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\heading{E601: Spin Orbit Interaction}

\auname{Chen Giladi, Yoav Pollack, Alon Oringn}

\noindent \textbf{}

\noindent \textbf{The problem:}

\noindent Electron with mass m and a charge q is placed on one dimensional ring with radius R. There is a another point charge Q which is placed in the center of the ring, so there is a spin -- orbit interaction

\noindent (1) Write the Hamiltonian $H(x,p,S)$ of the electron.

\noindent (2) What are the energies of the electron?

\noindent (3) What is the velocity of the electron in each energy level?

\noindent (4) Which magnetic Flux $\phi $ trough the ring will cancel out the effect of charge Q on the electron with spin up.

\noindent Use $\hbar =c=1$and write the represent the final answers with the provided data.

\noindent 

\noindent \textbf{The solution:}

\noindent (1)We know that the spin -- orbit term is $H_{spin-orbit} =-\frac{e}{2m^{2} } (\vec{\varepsilon }\times \vec{p})\cdot \vec{S}$  where $\vec{\varepsilon }=-\frac{V'(r)}{r} \stackrel{\rightharpoonup}{r}\Rightarrow \vec{\varepsilon }=\frac{Q}{R^{2} } \cdot \stackrel{\frown}{r}$and using $\vec{p}=p_{\varphi } \stackrel{\frown}{\varphi }$, So our Hamiltonian is

\[H=\frac{\vec{p}^{2} }{2m} -\frac{e}{2m^{2} } (\vec{\varepsilon }\times \vec{p})\cdot \vec{S}=\frac{1}{2m} (\frac{\partial ^{2} }{\partial r^{2} } +\frac{1}{r^{2} } \frac{\partial ^{2} }{\partial \varphi ^{2} } +\frac{\partial ^{2} }{\partial z^{2} } )-\frac{e}{2m^{2} } (\frac{Q}{R^{2} } \cdot \stackrel{\frown}{r}\times \vec{p})\cdot \vec{S}\] 
Putting our ring on the XY plane and we know that $r=R$ we get the following Hamiltonian

\[H=\frac{p^{2} _{\varphi } }{2mR^{2} } -\frac{eQ}{2m^{2} R^{2} } S_{z} \] 


\noindent 

\noindent (2) Using $[p_{\varphi } ^{2} ,S_{z} ]=[H,S_{z} ]=[p_{\varphi } ^{2} ,H]=0$ we can write the energies of the electron as

\[E_{n,l} =\frac{1}{2mR^{2} } n^{2} -\frac{eQ}{2m^{2} R^{2} } l\] 


\noindent (3) The velocity can be found by using the Heisenberg equation of motion

\[\frac{d\varphi }{dt} =i[H,\varphi ]=\frac{p_{\varphi } }{mR^{2} } +\frac{eQ}{2m^{2} R^{2} } S_{z} \] 
So the velocity for the Eigen stats is $V_{\varphi } =\frac{n}{mR^{2} } +\frac{eQ}{2m^{2} R^{2} } l$

\noindent 

\noindent (4) By using the next Hamiltonian $H=\frac{1}{2m} (p_{\varphi } -\frac{e\phi }{2\pi R} )^{2} -\frac{eQ}{2m^{2} R^{2} } p_{\varphi } S_{Z} $

\noindent we get the condition on $\phi $ for which  there is no effect of the interaction spin orbit

\[\phi =-\frac{Q}{2m} \frac{\pi }{R} \] 


\noindent 


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