^2$ to find the particle at north pole ($\theta=0$)
will be maximal, so we look for a solution of the following form 
$\Psi\sim 1+\cos \theta$
\[
\Psi=c( 1+\cos \theta)=c(\sqrt{3}Y_{00}+Y_{10})
\]
\[ |\Psi \rangle = c(\sqrt{3}|00 \rangle+|10 \rangle)\]
After Normalization: $\int |\bra\Psi|\Psi\ket|^2d\Omega=1$ we find that  $c=\frac{1}{2}$ and we get
\[\Psi=\frac{1}{2}(\sqrt{3}Y_{00}+Y_{10})
\]



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\Dn

(2)
The energy states of the particle of the shell $E_l=\frac{l(l+1)}{2 M R^2}$.
We are restricted to $l=0,1$ so the energies are $E_0=0$ and $E_1=\frac{1}{ M R^2}$.
The time depended wave function is
\[
\Psi(t)=\frac{\sqrt{3}}{2}\eexp{-i E_0 t}Y_{00}+
			\frac{1}{2}\eexp{-i E_1 t}Y_{10}
		=\frac{\sqrt{3}}{2}Y_{00}+
			\frac{1}{2}\eexp{-i \frac{1}{ M R^2} t}Y_{10}
\]
we can see the function is periodic with frequency $\omega=E_1=\frac{1}{ M R^2}$, 
so the periodic time of oscillations is 
\[T=2\pi M R^2\]

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

(3)
To find the polarization we use $z=Rcos\theta=\sqrt{\frac{3}{4\pi}}RY_{10}$
\[
\bra\Psi|\cos\theta|\Psi\ket=\bra\cos\theta\ket
=\left(
\frac{\sqrt{3}}{2}\langle 00|+\frac{1}{2}\eexp{i\omega t}\langle 10|
\right)\cos\theta
\left(
\frac{\sqrt{3}}{2}|00\rangle+\frac{1}{2}\eexp{-i\omega t}|10\rangle 
\right)
\]
\[
\cos\theta |00\rangle=\sqrt{\frac{4\pi}{3}}Y_{10}\frac{\sqrt{3}}{2}Y_{00}=\frac{1}{2}Y_{10}=|10\rangle
\]
\[
\bra\cos\theta\ket=\alpha\bra 00|10 \ket+\frac{1}{4}\eexp{i\omega t}\bra 10|10 \ket
+\frac{1}{4}\eexp{-i\omega t}\bra 10|10 \ket+\beta \bra 10|\cos\theta|10 \ket
\]
the first and the last parts are zero and we get
\[
\bra\cos\theta\ket=\frac{1}{2}\cos(\omega t)
\]
\[
P(t)=\langle\Psi|\hat{z}|\Psi\rangle=\frac{1}{2}R\cos(\omega t)
\]


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