(a) The density matrix for a spin-1/2 state is a two by two matrix which can be written
\rho=\left(\begin{array}{ll} c_{1} & c_{2} \\ c_{3} & c_{4} \end{array}\right) .
for complex numbers c_{1}, c_{2}, c_{3}, c_{4} . Because the matrix is Hermitian we know that c_{1} \text { and } c_{4} must be real and c_2 and c_3 must be complex conjugates. Thus
\rho=\left(\begin{array}{cc} b_{1} & b_{2}-i b_{3} \\ b_{2}+i b_{3} & b_{4} \end{array}\right) .
for real numbers b_{1}, b_{2}, b_{3}, b_{4} In addition, since the trace is 1 we know that b_{4}=1-b_{1} . \text { Defining } a_{1}=2 b_{2} , a_{2}=2 b_{3}, \text { and } a_{3}=2 b_{1}-1 then gives
\rho=\frac{1}{2}\left(\begin{array}{cc} 1+a_{3} & a_{1}-i a_{2} \\ a_{1}+i a_{2} & 1-a_{3} \end{array}\right)=\frac{1}{2}\left[\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)+a_{1}\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)+a_{2}\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)+a_{3}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\right] .
=\frac{1}{2}\left(1+a_{1} \sigma_{x}+a_{2} \sigma_{y}+a_{3} \sigma_{z}\right)=\frac{1}{2}(1+ a \cdot \sigma) .
(b)
\operatorname{Tr}\left(\rho^{2}\right)=\operatorname{Tr}\left[\frac{1}{2}\left(\begin{array}{cc} 1+a_{3} & a_{1}-i a_{2} \\ a_{1}+i a_{2} & 1-a_{3} \end{array}\right) \frac{1}{2}\left(\begin{array}{cc} 1+a_{3} & a_{1}-i a_{2} \\ a_{1}+i a_{2} & 1-a_{3} \end{array}\right)\right] .
=\frac{1}{4} \operatorname{Tr}\left[\left(\begin{array}{cc} a_{1}^{2}+a_{2}^{2}+\left(1+a_{3}\right)^{2} & 2\left(a_{1}-i a_{2}\right) \\ 2\left(a_{1}+i a_{2}\right) & a_{1}^{2}+a_{2}^{2}+\left(1-a_{3}\right)^{2} \end{array}\right)\right]=\frac{1}{2}\left(1+| a |^{2}\right) .
By virtue of Problem 12.6c we see that | a | \leq 1 in order for this to be a density matrix, and it describes a pure state if and only if | a | = 1 .
(c) We can calculate the expectation value of S_z and, since there are only two possibilities,
\left\langle S_{z}\right\rangle=P_{+} \frac{\hbar}{2}+P_{-}\left(-\frac{\hbar}{2}\right)=P_{+} \frac{\hbar}{2}+\left(1-P_{+}\right)\left(-\frac{\hbar}{2}\right)=\hbar\left(P_{+}-\frac{1}{2}\right) .
This tells us the probability that a measurement will return spin up along z. Now
\left\langle S_{z}\right\rangle=\operatorname{Tr}\left(\frac{\hbar}{2} \sigma_{z} \rho\right)=\frac{\hbar}{2} \operatorname{Tr}\left[\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \frac{1}{2}\left(\begin{array}{cc} 1+a_{3} & a_{1}-i a_{2} \\ a_{1}+i a_{2} & 1-a_{3} \end{array}\right)\right] .
=\frac{\hbar}{4} \operatorname{Tr}\left[\left(\begin{array}{cc} 1+a_{3} & a_{1}-i a_{2} \\ -a_{1}-i a_{2} & -1+a_{3} \end{array}\right)\right]=\frac{\hbar}{2} a_{3} .
Combining these two results we have
P_{+}=\frac{1+a_{3}}{2} .
For the given directions we have: \text { (i) } P_{+}=1 \text {, (ii) } P_{+}=1 / 2, \text { and (iii) } P_{+}=0 \text {. }
(d) For a point on the equator at azimuthal angle \phi \text {, we have } a_{1}=\cos \phi, a_{2}=\sin \phi, a_{3}=0 \text {, so }
\rho=\frac{1}{2}\left(\begin{array}{cc} 1 & e^{-i \phi} \\ e^{i \phi} & 1 \end{array}\right) .
For a spinor \chi=\left(\begin{array}{l} a \\ b \end{array}\right) the density matrix is
\rho=\chi \chi^{\dagger}=\left(\begin{array}{l} a \\ b \end{array}\right)\left(\begin{array}{ll} a^{*} b^{*} \end{array}\right)=\left(\begin{array}{ll} |a|^{2} & a b^{*} \\ b a^{*} & |b|^{2} \end{array}\right) .
so, up to an arbitrary overall phase,
\chi=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ e^{i \phi} \end{array}\right) .