Past exam of the laser science and quantum information processing course of the University of Oxford / 2021 / 3

Use the source's north-pole assignment $|1⟩\leftrightarrow (0,0,1)$. Fix the ordered basis as $(|1⟩,|0⟩)$ and use the standard Pauli matrices in that order, soA pure state has $|\mathbf{r}|=1$: examples are $|1⟩$ at the north pole and $(|1⟩+|0⟩)/\sqrt{2}$ at $(1,0,0)$. An incoherent mixed statehas $\mathbf{r}=(0,0,2p-1)$ inside the Bloch sphere. At $p=1/2$ it is the origin, distinct from an equatorial pure state despite equal populations. The distinction is the off-diagonal quantum coherence of the density matrix.

For the stated Hamiltonian operator, put $\mathbf{\Omega }=(V,0,-\delta )$ and $\Omega =\sqrt{V^2+{\delta }^2}$. The Bloch vector obeysThus it rotates right-handedly about the fixed axis $\mathbf{\Omega }$, preserving its length and its projection on that axis. For example, on resonance $\delta =0$, a north-pole Bloch vector follows $(0,-\sin Vt,\cos Vt)$ and reaches the south pole at $t=\pi /V$.

Take $V>0$, with the field phase chosen accordingly. The Hadamard gate is $H=({\sigma }_x+{\sigma }_z)/\sqrt{2}$. Align the axis with $(1,0,1)$ and rotate through $\pi$:The factor $-i$ is an irrelevant common global phase. This realizes the whole quantum gate, rather than merely its action on one input quantum state.

There is a basis-ordering convention to make explicit: this $H$ is the Hadamard gate matrix in the ordered basis $(|1⟩,|0⟩)$ chosen in part (a). If instead its definition is the usual ket rules $H_{01}|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H_{01}|1⟩=(|0⟩-|1⟩)/\sqrt{2}$, its matrix in our order is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$. For that definition choose $\boxed{\delta =+V,\tau =\pi /(\sqrt{2}V)}$. Both conventions use the same pulse duration; the sign of the Rabi detuning selects the desired axis.

At resonance, the Hamiltonian operator becomes $\hslash V{\sigma }_x/2$. ThereforeIt gives the requested positive rotation about the x-axis. In the ordered basis $(|1⟩,|0⟩)$, $|1⟩$ becomes $(|1⟩-i|0⟩)/\sqrt{2}$, whose Bloch vector is $(0,-1,0)$.

A resonant Rabi pi pulse givesIt swaps $|0⟩$ and $|1⟩$ with a common global phase, implementing the Pauli gate $X$, or NOT. Its Bloch sphere action is a half-turn about the $x$-axis.

Keep the Pauli matrix and north-pole conventions of part (a), and neglect Rabi detuning during each short pulse. Write $P=R_x(\pi /2)$ and $\varphi =\delta T$. During the first pulse the Bloch vector is $(0,-\sin \theta ,\cos \theta )$ for $0\le \theta \le \pi /2$. During free evolution,because the Hamiltonian operator generates $R_z(-\delta t)$. During the final pulse it isThese formulas specify every arc in the two Bloch sphere sketches. For $T=\pi /(2\delta )$ the path is north $\to -y\to -x$, and the last pulse leaves $-x$ fixed. For $T=\pi /\delta$ the path is north $\to -y\to +y\to$ north.

To test the complete quantum gate, multiply its unitary operators:It is proportional to the Hadamard gate when $\cos (\varphi /2)=-\sin (\varphi /2)$, hence the shortest positive waiting time isMore generally $\delta T=-\pi /2(\mathrm{mod}2\pi )$. With the alternative conventional ket definition $H_{01}$ specified in part (b)(i), the required matrix is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$, so $\boxed{T_{H_{01}}=\pi /(2\delta )}$ and $U_R=-i{H}_{01}$. The distinction concerns which basis order defines the named Hadamard gate; the evolution and the return population do not depend on that naming choice. Finally $r_z=-\cos \varphi$, so the actual return population isThe printed plus sign is inconsistent with the two identical pulses. At $T=0$, $U_R=P^2=-i{\sigma }_x$, so $|1⟩$ becomes $|0⟩$ and ${\rho }_{11}(0)=0$; the printed expression gives one. This counterexample is independent of Bloch sphere labeling. A plus sign would describe the complementary population ${\rho }_{00}$, or the return population with an opposite-sign final pulse. That changed sequence also changes the gate question and must not be silently substituted.

Interpret the unspecified Gaussian width $\Delta$ as the standard deviation, soThe characteristic function of this Gaussian distribution isAverage the free-evolution Bloch vectors from part (c):It starts at $-y$ and spirals towards the origin in the equatorial plane, with period $2\pi /{\delta }_0$ and shrinking length $e^{-{\Delta }^2{T}^2/2}$. Individual pure states retain unit-length Bloch vectors; only ensemble averaging produces the apparent loss of quantum coherence.

After the final pulse, the corrected return population isThe oscillations lie between $[1\pm e^{-{\Delta }^2{T}^2/2}]/2$, start at zero, and tend to $1/2$. The upper envelope has no implication of exponential energy relaxation: this is static inhomogeneous dephasing. If instead the intended Gaussian convention was $p\propto e^{-(\delta -{\delta }_0{)}^2/{\Delta }^2}$, every envelope above becomes $e^{-{\Delta }^2{T}^2/4}$. The printed-plus-sign convention would give the complementary fringe.

An ideal instantaneous NOT about $x$ reverses each atom's accumulated phase relative to its subsequent free precession. Algebraically, $X{\sigma }_{z}X=-{\sigma }_z$, so for a static Rabi detuning and $F=e^{i\delta T{\sigma }_z/4}$,The entire Ramsey interferometry sequence is consequently $PXP=-iI$, up to the additional common global phase if NOT is implemented as $-iX$. ThusThis spin echo restores the initial population even for the distribution of static Rabi detunings in part (d). It removes reversible inhomogeneous dephasing; fluctuating Rabi detuning, finite pulse errors and irreversible loss of quantum coherence need not refocus. With an opposite final pulse, the return population would instead be zero; this distinguishes that alternative from the sequence actually specified.