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Shadow Theory

Section 4 4 October 2026

Exact Gaussian returns with fixed interactions

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4 Exact Gaussian returns with fixed interactions

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.

4.1 Symplectic endpoint correction

Let K0=OΩ2OT>0K_0=O\Omega^2O^T>0, with Ω=diag⁡(ω1,…,ωd)\Omega=\diag(\omega_1,\ldots,\omega_d). First suppose the frequencies are positive integers. At T=2πT=2\pi the uncontrolled symplectic propagator is the identity. Define

S˙=A(K0+D(t))S,A(K)=(0I−K0),S(0)=I. \dot{\mathsf S}=\mathsf A(K_0+D(t))\mathsf S,\qquad \mathsf A(K)=\begin{pmatrix}0&I\\-K&0\end{pmatrix},\qquad \mathsf S(0)=I . (4.1)

Here S\mathsf 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\mathsf S(T)=I implies U(T)=eiθIU(T)=e^{i\theta}I 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∈DM\in\calD write M~=OTMO\widetilde M=O^TMO. The following finite-dimensional criterion will be arranged by engineering the holding traps.

Lemma 4.1 (Endpoint submersion)

Assume that

  1. (i)

    the positive numbers 2ωi2\omega_i, ωi+ωj\omega_i+\omega_j and ∣ωi−ωj∣|\omega_i-\omega_j| (i<ji<j) are pairwise distinct;

  2. (ii)

    for each i<ji<j some M∈DM\in\calD has M~ij≠0\widetilde M_{ij}\ne0;

  3. (iii)

    the vectors (M~11,…,M~dd)(\widetilde M_{11},\ldots,\widetilde M_{dd}), M∈DM\in\calD, span Rd\R^d.

Then the endpoint map from smooth, interior-supported physical quadratic controls to Sp⁡(2d,R)\Sp(2d,\R) is a submersion at zero control and time 2π2\pi. 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(00−D(t)0)S0(t) dt. D\mathcal E(0)[D]=\int_0^{2\pi} \mathsf S_0(t)^{-1} \begin{pmatrix}0&0\\-D(t)&0\end{pmatrix} \mathsf S_0(t)\,dt . (4.2)

In oscillator coordinates put qi(t)=(aie−iωit+aˉieiωit)/2ωiq_i(t)=(a_i e^{-i\omega_i t}+\bar a_i e^{i\omega_i t})/ \sqrt{2\omega_i}. A real quadratic qTMq/2q^TMq/2 then has the two real quadratures of aiaja_ia_j at ωi+ωj\omega_i+\omega_j, those of aiaˉja_i\bar a_j at ωi−ωj\omega_i-\omega_j, and the diagonal number terms at frequency zero. For i=ji=j the nonzero frequency is 2ωi2\omega_i.

If a real covector on sp(2d,R)\mathfrak{sp}(2d,\R) annihilates (4.2) for every interior-supported time waveform and every MM, its pairing with this quadratic expression vanishes identically in tt. 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)\mathfrak{sp}(2d,\R), so the differential is onto. Select d(2d+1)d(2d+1) smooth controls whose images are a basis. Their coefficients give the asserted finite-dimensional submersion.

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Choose two distinct positive integers ν,ζ\nu,\zeta larger than every frequency occurring in (i). For M,N∈DM,N\in\calD the trial control

Dtr(t,ϵ)=ϵ[(cos⁡νt−cos⁡ζt)M+sin⁡νt N] D_{\mathrm{tr}}(t,\epsilon)= \epsilon\bigl[(\cos\nu t-\cos\zeta t)M+\sin\nu t\,N\bigr] (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=1d(2d+1)cℓ(ϵ)Dℓ(t). D(t,\epsilon)=D_{\mathrm{tr}}(t,\epsilon) +\sum_{\ell=1}^{d(2d+1)}c_\ell(\epsilon)D_\ell(t). (4.4)

The ordinary analytic implicit-function theorem, in a chart at I∈Sp⁡(2d,R)I\in\Sp(2d,\R), gives cℓ(ϵ)=O(ϵ2)c_\ell(\epsilon)=O(\epsilon^2) and S(2π)=I\mathsf S(2\pi)=I exactly. For sufficiently small ϵ\epsilon, K0+D(t,ϵ)>0K_0+D(t,\epsilon)>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.

4.2 The actual Gaussian configuration curvature

Starting in ψ0\psi_0, a quadratic history remains Gaussian:

ψt(q)=ctexp⁡[−qTA(t)q/2+iqTB(t)q/2],A(t)>0,B(t)=B(t)T. \psi_t(q)=c_t\exp[-q^TA(t)q/2+iq^TB(t)q/2], \qquad A(t)>0,\quad B(t)=B(t)^T .

Substitution in the Schrödinger equation gives

A˙=−AB−BA,B˙=A2−B2−K,L˙=BL,L(0)=I. \dot A=-AB-BA,\qquad \dot B=A^2-B^2-K,\qquad \dot L=BL,\quad L(0)=I . (4.5)

The last equation is the actual guided map q(t)=L(t)q(0)q(t)=L(t)q(0). To see that these formulas are global for finite time, write

X=Sqq+iSqpA0,Y=Spq+iSppA0. X=\mathsf S_{qq}+i\mathsf S_{qp}A_0,\qquad Y=\mathsf S_{pq}+i\mathsf S_{pp}A_0 .

Symplecticity gives X∗Y−Y∗X=2iA0X^*Y-Y^*X=2iA_0. Hence XX is invertible, A−iB=−iYX−1A-iB=-iYX^{-1} is symmetric, and A=X−∗A0X−1>0A=X^{-*}A_0X^{-1}>0. The Gaussian has no nodes, BB is bounded on every finite time interval, and LL is a global linear diffeomorphism. Also

L(t)TA(t)L(t)=A0,det⁡L(t)=exp⁡(∫0ttr⁡B(s) ds)>0. L(t)^TA(t)L(t)=A_0,\qquad \det L(t)=\exp\left(\int_0^t\tr B(s)\,ds\right)>0 . (4.6)

Every Gaussian ray return therefore gives A01/2L(T)A0−1/2∈SO(d)A_0^{1/2}L(T)A_0^{-1/2}\in SO(d).

Proposition 4.2 (Curvature surviving exact repair)

For the corrected loops (4.4), in modal coordinates, L(T)=I+ϵ2L2+O(ϵ3)L(T)=I+\epsilon^2L_2+O(\epsilon^3) and

skew⁡L2=−πν[RνM~,RνN~],(RνQ)ij=Qij(ωi+ωj)2−ν2,Gij=2ωiωjωi+ωj(skew⁡L2)ij,Ω1/2L(T)Ω−1/2=I+ϵ2G+O(ϵ3).\begin{align}\skw L_2 &=-\pi\nu[\mathcal R_\nu\widetilde M, \mathcal R_\nu\widetilde N], &(\mathcal R_\nu Q)_{ij} &=\frac{Q_{ij}}{(\omega_i+\omega_j)^2-\nu^2},\tag{4.7}\\ G_{ij} &=\frac{2\sqrt{\omega_i\omega_j}}{\omega_i+\omega_j} (\skw L_2)_{ij}, &\Omega^{1/2}L(T)\Omega^{-1/2} &=I+\epsilon^2G+O(\epsilon^3). \tag{4.8}\end{align}

Matrix commutators in (4.7) have the convention [P,Q]=PQ−QP[P,Q]=PQ-QP.

Proof

Let A=Ω+ϵa+⋯A=\Omega+\epsilon a+\cdots and B=ϵb+⋯B=\epsilon b+\cdots. For σij=ωi+ωj\sigma_{ij}=\omega_i+\omega_j the linearized equations are a˙ij=−σijbij\dot a_{ij}=-\sigma_{ij}b_{ij} and b˙ij=σijaij−Dij\dot b_{ij}=\sigma_{ij}a_{ij}-D_{ij}. A forcing M~ijcos⁡νt\widetilde M_{ij}\cos\nu t contributes

bij(t)=M~ijνsin⁡νt−σijsin⁡σijtσij2−ν2; b_{ij}(t)=\widetilde M_{ij} \frac{\nu\sin\nu t-\sigma_{ij}\sin\sigma_{ij}t} {\sigma_{ij}^2-\nu^2};

a forcing N~ijsin⁡νt\widetilde N_{ij}\sin\nu t contributes

bij(t)=N~ijν(cos⁡σijt−cos⁡νt)σij2−ν2. b_{ij}(t)=\widetilde N_{ij} \frac{\nu(\cos\sigma_{ij}t-\cos\nu t)} {\sigma_{ij}^2-\nu^2}.

There is an analogous cosine expression at ζ\zeta. All these functions integrate to zero over T=2πT=2\pi. Expansion of L˙=BL\dot L=BL yields

skew⁡L2=12∫02π[b(t),∫0tb(s) ds]dt. \skw L_2=\frac12\int_0^{2\pi} \left[b(t),\int_0^t b(s)\,ds\right]dt . (4.9)

The direct integral of the second-order BB is symmetric, including the contribution of the endpoint repair.

Orthogonality removes unequal frequencies from (4.9). At a natural frequency σij\sigma_{ij} 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\nu: they are νRνM~\nu\mathcal R_\nu\widetilde M and −νRνN~-\nu\mathcal R_\nu\widetilde N. Their elementary integral is −πν[RνM~,RνN~]-\pi\nu[\mathcal R_\nu\widetilde M, \mathcal R_\nu\widetilde N], proving (4.7).

Exact return and (4.6) imply ΩL2+L2TΩ=0\Omega L_2+L_2^T\Omega=0. Solving this entrywise in terms of skew⁡L2\skw L_2 gives (4.8). This is why a symmetric contribution before whitening cannot be simply ignored after whitening: the covariance identity supplies the necessary relation.

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4.3 From return curves to the exact group

Lemma 4.3 (Exact group generation)

Suppose a subgroup G⊂SO(d)\mathscr G\subset SO(d) contains analytic curves Rj(ϵ)=I+ϵ2Gj+O(ϵ3)R_j(\epsilon)=I+\epsilon^2G_j+O(\epsilon^3) and their inverses. If the GjG_j generate so(d)\mathfrak{so}(d) as a Lie algebra, then G=SO(d)\mathscr G=SO(d).

Proof

For a sufficiently small fixed nonzero aa, the curves Cj(t)=Rj(a+t)Rj(a)−1C_j(t)=R_j(a+t)R_j(a)^{-1} pass through the identity and have velocities Xj=2aGj+O(a2)X_j=2aG_j+O(a^2). A finite list of Lie words in the GjG_j is a basis, so the corresponding determinant remains nonzero for Xj/(2a)X_j/(2a). Let VV be the span of Ad⁡hXj\operatorname{Ad}_hX_j for all hh in the subgroup generated by the CjC_j. It is invariant under these adjoint actions. Differentiating Ad⁡Cj(t)V=V\operatorname{Ad}_{C_j(t)}V=V gives [Xj,V]⊂V[X_j,V]\subset V. Thus V=so(d)V=\mathfrak{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\mathscr G is open. Since SO(d)SO(d) is connected, G=SO(d)\mathscr 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.

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