= Solution
The <Bloch vector> representation of a <qubit> <density matrix> is $\rho=\frac12(I+r_xX+r_yY+r_zZ)$, where $r_j=\operatorname{Tr}(\rho\sigma_j)$ and $|\boldsymbol r|\le1$. Pure states have $|\boldsymbol r|=1$, while the origin represents the <maximally mixed state> $I/2$, not a <pure state>.
For the two preparations, direct evaluation of the <Pauli matrices> gives
$$
\boxed{\boldsymbol r_1=(0,0,1),\qquad\boldsymbol r_2=(-\sqrt3/2,0,-1/2).}
$$
Their dot product is $-1/2$, so their <Bloch vectors> meet at angle $\boxed{2\pi/3}$, or $120$ degrees. This is the angle between Bloch vectors; the angle determined by the modulus of the Hilbert-space overlap is different.
Completeness of the <POVM> requires $E_3=I-E_1-E_2$. In the <computational basis>, subtraction gives
$$
E_3=\begin{pmatrix}\frac12&-\frac{\sqrt3}{6}\\-\frac{\sqrt3}{6}&\frac16\end{pmatrix}=\frac23|\phi_3\rangle\langle\phi_3|,\qquad\boxed{|\phi_3\rangle=\frac{\sqrt3}{2}|0\rangle-\frac12|1\rangle.}
$$
An overall phase of $\phi_3$ is immaterial. The matrix has <eigenvalues> $2/3$ and zero, so it is positive; together with $E_1,E_2$ it is a valid <trine qubit POVM>.
The <Born rule> gives $\Pr(j\mid\rho_i)=\operatorname{Tr}(E_j\rho_i)=\frac23|\langle\phi_j|\psi_i\rangle|^2$. The conditional probabilities are
$$
\begin{array}{c|ccc}&j=1&j=2&j=3\\\hline\rho_1&0&1/2&1/2\\\rho_2&1/2&0&1/2\end{array}.
$$
Outcome one identifies $\rho_2$, since $\phi_1$ is orthogonal to $\psi_1$; outcome two identifies $\rho_1$, since $\phi_2$ is orthogonal to $\psi_2$. Outcome three is inconclusive. If Alice's prior probability for $\rho_1$ is $q$, the unconditional outcome probabilities are $(1-q)/2$, $q/2$, and $1/2$ respectively. Because outcome three has the same likelihood for both preparations, it leaves the prior odds unchanged. \b[Bob's conclusive identification succeeds with probability one half and is never wrong; he does not identify the state on the inconclusive trials.] This is <unambiguous quantum state discrimination>.
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