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\heading{E4510: Operating rotation matrices on spin states (2002A2)}
\auname{Oded Betzalel}


{\bf The problem:}

\Dn

Find the state vector ${\Psi}_m$ in the standard base for a spin polarized in the XY plane at angle ${\varphi}= 60^\circ.$

Relate to the following cases:
\Dn

(1) Spin 1/2.

(2) Spin 1 with circular polarization.

(3) Spin 1 with linear polarization.
\Dn

You must express the elements of the state vector in your final answer with ${e^{i{\pi}/integer} }$, ${\sqrt{2}}$ etc.

In order to prevent losing all points because of an algebraic mistake, please write first the analytic formula and only after calculate it to receive your final answer.

\Dn\Dn

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{\bf The solution:}
\Dn

(1)
Any spin 1/2 polarization can be obtained by combining a rotation round the Y axis and a rotation round the Z axis on the "up" state:

\[
|\vec{e}_{\theta,\varphi}>=R_{z}(\varphi)R_{y}(\theta)|\uparrow>=e^{-i \varphi S_z}e^{-i \theta S_y}|\uparrow> \rightarrow \left(\amatrix{e^{-i \varphi/2}cos(\theta/2) \cr e^{i \varphi/2}sin(\theta/2)}\right)
\]
\Dn

In order to get the question's polarization one needs to rotate by $\theta=\pi/2$ round the Y axis and then rotate by $ \varphi = \pi/3 $ round the Z axis. By doing so, one gets:

\[
\Psi_m=\frac{1}{\sqrt{2}}\left(\amatrix{e^{-i(\pi/6)} \cr e^{i(\pi/6)}}\right)
\]

\Dn
%%%%%%%%%%%%%%%
\Dn
(2) Any right circular state can be obtained by combining a rotation round the Y axis and a rotation round the Z axis on the $ | \Uparrow > $ state :

\[
|\vec{e}_{\theta,\varphi}>=U(\varphi\vec{e}_z)U(\theta\vec{e}_y)|\Uparrow>=e^{-i \varphi S_z}e^{-i \theta S_y}|\Uparrow> \rightarrow
\left(\amatrix{\frac{1}{2} (1+cos\theta) e^{-i \varphi} \cr \frac{1}{\sqrt{2}} sin \theta \cr \frac{1}{2} (1-cos\theta) e^{i \varphi}}\right)
\]

\Dn

The rotation action that needs to be done is a rotation by $\theta=\pi/2$ round the Y axis and then a rotation by $ \varphi = \pi/3 $ round the Z axis. Therefore:

\[
\Psi_m= \frac{1}{2} \left(\amatrix{e^{-i \pi/3} \cr {\sqrt{2}} \cr e^{i \pi/3}}\right)
\]

\Dn

Any left circular state can be obtained by combining a rotation round the Y axis and a rotation round the Z axis on the $ | \Downarrow > $ state :

\[
|\vec{e}_{\theta,\varphi}>=U(\varphi\vec{e}_z)U(\theta\vec{e}_y)|\Downarrow>=e^{-i \varphi S_z}e^{-i \theta S_y}|\Downarrow> \rightarrow
\left(\amatrix{\frac{1}{2} (1-cos\theta) e^{-i \varphi} \cr -\frac{1}{\sqrt{2}} sin \theta \cr \frac{1}{2} (1+cos\theta) e^{i \varphi}}\right)
\]

\Dn
The rotation action that needs to be done is a rotation by $\theta=\pi/2$ round the Y axis and after that a rotation by $ \varphi = \pi/3 $ round the Z axis. Therefore:

\[
\Psi_m= \frac{1}{2} \left(\amatrix{e^{-i \pi/3} \cr {-\sqrt{2}} \cr e^{i \pi/3}}\right)
\]

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

(3)Using the same technique as in (2) this time with a rotation matrix that rotates the state $|\Updownarrow>$ in order to get other linearly polarized states:

\[
|\vec{e}_{\theta,\varphi}>=
U(\varphi\vec{e}_z)U(\theta\vec{e}_y)|\Updownarrow>=
e^{-i \varphi S_z}e^{-i \theta S_y}| \Updownarrow> \rightarrow
\left(\amatrix{-\frac{1}{\sqrt{2}} sin\theta e^{-i \varphi} \cr cos\theta \cr \frac{1}{\sqrt{2}}sin\theta e^{i \varphi}}\right)
\]

\Dn

by using the same rotation as before $\theta=\pi/2$ and $ \varphi = \pi/3 $  :

\[
\Psi_m=\frac{1}{\sqrt{2}} \left(\amatrix{-e^{-i \pi/3} \cr {0} \cr e^{i \pi/3}}\right)
\]

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