Eigenvectors of $S_y$ for Spin ${1\over 2}$

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

1. Write the eigenvalue equation $(S_y - \alpha)\chi=0$

2. Solve the characteristic equation for the eigenvalues $\alpha_\pm$

3. Substitute the eigenvalues back into the original equation

4. Solve this equation for the eigenvectors

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

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

The characteristic equation is $det\vert S_y-\alpha\vert=0$, or

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

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

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

Writing this out in components gives the pair of equations

$$\bgroup\color{black}-\chi_1 -i\chi_2=0\qquad and\qquad i\chi_1 - \chi_2=0\egroup$$

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

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

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