Eigenvectors of \bgroup\color{black}$S_u$\egroup

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

\begin{displaymath}\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\end{displaymath}

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

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

\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}\\

For the positive eigenvalue, we have \bgroup\color{black}$a=\sqrt{3}b$\egroup, giving the eigenvector \bgroup\color{black}$\chi^{(u)}_+=\pmatrix{\sqrt{3\over 4}\cr {1\over 2}}$\egroup. For the negative eigenvalue, we have \bgroup\color{black}$a=-\sqrt{1\over 3}b$\egroup, giving the eigenvector \bgroup\color{black}$\chi^{(u)}_-=\pmatrix{-{1\over 2}\cr \sqrt{3\over 4}}$\egroup. 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 \bgroup\color{black}$\chi^{(z)}_\pm$\egroup and rotate the axes so that the \bgroup\color{black}$u$\egroup axis is where the \bgroup\color{black}$z$\egroup axis was. We must think carefully about exacty what rotation to do. Clearly we need a rotation about the \bgroup\color{black}$y$\egroup axis. Thinking about the signs carefully, we see that a rotation of -60 degrees moves the \bgroup\color{black}$u$\egroup axis to the old \bgroup\color{black}$z$\egroup axis.

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}} \\

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

Jim Branson 2013-04-22