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\noindent Quantum Mechanics II, Ex 4730

\noindent 

\heading{Ex 4730 - spherical shell}
\auname{Roee Steiner}

\noindent 

\noindent Given a spherical shell with radius R and a particle with mass M and charge e.

\noindent Notice that the standard variables which show the particle are ($\theta $, $\varphi $, Lx, Ly, Lz)

\noindent In this question we have to assume that the particle can be excited from ground state to first energy level but not beyond so the state space is four dimensional \textbar 0$>$,\textbar ?$>$,\textbar ?$>$,\textbar ?$>$.

\noindent 

\noindent 1. What is the energy of every base.

\noindent 

\noindent We prepare the particle as ''concentrated'' as possible in the ''north pole'' of the shell.

\noindent 

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

\noindent 

\noindent What you have written is not stationary. The particle oscillates between the North and the South pole.

\noindent 

\noindent 3. What is the period of the oscillations?

\noindent 

\noindent We torn on a constant electric field\textbf{ ${\mathbf \varepsilon }$ }in the Z direction  

\noindent 

\noindent 4. Write the electric perturbation in formal way with  the standard variables.

\noindent 

\noindent 5. What is the period time T of the fluctuation?

\noindent 

\noindent We torn on a constant magnetic field\textbf{ B  }in the X direction  

\noindent 

\noindent 6. Write the magnetic perturbation in formal way with  the standard variables.

\noindent 

\noindent 7. Find a value for B so in the period of   T/2 where the particle will be in the first place.

\noindent 

\[Y_{00}=\frac{1}{\sqrt{4\pi }}{\rm \ ,}Y_{10}=\sqrt{\frac{3}{4\pi }}{\cos  \theta \ }{\rm \ ,}Y_{1,\pm 1}\ \ \alpha \ \ {\sin  \left(\theta \right)\ }\] 


\noindent Paragraph 2 and 7 demands ''thinking'', the answer is simple and do not demand ''algebra''.

\noindent 

\noindent \textbf{Solution:}

\noindent  

\noindent 1. The energy level for particle on sphere is $E_l=\frac{l\left(l+1\right)}{2MR^2}$.

\noindent So for the first state \textbar 0$>$,  $l=0$ so  $E_0=0$.

\noindent For the three others state $l=1$ so ${\rm \ }E_1=\frac{1}{MR^2}$.

\noindent 

\noindent 2 We want to find a solution that gives us the best probability to find the particle in the North

\noindent pole of the sphere. We need to use the spherical harmonics function.

\noindent Its means that the function will have a trigonometric variable, so the function will look like

\noindent $\psi =\frac{1}{2}\left(1+{\cos  \theta \ }\right)$ And after we use the spherical harmonics and normalization we will get:

\noindent 

\[\psi =\frac{1}{2}\left(\sqrt{3}Y_{00}+Y_{10}\right)\] 


\noindent 

\noindent 3. We will write a time depended wave function:

\[\psi \left(t\right)=\frac{1}{2}\left(\sqrt{3}Y_{00}+Y_{10}e^{-iE_1t}\right)\] 
From here we can see easily that the frequency is $E_1$,so the period is

\noindent 

\[T=\frac{2\pi }{E_1}=2\pi MR^2\] 


\[4.    V=e\varepsilon \hat{z}=e\varepsilon R{\cos  \theta \ }{\rm =}\sqrt{\frac{4\pi }{3}}e\varepsilon RY_{10}\] 


\noindent 5. We will find first the matrix elements of V that given by:

\[\left\langle l^',m^'\left|V\right|l,m\right\rangle =\left\langle l^',m^'\left|\sqrt{\frac{4\pi }{3}}e\varepsilon RY_{10}\right|l,m\right\rangle {\rm =}\frac{e\varepsilon R}{\sqrt{3}}{\delta }_{l1,m0}{\delta }_{l^'0,m^'0}\] 
So we can see that the system work like a two site system so the new Hamiltonian is:

\noindent 

\[H=\left( \begin{array}{cc}
0 & \frac{e\varepsilon R}{\sqrt{3}} \\ 
\frac{e\varepsilon R}{\sqrt{3}} & \frac{1}{mR^2} \end{array}
\right)=\frac{1}{2mR^2}\hat{I}-\frac{1}{2mR^2}{\sigma }_z+\frac{e\varepsilon R}{\sqrt{3}}{\sigma }_x=\frac{1}{2mR^2}\hat{I}+\Omega S\] 


\noindent Where $S=\frac{1}{2}{\sigma }_i$ and $\Omega =\left(\frac{2e\varepsilon R}{\sqrt{3}},0,\frac{-1}{mR^2}\right)$

\noindent According to "precession picture" (see lecture notes chapter 36), the frequency is:

\noindent 

\[\omega =\left|\Omega \right|=\sqrt{\frac{4}{3}{\left(e\varepsilon R\right)}^2+{\left(\frac{1}{mR^2}\right)}^2}\] 


\noindent So the period time is:

\noindent 

\[T=\frac{2\2pi }{\omega }=\frac{\2pi }{\sqrt{\frac{4}{3}{\left(e\varepsilon R\right)}^2+{\left(\frac{1}{mR^2}\right)}^2}}\] 


\noindent 6. From Zeeman we know that the impact of magnetic field is:

\noindent 

\[V_B=-\frac{eB}{2m}L_x\] 


noindent 7. We can see that $V_B$ react in the same way on the system as the electric perturbation  

\noindent So in analogy we can see that the wave function will bee

\noindent 

\[\Psi \left(t\right)=\frac{1}{2}\left(\sqrt{3}Y_{00}+e^{-i\Omega t}e^{?V_Bt}Y_{10}\right)\] 
Because the time is two time lease its means that 

\[\left|\Omega \right|+V_B=2\left|\Omega \right|\] 


\noindent So             $B=\frac{2m}{?}\sqrt{\frac{4}{3}{\left(e\varepsilon R\right)}^2+{\left(\frac{{\rm 1}}{4mR^2}\right)}^2}$

\noindent 


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