Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 58 / 3

For a pure qubit $a|0⟩+b|1⟩$, the Bloch vector is $(2\mathrm{Re}(a^{\ast }b),2\mathrm{Im}(a^{\ast }b),|a{|}^2-|b{|}^2)$. Substitution givesBoth lie on the unit Bloch sphere, at the same azimuth and opposite heights. They are reflections across its equatorial plane, not antipodal vectors. Choose $\theta =\pi /3$ and $\phi =\pi /4$, so their coordinates are $(\sqrt{6}/4,\sqrt{6}/4,\pm 1/2)$. The following original sketch shows those vectors and their common meridian:

Represent the pure states by density matrices ${\rho }_u=(I+u\cdot \sigma )/2$ and ${\rho }_v=(I+v\cdot \sigma )/2$. The Pauli matrices obey $\mathrm{Tr}{\sigma }_i=0$ and $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$. ConsequentlyOn the other hand, ${\rho }_u=|u⟩⟨u|$ and ${\rho }_v=|v⟩⟨v|$ make that trace equal to $⟨u|v⟩⟨v|u⟩$. Thus the pure-qubit overlap identity isFor mixed states the trace formula still holds, but its left side is then a Hilbert–Schmidt overlap rather than this squared pure-state overlap.

Under the prescribed computational-basis measurement and decision rule, the first state is correctly declared when outcome zero occurs, with probability ${\cos }^2(\theta /2)$. The second is correctly declared on outcome one, with the same probability. Averaging the two equal priors givesOne can also verify optimality directly: the two density matrices differ by $\cos \theta {\sigma }_z$. Since $\cos \theta >0$, choosing the positive eigenspace as the decision for the first state maximizes the trace contribution to success. This is the equal-prior discrimination of qubit states construction along the vertical axis.

Continue with equal prior probabilities as in part (c). The two Bloch vectors areTheir density-matrix difference is $\Delta =\sin \theta \cos \phi {\sigma }_x$. If a binary measurement effect $E$ declares the first state, its success isDiagonalize $\Delta$. Because $0\le E\le I$, the trace is maximized by taking $E$ to be its positive-eigenvalue projector. Thus perform a Pauli measurement of ${\sigma }_x$, equivalently measure in $(|0⟩\pm |1⟩)/\sqrt{2}$. If $\cos \phi >0$, identify the first state on the positive outcome and the second on the negative; reverse the labels if $\cos \phi <0$. The optimum isFor $\cos \phi =0$ the states coincide, so the best probability is $1/2$. This explicit positive-eigenspace optimization proves the Helstrom measurement for two pure states result in the present case, rather than only quoting its name.

Use the optimal three-to-one qubit random access code. For Alice's bits $(b_1,b_2,b_3)$, put $s_j=(-1{)}^{b_j}$ and send the pure state with Bloch vectorThese are the eight vertices of a cube inscribed in the Bloch sphere. If an explicit ket is wanted, choose $\cos {\theta }_b=s_3/\sqrt{3}$ and $e^{i{\phi }_b}=(s_1+i{s}_2)/\sqrt{2}$, and prepare $\cos ({\theta }_b/2)|0⟩+e^{i{\phi }_b}\sin ({\theta }_b/2)|1⟩$.

Bob wanting bit $j$ performs a Pauli measurement of ${\sigma }_x,{\sigma }_y,{\sigma }_z$ respectively, reporting zero for eigenvalue $+1$ and one for $-1$. Its success for every input and every requested bit isOnly one chosen bit is extracted; the protocol does not recover all three simultaneously and uses no shared entanglement.

To prove that no one-qubit protocol gives a greater guaranteed probability, allow Alice arbitrary mixed encoding vectors $|r_b|\le 1$ and Bob any binary positive operator-valued measure. Write his decision difference as $D_j=E_{j,0}-E_{j,1}=t_{j}I+v_j\cdot \sigma$. Its eigenvalues lie in $[-1,1]$, implying $|v_j|\le 1-|t_j|\le 1$. On input $b$, success for bit $j$ is $[1+s_j(t_j+v_j\cdot r_b)]/2$. Average over all eight strings and three choices. The terms in $t_j$ cancel, leavingThe Cauchy-Schwarz inequality and cancellation of cross terms over all sign choices giveHence $p_{\mathrm{avg}}\le p_{\ast }$. Any worst-case guarantee is no greater than this average; the cube protocol attains it uniformly, proving optimality both for guaranteed success and for the uniform-input average. Biased prior information about the bits would define a different optimization problem.