Eigenvectors of $S_u$

As an example, lets take the $u$ direction to be in the $xz$ plane, between the positive $x$ and $z$ axes, 30 degrees from the x axis. The unit vector is then $\hat{u}=(\cos(30),0,\sin(30))=(\sqrt{3\over 4},0,{1\over 2})$. We may simply calculate the matrix $S_u=\hat{u}\cdot\vec{S}$.

$$\bgroup\color{black}S_u=\sqrt{3\over 4}S_x+{1\over 2}S_z={\hb... ...\over 2}&\sqrt{3\over 4}\cr \sqrt{3\over 4}&-{1\over 2}}\egroup$$

We expect the eigenvalues to be $\pm{\hbar\over 2}$ as for all axes.

Factoring out the ${\hbar\over 2}$, the equation for the eigenvectors is.

$$\begin{eqnarray*} \pmatrix{{1\over 2}&\sqrt{3\over 4}\cr \sqrt{3\over 4}&-{1\ove... ...over 4}b\cr \sqrt{3\over 4}a-{1\over 2}b}=\pm\pmatrix{a\cr b}\\ \end{eqnarray*}$$

For the positive eigenvalue, we have $a=\sqrt{3}b$, giving the eigenvector $\chi^{(u)}_+=\pmatrix{\sqrt{3\over 4}\cr {1\over 2}}$. For the negative eigenvalue, we have $a=-\sqrt{1\over 3}b$, giving the eigenvector $\chi^{(u)}_-=\pmatrix{-{1\over 2}\cr \sqrt{3\over 4}}$. Of course each of these could be multiplied by an arbitrary phase factor.

There is an alternate way to solve the problem using rotation matrices. We take the states $\chi^{(z)}_\pm$ and rotate the axes so that the $u$ axis is where the $z$ axis was. We must think carefully about exacty what rotation to do. Clearly we need a rotation about the $y$ axis. Thinking about the signs carefully, we see that a rotation of -60 degrees moves the $u$ axis to the old $z$ axis.

$$\begin{eqnarray*} R_y=\pmatrix{\cos{\theta\over 2}&\sin{\theta\over 2}\cr -\sin{... ...4}}\pmatrix{0\cr 1} =\pmatrix{-{1\over 2}\cr\sqrt{3\over 4}} \\ \end{eqnarray*}$$

This gives the same answer. By using the rotation operator, the phase of the eigenvectors is consistent with the choice made for $\chi^{(z)}_\pm$. For most problems, this is not important but it is for some.