Question 5.P.4: (a) Find the eigenvalues and eigenstates of the spin operato...

(a) In this question we want to solve

\vec{n}.\vec{S}|\lambda〉 =\frac{\hbar }{2} \lambda |\lambda 〉,                   (5.221)

where \vec{n} is given by \vec{n}=(\sin \theta \vec{i}+\cos \theta \vec{k}), because it lies in the xz plane, with 0 ≤ θ ≤ π. We can thus write

\vec{n}.\vec{S}=(\sin \theta \vec{i}+\cos \theta \vec{k}).(S_{x}\vec{i}+S_{y}\vec{j}+S_{z}\vec{k} )=S_{x} \sin \theta +S_{z} \cos \theta .               (5.222)

Using the spin matrices

\vec{S}_{x}=\frac{\hbar }{2}\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right),       \vec{S}_{y}=\frac{\hbar }{2} \left(\begin{matrix} 0 & -i \\ i & 0 \end{matrix} \right),        \vec{S}_{z}=\frac{\hbar }{2}\left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right),         (5.223)

we can write (5.222) in the following matrix form:

\vec{n}.\vec{S}=\frac{\hbar }{2}\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)\sin \theta +\frac{\hbar }{2}\left(\begin {matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)\cos \theta =\frac{\hbar }{2}\left(\begin{matrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{matrix} \right).           (5.224)

The diagonalization of this matrix leads to the following secular equation:

-\frac{\hbar ^{2} }{4}(\cos \theta -\lambda )(\cos \theta +\lambda ) -\frac{\hbar ^{2} }{4}\sin ^{2} \theta =0,                (5.225)

which in turn leads as expected to the eigenvalues λ = ±1.

The eigenvector corresponding to λ = 1 can be obtained from

\frac{\hbar }{2}\left(\begin{matrix} \cos \theta & \sin \theta \\ \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).                     (5.226)

This matrix equation can be reduced to a single equation

a\sin \frac{1}{2} \theta =b\cos \frac{1}{2}\theta.                     (5.227)

Combining this equation with the normalization condition \left |a\right|^{2} +\left|b\right|^{2} =1, we infer that a= \cos \frac{1}{2}\theta and b=\sin \frac{1}{2} \theta ; hence the eigenvector corresponding to λ = 1 is

|\lambda _{+} 〉=\left(\begin{matrix} \cos (\theta /2) \\ \sin (\theta /2) \end{matrix} \right) .                      (5.228)

Proceeding in the same way, we can easily obtain the eigenvector for λ = -1:

|\lambda _{-} 〉=\left(\begin{matrix} -\sin (\theta /2) \\ \cos (\theta /2) \end{matrix} \right).                      (5.229)

(b) Let us write |\lambda _{\pm } 〉of (5.228) and (5.229) in terms of the spin-up and spin-down eigen-vectors, \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+\sin \frac{1}{2} \theta \left|{\frac{1}{2},-\frac{1}{2}}\right\rangle,                   (5.230)

|\lambda _{-} 〉=-\sin \frac{1}{2} \theta\left|{\frac{1}{2},\frac {1}{2}}\right\rangle+\cos \frac{1}{2}\theta\left|{\frac{1}{2},-\frac{1}{2}}\right\rangle.                 (5.231)

We see that the probability of measuring \vec{S}_{z} =+\hbar/2 is given by

\left|\left\langle \frac{1}{2},\frac{1}{2}| \lambda _{+} \right\rangle \right| ^{2} =\cos ^{2} \frac{1}{2}\theta .                        (5.232)