Homework 11

6.28

$$S_y=\frac{\hslash }{y}\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]$$ $$\left[\begin{array}{cc}-\lambda & -\frac{i\hslash }{2}\\ \frac{i\hslash }{2}& -\lambda \end{array}\right]={\lambda }^2-\frac{{\hslash }^2}{4}⟹\lambda =\pm \frac{\hslash }{2}$$

Eigenvalues:

$$\pm \frac{\hslash }{2}$$

Eigenvectors

$$\frac{\hslash }{2}\left[\begin{array}{cc}o& -i\\ i& 0\end{array}\right]\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]=\pm \frac{\hslash }{2}\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]$$ $$⟹-i\beta =\pm \alpha$$

and

$$|\alpha {|}^2+|\beta {|}^2=1$$

Solving above we get

$${\chi }_{+}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ i\end{array}\right]$$

for $\frac{\hslash }{2}$ and

$${\chi }_{-}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -i\end{array}\right]$$

for $-\frac{\hslash }{2}$

Probability for obtaining $\frac{\hslash }{y}=S_z$ is $\frac{1}{2}$