Theorem: Relational Enter

For the frozen data of Section sec:data, the following statements hold.

1. Complete instrument identity. The map ${\mathcal{M}}_C$ is completely positive and trace preserving, and

   $${\mathcal{M}}_C={\mathcal{J}}_C$$

   as linear maps on all clock–system operators. This equality remains valid after tensoring with ${\mathrm{id}}_Q$ for every untouched reference.

2. Record law and residual. On input ${\rho }_{CS}$, the probability of code $k$, hence of calibrated reading $t_k$, is

   $$p_k=\mathrm{Tr}\left[(P_k\otimes I_S){\rho }_{CS}\right].$$

   On a positive branch, the surviving system is exactly ${\rho }_S(t_k)$ of (eq:pw-normalized). Under Hypothesis hyp:enter, these are the probabilities and updates selected by the Enter characterization, not an additional measurement postulate introduced here.

3. Linear consumption. The readout consumes every measured quantum name in $C$. Its quantum output context contains $S$, not $C$. A continuation in the same process cannot use a consumed clock name as a quantum input.

4. Relational evolution under covariance. If ${\rho }_{CS}$ satisfies (eq:stationarity) and the reading basis satisfies (eq:clock-covariance), then

   $${\sigma }_{S,k}=U_S(t_k){\sigma }_{S,0}{U}_S(t_k{)}^{†},U_S(t)=e^{-i{H}_{S}t/\hslash }.$$

   In this complete covariant basis, $p_k=1/d$, and therefore

   $${\rho }_S(t_k)=U_S(t_k){\rho }_S(0)U_S(t_k{)}^{†}.$$

5. Complete operational realization. The record, surviving system, and reference extension realize precisely the declared relational extraction. Discarding the record yields

   $${\mathrm{Tr}}_R[({\mathcal{M}}_C\otimes {\mathrm{id}}_Q)(X)]={\mathrm{Tr}}_{C}X.$$

   No quantum clock output or spacetime identification is part of this channel.

Proof. Define Kraus operators from ${\mathcal{H}}_C\otimes {\mathcal{H}}_S$ to ${\mathcal{H}}_R\otimes {\mathcal{H}}_S$ by

Then ${\mathcal{M}}_C(X)={\sum }_k{A}_{k}X{A}_k^{†}$, which is completely positive. Orthonormality of the clock basis gives

Hence ${\mathcal{M}}_C$ is trace preserving.

For an arbitrary operator $X$, expand in the clock basis:

Both

equal $X_{kk}$. This proves (eq:exact-channel-identity) on all operators, not only on the frozen preparation. Replacing $I_S$ by $I_S\otimes I_Q$ proves the reference statement.

The $R$-marginal is

Cyclicity of the full trace and ${P_k}^2=P_k$ give

proving (eq:born-clock). Dividing the positive $k$-block by its trace gives exactly (eq:pw-normalized). Lemma lem:density-semantics supplies the semantic attribution under Hypothesis hyp:enter.

For consumption, the register-level readout has the first-order interface

It is implemented by the calibrated basis change and destructive Enter on the $m$ clock names. Each elementary readout removes its measured name from the quantum context. After all $m$ readouts, the surviving quantum context is $S$. Tensor permutation and classical packaging do not recreate a quantum name. This proves the linear-resource assertion. The system output in (eq:linear-interface) is unmeasured residual output; it is not a quantum return of any name in $C$.

To prove the dynamical clause, fix $t=t_k$, write $U_C=U_C(t)$, $U_S=U_S(t)$, and use stationarity:

Consequently,

For partial trace over $C$, the two-projector expression in (eq:pw-subnormalized) equals the corresponding single-projector expression. Using covariance and unitary invariance of the partial trace,

Thus all $p_k$ coincide. Their sum is one and there are $d$ outcomes, so $p_k=1/d$. Normalization proves (eq:relational-evolution).

Finally, the clock-basis expansion gives

The same calculation includes $Q$. The complete channel therefore returns exactly the record and residual specified above. Its codomain has no quantum $C$ factor, and no geometric object occurs in its definition.