Work out to analog to Equation 10.62 for a particle of spin I.

Geometric phase is a phase difference gained during the course of a cycle when a system is subjected to cyclic adiabatic processes, which derives from the geometrical features of the Hamiltonian's parameter space in both classical and quantum mechanics.

To work out the analog to the equation 10.62 using,

Here, the magnetic field is $B=B_0\left[\sin \left(\theta \right)\cos \left(\varphi \right)\hat{i}+\sin \left(\theta \right)\sin \left(\varphi \right)\hat{j}+\cos \left(\theta \right)\hat{K}\right]$, and the spin matrices are,

$$\begin{gathered} S_x=\frac{\sout{h}}{\sqrt{2}}\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right) \\ {S}_Y=\frac{i\sout{h}}{\sqrt{2}}\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& -1\\ 0& 1& 0\end{array}\right) \\ {S}_Z=h\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right) \end{gathered}$$

Substitute the value in the equation (2) to get:

$$\begin{gathered} \mathrm{H}=\frac{{\mathrm{eB}}_0}{\mathrm{m}}\frac{\sout{\mathrm{h}}}{\sqrt{2}}\left[\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\varphi }\right)\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right)+\sin \left(\mathrm{\theta }\right)\sin \left(\mathrm{\varphi }\right)\left(\begin{array}{ccc}0& -\mathrm{i}& 0\\ \mathrm{i}& 0& -\mathrm{i}\\ 0& \mathrm{i}& 0\end{array}\right)+\cos \left(\mathrm{\theta }\right)\left(\begin{array}{ccc}\sqrt{2}& 0& 0\\ 0& 0& 0\\ 0& 0& -\sqrt{2}\end{array}\right)\right] \\ =\frac{{\mathrm{eB}}_0}{\mathrm{m}}\frac{\sout{\mathrm{h}}}{\sqrt{2}}\sin \left(\mathrm{\theta }\right)\left(\begin{array}{ccc}\frac{\cos \left(\mathrm{\theta }\right)\sqrt{2}}{\sin \left(\mathrm{\theta }\right)}& \cos \left(\mathrm{\varphi }\right)-\mathrm{i}\sin \left(\mathrm{\theta }\right)& 0\\ \cos \left(\mathrm{\varphi }\right)+\sin \left(\mathrm{\theta }\right)& 0& \cos \left(\mathrm{\varphi }\right)-\mathrm{i}\sin \left(\mathrm{\theta }\right)\\ 0& \cos \left(\mathrm{\varphi }\right)+\mathrm{i}\sin \left(\mathrm{\theta }\right)& -\frac{\cos \left(\mathrm{\theta }\right)\sqrt{2}}{\sin \left(\mathrm{\theta }\right)}\end{array}\right) \\ \mathrm{Using}\mathrm{the}\mathrm{values}{\mathrm{e}}^{\mathrm{i\varphi }}=\cos \left(\mathrm{\varphi }\right)-\mathrm{i}\sin \left(\mathrm{\theta }\right),{\mathrm{e}}^{\mathrm{i\varphi }}=\cos \left(\mathrm{\varphi }\right)+\mathrm{i}\sin \left(\mathrm{\theta }\right),\mathrm{to}\mathrm{get}: \\ \mathrm{H}=\frac{{\mathrm{eB}}_0\sout{\mathrm{h}}}{\sqrt{2}\mathrm{m}}\left(\begin{array}{ccc}\sqrt{2}\cos \left(\mathrm{\theta }\right)& {\mathrm{e}}^{-\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)& 0\\ {\mathrm{e}}^{\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)& 0& {\mathrm{e}}^{\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)\\ 0& {\mathrm{e}}^{\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)& -\sqrt{2}\cos \left(\mathrm{\theta }\right)\end{array}\right)...\left(3\right) \end{gathered}$$

$$\begin{gathered} Lettheeigenvectorforthespin-upbe,x_{+}=\left(\begin{array}{c}a\\ b\\ c\end{array}\right). \\ Applythefollowingequation:H_{x_{+}}=\frac{e{B}_0}{m}{\sout{h}}_{x_{+}} \end{gathered}$$

$$\begin{gathered} \mathrm{As}\left(\begin{array}{ccc}\sqrt{2}\cos \left(\mathrm{\theta }\right)& {\mathrm{e}}^{-\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)& 0\\ {\mathrm{e}}^{\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)& 0& {\mathrm{e}}^{\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)\\ 0& {\mathrm{e}}^{\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)& -\sqrt{2}\cos \left(\mathrm{\theta }\right)\end{array}\right)\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\\ \mathrm{c}\end{array}\right)=\sqrt{2}\left(\begin{array}{c}\mathrm{a}\\ \mathrm{b}\\ \mathrm{c}\end{array}\right),\mathrm{so}\mathrm{the}\mathrm{following}\mathrm{equations}\mathrm{are}, \\ \sqrt{2}\cos \left(\mathrm{\theta }\right)\mathrm{a}+{\mathrm{e}}^{-\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)\mathrm{b}=\sqrt{2}\mathrm{a} \\ {\mathrm{e}}^{-\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)\mathrm{a}+{\mathrm{e}}^{-\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)\mathrm{c}=\sqrt{2\mathrm{b}}\begin{array}{}\end{array}...\left(4\right) \\ {\mathrm{e}}^{-\mathrm{i\varphi }}\sin \left(\mathrm{\theta }\right)\mathrm{b}-\sqrt{2}\cos \left(\mathrm{\theta }\right)\mathrm{c}=\sqrt{2}\mathrm{c} \end{gathered}$$

Solve the first equation for :

$b=\sqrt{2}{c}^{i\varphi }\left(\underset{-\tan (\theta /1)}{\underbrace{\frac{1-\cos \left(\theta \right)}{\sin \left(\theta \right)}}}\right)a$

$\mathrm{b}=\sqrt{2}{\mathrm{e}}^{\mathrm{i\varphi }}\tan \left(\frac{\mathrm{\theta }}{2}\right)\mathrm{a}$

From (3),

$$\begin{gathered} b=\sqrt{2}{e}^{i\theta }\left(\frac{1+\cos \left(\theta \right)}{\sin \left(\theta \right)}\right)c...\left(5\right) \\ =\sqrt{2}{e}^{i\theta }cot\left(\frac{\theta }{2}\right) \end{gathered}$$

Combine the above equation with (5) to get: $c=e^{i\varphi }{\tan }^2\left(\frac{\theta }{2}\right)a....\left(6\right)$

As the normalization condition appears, then ${\left|x+\right|}^2={\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2=1$.

Substitute the equation (5) and (6) to get

$$\begin{gathered} {\left|\mathrm{a}\right|}^2+2{\tan }^2\left(\frac{\mathrm{\theta }}{2}\right){\left|\mathrm{a}\right|}^2+{\tan }^4\left(\frac{\mathrm{\theta }}{2}\right){\left|\mathrm{a}\right|}^2=1 \\ \left[1+2{\tan }^2\left(\frac{\mathrm{\theta }}{2}\right)+{\tan }^4\left(\frac{\mathrm{\theta }}{2}\right)\right]{\left|\mathrm{a}\right|}^2=1 \\ {\left[1+{\tan }^2\left(\frac{\mathrm{\theta }}{2}\right)\right]}^2{\left|\mathrm{a}\right|}^2=1 \\ \mathrm{But},1+{\tan }^2\left(\frac{\mathrm{\theta }}{2}\right)=1+\frac{{\sin }^2\left({\displaystyle \frac{\mathrm{\theta }}{2}}\right)}{{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right)}=\frac{{\sin }^2\left(\frac{\mathrm{\theta }}{2}\right)+{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right)}{{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right)}=\frac{1}{{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right)} \end{gathered}$$

Thus,

localid="1656132153128" $$\begin{gathered} {\left|\mathrm{a}\right|}^2={\cos }^4\left(\frac{\mathrm{\theta }}{2}\right) \\ \mathrm{a}={\mathrm{e}}^{-\mathrm{i}0}{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right) \\ \mathrm{Then},\mathrm{c}=\sqrt{2}\sin \left(\frac{\mathrm{\theta }}{2}\right)\cos \left(\frac{\mathrm{\theta }}{2}\right)\mathrm{and}\mathrm{c}={\mathrm{e}}^{\mathrm{i\varphi }}{\sin }^2\left(\frac{\mathrm{\theta }}{2}\right). \\ \mathrm{Hence},\mathrm{the}\mathrm{eigenvector}\mathrm{for}\mathrm{the}\mathrm{spin}\mathrm{up}\mathrm{is}x+=\left(\begin{array}{c}{\mathrm{e}}^{-\mathrm{i\varphi }}{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right)\\ \sqrt{2}\sin \left(\frac{\mathrm{\theta }}{2}\right)\cos \left(\frac{\mathrm{\theta }}{2}\right)\\ {\mathrm{e}}^{-\mathrm{i\varphi }}{\sin }^2\left(\frac{\mathrm{\theta }}{2}\right)\end{array}\right). \end{gathered}$$

But,

$$\begin{gathered} \nabla \times \mathrm{A}=\frac{1}{\mathrm{r}\sin \mathrm{\theta }}\left(\frac{𝜕}{𝜕\mathrm{\theta }}({\mathrm{A}}_{\mathrm{\varphi }}\sin \mathrm{\theta })-\frac{𝜕{\mathrm{A}}_{\mathrm{\theta }}}{𝜕\mathrm{\varphi }}\right)\hat{\mathrm{r}} \\ +\frac{1}{\mathrm{r}}\left(\frac{1}{\sin \left(\mathrm{\theta }\right)}\frac{𝜕{\mathrm{A}}_{\mathrm{r}}}{𝜕\mathrm{\varphi }}-\frac{𝜕}{𝜕\mathrm{r}}\left({\mathrm{rA}}_{\mathrm{\varphi }}\right)\right)\hat{\mathrm{\theta }} \\ +\frac{1}{\mathrm{r}}\left(\frac{𝜕}{𝜕\mathrm{r}}\left({\mathrm{rA}}_{\mathrm{\varphi }}\right)-\frac{𝜕{\mathrm{A}}_{\mathrm{r}}}{𝜕\mathrm{\varphi }}\right)\hat{\mathrm{\varphi }} \\ \mathrm{Thus}, \end{gathered}$$