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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.

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.

Integrating the two vector fields gives their complete flow maps:The first is a simultaneous rotation about the z-axis through angle $t$ and a dilation by $e^t$ in the $z$-direction. The second is an isotropic dilation about the origin. Both satisfy the group law for all real $t$ and have inverse given by time $-t$. In particular ${\varphi }_t$ is not a pure rotation unless restricted to the plane $z=0$.

Since $A|x⟩=(-1{)}^{f(x)}|x⟩$, the spectral decomposition givesReplace the middle gate of the previous compute-phase-uncompute construction bya one-qubit rotation about the z-axis. ThenThe circuit uses the gates of $C$, their inverses, and one fixed rotation; it returns the quantum ancilla to $|0⟩$. The same construction implements $e^{itA}$ with middle gate $e^{itZ}$, an instance of simulation of a computable diagonal Hamiltonian.