Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 53 / 1

Write $C_U=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes U$ for the controlled unitary gate. For a normalized eigenvector $|u⟩$, the first Hadamard gate and quantum phase kickback giveThe final Hadamard gate therefore produces the unentangled outputThe auxiliary system is unchanged, while its eigenphase becomes accessible through interference of the control qubit. The special unitary group assumption is not needed for this calculation; unitarity and the normalized eigenvector suffice.

Apply the Born rule to the two amplitudes in the introductory calculation. In the computational basis, the probabilities areThe other basis printed in this question is $(|0⟩\pm i|1⟩)/\sqrt{2}$, the Pauli Y gate eigenbasis. It is not the real Hadamard basis. Write the output amplitudes as $a=(1+e^{i\varphi })/2$ and $b=(1-e^{i\varphi })/2$. The plus-outcome amplitude is $(a-ib)/\sqrt{2}$ and the minus-outcome amplitude is $(a+ib)/\sqrt{2}$. Since $\mathrm{Im}(\overline{a}b)=-\sin \varphi /2$, the Born rule givesIn particular, at $\varphi =\pi /2$ the output is the minus eigenstate, which checks the sign of the imaginary quadrature.

Use separate fresh runs for each measurement basis. If ${\hat{P}}_j$ denotes the empirical outcome frequency, thenThe two differences estimate $\cos \varphi$ and $\sin \varphi$, resolving the ambiguity that would remain from the first basis alone. Each Bernoulli frequency converges by the strong law of large numbers, with standard error of order the inverse square root of the number of runs. One may project the noisy estimate onto the unit circle if a phase estimate is desired, but that projection is not necessary for consistency.

The mixed-state expression in the PDF contains a mismatched bra in its second term. A probability mixture of the stated orthonormal eigenvectors must be interpreted as $\rho ={\sum }_k{p}_k|u_k⟩⟨u_k|$. Taken literally with a different bra, it need not be Hermitian or positive and is not generally a density operator.

Let $m={\sum }_k{p}_k{e}^{i{\varphi }_k}=\mathrm{Tr}(\rho U)$. A quantum channel and its measurement probabilities are linear in the density operator, so averaging the pure-state results givesTherefore the same frequency estimator yieldsMore generally, after the controlled operation and before the final Hadamard gate, the reduced control density operator isThis follows by a partial trace of the control blocks $\rho$, $\rho U^{†}$, $U\rho$ and $U\rho U^{†}$. It proves the mixed-state Hadamard-test quadratures for any auxiliary density operator, even one not diagonal in the eigenbasis. Conjugation by the final Hadamard gate maps $X$ to $Z$ and $Y$ to $-Y$, explaining the imaginary-part sign.

Prepare the auxiliary system in the maximally mixed state $\rho =I/N$. Then $\mathrm{Tr}(\rho U)=\mathrm{Tr}U/N$, so the two measurement bases giveThere is no need to know the eigenvectors of $U$. Choose a uniformly random member of any known orthonormal basis on each run; discarding the classical preparation label produces $I/N$. Alternatively, discard one half of a maximally entangled pair of $N$-level systems. This is trace estimation by a maximally mixed Hadamard test.

The circuit directly estimates the normalized trace. To demand fixed absolute accuracy $\epsilon$ in the unnormalized trace requires normalized accuracy $\epsilon /N$; straightforward independent frequency sampling consequently uses order $N^2/{\epsilon }^2$ runs at fixed confidence. Thus the construction should not be confused with a dimension-independent guarantee for arbitrary absolute trace precision.