Construct the matrix${\mathbf{S}}_{\mathbf{r}}$representing the component of spin angular momentum along an arbitrary direction$\stackrel{\mathbf{⏜}}{\mathbf{r}}$. Use spherical coordinates, for which

$\stackrel{\mathbf{⏜}}{\mathbf{r}}\mathbf{}\mathbf{sin\theta cos\Phi }\stackrel{\mathbf{⏜}}{\mathbf{ı}\mathbf{}}\mathbf{+}\mathbf{sin\theta sin\Phi }\stackrel{\mathbf{⏜}}{\mathbf{ø}}\mathbf{+}\mathbf{cos\theta }\stackrel{\mathbf{⏜}}{\mathbf{k}}\mathbf{}\mathbf{}\mathbf{}$ [4.154]

Find the eigenvalues and (normalized) eigen spinors of${\mathbf{S}}_{\mathbf{r}}$. Answer:

${\mathbf{x}}_{\mathbf{+}}^{\left(r\right)}\mathbf{=}\left(\begin{array}{c}\cos \left(\theta /2\right)\\ e^{\mathrm{i\varphi }}\sin \left(\theta /2\right)\end{array}\right)$; ${\mathbf{x}}_{\mathbf{+}}^{\left(r\right)}\mathbf{=}\mathbf{\left(}\begin{array}{c}{\mathbf{e}}^{\mathbf{i\varphi }}\mathbf{sin}\left(\theta /2\right)\\ \mathbf{-}\mathbf{cos}\left(\theta /2\right)\end{array}\mathbf{\right)}$ [4.155]

Note: You're always free to multiply by an arbitrary phase factor-say,${\mathit{e}}^{\mathbf{i}\mathbf{\varphi }}$-so your answer may not look exactly the same as mine.

Eigenvalues are a unique set of scalar values associated with a set of linear equations, most commonly found in matrix equations.

Characteristic roots are another name for eigenvectors. It's a non-zero vector that can only be altered by its scalar factor once linear transformations are applied.

In quantum physics, Eigen spinors are considered basis vectors that represent a particle's general spin state.

Use eigenvalues $|S_r-\lambda I|=0$and find the value the eigenvalues $S_r$,$\lambda$.

$$\begin{gathered} \left|\begin{array}{cc}\frac{ħ}{2}\cos \theta -\lambda & \frac{ħ}{2}{e}^{-i\varphi }\sin \theta \\ \frac{ħ}{2}{e}^{-i\varphi }\sin \theta & \frac{ħ}{2}\cos \theta -\lambda \end{array}\right|=0 \\ -\frac{ħ}{4}{\cos }^2\theta +{\lambda }^2-\frac{ħ}{4}{\sin }^2\theta =0 \\ {\lambda }^2=\frac{ħ}{4}\left({\sin }^2\theta +{\cos }^2\theta \right) \\ {\lambda }^2=\frac{-ħ}{4} \\ \lambda =\pm \frac{ħ}{2} \\ \mathrm{Thus},\mathrm{the}\mathrm{eigenvalues}\mathrm{of}{\mathrm{S}}_{\mathrm{r}}\mathrm{is}\pm \frac{\mathrm{ħ}}{2} \end{gathered}$$

Use the eigen-spinor as an example$\left(\begin{array}{c}a\\ b\end{array}\right)$.

$S_2\left(\begin{array}{c}a\\ b\end{array}\right)=\lambda \left(\begin{array}{c}a\\ b\end{array}\right)$

For $\lambda =\frac{ħ}{2}$,

localid="1659013250803" $$\begin{gathered} \frac{ħ}{2}\left(\begin{array}{cc}\cos \theta & e^{-i\varphi }\sin \theta \\ e^{-i\varphi }\sin \theta & -\cos \theta \end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\frac{ħ}{2}\right) \\ \left(\begin{array}{cc}\cos \theta & e^{-i\varphi }\sin \theta \\ e^{-i\varphi }\sin \theta & -\cos \theta \end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}a\\ b\end{array}\right) \end{gathered}$$

Compare the relevant entries on both sides of the matrices.

$$\begin{gathered} a\cos \theta +b{e}^{-i\varphi }\sin \theta =a \\ a{e}^{i\varphi }\sin \theta -b\cos \theta =b \end{gathered}$$

Solve the above two equations.

$$\begin{gathered} b=\frac{a\left(1-\cos \theta \right)}{e^{-i\varphi }\sin \theta } \\ =e^{-i\varphi }\frac{\left(1-\cos \theta \right)a}{\sin \theta } \\ =e^{-i\varphi }\frac{\left(2{\sin }^2{\displaystyle \frac{\theta }{2}}\right)}{2{\sin }^2{\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}a \\ =e^{-i\varphi }\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}}a \end{gathered}$$

For $\left(\begin{array}{c}a\\ b\end{array}\right)$become commonplace,

Apply ${\left|a\right|}^2+{\left|b\right|}^2=1$.

${\left|a\right|}^2+\frac{{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)}{\left|a\right|}^2=1$

Solve the above expression further.

$$\begin{gathered} {\left|a\right|}^2\left[\frac{{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)}{{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}+1\right]=1 \\ {\left|a\right|}^2\left[\frac{\sin \left({\displaystyle \frac{\theta }{2}}\right)+{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)}{{\cos }^2\left(\frac{\theta }{2}\right)}\right]=1 \\ {\left|a\right|}^2\left[\frac{1}{\cos {\displaystyle \frac{\theta }{2}}}\right]=1 \\ a=\cos \frac{\theta }{2} \end{gathered}$$

Substitute the above value in $b=e^{i\varphi }\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}}a$.

$$\begin{gathered} b=e^{i\varphi }\frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}}\cos \frac{\theta }{2} \\ =e^{i\varphi }\sin \frac{\theta }{2} \end{gathered}$$

Substitute the values of a, and b in $\left(\begin{array}{c}a\\ b\end{array}\right)$.

$\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}\cos \frac{\theta }{2}\\ e^{i\varphi }\sin \frac{\theta }{2}\end{array}\right)$

For $\lambda =\frac{-ħ}{2}$,

$$\begin{gathered} \frac{ħ}{2}\left(\begin{array}{cc}\cos \theta & e^{-i\varphi }\sin \theta \\ e^{-i\varphi }\sin \theta & -\cos \theta \end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=-\frac{ħ}{2}\left(\begin{array}{c}a\\ b\end{array}\right) \\ a\cos \theta +b{e}^{-i\varphi }\sin \theta =-a \\ a{e}^{i\varphi }\sin \theta -b\cos \theta =b \end{gathered}$$

Solve the above two equations.

$$\begin{gathered} b=\frac{-a\left(1+\cos \theta \right)}{\sin \theta }{e}^{i\varphi } \\ =-a{e}^{i\varphi }\frac{2{\cos }^2{\displaystyle \frac{\theta }{2}}}{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}} \\ =-a{e}^{i\varphi }\frac{2{\cos }^2{\displaystyle \frac{\theta }{2}}}{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}} \\ \mathrm{Apply}{\left|\mathrm{a}\right|}^2+{\left|\mathrm{b}\right|}^2 \end{gathered}$$

$$\begin{gathered} {\left|a\right|}^2+\frac{{\cos }^2{\displaystyle \frac{\theta }{2}}}{{\sin }^2{\displaystyle \frac{\theta }{2}}}{\left|a\right|}^2=1 \\ {\left|a\right|}^2\left[1+\frac{{\cos }^2{\displaystyle \frac{\theta }{2}}}{{\sin }^2{\displaystyle \frac{\theta }{2}}}\right]=1 \\ a=e^{-i\Phi }\sin \left(\frac{\theta }{2}\right) \end{gathered}$$

Substitute the above value in$b=-a{e}^{i\varphi }\frac{\cos {\displaystyle \frac{\theta }{2}}}{\sin {\displaystyle \frac{\theta }{2}}}$.

$$\begin{gathered} b=-\left(e^{-i\Phi }\sin \left(\frac{\theta }{2}\right)\right)e^{-i\varphi }\frac{\cos {\displaystyle \frac{\theta }{2}}}{\sin {\displaystyle \frac{\theta }{2}}} \\ =-e^{i\varphi }\cos \frac{\theta }{2} \end{gathered}$$

Substitute the values of a, and b in $\left(\begin{array}{c}a\\ b\end{array}\right)$.

$\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}{e}^{-i\Phi }\sin \frac{\theta }{2}\\ -\cos \frac{\theta }{2}\end{array}\right)$

Thus, the eigen values and the eigen spinors are role="math" localid="1659015532483" $\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}\cos \frac{\theta }{2}\\ e^{-i\Phi }\sin \frac{\theta }{2}\end{array}\right)$and

$\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}{e}^{-i\Phi }\sin \frac{\theta }{2}\\ -\cos \frac{\theta }{2}\end{array}\right)$.