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

From $A^2=I$, induction gives $A^{2n}=I$ and $A^{2n+1}=A$. The matrix exponential converges absolutely in finite dimension, so its even and odd terms may be separated:This involution exponential formula requires no Hermiticity assumption on $A$. Hermiticity is needed if the exponential is additionally asserted to be unitary.

The Pauli matrices obey ${\sigma }_x^2={\sigma }_y^2=I$ and ${\sigma }_x{\sigma }_y+{\sigma }_y{\sigma }_x=0$, so $A=\cos \varphi {\sigma }_x+\sin \varphi {\sigma }_y$ has $A^2=I$. Its entries are $A_{12}=\cos \varphi -i\sin \varphi =e^{-i\varphi }$ and $A_{21}=e^{i\varphi }$, with zero diagonal. Substitute this matrix into the involution exponential formula:The drive phase selects the transverse rotation axis, while the quantum pulse area selects its rotation angle.

For each fixed phase ${\varphi }_k$, the single-transition Quantum Hamiltonian is ${\Omega }_k(t)H_k({\varphi }_k)$, so its generators at different times commute. Its exact propagator depends only on ${\theta }_k(t)={\hslash }^{-1}{\int }_0^t{\Omega }_k(\tau )d\tau$. On the active two-level subspace, $H_k^2=I$; on the spectator state, $H_k=0$. Thus it is an embedded two-level quantum rotation, not an involution on the whole three-dimensional space. Explicitly,The spectator entries in both matrices are exactly one; the converted TeX damages the second matrix, so these entries use the PDF.

The proposed combined exponential is false in general for two independent envelopes. Direct multiplication givessoThe constant pulse-axis condition for removing time ordering holds if both envelopes are proportional to the same function, but not for generic independent controls. For example, turn on only the first transition, then only the second. The result is $e^{-i{\theta }_2{H}_2}{e}^{-i{\theta }_1{H}_1}$, whose Baker--Campbell--Hausdorff formula includes a nonzero commutator correction, rather than simply $e^{-i({\theta }_1{H}_1+{\theta }_2{H}_2)}$.

There is also a literal source defect: the PDF's right-hand exponent omits the factor $-i$ on its second term. With ${\theta }_1=0$ it would give $e^{{\theta }_2{H}_2}$, which has eigenvalues $e^{\pm {\theta }_2}$ and is not unitary for real nonzero ${\theta }_2$, whereas the true propagator is unitary. Even after repairing that missing factor, time ordering is still necessary unless the commutators vanish.

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.

For real nonnegative envelopes, put $\Omega =\sqrt{{\Omega }_1^2+{\Omega }_2^2}$ and choose a continuous $\theta =atan2({\Omega }_1,{\Omega }_2)$. The state $|D⟩=\cos \theta |1⟩-\sin \theta |3⟩$ is dark because its two amplitudes feeding $|2⟩$ cancel: ${\Omega }_1\cos \theta -{\Omega }_2\sin \theta =0$. The orthogonal bright combination $|B⟩=\sin \theta |1⟩+\cos \theta |3⟩$ couples to $|2⟩$ with strength $\Omega$. Thus the other eigenstates are $(|B⟩\pm |2⟩)/\sqrt{2}$ with energies $\pm \Omega$.

Start with ${\Omega }_1=0$ and ${\Omega }_2\ne 0$, so $\theta =0$ and $|D⟩=|1⟩$. Slowly increase ${\Omega }_1/{\Omega }_2$ to infinity while maintaining a nonzero gap; end with ${\Omega }_2=0$ and ${\Omega }_1\ne 0$, so $|D⟩=-|3⟩$. This target-to-intermediate pulse first is the counterintuitive ordering of stimulated Raman adiabatic passage. Its quantitative slowness condition isIt follows from $|⟨\pm |\dot{D}⟩|=|\dot{\theta }|/\sqrt{2}$ and the gap $\Omega$. Avoid switching both envelopes off while $\theta$ is changing; switch off together only after the transfer angle has become stationary. The dark eigenvalue is zero, so there is no dynamical phase; with this real eigenvector $⟨D|\dot{D}⟩=0$ as well. The ideal adiabatic quantum control therefore transfers $|1⟩$ to $-|3⟩$, physically the same target ray.

The adiabatic dark state of a driven Hamiltonian has exactly zero component on the lossy intermediate level throughout the ideal passage. It therefore suppresses spontaneous emission, whereas the two-pulse route explicitly transfers the population into $|2⟩$ before moving it to $|3⟩$. For an intermediate-level decay rate $\Gamma$, the leading loss probability is controlled by $\Gamma \int P_2(t)dt$. In stimulated Raman adiabatic passage, nonadiabatic leakage makes $P_2$ of order $(\hslash \dot{\theta }/\Omega {)}^2$, rather than the order-one intermediate population of the sequential swaps. Transfer is also less sensitive to precise pulse areas, provided the ordering, gap and slowness requirements hold. Making the process slower indefinitely is not always beneficial if the outer states also decohere; their coherence must survive the total passage duration.