(a) Find the eigenvalues and eigenspinors of S_y .

(b) If you measured S_yon a particle in the general state X(Equation 4.139), what values might you get, and what is the probability of each? Check that the probabilities add up to 1 . Note: a and b need not be real!

(c) If you measured${\mathbf{S}}_{\mathbf{y}}^{\mathbf{2}}$ , what values might you get, and with what probabilities?

The theoretical probability of an event is the number of possible outcomes divided by the number of possible outcomes, with the probability being the chance of an outcome or event.

As a result, probability refers to the possibility or frequency with which something occurs.

Determine the eigenvalues in the following way.

$\begin{array}{rcl}\left|\begin{array}{cc}-\mathrm{\lambda }& -\frac{\mathrm{i}\sout{\mathrm{h}}}{2}\\ -\frac{\mathrm{i}\sout{\mathrm{h}}}{2}& -\mathrm{\lambda }\end{array}\right|& =& {\mathrm{\lambda }}^2-\frac{{\mathrm{h}}^2}{4}\\ & =& 0\\ \mathrm{\lambda }& =& \pm \frac{\mathrm{h}}{2}\end{array}$

Determine the eigenspinors in the following way.

For$\mathrm{\lambda }=\frac{\mathrm{h}}{2}$,

$\begin{array}{rcl}\frac{\mathrm{h}}{2}\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right)\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)& =& \frac{\mathrm{h}}{2}\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)\\ \left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)& =& \left(\begin{array}{c}-\mathrm{ib}\\ \mathrm{ia}\end{array}\right)\end{array}$

So, $b=ia$and $\overline{)\mathrm{\chi }\widetilde{{\mathrm{n}}_{+}}=\left(\begin{array}{c}\mathrm{a}\\ \mathrm{ia}\end{array}\right)}$.

It is known that $\stackrel{\text{'}}{\mathrm{a}}\mathrm{\chi }\overline{)\mathrm{\chi }}\tilde{\mathrm{n}}=1$.

$\begin{array}{rcl}\left({\mathrm{a}}^{*}-{\mathrm{ia}}^{*}\right)\left(\begin{array}{c}\mathrm{a}\\ \mathrm{ia}\end{array}\right)& =& {\left|\mathrm{a}\right|}^2+{\left|\mathrm{a}\right|}^2\\ & =& 1\\ \mathrm{a}& =& \frac{1}{\sqrt{2}}\end{array}$

So, $\overline{)\chi \widetilde{n_{+}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ i\end{array}\right)}}$

Determine the eigenspinors in the following way.

For$\lambda =-\frac{h}{2}$,

$\begin{array}{rcl}\frac{\mathrm{h}}{2}\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right)\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)& =& -\frac{\mathrm{h}}{2}\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)\\ -\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right)& =& \left(\begin{array}{c}-\mathrm{ib}\\ \mathrm{ia}\end{array}\right)\end{array}$

So, b=-ia, and $|\chi \widetilde{n_{+}}=\left(\begin{array}{c}a\\ -ia\end{array}\right)$.

It is known that $\stackrel{\text{'}}{a}\chi \overline{)x\tilde{n}=1}$.

$\begin{array}{rcl}\left({\mathrm{a}}^{*}-{\mathrm{ia}}^{*}\right)\left(\begin{array}{c}\mathrm{a}\\ -\mathrm{ia}\end{array}\right)& =& {\left|\mathrm{a}\right|}^2+{\left|\mathrm{a}\right|}^2\\ & =& 1\\ \mathrm{a}& =& \frac{1}{\sqrt{2}}\end{array}$

So,$|x\tilde{n}-\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -i\end{array}\right)$.

Thus, the required eigenvalues are $\pm \frac{h}{2}$and eigenspinors of S_y are $\frac{1}{\sqrt{2}}i,\frac{1}{\sqrt{2}}-i$.

The likelihood of measuring $\pm \frac{h}{2}$ of S_y is obtained as follows,

Add the probabilities.

$\begin{array}{rcl}{\mathrm{P}}_{+}^{\mathrm{y}}+{\mathrm{P}}_{-}^{\mathrm{y}}& =& {\left|\mathrm{a}\right|}^2+{\left|\mathrm{b}\right|}^2\\ & =& \stackrel{\text{'}}{\mathrm{a}}\mathrm{\chi }|\mathrm{\chi }\tilde{\mathrm{n}}\\ & =& 1\end{array}$

Thus, the probability of each is $\frac{1}{2}\left({\left|\mathrm{a}\right|}^2+{\left|\mathrm{b}\right|}^2+\mathrm{i}\left({\mathrm{ab}}^{*}-{\mathrm{ba}}^{*}\right)\right)$and $\frac{1}{2}\left({\left|\mathrm{a}\right|}^2+{\left|\mathrm{b}\right|}^2-\mathrm{i}\left({\mathrm{ab}}^{*}-{\mathrm{ba}}^{*}\right)\right)$.

The anticipated value of${\mathrm{S}}_{\mathrm{y}}^2$is obtained in the following way,

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{y}}^2& =& \frac{{\mathrm{h}}^2}{4}\\ \begin{array}{cc}1& 0\end{array}& =& \frac{{\mathrm{h}}^2}{4}1\\ \mathrm{\chi }\left|{\mathrm{S}}_{\mathrm{y}}^2\right|\mathrm{\chi }& =& \frac{{\mathrm{h}}^2}{4}\stackrel{\text{'}}{\mathrm{a}}\mathrm{\chi }|\mathrm{\chi }\tilde{\mathrm{n}}\\ & =& \frac{{\mathrm{h}}^2}{4}\end{array}$

Thus, the value of ${\mathrm{S}}_{\mathrm{y}}^2$ is $\frac{{\mathrm{h}}^2}{4}$and the probability is 1.