Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 61 / 4

The feedback law printed in the PDF is not sufficient for the asserted monotonicity. We first derive the exact identity, then give a counterexample and the corrected law. Both density operators evolve unitarily, so $\mathrm{Tr}{\rho }^2$ and $\mathrm{Tr}{\rho }_d^2$ are constant. ThusCyclicity of the trace gives $\mathrm{Tr}([-i{H}_0,{\rho }_d]\rho )=-\mathrm{Tr}({\rho }_d[-i{H}_0,\rho ])$. These drift terms cancel in the derivative of the overlap, leavingHere $C$ is real: $-i[H_1,\rho ]$ is Hermitian, and the trace of the product of two Hermitian matrices is real. This is the Hilbert-Schmidt Lyapunov feedback identity.

For a direct counterexample to the source's overlap law, take $H_0=0$, $H_1={\sigma }_y/2$, ${\rho }_d=(I+{\sigma }_z)/2$ and $\rho =(I+{\sigma }_x)/2$. All are admissible qubit operators and states. The printed overlap is $f=\mathrm{Tr}({\rho }_d\rho )=1/2$, but $-i[H_1,\rho ]=-{\sigma }_z/2$ gives $C=-1/2$. Hence $\dot{V}=1/2>0$: the distance initially increases. This proves that feedback based on state overlap need not be stabilizing.

The intended Lyapunov quantum control replaces the overlap by its control-direction derivative:The conclusion is nonincrease, not strict decrease or automatic convergence. For example, opposite computational-basis pure states, diagonal $H_0$, and $H_1={\sigma }_x$ give $C=0$ and zero control even though $V=2$. Target convergence requires further invariant-set and reachability conditions. A target with a different spectrum is also inaccessible under Quantum Hamiltonian evolution. The density-operator derivatives in the governing equations use the dots visible in the PDF, which the TeX loses.

The deterministic law cannot be transplanted unchanged into measurement-based quantum feedback. Evaluating its state-dependent scalar requires knowledge of the current quantum state. In fact $C=\mathrm{Tr}(i[H_1,{\rho }_d]\rho )$ is the expectation value of a Hermitian observable, but that expectation cannot be obtained exactly from a single system without measurement backaction. Tomography uses many independently prepared copies, not a nondisturbing readout of one evolving state.

Continuous weak measurements can supply a noisy record and a conditional state estimate, but then the state follows stochastic measurement dynamics with additional disturbance and decoherence, rather than the closed unitary equation used in part (a). A new stochastic stability analysis is required. One may instead simulate the corrected Lyapunov quantum control from known initial data and apply its waveform as open-loop control, or learn a pulse across repeated experiments as in Question 3. Those procedures do not constitute real-time state feedback on a single quantum trajectory. The separate source error in part (a) remains: even ideal knowledge of the state would not make its printed overlap law monotone.

Assume $\gamma >0$ and zero initial cavity amplitude. Taking Laplace transforms givesEliminating the mode yields the passive cavity input-output transfer functionIts unique pole is $s=-\gamma /2$, so the cavity's internal mode decays and its causal input-output response is stable. A nonzero initial amplitude would add $\sqrt{\gamma }a(0)/(s+\gamma /2)$ to the output, a decaying transient.

For a precise Nyquist stability criterion interpretation of this open cavity, rewrite the mode equation as an integrator with negative state feedback. Its return ratio is $L(s)=(\gamma /2)/s$, and its characteristic equation is $1+L(s)=0$. The frequency locus $L(i\omega )=-i\gamma /(2\omega )$ lies on the imaginary axis. The Nyquist indentation excluding the pole at the origin maps to a large semicircle in the right half-plane; the whole contour has no winding around $-1$. There are no right-half-plane open-loop poles within that indented contour, so there are no unstable characteristic zeros. There is also no zero-frequency characteristic root, since the physical characteristic polynomial is $s+\gamma /2$. This recovers asymptotic stability.

Do not use the plant $G$ as a return ratio without specifying an additional feedback connection. Indeedand its locus passes through $-1$ at $\omega =0$. That would indicate a marginal mode for a new unity-negative-feedback loop with characteristic factor $1+G$, but not instability of the original cavity. Part (e) supplies the actual return ratio $\alpha G$ for the beam-splitter network.

The beam splitter relation and the cavity output equation giveFor $\alpha \ne -1$, solve this scalar algebraic relation and substitute back:The cavity equation becomesFinally, use the other beam-splitter output and ${\alpha }^2+{\beta }^2=1$:These derivations retain the printed input $b_{min}$ in both beam-splitter equations; the converted TeX changes one of these subscripts. Define ${\gamma }_{meff}=\gamma (1-\alpha )/(1+\alpha )$ and $c=\beta \sqrt{\gamma }/(1+\alpha )$, so $\dot{a}=-{\gamma }_{meff}a/2-c{b}_{min}$ and $b_2=b_{min}+ca$. In the nondegenerate range $|\alpha |<1$, ${\gamma }_{meff}>0$ and $c^2={\gamma }_{meff}$, the expected passive input-output relation. This is instantaneous coherent cavity feedback through a beam splitter; it uses the zero-delay interconnection assumed by the equations.

The stated $M$ is specifically the transfer from the external input to the internal field $b_1$. Combining $b_1=G{b}_0$ with $b_0=\beta b_{min}-\alpha b_1$ givesFor $\gamma >0$ and a nondegenerate beam splitter $|\alpha |<1$, the closed-loop pole is $-{\gamma }_{meff}/2<0$. The return ratio for the Nyquist stability criterion is $L=\alpha G$. Its frequency locus is a circle of radius $|\alpha |<1$, so it cannot reach or encircle $-1$. The open cavity has no unstable poles; hence the closed loop has none. This agrees with the directly derived state damping rate.

The externally accessible output is different. From $b_2=\alpha b_{min}+\beta b_1$,It is again an all-pass cavity response. Confusing $b_1$ and $b_2$ would give the wrong output steady state.

For a constant coherent input mean $B=⟨b_{min}⟩$, the steady mean cavity amplitude follows by setting its derivative to zero:The input-output values also follow from $G(0)=-1$ and $M(0)=-\beta /(1-\alpha )$. For any initial mean amplitude, its difference from the stationary mean decays as $e^{-{\gamma }_{meff}t/2}$.

The stability statement needs endpoint qualifications. At $\alpha =1$, normalization forces $\beta =0$, and the cavity obeys $\dot{a}=0$: it is isolated and undamped, not asymptotically stable. Its external output is nevertheless the decoupled input. At $\alpha =-1$, the algebraic elimination divides by zero; the instantaneous feedback connection is singular and is not a well-posed instance of these reduced equations. Thus strict cavity stability holds for $|\alpha |<1$, not for every real pair satisfying only ${\alpha }^2+{\beta }^2=1$.

Finally, $b_{min}(t)$ is a quantum stochastic field, so arbitrary inputs do not approach a literal time-independent operator amplitude. The stationary expressions above describe a constant coherent drive's means, or the formal zero-frequency response. The full solution contains the filtered input-noise convolutionThe stationary cavity density operator also depends on the input noise statistics. For this ideal passive cavity with a coherent drive and vacuum noise, it is the coherent state with the amplitude just calculated; without such input statistics, the gain function alone does not specify a complete stationary quantum state.