Past exam of the computer science course of the University of Oxford / 2020 / Quantum information / 1 / b

For an orthonormal basis, choosing coefficients with magnitudes $\sqrt{p_j}$ gives the required Born rule probabilities. More generally, ${\sum }_j\sqrt{p_j}{e}^{i{\theta }_j}|{\varphi }_j⟩$ realizes the same distribution for arbitrary phases. The following are convenient examples of pure-state realization of a basis measurement distribution.

Choose $|\psi ⟩=|0⟩$. Since $|0⟩=(|{\varphi }_0⟩+|{\varphi }_2⟩)/\sqrt{2}$, the probability amplitudes in the measurement basis are $(1/\sqrt{2},0,1/\sqrt{2})$. Their squared moduli give $(1/2,0,1/2)$ by the Born rule.

Choose $|\psi ⟩=|{\varphi }_2⟩$. Its inner products with the three measurement vectors are $(0,0,1)$, so the Born rule gives certainty for the outcome labelled 2 and zero probability for the other two outcomes. This uses the measurement basis, rather than assuming that its label 2 means the initial state must be the computational vector $|2⟩$.

ChooseOrthonormality makes this a unit vector. Its probability amplitudes in the given measurement basis are $(1/\sqrt{2},1/\sqrt{2},0)$, so the required probabilities are $(1/2,1/2,0)$.