Eigenvectors of \bgroup\color{black}$S_y$\egroup for Spin \bgroup\color{black}${1\over 2}$\egroup

To find the eigenvectors of the operator \bgroup\color{black}$S_y$\egroup we follow precisely the same procedure as we did for \bgroup\color{black}$S_x$\egroup (see previous example for details). The steps are:

1. Write the eigenvalue equation \bgroup\color{black}$(S_y - \alpha)\chi=0$\egroup

2. Solve the characteristic equation for the eigenvalues \bgroup\color{black}$\alpha_\pm$\egroup

3. Substitute the eigenvalues back into the original equation

4. Solve this equation for the eigenvectors

Here we go! The operator \bgroup\color{black}$S_y={\hbar\over 2} \left(\matrix{0&-i \cr
i&0}\right)$\egroup, so that, in matrix notation the eigenvalue equation becomes

\begin{displaymath}\bgroup\color{black}\left(\matrix{-\alpha&-i\hbar/ 2 \cr
{i...
...pha}\right)
\left(\matrix{\chi_1 \cr \chi_2}\right )=0 \egroup\end{displaymath}

The characteristic equation is \bgroup\color{black}$det\vert S_y-\alpha\vert=0$\egroup, or

\begin{displaymath}\bgroup\color{black}\alpha^2-{\hbar^2\over 4}=0\qquad\Rightarrow\qquad\alpha=\pm{\hbar\over
2}\egroup\end{displaymath}

These are the same eigenvalues we found for \bgroup\color{black}$S_x$\egroup (no surprise!) Plugging \bgroup\color{black}$\alpha_+$\egroup back into the equation, we obtain

\begin{displaymath}\bgroup\color{black}\left(\matrix{-\alpha_+&-i\hbar/ 2 \cr
...
...1}\right)
\left(\matrix{\chi_1 \cr \chi_2}\right )
=0 \egroup\end{displaymath}

Writing this out in components gives the pair of equations

\begin{displaymath}\bgroup\color{black}-\chi_1 -i\chi_2=0\qquad and\qquad i\chi_1 - \chi_2=0\egroup\end{displaymath}

which are both equivalent to \bgroup\color{black}$\chi_2=i\chi_1$\egroup. Repeating the process for \bgroup\color{black}$\alpha_-$\egroup, we find that \bgroup\color{black}$\chi_2=-i\chi_1$\egroup. Hence the two eigenvalues and their corresponding normalized eigenvectors are

\begin{displaymath}\bgroup\color{black}\alpha_+=+\hbar/2 \qquad
\chi^{(y)}_{+}={1\over\sqrt{2}}\left(\matrix{1\cr i}\right)\egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black}\alpha_-=-\hbar/2 \qquad
\chi^{(y)}_{-}={1\over\sqrt{2}}\left(\matrix{1\cr -i}\right)\egroup\end{displaymath}

Jim Branson 2013-04-22