Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 3

Since $A$ is a bounded operator, its exponential power series converges in operator norm. The identity $A^2=I$ gives $A^{2k}=I$ and $A^{2k+1}=A$. Separating the even and odd powers therefore givesThe algebraic identity does not require $A$ to be Hermitian. Hermiticity is additionally needed for this exponential to be a unitary operator for real $\theta$.

A general single-qubit quantum gate belongs to $U(2)$. Remove its global phase to obtain $V\in SU(2)$; the displayed expression in the question is literally a special-unitary gate and represents a general physical gate up to that phase. Every $V\in SU(2)$ can be writtenand consequentlyChoose $\theta$ with $q_0=\cos (\theta /2)$ and $\Vert q\Vert =\sin (\theta /2)$, and set $n=q/\Vert q\Vert$ when $q\ne 0$. The Pauli matrix multiplication law shows $(n\cdot \sigma {)}^2=\Vert n{\Vert }^{2}I=I$. Applying part (a) provesFor $q=0$, $V=\pm I$ and any axis can be used with angle zero or $2\pi$. Thus the phase-free gate is a rotation gate by angle $\theta$ about the unit axis $n$. An arbitrary exact $U(2)$ operator also needs the scalar phase $e^{i\chi }$, which cannot in general be produced by a traceless Quantum Hamiltonian.

With $B_y=0$, a pulse of the $x$ field realizes $R_x(\theta )=e^{-i\theta {\sigma }_x/2}$ when $\int B_x(t)dt=\theta /2$. Similarly the $y$ field realizes $R_y(\theta )$ with area $\theta /2$. These are instances of the quantum pulse area rule. A rotation about the z-axis can be synthesized from the two available axes:because $R_x(\pi /2){\sigma }_y{R}_x(-\pi /2)={\sigma }_z$.

For a target $V=\left(\begin{array}{cc}u& v\\ -v^{\ast }& u^{\ast }\end{array}\right)$, one convenient construction isTake $\beta =2atan2(|v|,|u|)$ and choose $\alpha +\gamma =-2\arg u$, $\alpha -\gamma =-2\arg (-v)$ when both entries are nonzero; if an entry vanishes its phase constraint can be omitted. Substitute the three-pulse construction for each $z$ rotation. Apply the factors from right to left in time. Pulses along the available $x,y$ axes therefore implement every single-qubit gate up to global phase. Negative angles use negative field area, or an equivalent full-period rotation.

Write $X={\sigma }_x$, $Y={\sigma }_y$, $Z={\sigma }_z$. The specified diagonal two-qubit gate has the exact factorizationThus one exact implementation is to set the controllable Ising coupling to zero and perform the two local rotations about the z-axis, synthesized from the available $x,y$ pulses as in part (c). Alternatively, turn off the local fields and choose the interaction area $\int J_{12}(t)dt=\pi /2$. Thenwhich implements the same physical gate up to global phase. The printed gate is local, not entangling; the pulse construction does not change that fact.

The claimed universality is false for the particular gate specified in part (d). Every allowed single-qubit gate is local, and $C_{\mathrm{phase}}=-Z\otimes Z$ is local too. Products of these gates remain of the form $e^{i\chi }A\otimes B$. They preserve all product states and cannot, for example, implement the CNOT gate that sends $|+⟩\otimes |0⟩$ to $(|00⟩+|11⟩)/\sqrt{2}$. Equivalently, its diagonal two-qubit phase entanglement criterion gives the alternating phase $\pi -0-0+\pi =0$ modulo $2\pi$. No construction with only the printed phase gate and local rotations implements arbitrary two-qubit gates.

The intended universality argument works after replacing that local gate by an entangling controlled phase. The same tunable Ising coupling suppliesThus this quarter-area Ising pulse, with local corrections, realizes the Controlled-Z gate. Conjugating it by the Hadamard gate on the second qubit gives $\mathrm{CNOT}=(I\otimes H)\mathrm{CZ}(I\otimes H)$, with harmless scalar phases if the Hadamards are implemented using traceless control Hamiltonians.

The Cartan decomposition of a two-qubit gate writes a target aswith local $SU(2)$ factors. The three central Pauli-product operators commute, so their exponential is a product of three variable interaction gates. From two fixed CNOT gates and a variable local rotation about the z-axis,because $\mathrm{CNOT}(I\otimes Z)\mathrm{CNOT}=Z\otimes Z$. Conjugating this interaction by $R_y(\pi /2)$ on both qubits yields $e^{-icX\otimes X}$; conjugating by $R_x(-\pi /2)$ on both yields $e^{-icY\otimes Y}$. Part (c) implements all outer local factors. This gives the full requested Cartan construction with an entangling phase gate, and pinpoints the defect in the printed version. Exact arbitrary scalar phase requires an identity Hamiltonian term; the physical gate construction is up to global phase.

An always-on Ising coupling produces evolution during every nominally local pulse, and generally does not commute with the applied $x,y$ Quantum Hamiltonians. A naive product of single-qubit rotations therefore accumulates unwanted conditional phases and can entangle the qubits. Idling is also no longer an identity operation.

For a pulse of duration $\tau$, neglecting the coupling requires $|J_{12}|\tau \ll 1$ in units $\hslash =1$, or $|J_{12}|\tau /\hslash \ll 1$ in ordinary units. For order-one rotation angles, this is the strong-local-drive regime $|B|\gg |J_{12}|$. Large drive amplitude alone does not justify neglect over a long total sequence: the accumulated interaction time or an appropriate error bound must also be small.

If it cannot be neglected, include it in the quantum optimal control model, or refocus it. A $\pi$ pulse $P=R_x^{(1)}(\pi )$ obeys $P(Z\otimes Z)P^{†}=-Z\otimes Z$, giving the ideal Ising spin echoThis cancellation assumes the refocusing pulses are instantaneous relative to $1/|J_{12}|$, or that the coupling during finite pulses is compensated. It explains how fast local controls can suppress the fixed coupling without mistaking it for a controllable zero interaction.