Exact apparatus energy leaves a receiver-sign ambiguity
Knowing both initial receiver energy and initial apparatus energy still permits two distinct receiver states with identical complete final apparatus records. For every sufficiently small positive coupling, a backward-flow construction gives an antipodal receiver pair inside the preparation box. The clock and both energy records agree exactly; both canonical reconstruction risks stay positive as coupling decreases at fixed preparation width.
R26, 2026-09-12. Retain R06/R25’s finite apparatus and smooth coordinate interaction, fixed positive receiver energy E and observation time T. Let H_0=M_c v_0^2/2 be the nominal initial apparatus kinetic energy. All incoming apparatus coordinates are unknown within a fixed positive box about nominal clock phase and zero probe phase. Require exactly H_app=H_0, in addition to H_s=E. Pulse supports are absent at both endpoints. The record consists of all ten final apparatus coordinates and the two initial energies. Source coverage is in B57.
1. Reverse a terminal state with zero probe phase
Let z_* be any receiver shell point and Phi_T its free evolution. Prescribe terminal data
\[z(T)=r\Phi_Tz_*,\quad q(T)=\pi(T)=0,\quad s(T)=s_0+v_0T,\] \[p_s(T)=\sqrt{2M_c[H_0+E(1-r^2)]}. \tag{1}\]
For r near 1 the square root is positive. Terminal total energy is exactly E+H_0 and the interaction is zero. Evolve the exact autonomous Hamiltonian backward for time T and denote its initial receiver state Z_lambda(r). Smooth finite-time dependence gives Z_0(r)=r z_*. Hence
\[\partial_r[H_s(Z_0(r))-E]_{r=1}=2E>0.\]
The implicit-function theorem provides a smooth r_lambda near 1 such that
\[H_s(Z_\lambda(r_\lambda))=E. \tag{2}\]
At the resulting initial point all pulse supports are absent, by continuity of the strict free endpoint margin. Conservation of the total energy E+H_0 then gives H_app=H_0 exactly. Thus (2) imposes one energy and (1) plus conservation imposes the other; neither is merely an asymptotic equality.
2. Bounds and preparation margins
On the backward interval, terminal zero probe coordinates and momenta imply q,pi=O(lambda). The receiver and clock reaction forces contain lambda q, so their deviations from the corresponding free paths are O(lambda squared). These bounds hold with r derivatives on a fixed small r interval, by the smooth variational equations and finite-time estimates. Therefore
\[r_\lambda=1+O(\lambda^2),\qquad Z_\lambda(r_\lambda)=z_*+O(\lambda^2),\] \[p_s(T)=M_cv_0+O(\lambda^2). \tag{3}\]
Initial probe phase differs from zero by O(lambda) and initial clock phase differs from (s_0,M_c v_0) by O(lambda squared). In fixed component units there is C with ||a(0)-a_*||<=C lambda. For any sufficiently small fixed box half-width b>0, take lambda_0 small enough that C lambda_0<=b/2. Both the box margin and the linear cutoff regions then hold throughout the whole construction. All masses, durations and pulse widths stay fixed.
The energy-constrained preparation is the shell times the intersection of the apparatus box with H_app=H_0. Interior margin refers to the box, not an open neighbourhood in the full ten-dimensional apparatus space.
3. An exact symmetry makes a second preparation with the same record
Inside the common linear cutoff regions, the involution
\[\mathcal I:(z,q,\pi,s,p_s)\mapsto(-z,-q,-\pi,s,p_s) \tag{4}\]
preserves the Hamiltonian and its equations: receiver and probe forces change sign, while the clock force contains the invariant products x q_j. The free energies are quadratic. Global oddness of the cutoff functions is unnecessary; both trajectories stay in the regions where X(x)=x and R(q)=q.
Apply (4) to the trajectory constructed above. At T its probes are again exactly zero and its clock phase is unchanged. The complete final apparatus record is therefore identical. At the initial time the receiver states are exactly Z_lambda and -Z_lambda, both with energy E. Probe phases reverse sign, clock phase is unchanged, and apparatus kinetic energy is exactly H_0 for both. The symmetric preparation box admits both points with the same b/2 margin. This proves an actual pair rather than a dimension or tangent count.
4. Canonical risk and limit
Put k_x=a-g^2/d>0 and choose the physical shell point
\[z_*=(x_*,P_*,(g/d)x_*,0),\qquad x_*=\sqrt{E/k_x},\quad P_*=\sqrt{\mu E}. \tag{5}\]
Here x_,P_ denote state components, not the fixed component units. Its potential and kinetic energies are each E/2. Identical records for the antipodal pair force every deterministic estimator’s coordinate risks to obey
\[\epsilon_x\ge |(Z_\lambda)_x|,\qquad \epsilon_P\ge |(Z_\lambda)_P|,\] \[\epsilon_x\epsilon_P\ge |(Z_\lambda)_x(Z_\lambda)_P| \longrightarrow E\sqrt{\mu/k_x}>0 \quad(\lambda\downarrow0). \tag{6}\]
The arrow concerns the displayed lower bound. In particular it is bounded below by E sqrt(mu/k_x)/4 for sufficiently small positive coupling, by keeping each coordinate at least half its limiting magnitude. The constant has action units; no 2 pi normalization is introduced. Neither exact minimax optimality nor a positive projected area is claimed for this two-point construction.
The scale is set by receiver energy and stiffness/mass. The allowed weak- coupling regime depends on the positive preparation width. Exact energy information removes neither the symmetric preparation freedom nor the resulting global ambiguity. The common record was chosen by terminal zero probe phase; it need not be R25’s originally chosen nominal-record fibre.
Next R27: test whether a known nonzero incoming probe displacement breaks this sign ambiguity while exact initial energies and full final records remain supplied. Separate breaking this particular involution from proving global recovery, and keep the admissible preparation-energy intersection explicit.