Past exam of the laser science and quantum information processing course of the University of Oxford / 2008 / 5 (DPHY 2752, DPHP 2752)

Source extract: users.physics.ox.ac.uk/~Foot/C2/pdf/C22008Q5.pdf

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TODO: The remainder of this historical paper is not held. Only the supplied question is covered.

For a normalized pure state, expand $|\psi ⟩={\sum }_m{c}_m|m⟩$ in an orthonormal basis of eigenvectors of the measured Hermitian operator. The Born rule assigns outcome $m$ the probability $p_m=|c_m{|}^2$. Under the ideal projective measurement update, a nondegenerate outcome leaves $|m⟩$, up to a global phase. The sum of the outcome probabilities is one, and the expectation value is ${\sum }_{m}m|c_m{|}^2$; a single trial reports one eigenvalue, not that average.

If $M|m⟩=m|m⟩$, thensince $M=M^{†}$. A nonzero eigenvector has positive squared norm, so $m$ is real. For two eigenvectors,Thus $(m-n)⟨n|m⟩=0$: distinct eigenvalues have orthogonal eigenvectors.

The overlap $⟨0|+⟩=1/\sqrt{2}$ is nonzero, whereas perfect discrimination of pure states requires orthogonality. Consequently no single-copy quantum measurement can identify both preparations with certainty. After a nondegenerate projective measurement, the loss of preparation information after a nondegenerate projective measurement prevents further measurements from recovering the missing label: conditional on the first outcome, the post-measurement state is the same for every compatible initial hypothesis. Repeating the same basis gives that outcome again; changing basis probes the collapsed state. Fresh, independently prepared copies would be additional resources.

In the computational basis, input $|0⟩$ always gives zero and remains $|0⟩$. Input $|+⟩$ gives zero or one with probability $1/2$, leaving the corresponding basis state. Combining these likelihoods with equal prior probabilities, the Bayesian update after a projective measurement givesFor example, Bayes' theorem gives $P(0\text{ input}\mid 0\text{ outcome})=(1/2)/(3/4)=2/3$. Outcome one certifies input $|+⟩$; outcome zero is ambiguous. Both inputs that give outcome zero leave exactly the same post-measurement state $|0⟩$.

In the Hadamard basis, input $|+⟩$ always gives $+$, while input $|0⟩$ gives $+$ or $-$ equally. ThereforeOutcome minus certifies input $|0⟩$; outcome plus is ambiguous. Guessing the more likely input after the ambiguous outcome gives success probability $3/4$ for either of these two fixed-basis measurements.

Put $q={\cos }^2(\pi /8)=(2+\sqrt{2})/4$ and $1-q={\sin }^2(\pi /8)$. In the computational basis, input $|a⟩$ has outcome probabilities $(q,1-q)$, while $|b⟩$ has $(1-q,q)$. The marginal outcomes are equally likely; the Bayesian posterior probabilities areGuess $a$ after zero and $b$ after one. Each guess is correct with probability $q\simeq 0.853553$ and wrong with probability $1-q\simeq 0.146447$. Neither outcome is certain evidence for one input.

For a real state $|{\psi }_{\theta }⟩=\cos \theta |0⟩+\sin \theta |1⟩$, the Born rule in the Hadamard basis gives $P(\pm \mid \theta )=(1\pm \sin 2\theta )/2$. Both angles $\pi /8$ and $3\pi /8$ have $\sin 2\theta =1/\sqrt{2}$. Thus both preparations give $P(+)=q$ and $P(-)=1-q$. By Bayes' theorem, either outcome leaves the two input hypotheses equally likely. The biased outcome frequencies carry no information about which state was prepared; the best success probability with this measurement is $1/2$.

The Bloch vectors of $|a⟩$ and $|b⟩$ areThey share the same $x$ component and differ along $z$. The equal-prior discrimination of qubit states therefore selects the $Z$ axis: it maximizes the separation of their projected outcome probabilities. This also explains why the $X$ measurement in part 3 is uninformative.

For $|0⟩$ and $|+⟩$, the Bloch vectors are $(0,0,1)$ and $(1,0,0)$. Their difference points along $\mathbf{n}=(-1,0,1)/\sqrt{2}$. Hence Carl W. Helstrom's Helstrom measurement for the zero and plus states measuresGuess $|0⟩$ after $P_{+}$ and $|+⟩$ after $P_{-}$. An explicit orthonormal basis for these projectors isThe positive measurement axis is tilted by $\pi /4$ from $+z$ towards $-x$ on the Bloch sphere; the corresponding state-vector rotation angle is $-\pi /8$.

For completeness, if a projective measurement guesses $0$ along a unit Bloch axis $\mathbf{n}$, its average success is $1/2+(n_z-n_x)/4$. The Cauchy-Schwarz inequality bounds $n_z-n_x\le \sqrt{2}$, with equality at the axis above. The Holevo–Helstrom theorem shows this optimum also holds among general POVMs. Thus