Past exam of the laser science and quantum information processing course of the University of Oxford / 2022 / 5

Assume $|\alpha {|}^2+|\beta {|}^2=1$. Let Alice's first recorded quantum measurement outcome be $a\in \{0,1\}$. Immediately afterward she assigns the conditional pure quantum state $|a⟩$, with probabilities $p_0=|\alpha {|}^2$, $p_1=|\beta {|}^2$. Bob, who has no record, assigns the density operatorThis is a nonselective quantum measurement, which removes the initial computational basis coherence. It is not generally the original pure quantum state.

After the Hadamard gate, Alice assigns $H|a⟩=|+⟩$ for $a=0$, or $|-⟩$ for $a=1$. Bob assignsBoth $|+⟩$ and $|-⟩$ give either final computational basis outcome with probability $1/2$. Thus, after Alice records the second outcome $b$, she assigns $|b⟩$, whereas Bob assignsAlice's conditional quantum states are pure; Bob's are averages over her unobserved records. They agree on all probabilities when supplied the same information. Their first descriptions can coincide if the initial quantum state is a computational basis state; after the second quantum measurement, Bob's maximally mixed state differs from each conditional pure quantum state.

Expand the possibly unnormalized second-qubit kets as $|\alpha ⟩=a_0|0⟩+a_1|1⟩$ and $|\beta ⟩=b_0|0⟩+b_1|1⟩$. The joint quantum state isThe Born rule in the orthonormal computational basis gives $P_{00}=|a_0{|}^2$, $P_{01}=|a_1{|}^2$, and similarly for the two outcomes beginning with one. Ignoring the second outcome therefore givesEquivalently these are the squared norms of $(|0⟩⟨0|\otimes I)|\psi ⟩$ and $(|1⟩⟨1|\otimes I)|\psi ⟩$. Orthogonality of the first qubit's basis gives $P_0+P_1=⟨\psi |\psi ⟩=1$, even if $⟨\alpha |\beta ⟩$ is nonzero. For a nonzero-probability outcome, the conditional second-qubit quantum state is the corresponding ket divided by $\sqrt{P_i}$.

Read the source network as a Hadamard test: a Hadamard gate on the first qubit, a controlled unitary gate with the first qubit as control, and another Hadamard gate on that control. Starting from $|0⟩|\varphi ⟩$, the successive quantum states areApply b to the last line. Since $U^{†}U=I$ by unitarity,For the Pauli Z gate observable on the first qubit, the two eigenvalues are $+1$ and $-1$. ThusAs a check, $U=I$ returns outcome zero certainly, while $U=-I$ returns one certainly. The controlled unitary gate makes this relative phase measurable even though multiplying an isolated system's quantum state by $-1$ is only a global phase.

For second-qubit inputs $|0⟩$ and $|1⟩$, c gives control expectation values $\mathrm{Re}{U}_{00}$ and $\mathrm{Re}{U}_{11}$. Preparing the maximally mixed state $I/2$ is equivalent to choosing these inputs independently with equal probability and discarding the choice. Linearity of the density operator evolution and quantum measurement probabilities therefore givesHenceThe factor two is essential: the Hadamard test directly estimates the normalized trace. More generally, a target density operator $\rho$ gives $⟨Z⟩=\mathrm{Re}\mathrm{Tr}(\rho U)$. This argument never requires knowledge of the eigenvectors of $U$.

Yes. Replace the target qubit by a register of $k$ qubits and implement the controlled unitary gateThe algebra in c uses only normalization and unitarity, so for an arbitrary pure register quantum state $|\varphi ⟩$ it still gives $⟨Z⟩=\mathrm{Re}⟨\varphi |U|\varphi ⟩$. For a mixed register quantum state $\rho$, linearity gives $\mathrm{Re}\mathrm{Tr}(\rho U)$. With $\rho =I/2^k$,Only the control qubit needs to be pure. A uniformly randomized computational basis preparation supplies the target maximally mixed state. This is the principle of trace estimation by a maximally mixed Hadamard test. The extension is algebraic; efficiently implementing controlled-$U$ still requires a known suitable gate construction. Moreover, estimating the unnormalized trace to fixed additive precision requires increasingly precise estimation of the normalized signal as $k$ grows. There is no automatic efficient reconstruction of all entries of $U$.