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\heading{E6010: The spin-orbit interaction} 
\auname{Rybka Yaroslav, Idan Barak}


{\bf The problem:}

\Dn

An electron with mass ${M}$ and charge ${e}$ is free 
to move on a single dimensional ring of radius ${R}$.
The electron has a spin-orbit interaction with the electric 
field created by a charge ${Q}$ located in the center of the ring.
It's recommended to use the units  \hbar${=c=1}$

\Dn

(1) Write the Hamiltonian ${H(x,p,s)}$.

(2) Find the eigenenergies of the electron.

(3) Find the speed of the electron for each energy state.

(4) What magnetic flux $(\Phi)$ through the ring would negate the effect the charge ${Q}$ has over en electron with spin up? 
\Dn

Give the answer in terms of the given parameters only. 

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
The Hamiltonian that includes the spin-orbit interaction is as following

\[(a) 
\mathcal{H} 
= \frac{1}{2m}p^{2}-\frac{e}{2m^{2}}\left(\vec{\epsilon}\times\vec{p} \right)\cdot\vec{s}
\]


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

(2)
The electtric field created by a point charge is given by:

\[(b)
\mathcal{\vec{\epsilon}} 
= \frac{Q}{R^{2}}\widehat{r}
\]


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


We'll write the momentum as: 

\[(c) 
\vec{p}}= p\widehat{\phi}
\]


The spin of a electron is: 

\[(d) 
\vec{s}}= s\widehat{z}
\]

Substituting (d), (c) and (b) into (a) we get:

\[(e) 
\mathcal{H} 
= \frac{1}{2m}p^{2}-\frac{Qes}{2m^{2}R^{2}}p\]

\Dn

Notice that the Hamiltonian can be rewritten as: 
\[
\mathcal{H} 
= \frac{1}{2m}\left(p-\frac{Qes}{2mR^{2}}\right)^{2}
-\frac{Q^{2}e^{2}s^{2}}{8m^{3}R^{4}}
\]

\Dn

As seen in the Aharonov-Bohm effect, if we have shifted momentum, the eigentsates will be as those as of a regular momentum. Addition of a constant won't affect the eigenstates. The eigenvalues of a Hamilitonian with momentum shifted by +${a}$ will be as following:

\[(f) 
E_{n}=\frac{1}{2m}\left(\frac{2\pi\hbar}{L}n+a\right)^{2}
\]

\Dn

The constant addition to the Hamiltonian will appear in eigenvalues as a constant addition. Substituting (e) into (f) and adding the contant we get:

\[(g) 
E_{n}=\frac{1}{2m}\left(\frac{2\pi\hbar}{L}n+\frac{Qes}{2mR^{2}}\right)^{2}-\frac{Q^{2}e^{2}s^{2}}{8m^{3}R^{4}}
\]

\Dn

Finally, substituting ${L}$ with 2${\pi}$${R}$ and opening the parentheses we end up with:

\[ 
E_{n}=\frac{1}{2mR^{2}}\left(n^{2}}+\frac{Qes}{2mR}\right)
\]

\Dn

(3)
Because the eigenstates of the energies are the eigenstates of momentum:

\[ 
v=\frac{1}{mR}\left(n+\frac{Qes}{2mR}\right)
\]

(4)
We know that flux $\Phi$ will appear in the Hamiltonian as shift in momentum. All we need is for two shifts to cancel each other:

\[ 
\frac{Qes}{2mR^{2}}-\frac{e\Phi}{2R\pi}=0
\]

\[ 
\Phi=\frac{\pi sQ}{mR}
\]

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