Question 5.P.5: (a) Find the eigenvalues and eigenstates of the spin operato...

(a)We need to solve

\vec{n}.\vec{S}|\lambda〉 =\frac{\hbar }{2} \lambda |\lambda 〉,                    (5.233)

where \vec{n}, a unit vector pointing along an arbitrary direction, is given in spherical coordinates by

\vec{n}=(\sin \theta \cos \varphi )\vec{i}+(\sin \theta \sin \varphi )\vec{j}+(\cos \theta )\vec{k},                 (5.234)

with 0\leq \theta \leq \pi and 0\leq \varphi \leq 2\pi . We can thus write

=S_{x}\sin \theta \cos \varphi +S_{y}\sin \theta \sin \varphi +S_{z}\cos \theta .                                     (5.235)

Using the spin matrices, we can write this equation in the following matrix form:

=\frac{\hbar }{2}\left(\begin{matrix} \cos \theta & e^{-i\varphi }\sin \theta \\ e^{i\varphi }\sin \theta & -\cos \theta \end{matrix} \right).                                                      (5.236)

Diagonalization of this matrix leads to the secular equation

-\frac{\hbar ^{2} }{4} (\cos \theta -\lambda )(\cos \theta +\lambda )-\frac{\hbar ^{2} }{4}\sin ^{2} \theta =0,                (5.237)

which in turn leads to the eigenvalues λ = ±1.
The eigenvector corresponding to λ = 1 can be obtained from

\frac{\hbar }{2}\left(\begin{matrix} \cos \theta & e^{-i\varphi }\sin \theta \\ e^{i\varphi }\sin \theta & -\cos \theta \end{matrix} \right)\left(\begin{matrix} a \\ b \end{matrix} \right) =\frac{\hbar }{2}\left(\begin{matrix} a \\ b \end{matrix} \right),           (5.238)

which leads to

a\cos \theta +be^{-i\varphi }\sin \theta =a                           (5.239)

or

a(1-\cos \theta )=be^{-i\varphi }\sin \theta.                    (5.240)

Using the relations 1-\cos \theta =2\sin ^{2} \frac{1}{2}\theta and \sin \theta =2\cos \frac{1}{2}\theta \sin \frac{1}{2}\theta, we have

b=a\tan \frac{1}{2}\theta e^{i\varphi }.                        (5.241)

Combining this equation with the normalization condition \left| a\right| ^{2}+\left|b\right| ^{2} =1, we obtain a=\cos \frac{1}{2}\theta and b=e^{i\varphi }\sin \frac{1}{2}\theta. Thus, the eigenvector corresponding to λ = 1 is

|\lambda _{+} 〉=\left(\begin{matrix} \cos (\theta /2) \\ e^{i\varphi }\sin (\theta /2) \end{matrix} \right) .               (5.242)

A similar treatment leads to the eigenvector for λ = -1:

|\lambda _{-} 〉=\left(\begin{matrix} -\sin (\theta /2) \\ e^{i\varphi }\cos (\theta /2) \end{matrix} \right) .                    (5.243)

(b) Write |\lambda _{-} 〉 of (5.243) in terms of \left|{\frac{1}{2},\frac{1}{2}}\right\rangle =\left(\begin{matrix} 1 \\ 0 \end{matrix} \right) and \left|{\frac{1}{2},-\frac{1}{2}}\right\rangle=\left(\begin{matrix} 0 \\ 1 \end{matrix} \right):

|\lambda _{+} 〉=\cos \frac{1}{2}\theta \left|{\frac{1}{2},\frac {1}{2}}\right\rangle+e^{i\varphi } \sin \frac{1}{2} \theta \left|{\frac {1}{2},-\frac{1}{2}}\right\rangle,               (5.244)

|\lambda _{-} 〉=-\sin \frac{1}{2}\theta \left|{\frac{1}{2},\frac {1}{2}}\right\rangle+e^{i\varphi }\cos \frac{1}{2}\theta \left|{\frac{1}{2},-\frac{1}{2}}\right\rangle.                    (5.245)

We can then obtain the probability of measuring \hat{S}_{z} =-\hbar /2:

\left|\left\langle \frac{1}{2},-\frac{1}{2}| \lambda _{-} \right \rangle \right| ^{2}=\cos ^{2} \frac{1}{2}\theta .          (5.246)

(c) The spin’s eigenstates at time t are given by

|\lambda _{+}(t) 〉=e^{-iE_{+}t/\hbar }\cos \frac{1}{2}\theta \left|{\frac{1}{2},\frac{1}{2}}\right\rangle+e^{i(\varphi -E-t/\hbar )} \sin \frac{1}{2} \theta \left|{\frac{1}{2},-\frac{1}{2}}\right\rangle ,             (5.247)

|\lambda _{-}(t) 〉=-e^{-iE_{+}t/\hbar }\sin \frac{1}{2} \theta \left|{\frac{1}{2},\frac{1}{2}}\right\rangle+e^{i(\varphi -E-t/\hbar )}\cos \frac{1}{2}\theta\left|{\frac{1}{2},-\frac{1}{2}}\right\rangle ,              (5.248)

where E_{\pm }are the energy eigenvalues corresponding to the spin-up and spin-down states, respectively.