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\begin{document} 
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\heading{E472: Motion on a spherical shell} 
\auname{Andrey Gouternakht}


{\bf The problem:}

\Dn

Given a spherical shell with radius $R$ ;and a particle with mass $M$ and a charge $e$.
In $t=-\infty$ the particle is in the ground state.

\Dn

(1) Find the wave function: $\Psi(\theta,\varphi)$\\

Given an electric pulse: 
	\[\varepsilon(t)=\varepsilon_{0}e^{ - \frac{1}{2}(t / r)^2}
\]

(2) Which energy levels can the particle pass to?

(3) What is the probability to find the particle not in the ground 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 $R$ ( lets us assume for simplisity of the solution that $R$=1 ) is given by:
	
	\[\hat{H}_0=\frac{\hat{p}^2}{2M}=\frac{\hat{L}^2}{2MR^2}=\frac{\hat{L}^2}{2M}
\]
the system is found at the ground state, therefore $l$=0
and the wave function of the system is:
	\[\Psi($r$,\theta,\varphi)=F(R)Y_{00}\ \ \ \ \ \ \ \ \ \ \  E_{lm}=\frac{\hbar^2l(l+1)}{2M}
\]
%%%%%%%%%%%%%%%
\Dn

(2) 
If the electric pulse is turn on the Hamiltonian of the system is parturbated by the  $\hat{H}^{'}$:
\begin{center}
	\[\hat{H}=\hat{H}_{0}+\hat{H}^{'}
\]
\\ 
	\[\hat{H}^{'}=-q\varepsilon(t)\hat{z}=-q\varepsilon(t)\cos(\theta)=-q\varepsilon(t)\sqrt{\frac{4\pi}{3}}Y_{10}
\]
\end{center}
the correction matrix is given by:\\

	\[\bra 00\left|\hat{H}^{'}\right|lm\ket=-q\varepsilon_{0}e^{ - \frac{1}{2}(t / r)^2}\sqrt{\frac{4\pi}{3}}\int^{2\pi}_{0}d\varphi\int^{\pi}_{0}Y_{00}Y_{10}Y_{lm}\sin\theta d\theta 
  \]
\\\
the integral vanishes for all $\left|lm\right\rangle$ except $\left|{lm}\right\rangle$=$\left|{10}\right\rangle$, that is to say that system can pass to the first excited level: ($l$=1, $m$=0)\\\\
	\[\bra 00\left|\hat{H}^{'}\right|01\ket = \frac{-q\varepsilon(t)}{\sqrt{3}} \Rightarrow \hat{V}=\frac{-q\varepsilon_{0}e^{ - \frac{1}{2}(t / r)^2}}{\sqrt{3}}\delta_{l,1;m,0}
\]So, only $E_{00}\rightarrow E_{10}$, that is to say $0\rightarrow\frac{\hbar^2}{M}$, transition is possible!\\\\\\
%%%%%%%%%%%%%%%
\Dn

(3)
The particle can be found only at two different states: $\left|00\right\rangle$ and $\left|10\right\rangle$, so the probability of the state $\left|10\right\rangle$ is:\
	\[
	P_{10}=\frac{1}{\hbar^2}\left|\int_{-\infty}^\infty V_{01}e^{i\Omega_{10}t}dt\right|^2 = 
	\frac{1}{\hbar^2}\left|\int_{-\infty}^\infty \frac{-q\varepsilon_{0}e^{ - \frac{1}{2}(t /
	r)^2}}{\sqrt{3}}e^{i\Omega_{10}t}dt\right|^2 = 
	\frac{{(q\varepsilon_{0})}^2}{3\hbar^2}\left|\int_{-\infty}^\infty {e^{ - \frac{1}{2}(t /
	r)^2}}e^{i\frac{\hbar}{M}t}dt\right|^2
	\]
	in last transition we used: 
	\[\Omega_{10}=\frac{1}{\hbar}(E_1-E_0)=\frac{\hbar}{M}
\]
in order to solve the aforementioned integral one may use the solved integral (Schaum's handbook 15.75):
	\[
	\int_{-\infty}^{\infty}e^{-(ax^2+bx+c)}dx=\sqrt{\frac{\pi}{a}}e^{(b^2-4ac)/{4a}}
\]

	\[
	P_{10}=\frac{{(q\varepsilon_{0})}^2}{3\hbar^2}(2\tau^2\pi)e^{-\frac{\hbar^2 \tau^2}{M^2}}
\]


\Dn
\Dn
\Dn
\Dn
\begin{verbatim}

Do not assume that R=1.
The exercise shold be solved using known results from perturbation theory.

\end{verbatim}



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\end{document} 
