Question

For the most general normalized spinor (Equation 4.139),

compute$\left\{S_x\right\}\mathbf{,}\left\{S_y\right\}\mathbf{,}\left\{S_z\right\}\mathbf{,}\left\{S_x^2\right\}\mathbf{,}\left\{S_y^2\right\}\mathbf{,}\mathbf{}\mathbf{and}\left\{S_x^2\right\}\mathbf{.}\mathbf{}\mathbf{check}\mathbf{}\mathbf{that}\left\{S_x^2\right\}\mathbf{+}\left\{S_y^2\right\}\mathbf{+}\left\{S_z^2\right\}\mathbf{=}\left\{S^2\right\}\mathbf{.}$

$$\begin{gathered} \mathcal{X}\mathbf{=}\left(\begin{array}{c}a\\ b\end{array}\right)\mathbf{=}{{\mathbf{a}}_{\mathbf{X}}}_{\mathbf{+}}\mathbf{+}{\mathbf{b}}_{\mathbf{X}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(4.139\right)\mathbf{.} \end{gathered}$$

Short answer

By solving we find the above value to be.

Step-by-step solution

Calculating

Computing

$=\frac{\sout{h}}{2}\left(-i{a}^{*}b+ia{b}^{*}\right)=\frac{\sout{h}}{2}i\left(a{b}^{*}-a^{*}b\right)=-\sout{h}lm\left(a{b}^{*}\right).$

$S_x^2=\frac{{\sout{h}}^2}{4}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\frac{{\sout{h}}^2}{4}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\frac{{\sout{h}}^2}{4}$

$S_y^2=\frac{{\sout{h}}^2}{4}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\frac{{\sout{h}}^2}{4}$

localid="1655968926137"

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