This section separates two endpoint problems. The classical symplectic propagator determines exact quantum return for a quadratic Hamiltonian. The configuration propagator is computed from its evolving Gaussian phase. They are different matrices.
Let K0=OΩ2OT>0, with Ω=diag(ω1,…,ωd). First suppose the frequencies are positive integers. At T=2π the uncontrolled symplectic propagator is the identity. Define
S˙=A(K0+D(t))S,A(K)=(0−KI0),S(0)=I.(4.1)
Here S acts on classical position and momentum, used to represent the quadratic quantum evolution. It is not a guidance flow. The metaplectic representation, or equivalently the exact Heisenberg equations for Weyl operators, shows that S(T)=I implies U(T)=eiθI on the full quantum Hilbert space [5, Chapter 4]. Irreducibility of the Weyl representation gives the same conclusion directly. There is no spectral truncation in this implication.
For a physical matrix M∈D write M=OTMO. The following finite-dimensional criterion will be arranged by engineering the holding traps.
Lemma 4.1 (Endpoint submersion)
Assume that
(i)
the positive numbers 2ωi, ωi+ωj and ∣ωi−ωj∣ (i<j) are pairwise distinct;
(ii)
for each i<j some M∈D has Mij=0;
(iii)
the vectors (M11,…,Mdd), M∈D, span Rd.
Then the endpoint map from smooth, interior-supported physical quadratic controls to Sp(2d,R) is a submersion at zero control and time 2π. A finite-dimensional family of such controls already has this property.
Proof
At the return time the differential is
DE(0)[D]=∫02πS0(t)−1(0−D(t)00)S0(t)dt.(4.2)
In oscillator coordinates put qi(t)=(aie−iωit+aˉieiωit)/2ωi. A real quadratic qTMq/2 then has the two real quadratures of aiaj at ωi+ωj, those of aiaˉj at ωi−ωj, and the diagonal number terms at frequency zero. For i=j the nonzero frequency is 2ωi.
If a real covector on sp(2d,R) annihilates (4.2) for every interior-supported time waveform and every M, its pairing with this quadratic expression vanishes identically in t. Independence of the distinct trigonometric frequencies, (ii), and (iii) force its coefficients on every root quadrature and on every diagonal number term to vanish. These form a real basis of sp(2d,R), so the differential is onto. Select d(2d+1) smooth controls whose images are a basis. Their coefficients give the asserted finite-dimensional submersion.
□
Choose two distinct positive integers ν,ζ larger than every frequency occurring in (i). For M,N∈D the trial control
Dtr(t,ϵ)=ϵ[(cosνt−cosζt)M+sinνtN](4.3)
vanishes at both endpoints. Fourier orthogonality makes its first endpoint derivative zero. Add a linear combination of the finite controls from lemma 4.1,
D(t,ϵ)=Dtr(t,ϵ)+ℓ=1∑d(2d+1)cℓ(ϵ)Dℓ(t).(4.4)
The ordinary analytic implicit-function theorem, in a chart at I∈Sp(2d,R), gives cℓ(ϵ)=O(ϵ2) and S(2π)=I exactly. For sufficiently small ϵ, K0+D(t,ϵ)>0. Thus these are finite-duration exact quantum returns with local controls and fixed off-block entries. Only a finite-dimensional endpoint equation has been inverted.
The last equation is the actual guided map q(t)=L(t)q(0). To see that these formulas are global for finite time, write
X=Sqq+iSqpA0,Y=Spq+iSppA0.
Symplecticity gives X∗Y−Y∗X=2iA0. Hence X is invertible, A−iB=−iYX−1 is symmetric, and A=X−∗A0X−1>0. The Gaussian has no nodes, B is bounded on every finite time interval, and L is a global linear diffeomorphism. Also
Matrix commutators in (4.7) have the convention [P,Q]=PQ−QP.
Proof
Let A=Ω+ϵa+⋯ and B=ϵb+⋯. For σij=ωi+ωj the linearized equations are a˙ij=−σijbij and b˙ij=σijaij−Dij. A forcing Mijcosνt contributes
bij(t)=Mijσij2−ν2νsinνt−σijsinσijt;
a forcing Nijsinνt contributes
bij(t)=Nijσij2−ν2ν(cosσijt−cosνt).
There is an analogous cosine expression at ζ. All these functions integrate to zero over T=2π. Expansion of L˙=BL yields
skewL2=21∫02π[b(t),∫0tb(s)ds]dt.(4.9)
The direct integral of the second-order B is symmetric, including the contribution of the endpoint repair.
Orthogonality removes unequal frequencies from (4.9). At a natural frequency σij all matrices are multiples of the same symmetric matrix unit, and their commutator is zero. The only nonzero paired sine and cosine matrices are at ν: they are νRνM and −νRνN. Their elementary integral is −πν[RνM,RνN], proving (4.7).
Exact return and (4.6) imply ΩL2+L2TΩ=0. Solving this entrywise in terms of skewL2 gives (4.8). This is why a symmetric contribution before whitening cannot be simply ignored after whitening: the covariance identity supplies the necessary relation.
Suppose a subgroup G⊂SO(d) contains analytic curves Rj(ϵ)=I+ϵ2Gj+O(ϵ3) and their inverses. If the Gj generate so(d) as a Lie algebra, then G=SO(d).
Proof
For a sufficiently small fixed nonzero a, the curves Cj(t)=Rj(a+t)Rj(a)−1 pass through the identity and have velocities Xj=2aGj+O(a2). A finite list of Lie words in the Gj is a basis, so the corresponding determinant remains nonzero for Xj/(2a). Let V be the span of AdhXj for all h in the subgroup generated by the Cj. It is invariant under these adjoint actions. Differentiating AdCj(t)V=V gives [Xj,V]⊂V. Thus V=so(d). Select finitely many adjoint velocities forming a basis. The product of their conjugated curves has surjective differential at the origin, so its image contains an identity neighborhood, by the finite-dimensional inverse-function theorem. Consequently G is open. Since SO(d) is connected, G=SO(d). Every element obtained this way is a finite product of actual curves and their inverses, rather than only a limit of such products.