Question 5.P.8: Consider a particle of total angular momentum j = 1. Find th...

Since \vec{J}=J_{x} \vec{i}+J_{y} \vec{j}+J_{z} \vec{k} and \vec{n}=(\sin \theta \cos \varphi )\vec{i}+(\sin \theta \sin \varphi )\vec{j}+(\cos \theta )\vec{k}, the component of \vec{J} along \vec{n} is

\vec{n}.\vec{J}=J_{x}\sin \theta \cos \varphi+J_{y}\sin \theta \sin \varphi+J_{z}\cos \theta,                (5.271)

with 0\leq \theta \leq \pi and 0\leq \varphi \leq 2\pi ; the matrices of \hat{J}_{x} ,\hat{J}_{y}, and \hat{J}_{z} are given by (5.259).

\hat{J}_{x}=\frac{\hbar }{\sqrt{2} } \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right) , \hat{J}_{y}= \frac {\hbar }{\sqrt{2} }\left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{matrix} \right),\hat{J}_{z}=\hbar \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right).

We can therefore write this equation in the following matrix form:

+\hbar \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right)\cos \theta =\frac{\hbar }{\sqrt{2} }\left(\begin {matrix} \sqrt{2}\cos \theta & e^{-i\varphi }\sin \theta & 0 \\ e^{i\varphi }\sin \theta & 0 & e^{-i\varphi }\sin \theta \\ 0 & e^{i\varphi }\sin \theta & -\sqrt{2}\cos \theta \end{matrix} \right) .       (5.272)

The diagonalization of this matrix leads to the eigenvalues \lambda _{1} =-\hbar ,\lambda _{2} =0, and \lambda _{3} =\hbar; the corresponding eigenvectors are given by

|\lambda _{1}〉=\frac{1}{2}\left(\begin{matrix} (1-\cos \theta )e^{-i\varphi } \\ -\frac{2}{\sqrt{2} }\sin \theta \\ (1+\cos \theta )e^{i\varphi } \end{matrix} \right) ,          |\lambda _{2}〉=\frac{1}{\sqrt{2} } \left(\begin{matrix} -e^{-i\varphi }\sin \theta \\ \sqrt{2}\cos \theta \\ e^{i\varphi }\sin \theta \end{matrix} \right),          (5.273)

|\lambda _{3}〉=\frac{1}{2}\left(\begin{matrix} (1+\cos \theta )e^{-i\varphi } \\ \frac{2}{\sqrt{2} }\sin \theta \\ (1-\cos \theta )e^{i\varphi } \end{matrix} \right) .                                 (5.274)