Finite receptor response, retained nulls and delayed records

The first checkpoint [C01] contains a distinct finite detector whose coherent excitation is followed by an assumed intrinsic latch. This chapter expands its exact null calculation and its controlled fresh-cell limit into complete theorems. The comparison includes classical event times, outcomes, the source and inaccessible references. It excludes the future return of receptors that have been discarded. That restriction is essential: the exact complete null contains source–receptor correlations which a reduced source instrument omits.

22.1 The finite receptor and its stochastic premise

Let $P_i$ be a finite projective resolution on the carried system. The apparatus has a ready vector $|A_0\rangle$, orthogonal excited vectors $|A_i^*\rangle$ and mutually distinguished terminal recorded/spent states $|R_i\rangle$. During an exposure set the source Hamiltonian to zero and use

Units have $\hbar=1$. The $g_i,\Gamma_i$ are positive apparatus parameters. $H_{\mathrm{int}}$ acts identically within each possibly degenerate $P_i$ sector and trivially on the reference.

Assumption 22.1 (Intrinsic apparatus latch)

An actual latch $i$ has conditional intensity $\|C_i\Psi\|^2$, normalized daughter $C_i\Psi/\|C_i\Psi\|$, and the associated no-event evolution generated by $H_{\mathrm{eff}}=H_{\mathrm{int}}-\tfrac i2\sum_iC_i^\dagger C_i$. An event creates its actual record and consumes this cell's readiness; there is at most one event per cell. Future event randomness is the specified Markov latch law. No separate physical sampling tape is available. All terminal products are specified, and failed readiness produces its declared failed or blocked branch.

This is a disclosed squared-norm statistical primitive. The following theorems derive the effective detector response from it and the coherent interaction. They do not derive this primitive from the Hamiltonian Bell current. The same-charge assignments can account for one readiness unit passing through excitation into the terminal flag; an energetic reservoir model would require additional dynamics.

Theorem 22.2 (Exact fresh-cell event and complete null)

Prepare the cell independently in $|A_0\rangle$ and allow one uninterrupted exposure. Define

$$\dot a_i=-ig_ib_i,\qquad \dot b_i=-ig_ia_i-\tfrac{\Gamma_i}{2}b_i, \quad a_i(0)=1,\quad b_i(0)=0.$$(22.2)

For $\rho_{ij}=(P_i\otimes I_R)\rho_{SR}(P_j\otimes I_R)$ and $\eta_i(t)=a_i(t)|A_0\rangle+b_i(t)|A_i^*\rangle$, the complete unnormalized null is

$$\widetilde\rho_{SRA}(t)=\sum_{i,j}\rho_{ij}\otimes |\eta_i(t)\rangle\langle\eta_j(t)|.$$(22.3)

The timed event map on source and reference is

$$\mathcal J_i(dt)(\rho)=\Gamma_i|b_i(t)|^2\rho_{ii}\,dt, \quad S(t)=\sum_iw_i(|a_i(t)|^2+|b_i(t)|^2), \quad w_i=\operatorname{tr}\rho_{ii}.$$(22.4)

For a nonzero event the retained source/reference state is $\rho_{ii}/w_i$. If all channels have the same $g,\Gamma$, then the receptor-discarded null is exactly

$$\mathcal N_\Gamma^T(\rho)=|a(T)|^2\rho+ |b(T)|^2\mathcal D(\rho), \qquad \mathcal D(\rho)=\sum_i\rho_{ii}.$$(22.5)

It is generally different from $S_\Gamma(T)\rho$.

Proof

The no-event Hamiltonian leaves each source sector invariant and acts on its ready/excited span by the two-by-two matrix in (22.2). Its propagator applied to a ready vector is $\eta_i$. Bilinearity gives (22.3) for every mixed input and reference. Application of $C_i$ annihilates every excited vector except $A_i^*$ and maps that one to the fixed terminal $R_i$, giving the event map. Direct differentiation gives

$$\frac{d}{dt}(|a_i|^2+|b_i|^2)=-\Gamma_i|b_i|^2.$$

Thus event integration and null trace sum to one. The conditional hazard is $w_i\Gamma_i|b_i(t)|^2/S(t)$, not the event density itself. For equal channels, $\langle\eta_j|\eta_i\rangle=|a|^2+\delta_{ij}|b|^2$; tracing the receptor gives (22.5). The normalized event formula follows by its trace. Degenerate internal coherence is preserved because neither operator in (22.1) acts on it.

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Both roots of $a_i''+(\Gamma_i/2)a_i'+g_i^2a_i=0$ have negative real part. Hence the cell eventually latches with probability one on each populated sector. Equal channels produce integrated mark weights $w_i$, but a finite response with

The density has quadratic onset and the cumulative probability cubic onset. No finite response is exactly the instantaneous constant-rate detector at its start.

22.2 Null recovery and a retained-excitation counterexperiment

A receptor-only unitary cannot send the different normalized $\eta_i$ to one common ready vector: it preserves their inner products. A source-controlled operation can do so if separately admitted. Choose $W_i\eta_i(T)=\sqrt{n_i(T)}A_0$, where $n_i=|a_i|^2+|b_i|^2$, and apply $\sum_iP_i\otimes W_i$ with the latch disabled. The factors $\sqrt{n_i}$ are forced by unitarity. For equal channels the recovered null is a scalar multiple of the original source; unequal channels retain sector-dependent filtering. This operation uses a shutter and source-controlled access. It is not a source-blind reset and is not part of the fresh-cell theorem below.

22.2.1 Finite pre-latch pointer contact and two overlap scales

The mechanical contact of [C01, §14] acts on the complete null of Theorem 22.2. It requires a shuttered interval with both exchange and latching disabled. Take an independent pointer with position variance $\sigma^2>0$ and wavefunction

In units $\hbar=1$, the only active Hamiltonian during the pulse is

The real pulse profiles have finite integrals; any pointer free evolution is absent or compensated as part of the declared control.

Theorem 22.3 (Complete contact state and distinct overlaps)

For every source state with an arbitrary inaccessible reference, let $a_i=a_i(T)$, $b_i=b_i(T)$ and $S(T)>0$ be as in Theorem 22.2. After the contact, its complete unnormalized null is

$$\widetilde\rho_{SRAY} =\sum_{i,j}\rho_{ij}\otimes|\Xi_i\rangle\langle\Xi_j|, \qquad |\Xi_i\rangle=a_i|A_0,\varphi_0\rangle +b_i|A_i^*,\varphi_{d_i}\rangle.$$(22.7)

Tracing only $Y$ multiplies the excited–excited receptor coherence $|A_i^*\rangle\langle A_j^*|$ by $m_{ij}$ and each ready–excited coherence involving $A_i^*$ by $m_i$, where

$$m_{ij}=e^{-(d_i-d_j)^2/(8\sigma^2)},\qquad m_i=e^{-d_i^2/(8\sigma^2)}.$$(22.8)

If a pointer-position acquisition with its usual squared-amplitude law is additionally supplied, its density conditioned on the first null is

$$p(y\mid\varnothing)=\frac{1}{S(T)}\sum_iw_i \left(|a_i|^2|\varphi_0(y)|^2 +|b_i|^2|\varphi_{d_i}(y)|^2\right).$$(22.9)

The contact by itself is unitary and supplies no acquisition or new stochastic latch law.

Proof

The excited projectors commute and the pulse translates only their pointer factors. Applying this unitary to (22.3) gives (22.7); all maps are the identity on $R$. Completing the square in $\int\varphi_{d_j}(y)^*\varphi_{d_i}(y)\,dy$ gives $m_{ij}$; setting one displacement to zero gives $m_i$. Expansion of each $|\Xi_i\rangle\langle\Xi_j|$ then gives the stated trace factors. For the additional acquisition, replace $a_i,b_i$ in the complete branch by $a_i\varphi_0(y),b_i\varphi_{d_i}(y)$ and take its trace. Distinct source sectors have zero off-diagonal trace, and the ready and excited receptor vectors are orthogonal, giving (22.9). Its integral is one by the definition of $S(T)$.

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Orthogonal receptor labels already eliminate some interference: tracing both $A$ and $Y$ gives the same source/reference marginal as tracing $A$ before the contact. One must not attach $m_{ij}$ to a source cross term that this receptor trace has already removed. If the pointer can return in the future programme, retain (22.7), rather than only its overlap-reduced state. A local later response with the pointer idle can be calculated from either that full state or its exact pointer trace.

For an explicit later effect, take two equal channels with $d_0=d_1=d\ne0$ and resume the same exchange and latch without source control, leaving the pointer idle. Then $m_{01}=1$ but $m_0=m_1=m=e^{-d^2/(8\sigma^2)}<1$. Define

The excited pointer amplitude in either populated sector is $v(t)a\varphi_0+z(t)b\varphi_d$. Summing the two record labels, the resumed first-event density, including the probability of the original null, is therefore

Choose a sufficiently short original exposure $T>0$, so that $a>0$ and $b=-i\beta$ with $\beta>0$. Since $v(t)=-igt+O(t^2)$ and $z(t)=1-\Gamma t/2+O(t^2)$, comparison with zero displacement ($m=1$) gives the finite-window joint-record difference

Dividing by $S(T)$ gives the difference conditioned on the original null. The equal instantaneous densities at resumption do not remove this later timing effect: equal excited displacements leave ready–excited interference suppressed. The resumed event probabilities still use Assumption 22.1; the contact calculation does not select that event law.

Counterexample 22.4 (A current event can report an old excitation)

Start with source $|0\rangle$ and a fresh binary cell. After a null exposure $T$ with $b(T)\ne0$, the unnormalized state is $|0\rangle(a|A_0\rangle+b|A_0^*\rangle)$. Apply a Hadamard to the source only. The old-excited component is now $b|+\rangle|A_0^*\rangle$. On resuming the latch, its instantaneous record-$0$ density is $\Gamma|b|^2$, and that contribution leaves $|+\rangle$, not $|0\rangle$. By continuity the discrepancy persists over a sufficiently short finite resumption window. A subsequent fresh finite $X$-basis detector, with success probability $r>0$, declares $+$ with conditional probability approaching $r$ on this contribution, whereas an erroneously inserted $|0\rangle$ daughter would give $r/2$. The corresponding unconditioned finite joint-record gap is $\tfrac12r\Gamma|b|^2\delta+o(\delta)$ for resumption duration $\delta$. The pre-null probability is already included by the unnormalized $b$.

The exact finite propagator after a control must act on the complete source–receptor bank. An excited $A_k^*$ paired with a different source sector has no coherent return through its stated $P_k$ coupling, but it can still latch. This is the mathematical reason that arbitrary controls during a retained exposure are outside the simple projective daughter claim.

22.3 A quantitative finite-response instrument theorem

Set $g=\tfrac12\sqrt{\kappa\Gamma}$, hold $\kappa>0$ fixed and put $\varepsilon=\kappa/\Gamma\le1/8$ and $d=\sqrt{1-4\varepsilon}$. Let $\mathfrak I_\Gamma^T$ be the finite instrument with timed events (22.4), null (22.5), and the old receptor permanently excluded from later use. Its comparator is the same output space with

Clock conventions agree exactly: both times are continuous physical times censored at the same $T$. Known terminal receptor states can be retained on an event in both comparators. The no-event receptor is discarded in both; its later coherent return is excluded.

Theorem 22.5 (Uniform fresh-cell error with the null retained correctly)

Under the stated finite receptor and primitive latch,

$$\tfrac12\|\mathfrak I_\Gamma^T- \mathfrak I_\infty^T\|_\diamond \le\min\{1,13\kappa/\Gamma\}$$(22.13)

uniformly for $T\ge0$, on one unknown input with arbitrary inaccessible reference. For at most $M$ fresh exposures, arbitrary finite durations and the same adaptive quantum/classical controls between exposures, the complete record/source output distance, and therefore record-history total variation, is at most

$$\min\{1,13M\kappa/\Gamma\}.$$(22.14)

The assertion includes stopping and censoring, but no source control during a retained finite exposure and no return of discarded receptors.

Proof

The two amplitude decay rates are $\Gamma(1\mp d)/4$. Solving (22.2) gives

$$b(t)=-i\frac{2g}{\Gamma d} \left(e^{-\Gamma(1-d)t/4}-e^{-\Gamma(1+d)t/4}\right).$$

Consequently, with $\lambda_\pm=\Gamma(1\pm d)/2$,

$$q_\Gamma(t)=\frac{\kappa}{d^2} \left(e^{-\lambda_-t}-2e^{-\Gamma t/2} +e^{-\lambda_+t}\right).$$(22.15)

The density is nonnegative and integrates to one by the survival identity and complete decay. Also $\lambda_-\ge\kappa$ and $\kappa/\lambda_-=(1+d)/2$. Split the density difference into its leading exponential coefficient, leading exponent and two fast terms. Integrating absolute values yields

$$\begin{align}\int_0^\infty|q_\Gamma(t)-\kappa e^{-\kappa t}|dt &\le(d^{-2}-1)\frac{\kappa}{\lambda_-} +1-\frac{\kappa}{\lambda_-} +\frac{\kappa}{d^2} \left(\frac4\Gamma+\frac1{\lambda_+}\right). \tag{22.16}\end{align}$$

To check the constants explicitly, $d\ge1/\sqrt2$, so the four terms on the right are at most $8\varepsilon$, $2\varepsilon/(1+1/\sqrt2)$, $8\varepsilon$ and $4\varepsilon/(1+1/\sqrt2)$, respectively. Their sum is less than $20\varepsilon$, and in particular less than the checkpoint's retained conservative $22\varepsilon$ bound. Censoring a probability law cannot increase total variation. Thus

$$\delta_\Gamma(T):=\tfrac12\int_0^T |q_\Gamma(t)-\kappa e^{-\kappa t}|dt +\tfrac12|S_\Gamma(T)-e^{-\kappa T}| \le11\varepsilon.$$(22.17)

Insert an intermediate normalized instrument with the same timed events $q_\Gamma(t)P_i\rho P_i\,dt$ but scalar null $S_\Gamma(T)\rho$. For every referenced positive input, the trace norm of its difference from (22.12) is exactly $2\delta_\Gamma(T)$: the event blocks have traces summing to $\operatorname{tr}\rho=1$ at each time, and the null block is a scalar multiple of the same state. This proves the corresponding half-diamond bound, since a Hermiticity-preserving channel difference can be optimized over states with a reference.

The actual null differs from that scalar surrogate by $|b(T)|^2(\mathcal D-\operatorname{id})(\rho)$. From the square representation of (22.15), $q_\Gamma(T)\le\kappa/d^2\le2\kappa$, whence $|b(T)|^2\le2\varepsilon$. The half-diamond distance between two channels is at most one, so this missing null term costs at most $2\varepsilon$. Adding it to (22.17) proves (22.13). The coefficient $13$ is retained as a simple conservative bound; the intermediate estimates show it is not optimal.

For the adaptive programme, both local instruments are normalized CP maps on the same retained output domain by direct calculation. The local bound is uniform in $T$, in the input reference and in all between-exposure controls. Pad stopping by identities, replace the $M$ exposure slots one at a time, and contract through each common suffix. This gives (22.14), including actual null continuations at every intermediate slot. Rare conditional outputs obey (21.13); they have no uniform error independent of their success probabilities.

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This establishes the checkpoint's stated constants with a complete proof and its correct scope. The convergence is integrated over event times, not uniform pointwise in the event density at zero. It improves useful throughput without taking $\kappa$ to zero: $\Gamma$ and $g=\tfrac12\sqrt{\kappa\Gamma}$ increase while the effective click rate stays fixed. The required latch strength and coherent excitation are therefore explicit growing resources. No theorem here obtains those resources from an unmodeled reservoir or proves complete microscopic output convergence after permitting old receptors to return.

22.4 A three-stage null and a completed loss branch

The distinction persists when excitation, capture and loss are separate. For a unit bright input, let the no-latch/no-loss amplitudes $x,y,z$ denote respectively the initial carrier, an emitted product and a receptor excitation. Specify

The primitive latch and loss channels give record density $\Gamma|z|^2$ and loss density $\ell|y|^2$. The norm identity

proves normalization, but an unobserved loss belongs in the observed no-record branch. For a source with bright population $a$ and a dark spectator component, put $D(T)=\int_0^T\Gamma|z|^2dt$. The observed record survival is $1-aD(T)$, and its observed-history hazard is

Dividing instead by $|x|^2+|y|^2+|z|^2$ would condition on both no latch and no loss, a different experiment. Since $z(t)=-gh t^2/2+O(t^3)$, the physical record density is $a\Gamma g^2h^2t^4/4+O(t^5)$. This quartic onset is an exact finite-response prediction. A new attempt must retain the dark component, the surviving three amplitudes and the lost branch; it cannot restart with a newly idealized bright input.

22.5 Delayed classical acquisition: a complete finite filter

The following result consolidates the checkpoint's delayed-record formulas, while keeping their separate statistical constitution explicit. A native classical source generator is supplied in advance. An event can create a pending daughter that later captures a finite ready site or is lost. This can be a useful benchmark for delayed recording, but it is not an admitted passive quantum current meter without a separate material compatibility theorem.

Theorem 22.6 (Finite hidden-state acquisition and continuation)

Let $Z_t$ be a finite-state Markov process whose complete states contain source state, pending daughters, readiness, fuel and every retained classical memory. On each record-adapted control segment let its generator, acting on row laws, be

$$Q(t)=A(t)+\sum_{r\in\mathcal R}B_r(t).$$(22.19)

Each $B_r$ is an entrywise nonnegative kernel of transitions that write observed mark $r$; self-transitions may represent physical marked events. $A$ has nonnegative off-diagonal entries and $A\mathbf1=-\sum_rB_r\mathbf1$. Its diagonal retains the killing rate for every observed event, while all unobserved transitions, including losses, stay in $A$. Rates are bounded and controls depend only on admitted previous observations.

Starting from row law $\alpha$, the unnormalized no-record law solves

$$\dot\alpha_t=\alpha_tA(t).$$(22.20)

Its trace $\alpha_t\mathbf1$ is the no-record probability. A record $r$ at $t$ has density $\alpha_tB_r(t)\mathbf1$, and the exact retained-state posterior is

$$\frac{\alpha_tB_r(t)}{\alpha_tB_r(t)\mathbf1}.$$(22.21)

The normalized update is asserted only at records of positive density, for almost every actual event time; zero-density histories carry no conditional-state claim. These rules iterate to every finite acquired history, including adaptive replenishment and selective stopping, using the actual posterior resources rather than a reset source state.

Proof

Over $dt$, retain every unobserved transition and remove every observed inflow from the no-record law. Its balance is exactly $\alpha_{t+dt}=\alpha_t+\alpha_tA(t)dt+o(dt)$. For a prospective record $r$, sum the marked transition probabilities from each incoming hidden state: the unnormalized post-event law is $\alpha_tB_r(t)dt+o(dt)$. Its trace is the density, and ordinary conditional probability gives (22.21). Bounded finite rates control the multiple-event remainder and ensure the finite propagator exists. Induct over record times; between them propagate by the appropriate $A$, and at them multiply by the appropriate $B_r$. The trace of this product is the joint density of that finite history. Summing all histories and the final no-record branch recovers total probability one from $Q\mathbf1=0$. Adaptive control chooses subsequent matrices from the actually observed history; the same conditional argument applies on each branch. No quantum measurement or normalized linear extraction was used in this classical filtering proof.

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If the extended physical reactions project to a native source update with exactly its original total propensities, then for source projection $\pi$ one has $\mathcal L_{\mathrm{ext}}(f\circ\pi) =(\mathcal L_{\mathrm{src}}f)\circ\pi$. Indeed apparatus-only transitions vanish on $f\circ\pi$, and the remaining projected transition sums agree term by term. Uniqueness of the finite-state martingale problem gives the native projected path law. This is a sufficient neutrality equation, not a derivation of why an added physical daughter leaves every source propensity unchanged. Multiplying source rates by a depleted readiness variable instead blocks the source and changes this equality.

For a concrete complete example, let the source have transitions $r\to q\to s$ at rates $a_1,a_2>0$, emitting labeled daughters $X_1,X_2$. Each pending daughter captures a single ready site at rate $\beta_R$ or is lost at rate $\beta_M$; the first capture exhausts the site. Set $k=\beta_R+\beta_M$. A daughter's probability of no capture by age $v$, allowing prior loss, is

For first acquired mark $R_1$ at $t$, define

Condition first on the emission of $X_1$ at $s<t$. Its survival to capture supplies $h_t(s)ds$; $B(t-s)$ sums no second source event and a second event whose daughter has not captured first. Hence

The actual source has already reached $s$ with conditional probability

This is strictly positive at $t>0$; an acquired $R_1$ cannot be interpreted as a present-state projection onto its historical destination $q$.

If a fresh site is supplied immediately without removing old daughters, the instantaneous next-record intensity is

The inner exponential now requires $X_2$ still to exist, rather than merely to have avoided capture through loss. Theorem 22.6 derives the same expression by retaining pending daughters in the posterior. Zero-current source holds can therefore contain new acquired historical records without new native source events.

22.6 What this part establishes for the measurement architecture

The strongest statements now have one proof each. Joint dark-channel and finite-order tests select held capture and its scalar rate under their stated source premises. A conserved finite converter plus a primitive actual diffusive reader yields a complete finite instrument, with explicit reference, archive, resource and network bounds. The finite receptor theorem establishes a controlled constant-rate instrument limit only after its intrinsic latch has been supplied, and the delayed-state theorem correctly propagates pending classical records. These are compatible mathematical tools where their state, output and resource hypotheses coincide.

These results do not identify three different null constitutions. A frozen intrinsic null, a native diffusive conditional multiplier and an unobserved finite receptor excitation are different physical states. None of the rate constants $\gamma,\nu,\kappa$ becomes the Hamiltonian Bell intensity merely by being an event rate. The detector theorems remain conditional on their own actualization laws. The later pilot and massive constructions provide separate complete measurement implementations; they do not derive every Gaussian or intrinsic-latch primitive in this part. Their source, record and continuation laws must be connected by the explicit state-space and interaction comparisons stated there.