Time ordering and the complete memory response

3 Conditional evolution and time ordering

Fix a realized $m(t)$ and put $c=\gamma+i\alpha$. The analysis here needs only bounded piecewise-continuous $K(t)\ge0$; the source supplies smoothness and $\mu I$ as a lower bound.

$$\partial_t U(t,s)=-cK(t)U(t,s),\quad U(s,s)=I,\tag{15}$$ $$\partial_t D(t,s)=-\gamma K(t)D(t,s),\quad D(s,s)=I,\tag{16}$$ $$\partial_t V(t,s)=-i\alpha K(t)V(t,s),\quad V(s,s)=I.\tag{17}$$

Evolution-family composition holds. Differentiating squared norms gives contraction of $U,D$ and unitarity of $V$. With $K\ge\mu I$ the first two norms are bounded by $e^{-\gamma\mu(t-s)}$. Shifting initial time changes the history, so the propagators need not be one-parameter semigroups or groups. Write $U(t)=U(t,0)$, and likewise $D,V$.

Theorem 3.1 (Exact interaction picture).

$$U(t)=V(t)R(t),\qquad \dot R=-\gamma V^*KV R,\qquad R(0)=I.\tag{18}$$

Proof. Set $R=V^*U$. Unitarity and $\partial_tV^*=i\alpha V^*K$ cancel the reactive term, giving the equation. The rotated carrier remains positive by unitary congruence. Generally $R\ne D$. Instantaneous commutation $[\gamma K(t),\alpha K(t)]=0$ leaves cross-time commutators intact.

Theorem 3.2 (Gap-free factorization estimate).

$$\|U(T)-D(T)V(T)\|\le\gamma|\alpha|\int_0^T\!\int_0^t\|[K(t),K(s)]\|\,ds\,dt.\tag{19}$$

For continuously differentiable $K$ with $\|K\|\le M,\|\dot K\|\le L$, the bound is at most $\gamma|\alpha|MLT^3/3$. For $K=K_0+tA$ it is $\gamma|\alpha|\|[A,K_0]\|T^3/6$.

Proof. Fix $t$, differentiate $D(t,s)K(t)D(s,0)$ in $s$, and integrate:

$$[K(t),D(t)]=-\gamma\int_0^t D(t,s)[K(t),K(s)]D(s,0)\,ds.\tag{20}$$

With $W=DV$, differentiation in that order gives

$$\dot W+cK(t)W=i\alpha[K(t),D(t)]V(t).\tag{21}$$

The difference $U-W$ starts at zero. Variation of constants with the contractive $U(T,t)$, both contractive $D$ factors and unitary $V$ gives (19). In the differentiable case $\|[K(t),K(s)]\|=\|[K(t)-K(s),K(s)]\|\le2ML(t-s)$; the double integral of $t-s$ is $T^3/6$. In the affine case the commutator is $(t-s)[A,K_0]$.

No gap is needed at a degeneracy or crossing. Pairwise commuting histories factor through exponentials of the integrated generators; one endpoint equality alone does not establish commutation. If a differentiable unitary basis $Q$ diagonalizes $K$, $u=Q^*z$ satisfies

$$\dot u=-c\Lambda(t)u-Q^*\dot Q\,u.\tag{22}$$

The skew-Hermitian connection can mix modes. A unique smooth eigenvalue-ordered basis need not exist at crossings, and fixed physical coordinates avoid that hypothesis.

3.1 Two pulses and asymptotic sharpness

Apply $K_1$ then $K_2$ for duration $\tau$ each:

$$U=e^{-c\tau K_2}e^{-c\tau K_1},\quad D=e^{-\gamma\tau K_2}e^{-\gamma\tau K_1},\quad V=e^{-i\alpha\tau K_2}e^{-i\alpha\tau K_1}.\tag{23}$$ $$U-DV=i\gamma\alpha\tau^2[K_2,K_1]+O(\tau^3),\tag{24}$$ $$U-U_{\rm rev}=c^2\tau^2[K_2,K_1]+O(\tau^3),\quad U_{\rm rev}=e^{-c\tau K_1}e^{-c\tau K_2}.\tag{25}$$

The operator-norm remainders hold for fixed matrices and coefficients as $\tau\downarrow0$. In $U$ the ordered mixed term is $c^2K_2K_1$ times $\tau^2$; subtract the mixed terms in $DV$ to obtain (24). Subtracting the reverse order yields (25). Here (19) becomes

$$\|U-DV\|\le\gamma|\alpha|\tau^2\|[K_2,K_1]\|.\tag{26}$$

The leading constant is attained asymptotically for nonzero commutator. Reversal preserves the pulse spectra and $\int_0^{2\tau}K(t)\,dt$ but changes the response. The naive integrated exponential differs from $U$ by the ordered correction $U-e^{-c\tau(K_1+K_2)}=(c^2\tau^2/2)[K_2,K_1]+O(\tau^3)$.

3.2 A moving but commuting control

For the three-node path $b_1=(1,-1,0)^\top,b_2=(0,1,-1)^\top$, write $w_e=e^{m_e}$.

$$[K(t),K(s)]=[w_1(t)w_2(s)-w_2(t)w_1(s)][B_1,B_2].\tag{27}$$

Multiplying the rank-one edge matrices gives a real skew commutator with singular values $\sqrt3,\sqrt3,0$. Reflection swaps nodes one and three and swaps memories; equivariance and uniqueness preserve symmetric data $z_1=z_3,m_1=m_2$. In that sector,

$$K(t)=\mu I+w(t)(B_1+B_2).\tag{28}$$

This moving history commutes exactly. More generally, a constant ratio $w_1/w_2$ is sufficient; the asymmetric comparison below changes the preparation, not the mobility law.

4 Full feedback, memory and changing coordinates

Along a trajectory let $B$ be a $2n\times p$ matrix with columns distinct from edge matrices $B_e=b_eb_e^\top$.

$$B_e^{\rm col}=(\partial_{m_e}K_R)q,\qquad H_m=\kappa I+\operatorname{diag}\big(\tfrac12e^{m_e}|b_e^\top z|^2\big).\tag{29}$$ $$H(t)=\begin{pmatrix}K_R&B\\B^\top&H_m\end{pmatrix}.\tag{30}$$

Its positive diagonal blocks do not guarantee nonnegative $H$.

Proposition 4.1 (Complete tangent and exact memory). For $(\xi,\nu)=(\delta q,\delta m)$,

$$\dot\xi=-GK_R\xi-GB\nu,\tag{31}$$ $$\dot\nu=-\eta B^\top\xi-\eta H_m\nu.\tag{32}$$

Let $\partial_tR_m(t,s)=-\eta H_m(t)R_m(t,s)$, $R_m(s,s)=I$. Then

$$\dot\xi(t)=-GK_R(t)\xi(t)-GB(t)R_m(t,0)\nu(0)+\eta\int_0^t GB(t)R_m(t,s)B(s)^\top\xi(s)\,ds.\tag{33}$$

Proof. Constant mobilities make the derivative of (4) the displayed Hessian-block system. Variation of constants gives

$$\nu(t)=R_m(t,0)\nu(0)-\eta\int_0^t R_m(t,s)B(s)^\top\xi(s)\,ds.$$

Substitute to obtain the positive kernel and negative initial-memory term. No adiabatic or Markov approximation enters.

Here diagonal $H_m$ gives an exponential of its time integral and $\|R_m(t,s)\|\le e^{-\eta\kappa(t-s)}$. The full product need not be symmetric, a scalar function of the amplitude carrier, or a stationary convolution. A clamp constrains $\nu=0$ throughout and gives the realification of $U$. Free memory with $\nu(0)=0$ still responds through $B^\top\xi$. They coincide under decoupling, such as $B=0$ on the trajectory. A physical clamp is an additional apparatus resource.

4.1 Exact transient expansion despite partial positivity

Take a scalar complex amplitude, one memory, $\mu=2/5,\kappa=1,m=0,z=3$ real. The $(\Re z,m)$ block and its determinant are

$$\begin{pmatrix}7/5&3\\3&11/2\end{pmatrix},\qquad\det=-13/10.\tag{34}$$

The imaginary diagonal is $7/5$, with no memory cross term. For $v=(1,0,-1/2)$, $\gamma=\eta=1$, any $\alpha$,

$$v^\top Hv=-9/40,\qquad\left.\frac{d}{dt}(|\xi|^2+|\nu|^2)\right|_{t=0}=9/20.\tag{35}$$

The reactive norm derivative vanishes on this perturbation; the dissipative contribution is $-2v^\top Hv$. Positive partial curvature therefore permits full tangent expansion. Sensitivity between nonstationary trajectories is compatible with energy decrease and global relaxation.

Even joint positivity does not restore complex alignment. At $z=1/10,m=0$,

$$H=\begin{pmatrix}7/5&0&1/10\\0&7/5&0\\1/10&0&201/200\end{pmatrix}>0.\tag{36}$$

Its commutator with $\operatorname{diag}(J,0)$ has nonzero entries of magnitude $1/10$. This skew operator is degenerate on real memory and is not a complex structure for the entire state. Exact rational checks accompany both examples.

4.2 The connection for a changing mobility metric

For $C^1$ positive $M(t)$, continuous Hermitian $H(t)$, consider the separate prescribed-path model

$$\dot x=-cM(t)H(t)x.\tag{37}$$

Put $S=M^{1/2}, y=S^{-1}x$. Differentiating $x=Sy$ gives

$$\dot y=-c\underbrace{SHS}_{K_y(t)}y-\underbrace{S^{-1}\dot S}_{A_{\rm conn}(t)}y.\tag{38}$$ $$\frac{d}{dt}\|y\|^2=-2\gamma y^*K_y y-y^*(A_{\rm conn}+A_{\rm conn}^*)y.\tag{39}$$

Positive $H$ gives positive-semidefinite $K_y$, but neither positivity nor whitening removes the connection. It need not be skew-Hermitian or a function of $K_y$. State-dependent metrics may contribute further tangent derivatives; (38) concerns a prescribed path. With $M=\operatorname{diag}(e^{2rt},e^{-2rt})$, $H=\operatorname{diag}(1,2)$, $r=1,\gamma=0.2,\alpha=0.4,x(0)=(0,1)^\top$,

$$x_2(t)=\exp[-c(1-e^{-2rt})/r],\qquad y_2(t)=e^{rt}x_2(t).\tag{40}$$

At $t=0.1$ the respective norms are $0.964395$ and $1.065822$; omitting the connection changes the result by $0.101426$. A moving coordinate norm is not physical active gain. The original network has fixed metric.

4.3 Why the autonomous closure certificate changes

For fixed selfadjoint $K\ge0$ and retained isometry $I$, the Gram defect is

$$I^*e^{-2tK}I-(I^*e^{-tK}I)^2=(Qe^{-tK}I)^*(Qe^{-tK}I),\quad Q=1-II^*.$$

For nonautonomous $D$, insert $1=II^*+Q$ in evolution-family composition:

$$A(t,r)-A(t,s)A(s,r)=I^*D(t,s)QD(s,r)I,\qquad A(t,s)=I^*D(t,s)I.\tag{41}$$

The product need not be a Gram matrix. With scalar $\gamma=1,K=1$ on $[0,1]$, $K=2$ on $(1,2]$, $D(2,0)-D(1,0)^2=e^{-3}-e^{-2}<0$ despite no hidden sector. For a distinct fixture take

$$K_1=\begin{pmatrix}2&1\\1&2\end{pmatrix},\quad K_2=\begin{pmatrix}2&-1\\-1&2\end{pmatrix},\quad P=\operatorname{diag}(1,0).\tag{42}$$

The positive matrices commute and sum to $4I$, so unit-duration pulses give $D(2,0)=e^{-4}I$ and zero endpoint leakage $(I-P)D(2,0)P$, while $[P,K_1]\ne0$. Endpoint invariance fails to imply invariance through the history, even here. Both examples preserve the autonomous certificate within its hypotheses.