Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 5

Use units $\hslash =1$. A Hermitian quadrature coupling of a cavity mode to a continuum of external modes can be writtenwith real coupling amplitude $\kappa$. The free Quantum Hamiltonian is $H_0={\omega }_c{a}^{†}a+\int \omega b_{\omega }^{†}{b}_{\omega }d\omega$. In the interaction picture, the two photon-exchange products $a{b}_{\omega }^{†}$ and $a^{†}{b}_{\omega }$ acquire phases with frequency $\omega -{\omega }_c$, whereas $a^{†}{b}_{\omega }^{†}$ and $a{b}_{\omega }$ oscillate at $\omega +{\omega }_c$.

For weak coupling near resonance, the rotating-wave approximation drops those rapidly oscillating pair-creation and pair-annihilation terms and retainsFor a broadband flat coupling $\kappa =\sqrt{\gamma /(2\pi )}$, define the envelope annihilation operator $b(t)=(2\pi {)}^{-1/2}\int b_{\omega }{e}^{-i(\omega -{\omega }_c)t}d\omega$. The Markov approximation extends the detuning integral over the full real line and gives $[b(t),b^{†}(s)]=\delta (t-s)$. Then the cavity-reservoir interaction isThe phases and normalization of the external field fix the sign convention. The creation operators and annihilation operators exchange one excitation between cavity and field, rather than creating or destroying a pair.

Under the Markov approximation, a short incoming time bin is independent of the earlier input and commutes with adapted cavity operators. Write its quantum noise increment as $d{B}_{\mathrm{in}}(t)={\int }_t^{t+dt}b(\tau )d\tau$; this is the increment denoted $B_{\mathrm{in}}(t)$ in the short-time definition, not an accumulated integral from zero.

During this bin, use the cavity operator at its left endpoint. Integrating the cavity-reservoir interaction and applying $U=e^{-i\int H_{I}d\tau }$ yields the bin propagatorThe exponent is anti-Hermitian, so the bin evolution is unitary. This is the short-time stochastic form: the time-bin limit, or equivalently the time-ordered product of these adapted propagators, gives continuous evolution. The quantum noise increments scale as $\sqrt{dt}$; therefore one must retain the square of this exponent, rather than treating its integral as an ordinary order-$dt$ bounded Hamiltonian.

Put $K=\sqrt{\gamma }(a\otimes d{B}^{†}-a^{†}\otimes dB)$, with the time argument suppressed. Since the incoming field commutes with the adapted cavity operators,Applying the Gaussian quantum noise Ito table givesThe power series expansion $e^K=I+K+K^2/2+\cdots$ consequently becomes, through Ito order $dt$,This has precisely the required quadratic drift correction. The short-time ordering is essential: the linear noise term is order $\sqrt{\gamma dt}$ and the drift is order $\gamma dt$; higher stochastic orders are discarded in the differential limit. A remainder written merely as a power of $\gamma$ is a formal coupling expansion, not a dimensionally complete statement of the short-time error.

As a sign check, $K^{†}=-K$ and $Q^{†}=Q$. Thus $U_I^{†}{U}_I=I-\gamma Qdt-K^2=I$ through this order. The cross term is needed for stochastic unitarity.

The Heisenberg update is $ds=U^{†}sU-s$; the left side of the printed first update relation must be this increment, rather than $s(t)$ itself. For $s=a$, the linear noise term in the supplied stochastic equation isby the bosonic canonical commutation relation $[a,a^{†}]=I$.

The two thermal drift expressions simplify separately toThe anomalous terms vanish because $[a^{†},[a^{†},a]]=0$ and $[a,[a,a]]=0$. Hence the factor inside the drift braces is $-(N+1)a+Na=-a$, independent of $N$ and $M$. ThereforeEquivalently, in the formal white-noise notation of the cavity-reservoir interaction,The dot is a derivative notation for this stochastic differential equation, not a claim that the white-noise field is an ordinary differentiable function.

Take unilateral Laplace transforms of the cavity equation, keeping the initial operator:The input-output relation givesA transfer function describes the forced zero-state part of the response. Thus the passive cavity input-output transfer function isThe omitted initial-state term is a decaying transient $\sqrt{\gamma }{e}^{-\gamma t/2}a(0)$, and is present for a general initial cavity state. This distinction avoids silently setting a physical annihilation operator equal to zero.

For a specified negative-feedback loop with return ratio $L(s)$, let $P$ be its number of open-right-half-plane poles. Let $N_{\mathrm{cw}}$ be the signed clockwise encirclement count of $-1$ by the standard Nyquist stability criterion contour. Provided the contour does not hit a singularity or the critical point, the number of right-half-plane closed-loop poles is $Z=P+N_{\mathrm{cw}}$. Thus closed-loop stability requires $N_{\mathrm{cw}}=-P$ and no closed-loop pole on the imaginary axis, with the usual well-posedness and absence of hidden unstable cancellations. For an already stable return ratio, there must be no encirclement of $-1$ and no passage through it.

The open cavity itself, for $\gamma >0$, has state eigenvalue and transfer-function pole $-\gamma /2$. Its initial transient decays and its impulse response is a direct-feedthrough term plus a decaying exponential. Hence the open-loop cavity is asymptotically stable and its transfer function is BIBO stable. The right-half-plane zero at $+\gamma /2$ does not make the open cavity unstable. On the frequency axis,Its unit-circle plot passes through $-1$ at $\omega =0$. A Nyquist stability criterion is a feedback criterion; this passage is not by itself a test of open-plant stability.

If the intended interpretation is a unit negative-feedback loop with $L=G$, thenThat loop has a pole at zero, so unit negative feedback is marginal rather than asymptotically stable, and the critical-point passage correctly prevents a strict Nyquist stability conclusion. Without a specified feedback interconnection, the unambiguous conclusion is the stable open cavity, not that particular closed loop.