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\heading{E5668: Spin precession due to spin-orbit interaction} 
\auname{Koby Yavilberg and Igor Plochotnikov}, Edited by Noam Riemer


{\bf The problem:}

\Dn

An electron, with mass ${M}$, charge ${e}$ and gyromagnetic constant ${g}$,
launched with energy ${E}$ in a one dimensional conductor, in the direction of the ${X}$ axis.
The conductor passes through capacitor plates of length ${L}$. The capacitor creates an
electric field ${\mathcal{E}}$ in the ${Y}$ direction.
Likewise, there's a magnetic field ${B}$ in that area in the ${Z}$ direction. When the 
electron enters the region of interaction, its spin is being polarized in the
direction of its motion.

\Dn

Throughout the problem we may ignore the possibility of the particle returning from the interaction region.
And also ${\hbar=1}$.

\Dn

In the first two paragraphs ${v \approx (2E/M)^{1/2}}$

\Dn

(1) What will be the direction ${\phi}$ of the spin, when the electron escapes the interaction region?

(2) What does the magnetic field has to be, for the spin not to rotate?

\Dn

In the next paragraphs ${x=0}$ will be set as the entering point.
Also giving an exact solution on the basis of the launching energy.

\Dn

(3) Write the particle's state in the region where the interaction takes place,
in the standart basis ${| x,m \rangle}$, where ${m = \uparrow, \downarrow}$.

(4) Give the exact answer to paragraph (1).

\Dn\Dn


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{\bf The solution:}

\Dn

(1) 
The Hamiltonian of the motion is as follows: 

\[\mathcal{H} = \frac{p^2}{2M} - g\frac{e}{2M}{\bf{B\cdot S}}-\frac{e}{2M^2}{(\bf{\mathcal{E}}\times p)\cdot S}\]
\[\mathcal{H} = \frac{p^2}{2M} - g\frac{eB_z}{2M}S_{z}+\frac{ev\mathcal{E}_y}{2M}S_{z} = \frac{p^2}{2M} - \frac{eS_{z}}{2M}(B_zg-v\mathcal{E}_y)\]



The time that it takes the electron, to get out of the capacitor is:
\[t=\frac{L}{v} \approx L \left(\frac{M}{2E}\right)^\frac{1}{2} \]
Also we know that since ${S_{z}}$ is the rotations generator around the ${z}$ direction,
it has to satisfy:
\[R(\bar{e}_{z}\phi) = e^{-i\phi S_{z}}=e^{-i\mathcal{H}t}=U(t)\]
The spin angle is calculated as follows: 
\[\phi=(k_\uparrow-k_\downarrow)L\]
Therefore, with respect to time $ t$ ,The angle with which it escapes the interaction region is:
\[\phi = -\frac{e}{2M}(B_zg-v\mathcal{E}_y)\times \frac {L}{v}\]



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\Dn

(2)
For the spin not to change, we must have ${\phi=0}$ 
\[\phi = -\frac{e}{2M}(B_zg-v\mathcal{E}_y)\times \frac {L}{v}=0\] 
\[\rightarrow B_z=\frac{v\varepsilon}{g}\]


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\Dn

(3)
With the addition of angular momentum we get:
\[ |\psi\rangle = \sum{\psi_{x,m}|x,m\rangle} \]
Where \[{|x,m\rangle=|x\rangle\otimes|m\rangle}$ and ${m=\uparrow,\downarrow}\]
\Dn
So we get:
\[ |\psi\rangle = e^{ik_{\uparrow} x} \otimes |\uparrow\rangle + e^{ik_{\downarrow} x} \otimes |\downarrow\rangle\]
\[ E = \frac{k^2}{2M} - \frac{eS_{z}}{2M}(B_zg-v\mathcal{E}_y)\]
\[ \Rightarrow k_{m} = \sqrt{2ME\pm\frac{1}{2}e(B_zg-v \mathcal{E}_y)}\]
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(4)
By solving the equation received in (3)
\[ k_{\uparrow} - k_{\downarrow} =\sqrt{2ME+\frac{1}{2}e(B_zg-v \mathcal{E}_y)}-\sqrt{2ME-\frac{1}{2}e(B_zg-v \mathcal{E}_y)}\]
\[\phi= -(k_{\uparrow} - k_{\downarrow})\cdot L\]
\Dn
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