Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 57 / 4

Label the control qubits by binary weight: the bit $x_n$ contributes $2^n$ to $x={\sum }_{n=0}^{N-1}{2}^n{x}_n$. For each $n$, use that qubit as the control of the supplied controlled-$V^{2^n}$, with the same target eigenstate $|v_j⟩$. The controlled unitary gate acts bysince $V^{2^n}|v_j⟩=e^{i{2}^n{\varphi }_j}|v_j⟩$. The target remains unchanged after every application. Multiplying these factors givesThe ordering of the tensor factors is chosen to match the binary place values. Discarding the unchanged target leaves the required phase register. This uses quantum phase kickback in the phase accumulation step of quantum phase estimation.

Set $L=2^N$. Applying the inverse quantum Fourier transform to the register in part (a) gives probability amplitude at label $m$ equal toChoose $m$ nearest to $\eta =L{\varphi }_j/(2\pi )$, with $m$ reduced modulo $L$ at the phase boundary. Then the difference can be represented with $|\delta |\le \pi /L$. For $\delta \ne 0$, sum the finite geometric series to obtainLet $z=L|\delta |/2\le \pi /2$. On $[0,\pi /2]$, concavity of sine gives $\sin z\ge 2z/\pi$, the line through its endpoint values. Also $|\sin (\delta /2)|\le |\delta |/2$. Thereforeand the Born rule proves the nearest-integer success bound for quantum phase estimation:If $\delta =0$, every summand is one and the probability is exactly one. If the scaled phase is halfway between two integers, either nearest label separately satisfies the displayed bound. Interpreting the measured integer as $m/L$ gives the closest available $N$-bit phase estimate modulo one.

Choose an orthonormal eigenbasis of $V$, with eigenphases ${\varphi }_0,{\varphi }_1$, counting multiplicity. The maximally mixed state is basis independent, soLinearity means that the experiment is the equal mixture of the two eigenstate experiments. After the inverse quantum Fourier transform, the joint state is $\frac{1}{2}{\sum }_{j=0}^1|{\chi }_j⟩⟨{\chi }_j|\otimes |v_j⟩⟨v_j|$, where $|{\chi }_j⟩={\sum }_m{a}_m({\varphi }_j)|m⟩$ andThus the exact output distribution for quantum phase estimation with a maximally mixed target isFor a nonzero phase mismatch, the squared probability amplitude is ${\sin }^2[L(\varphi -2\pi m/L)/2]/\{L^2{\sin }^2[(\varphi -2\pi m/L)/2]\}$, with value one at an exact match. If both eigenphases are exactly representable on the $L$-point grid, their two labels occur with probability $1/2$ each; coincident labels have their probabilities added. For general eigenphases there can be nonzero probabilities at other labels, so it would not be exact to list only the two best approximations with probability $1/2$ each. Each eigenphase is sampled with spectral weight $1/2$, and its nearest-label event has conditional probability at least $4/{\pi }^2$. In particular a nearest label for one eigenphase has unconditional probability at least $2/{\pi }^2$, with additional contributions possible from the other phase.

With access to the uncontrolled physical gate, use single-qubit process tomography. Prepare the positive eigenstate of each of $X,Y,Z$, apply $V$, and on independent copies measure each of $X,Y,Z$. Arbitrary single-qubit rotation gates let one prepare these inputs and rotate each measurement into the computational basis. The nine measured means areThey determine the real rotation matrix $R$ of the Bloch vector, and hence the action of $V$ on every qubit state. To see precisely what this identifies, suppose another unitary $U$ has the same action. Then $V^{†}U$ commutes with every Pauli matrix. Commuting with $Z$ makes it diagonal; commuting with $X$ makes its diagonal entries equal. Therefore $U=e^{i\gamma }V$.

Consequently tomography determines $V$, and its eigenvalues, up to a common phase. More explicitly, write a representative as $V=e^{i\gamma }(\cos aI-i\sin a\mathbf{n}\cdot \mathbf{\sigma })$. Its eigenvalues are $e^{i(\gamma -a)}$ and $e^{i(\gamma +a)}$, and its Bloch vector action is rotation through $2a$ about $\mathbf{n}$. The measured rotation identifies the unordered relative eigenphase information; for example $\mathrm{Tr}R=1+2\cos (2a)$. Once a representative with a specified overall phase is supplied, diagonalizing that representative gives the corresponding two eigenvalues.

The common phase itself cannot be recovered with the stated uncontrolled access. Indeed, $(e^{i\gamma }V)\rho (e^{i\gamma }V{)}^{†}=V\rho V^{†}$ for every input, including inputs entangled with ancillas. Replacing $V$ by $e^{i\gamma }V$ therefore leaves every such preparation-and-measurement experiment unchanged, whereas it multiplies both eigenvalues by $e^{i\gamma }$. For example, $I$ and $e^{i\gamma }I$ are indistinguishable physical gates under these operations. This proves that the request for absolute eigenvalue phases needs an additional phase reference or controlled-gate assumption: the global phase of an uncontrolled quantum gate is unobservable. The controlled powers in the earlier parts supply that extra resource, since $\mathrm{diag}(I,e^{i\gamma }V)$ changes a relative phase between the two control branches.

The practical advantages are that no unknown eigenstate preparation, controlled powers, entangling gates or inverse quantum Fourier transform are needed: a fixed set of single-qubit preparations and measurements suffices. The disadvantages are statistical estimation and the phase ambiguity just proved. Estimating the bounded Pauli measurement means to additive accuracy $ϵ$ at fixed confidence requires order ${ϵ}^{-2}$ independent repetitions, giving the usual sampling cost for this direct procedure. The earlier quantum phase estimation circuit uses $N$ phase qubits and high controlled powers to resolve scale $2^{-N}$ in one coherent experiment with a constant success bound; constructing those powers from controlled uses of $V$ costs $2^N-1$ gate calls, so its exponential-looking precision is not obtained with only $N$ uses of $V$. Repetitions can raise confidence. Finally, full process tomography grows exponentially with the number of system qubits, whereas the eigenstate-based phase procedure can estimate a selected eigenphase without reconstructing the entire operator.