Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 61 / 2 / d / Solution

Use pulse areas in the convention ${\theta }_k=\int {\Omega }_{k}dt/\hslash$, with the other channel switched off for a single-channel pulse. Substituting $\theta =\pi /2$ into part (c) showsHere the arguments denote area and phase. For $W_1$, the off-diagonal factors are $-i{e}^{-i\pi /2}=-1$ and $-i{e}^{i\pi /2}=1$. For $W_2$, the phase $-\pi /2$ reverses those two signs. Thus these choices implement the gates exactly on all basis states, not only on one population.

To implement $W_4$, let $P_1=U_1(\pi /2,0)$ and $P_2=U_2(\pi /2,0)$ and apply channel 1, channel 2, then channel 1 again. Each active swap contributes $-i$, so direct tracking givesHence $\boxed{W_4=P_1{P}_2{P}_1}$, including the overall minus sign and middle-state phase actually printed in the PDF. Alternatively equal zero-phase envelopes give $H(t)=\Omega (t)C$, where $C=E_{12}+E_{21}+E_{23}+E_{32}$. Since $C^3=2C$, its exponential is $I+[(\cos \sqrt{2}\theta -1)/2]C^2-i(\sin \sqrt{2}\theta /\sqrt{2})C$. One simultaneous pulse with $\theta =\pi /\sqrt{2}$ gives $I-C^2=-P_{13}=W_4$.

For population transfer alone, the two zero-phase pulses $P_1$ followed by $P_2$ take $|1⟩\mapsto -i|2⟩\mapsto -|3⟩$, so the target population is exactly one. If a positive final amplitude is desired, phases ${\varphi }_1={\varphi }_2=\pi /2$ give $|1⟩\mapsto |2⟩\mapsto |3⟩$. Both choices require area $\pi /2$ on each channel.