% En07-A.tex % Standalone source for pdfLaTeX; no external bibliography or figures. % Run pdflatex twice to resolve citations and cross-references. % Successful compilation, numerical execution, and machine verification % have not been asserted. \documentclass[11pt]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern} \usepackage[a4paper,margin=1in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{microtype} \usepackage{booktabs,tabularx,longtable,array} \usepackage{enumitem} \usepackage[hidelinks]{hyperref} \hypersetup{ pdftitle={En07-A: Enter, Reversible Recycling, and Thermodynamic Access}, pdfauthor={Ngo Cheung}, pdfsubject={Quantum recovery, destructive readout, complete residuals, and the separation of channel and thermodynamic ledgers}, pdfkeywords={quantum recovery, complementary channel, no-hiding, Landauer principle, reversible computation, destructive measurement, operational discard, holography} } \setlength{\emergencystretch}{3em} \setlength{\tabcolsep}{5pt} \setlist[enumerate]{leftmargin=*,itemsep=3pt} \setlist[itemize]{leftmargin=*,itemsep=3pt} \newcolumntype{L}[1]{>{\raggedright\arraybackslash}p{#1}} \newcolumntype{Y}{>{\raggedright\arraybackslash}X} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \numberwithin{equation}{section} \newcommand{\C}{\mathbb C} \newcommand{\R}{\mathbb R} \newcommand{\id}{\mathrm{id}} \newcommand{\ii}{\mathrm i} \newcommand{\dd}{\mathrm d} \newcommand{\Hil}{\mathcal H} \newcommand{\VV}{\mathcal V} \newcommand{\NN}{\mathcal N} \newcommand{\CC}{\mathcal C} \newcommand{\DD}{\mathcal D} \newcommand{\TT}{\mathcal T} \newcommand{\MM}{\mathcal M} \newcommand{\dtr}{d_{\mathrm{tr}}} \newcommand{\ddia}{d_{\diamond}} \newcommand{\erec}{\varepsilon_{\mathrm{rec}}} \newcommand{\dleak}{\delta_{\mathrm{full}}} \newcommand{\Qtype}{\mathsf Q} \newcommand{\Bit}{\mathsf{Bit}} \newcommand{\Itype}{\mathsf I} \newcommand{\New}{\mathsf{New}} \newcommand{\Enter}{\mathsf{Enter}} \newcommand{\Drop}{\mathsf{Drop}} \newcommand{\Reset}{\mathsf{Reset}} \newcommand{\ReturnQ}{\mathsf{returnQ}} \newcommand{\ReturnBit}{\mathsf{returnBit}} \newcommand{\ket}[1]{\lvert #1\rangle} \newcommand{\bra}[1]{\langle #1\rvert} \newcommand{\kb}[2]{\ket{#1}\bra{#2}} \newcommand{\norm}[1]{\left\lVert #1\right\rVert} \newcommand{\abs}[1]{\left\lvert #1\right\rvert} \DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\Ad}{Ad} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\supp}{supp} \title{ En07-A\\[0.3em] Enter, Reversible Recycling, and Thermodynamic Access:\\ Complete Residuals Without Holographic Identification } \author{Ngo Cheung} \date{} \begin{document} \maketitle \begin{abstract} This paper is a framing and separation note, not a new quantum recovery theorem. It places reversible recycling, operational forgetting, Landauer erasure, and holographic terminology within the reference-preserving access formulation of Hawk04-B. For a complete isometry \(V:C\to R\otimes B\), we locally restate and prove complete-access restoration and the equivalence between exact recovery from \(R\) and constancy of the complete complementary channel to \(B\). Destructive computational-basis readout returns a classical record, not an arbitrary unknown quantum code. Readout followed by classical discard is exactly partial trace, including external references. Its coherent dilation remains isometric, but this mathematical dilation is not a thermodynamic implementation certificate. Three ledgers remain distinct: conservation of a declared complete Hamiltonian, the heat and work of a specified erasure implementation, and information-theoretic residual leakage. Channel equality alone does not determine implementation cost. If radiation is within half-diamond distance \(\eta\) of a replacement channel, its optimal deterministic recovery error is at least \(\max\{0,1-d^{-2}-\eta\}\). An exact flagged-transfer example and a two-state control make the information destination explicit without assigning a temperature. No new Landauer bound, thermodynamic advantage, area law, bulk--boundary dictionary, Page curve, physical receiver, or gravitational evaporation endpoint is constructed. Companion manuscripts and supplied catalogue descriptions are documentary comparisons, not premises of the local channel proofs. No inherited interacting gravitational quantum coefficient is admitted; XV-B's admission-dependent exit is not opened, XVI-N's separate obligations are not discharged, and O.1 remains \texttt{NOT\_CLOSED}. \end{abstract} \begin{center} \fbox{\begin{minipage}{0.93\linewidth} \textbf{Governing distinctions.} \[ \begin{gathered} \text{complete unitary preservation} \ \not\Longrightarrow\ \text{recovery from a chosen subsystem},\\ \text{quantum recovery} \ \ne\ \text{destructive classical observation},\\ \text{operational discard} \ \ne\ \text{proved physical annihilation},\\ \text{finite-program termination} \ \ne\ \text{a physical evaporation endpoint}. \end{gathered} \[ \text{complete energy} \ \ne\ \text{erasure cost} \ \ne\ \text{residual leakage}. \] The residual is an output ledger, not a deletion mechanism. A channel interface does not, by itself, specify a thermal machine. \end{minipage}} \end{center} \tableofcontents %===================================================================== \section{Introduction and governing stance} \label{sec:introduction} The terms ``recycling,'' ``erasure,'' and ``holography'' can refer to different mathematical and physical tasks. This note fixes their interfaces before comparing them. Its immediate source is the supplied Hawk04-B version 2 manuscript \cite{CheungHawk04B}, with the predecessor Hawk04-A providing the complete-complement formulation and featureless-radiation comparison \cite{CheungHawk04A}. The present task is not to derive a new radiation decoder. It is to prevent three changes of meaning: \begin{enumerate}[label=\textup{(\roman*)}] \item replacing a quantum-return decoder by a classical readout; \item replacing a declared marginal by a claim of physical deletion or thermodynamic reset; \item replacing a channel-correction criterion by an area law or a gravitational bulk--boundary identification. \end{enumerate} Throughout, \(\Enter\) means destructive computational-basis readout. Its measured quantum names are consumed and its observable output is classical. A decoder instead has a quantum-return interface \[ R\xrightarrow{\DD}C_{\mathrm{out}}. \] A prescribed observation may subsequently consume \(C_{\mathrm{out}}\), but that subsequent operation has a measurement channel, not \(\id_C\), as its target. Unitary uncomputation, destructive readout-and-drop, and residual leakage are not three versions of the same operation. The first can coherently reverse a specified computation when the required registers are available. The second is generally many-to-one at its declared observable interface. The third is a numerical channel diagnostic, not an operation at all. \subsection{What is proved locally and what is documentary} Sections~\ref{sec:channels}--\ref{sec:examples} provide local proofs of the channel facts needed by the framing. These facts are standard exact-correction, no-hiding, and entanglement-breaking structures \cite{KnillLaflamme,BraunsteinPati,HorodeckiShorRuskai}. Their inclusion makes the present arguments independent of unavailable companion proofs. The thermodynamic inequality discussed in Section~\ref{subsec:landauer} is an explicitly external result, with additional physical assumptions \cite{Landauer,Bennett,ReebWolf}. It is not deduced from the readout-and-drop identity. The microscopic energy statements from Hawk04-B and Em03-E are recorded with their source-model restrictions. No new Hamiltonian, field-to-memory receiver, or reservoir regulator is constructed here. The companion comparisons in Section~\ref{sec:companions} and Appendix~\ref{app:analytic} are documentary, not theorem premises. Published references are distinguished from supplied unpublished manuscripts and excerpts. The supplied records are not represented as independently verified journal articles, and the bibliography does not invent missing publication or archival metadata. \subsection{The conceptual map} \begin{center} \small \begin{tabularx}{\linewidth}{@{}L{0.22\linewidth}Y Y@{}} \toprule Language & Named interface & Present status\\ \midrule Reversible recycling & Complete coherent output and its inverse & Restated channel fact; complete access required.\\[0.4em] Readout and forgetting & \(\Drop\circ\Enter\) & Exactly partial trace, not a physical reset specification.\\[0.4em] Landauer erasure & Specified memory--bath--control process & External thermal framework; no cost inferred from typing.\\[0.4em] Residual leakage & Distance of \(\CC\) from replacement channels & Information-theoretic diagnostic, not heat or work.\\[0.4em] Holographic reconstruction & A separately justified gravitational dictionary & Not identified with an arbitrary output partition.\\[0.4em] Recoverable ``bin'' & Declared inaccessible output \(B\) & May carry recoverable information; need not recover the whole code by itself.\\ \bottomrule \end{tabularx} \end{center} %===================================================================== \section{Complete channels, access, and local recovery facts} \label{sec:channels} Let \(C\simeq\C^d\), \(d\ge2\), be the logical code. Output Hilbert spaces are separable. States are normal density operators, and channels are normal completely positive trace-preserving maps on trace-class operators, abbreviated CPTP. The finite programs below use finite-dimensional encodings. Define \[ \dtr(\rho,\sigma)=\frac12\norm{\rho-\sigma}_1, \qquad \ddia(\Phi,\Psi)=\frac12\norm{\Phi-\Psi}_{\diamond}. \] The diamond distance retains arbitrary external references. For finite input dimension \(d\), a reference of dimension \(d\) suffices in channel discrimination \cite{Watrous}. For an isometry \[ V:C\longrightarrow R\otimes B, \] write \[ \VV(X)=VXV^\dagger,\qquad \NN(X)=\Tr_B\VV(X),\qquad \CC(X)=\Tr_R\VV(X). \] The output \(R\) is declared accessible. The residual \(B\) is complete relative to that declaration: it contains every inaccessible output of the specified pure dilation, including preparation purifiers and apparatus factors when present. For a state \(\sigma\), define \[ \TT_\sigma(X)=\Tr(X)\sigma. \] The two diagnostics are \begin{align} \erec(\NN) &=\inf_{\DD\ {\rm CPTP}}\ddia(\DD\NN,\id_C), \label{eq:rec}\\ \dleak(\CC) &=\inf_{\sigma}\ddia(\CC,\TT_\sigma). \label{eq:leak} \end{align} Decoders are deterministic. A normalized postselected branch is not, by itself, a candidate in \eqref{eq:rec}. For a fixed code basis, put \[ E_{ij}=\kb{i}{j}, \qquad \ket{\Phi_d}=\frac1{\sqrt d}\sum_{i=0}^{d-1} \ket{i}_Q\ket{i}_C, \] and define the normalized Choi state \[ J_{\mathrm n}(\Phi) =(\id_Q\otimes\Phi)(\kb{\Phi_d}{\Phi_d}). \] Agreement on the populations \(E_{ii}\) alone is not agreement on the quantum code. \begin{lemma}[Contraction and recovery stability] \label{lem:stability} CPTP preprocessing and postprocessing contract \(\ddia\). On fixed accessible output spaces, \[ \abs{\erec(\NN_1)-\erec(\NN_2)} \le\ddia(\NN_1,\NN_2). \] Replacement channels satisfy \[ \ddia(\TT_\rho,\TT_\sigma)=\dtr(\rho,\sigma). \] Moreover, \(\dleak(\CC)=0\) if and only if \(\CC\) is constant. \end{lemma} \begin{proof} CPTP maps have completely bounded trace norm one. The contraction statements follow by composing the reference-stable norm comparison. For any fixed decoder, the triangle inequality changes its error by at most \(\ddia(\NN_1,\NN_2)\). Taking infima in both directions proves recovery stability. On a referenced input state, a replacement channel returns the reference marginal tensored with its replacement state. This proves the replacement-channel equality; a product input already attains it. If leakage is zero, choose states \(\sigma_n\) with \(\ddia(\CC,\TT_{\sigma_n})\to0\). Evaluation at \(\tau_C=I_C/d\) gives \(\sigma_n\to\CC(\tau_C)\) in trace norm. The replacement-channel equality and the triangle inequality then give \[ \CC=\TT_{\CC(\tau_C)}. \] The converse is immediate. \end{proof} \begin{lemma}[Orthogonal unitary mixtures and constant reconstruction] \label{lem:unitary-mixture} Suppose the unitary channels \(\Ad_{U_j}\) have mutually orthogonal Bell outputs. For real coefficients \(a_j\), \[ \frac12\left\|\sum_j a_j\Ad_{U_j}\right\|_\diamond =\frac12\sum_j\abs{a_j}. \] In particular, \begin{equation} \inf_\sigma\ddia(\id_C,\TT_\sigma) =1-d^{-2}=:a_d. \label{eq:ad} \end{equation} Every channel from \(C\) has recovery error at most \(a_d\). \end{lemma} \begin{proof} A Bell input gives a diagonal signed output with eigenvalues \(a_j\), proving the lower bound. Each unitary channel has diamond norm one, so the triangle inequality gives the upper bound. For any state \(\sigma\), the Bell output of \(\TT_\sigma\) is \(I_Q/d\otimes\sigma\). Its overlap with the target Bell projector is \(d^{-2}\). Testing that projector therefore gives \[ \ddia(\id_C,\TT_\sigma)\ge1-d^{-2}. \] The \(d^2\) Weyl unitaries have orthogonal Bell outputs and uniform twirl \(\TT_{\tau_C}\). The first part gives the matching upper bound. A decoder that ignores its input and prepares \(\tau_C\) attains the same error after any channel. \end{proof} \subsection{Complete-access restoration is a decoder with a larger input} The following restatement is the restore structure in Hawk04-A and Hawk04-B \cite[\texttt{thm:global}]{CheungHawk04A} \cite[\texttt{thm:global}]{CheungHawk04B}. \begin{theorem}[Complete preservation and restoration] \label{thm:global} For an isometry \(V:C\to R\otimes B\), the map \begin{equation} \mathcal L(Y) = V^\dagger YV+ \Tr((I-VV^\dagger)Y)\tau_C \label{eq:restore} \end{equation} is CPTP and satisfies \[ \mathcal L\VV=\id_C. \] For every code--reference trace-class operator \(X\), \[ \norm{(V\otimes I)X(V^\dagger\otimes I)}_1=\norm X_1. \] In particular, if \[ V\ket\psi=U(J\ket\psi\otimes\ket{\Omega}), \] where \(J\) is an isometry, \(\ket{\Omega}\) is fixed and normalized, and \(U\) is a complete unitary, the entire code is preserved. \end{theorem} \begin{proof} The first term of \eqref{eq:restore} is completely positive. The second is a positive-effect measurement followed by fixed preparation and is also completely positive. Their traces sum to \(\Tr Y\). On the range of \(V\), the second term vanishes and the first returns \(X\). An isometric embedding preserves all nonzero singular values, including after tensoring with a reference identity. This proves the trace-norm equality. The specified preparation and unitary compose to an isometry. \end{proof} The map \(\mathcal L:R B\to C\) is a \emph{complete-access quantum decoder}. It is not generally a radiation-only decoder \(R\to C\). It is also not a unitary inverse on every state of \(R B\): the second term is an off-code completion. On the specified encoded range, complete coherent access allows reversal of the preparation and evolution. This corrects the possible shorthand ``the restore map is not a decoder.'' Its mathematical task is decoding; the issue is which input interface it requires. \subsection{Exact accessible recovery and the complete complement} \begin{theorem}[Exact recovery criterion, locally restated] \label{thm:exact} For an isometry \(V:C\to R\otimes B\), the following are equivalent: \begin{enumerate}[label=\textup{(\roman*)}] \item there is a CPTP map \(\DD:R\to C\) with \(\DD\NN=\id_C\); \item \(\CC=\TT_\sigma\) for some state \(\sigma\); \item \(\CC(E_{ij})=\delta_{ij}\sigma\) for every matrix unit; \item \[ J_{\mathrm n}(\CC)=\frac{I_Q}{d}\otimes\sigma. \] \end{enumerate} When these conditions hold, there are a separable Hilbert space \(K\), a state \(\sigma_K\), and an isometry \(A:C\otimes K\to R\) such that \begin{equation} \NN(X)=A(X\otimes\sigma_K)A^\dagger. \label{eq:normalform} \end{equation} An exact decoder is \begin{equation} \DD(Y) = \Tr_K(A^\dagger YA) +\Tr((I-AA^\dagger)Y)\tau_C. \label{eq:decoder} \end{equation} \end{theorem} \begin{proof} Linearity gives equivalence of (ii) and (iii). Expanding the Bell projector gives equivalence of (iii) and (iv). Assume (ii), and write the nonzero spectral decomposition \[ \sigma=\sum_\alpha p_\alpha\kb{b_\alpha}{b_\alpha}, \qquad p_\alpha>0,\qquad \sum_\alpha p_\alpha=1. \] The index set is finite or countable. The constancy condition puts every \(V\ket i\) in \(R\otimes\supp\sigma\). Write \[ V\ket i = \sum_\alpha\sqrt{p_\alpha} \ket{r_{i\alpha}}_R\ket{b_\alpha}_B. \] The complementary matrix-unit identities imply \[ \langle r_{j\beta},r_{i\alpha}\rangle =\delta_{ij}\delta_{\alpha\beta}. \] Consequently \[ A(\ket i\ket\alpha)=\ket{r_{i\alpha}} \] extends to an isometry. With \(\sigma_K=\sum_\alpha p_\alpha\kb{\alpha}{\alpha}\), tracing \(B\) gives \eqref{eq:normalform}. Both terms of \eqref{eq:decoder} are completely positive, and their traces sum to \(\Tr Y\). Substitution into \eqref{eq:normalform} gives exact recovery of every matrix unit. Conversely, dilate an exact decoder by a Stinespring isometry \(L:R\to C\otimes G\). The normal channel dilation is the standard mathematical structure theorem \cite{Stinespring,Watrous}. Put \(Z=(L\otimes I_B)V\). Since the decoded marginal for basis input \(\ket i\) is pure, \[ Z\ket i=\ket i\otimes\ket{\eta_i}. \] Exact recovery of \(E_{ij}\) forces \(\langle\eta_j,\eta_i\rangle=1\). All the normalized vectors \(\eta_i\) therefore coincide. Tracing the decoded register and \(G\) gives \[ \CC(E_{ij}) =\delta_{ij}\Tr_G\kb{\eta}{\eta}, \] which is (iii). \end{proof} These are the standard exact-correction/no-hiding conditions \cite{KnillLaflamme,BraunsteinPati}, restated in the complete output convention of Hawk04-A and Hawk04-B \cite[\texttt{thm:exact}]{CheungHawk04A} \cite[\texttt{thm:exact}]{CheungHawk04B}. No priority claim is made. \begin{corollary}[The precise no-hiding conclusion] \label{cor:nohiding} If \(\NN\) is constant on the entire code, \(B\) exactly recovers that code. If \(\CC\) is constant, \(R\) exactly recovers it. For \(d\ge2\), both complementary outputs cannot independently recover the entire unknown code. \end{corollary} \begin{proof} Interchange \(R\) and \(B\) in Theorem~\ref{thm:exact}. An exactly recoverable output has a constant complement, and a constant channel cannot recover two distinct inputs. \end{proof} \begin{remark}[Nonconstant is not independently recoverable] \label{rem:nonconstant} A nonconstant \(\CC\) distinguishes some logical inputs. It need not contain an independently recoverable copy of the entire code. Joint access to \(R B\) can be necessary. The examples in Section~\ref{sec:examples} exhibit this without changing the no-hiding theorem's hypotheses. \end{remark} \begin{remark}[Reversible recycling] \label{rem:recycling} A specified complete unitary may be uncomputed when its required registers remain available. This is reversible recycling of a coherent computation, not deletion of its unknown input. A local operation on \(B\), with all outputs ignored, cannot improve the pre-existing channel \(\NN\). Exact recovery from \(R\) requires a constant complete complement. A fixed mixed complement gives the syndrome form \eqref{eq:normalform}; it does not assert that every physical work register has already been reset to a pure blank state. These channel facts are compatible with irreversible erasure costs. They establish neither those costs nor a holographic encoding. \end{remark} %===================================================================== \section{Enter, Drop, and the complete apparatus ledger} \label{sec:enter} \subsection{Quantum and classical interfaces} For the finite fragment, use types \[ T::=\Itype\mid\Qtype\mid\Bit\mid T\otimes T \] with observable algebras \[ \mathcal A_{\Itype}=\C,\qquad \mathcal A_{\Qtype}=M_2(\C),\qquad \mathcal A_{\Bit}=\C\oplus\C. \] A hybrid state is block diagonal on its classical factors. A bit is not an unrestricted qubit under another name. The primitives are pure allocation, specified unitary constants, destructive computational readout, and classical discard: \[ \New(1)=\kb0{0}, \qquad U(\rho)=U\rho U^\dagger, \] \[ \Enter(\rho) =\sum_{x=0}^1\bra x\rho\ket x\,\kb x x, \qquad \Drop(p_0,p_1)=p_0+p_1. \] Return commands designate the retained interface; they are not additional dynamics. Every live quantum wire is returned or explicitly consumed. It is never duplicated. Classical control acts separately on classical blocks. Only finite acyclic instruction trees are considered. The fragment is defined here directly. The supplied Landmark I characterization concerns Born probabilities and positive-branch destructive residual rays within its own declared strategy class \cite[\texttt{thm:main}]{CheungI}. That full characterization is not reconstructed or imported. Likewise, the En03-A interface and scheduled-locality material is documentary \cite{CheungEn03A}. \subsection{Destructive partial readout} Let \(S\) be \(s\) measured qubits and \(T\) the surviving quantum register. Put \[ M_x=\bra{x}_S\otimes I_T, \qquad x\in\{0,1\}^s. \] The observable instrument is \begin{equation} \mathcal E_S(X) = \sum_x\kb x x_K\otimes M_xXM_x^\dagger. \label{eq:instrument} \end{equation} The names in \(S\) are consumed; \(T\) survives. For a pure input \(\psi\), the branch probability and positive-probability residual are \[ p_x=\norm{M_x\psi}^2, \qquad \psi_x=\frac{M_x\psi}{\sqrt{p_x}}\quad(p_x>0). \] No normalized residual is assigned to a zero-probability branch. Complete-channel calculations use unnormalized branches and retain every outcome. \begin{lemma}[Erasure-interface identity and coherent dilation] \label{lem:erasure-interface} For a finite encoded residual \(B\), destructive readout followed by discard equals partial trace: \begin{equation} (\id_{QR}\otimes(\Drop\circ\Enter))(\rho_{QRB}) =\Tr_B\rho_{QRB}. \label{eq:drop} \end{equation} Consequently, \[ (\id_R\otimes(\Drop\circ\Enter))\VV=\NN. \] For the instrument \eqref{eq:instrument}, the map \begin{equation} W_{\mathrm{read}}\ket\psi = \sum_x\ket x_K\ket x_E\otimes M_x\ket\psi \label{eq:read-dilation} \end{equation} is an isometry. Tracing \(E\) gives the observable instrument. Its complete coherent output \(K E T\) preserves the input, whereas the classical interface alone generally does not. \end{lemma} \begin{proof} The diagonal \(B\)-block sum is exactly partial trace: \[ \sum_x (I_{QR}\otimes\bra x)\rho_{QRB} (I_{QR}\otimes\ket x) =\Tr_B\rho_{QRB}. \] This identity holds for every joint state and therefore with arbitrary references. For the dilation, \[ W_{\mathrm{read}}^\dagger W_{\mathrm{read}} =\sum_xM_x^\dagger M_x=I. \] Tracing the orthogonal \(E\)-labels removes cross-outcome terms and gives \eqref{eq:instrument}. Complete restoration follows from Theorem~\ref{thm:global}. \end{proof} This is the local restatement of Hawk04-B's partial-trace and apparatus results \cite[\texttt{thm:drop}, \texttt{prop:measurement-dilation}] {CheungHawk04B}. The physical carrier \(K\) in \eqref{eq:read-dilation} is quantum. It becomes a commutative observable interface after the omitted factor \(E\) is traced out. The omitted information is preserved in the \emph{complete coherent output}; it is not claimed to be independently recoverable from \(E\) alone. For example, \[ a\ket0+b\ket1 \longmapsto a\ket0_K\ket0_E+b\ket1_K\ket1_E. \] Neither marginal generally retains the unknown phase. Joint \(K E\) access does. Neither marginal channel is constant on the entire qubit code, so this is not a counterexample to Corollary~\ref{cor:nohiding}. \begin{proposition}[Ignored local operations do not change untouched outputs] \label{prop:local} Let \(\Lambda:B\to B'\) be any normal CPTP map. For every joint state, \[ \Tr_{B'}((\id_{QR}\otimes\Lambda)(\rho_{QRB})) =\Tr_B\rho_{QRB}. \] In particular, unobserved residual readout-and-drop leaves \(\erec(\NN)\) exactly unchanged. \end{proposition} \begin{proof} The Heisenberg adjoint \(\Lambda^*\) is unital. For every bounded observable \(A\) on \(Q R\), \[ (\id_{QR}\otimes\Lambda^*)(A\otimes I_{B'}) =A\otimes I_B. \] Thus all expectations on \(Q R\) are unchanged. The recovery task is determined by that unchanged channel. \end{proof} This is an unconditional statement. Supplying a measurement outcome and conditioning on it changes the access task. \subsection{Discard is not reset} A physical-memory reset has a different output type. For a declared memory \(M\), define \[ \Reset_M(X)=\Tr(X)\kb0{0}_M. \] On a joint state, \begin{equation} (\id_{QR}\otimes\Reset_M)(\rho_{QRM}) =\Tr_M\rho_{QRM}\otimes\kb0{0}_M. \label{eq:reset} \end{equation} By contrast, discard returns no memory register at all. One may model ``discard, then allocate a blank'' as an observable implementation of \eqref{eq:reset}. That syntax still does not say how a reusable physical memory, its blank-state preparation, its controllers, or its environment are restored between runs. \begin{remark}[Completed records and later reset] \label{rem:record-reset} Resetting one apparatus factor cannot retroactively change a completed record already delivered to an untouched register. This follows from Proposition~\ref{prop:local}. Resetting the very register that holds the only retained record, however, removes that retained record from the subsequent interface. The matching distinction is recorded in Em03-B \cite{CheungEm03B}. It is a statement about specified interfaces, not a thermodynamic cost theorem. \end{remark} \subsection{Finite programs and quantum return} \begin{proposition}[Finite semantic realization and termination] \label{prop:finite} Every finite-dimensional CPTP map admits a semantic implementation by pure allocation, a specified unitary, destructive readout of unused work qubits, discard of their outcomes, and quantum return of its designated output. Every well-typed finite acyclic instruction tree in the declared fragment denotes a CPTP hybrid channel, including untouched references, and every evaluation branch terminates. \end{proposition} \begin{proof} For Kraus operators \(D_j\) of a finite channel, define \[ L\ket\psi=\sum_jD_j\ket\psi\ket j_G. \] Completeness of the Kraus family makes \(L\) an isometry. After qubit encoding and sufficient pure allocation, its orthonormal columns extend to a unitary. Readout and discard of \(G\) give \[ X\longmapsto\sum_jD_jXD_j^\dagger, \] while the designated logical output remains quantum. For a finite instruction tree, allocation and unitary nodes are complete channels, and readout satisfies \(\sum_xM_x^\dagger M_x=I\). Classical discard sums unnormalized blocks. Induction gives path maps \(K_h\) with \[ \sum_hK_h^\dagger K_h=I. \] Grouping paths by reported outputs gives the hybrid denotation. Tensoring every path map with a reference identity proves reference preservation. A finite acyclic tree has finite maximum branch length. \end{proof} This is a semantic existence statement, not finite-gate compilation, efficient synthesis, laboratory realization, or a proof about an unspecified larger calculus. An infinite-dimensional decoder supplied by Theorem~\ref{thm:exact} need not have a finite program. For finite encodings, the interfaces are: \[ P_{\mathrm{coh}}:Q_C\to Q_R\otimes Q_B, \qquad P_{\DD}:Q_R\to Q_C^{\mathrm{out}}. \] The local decoder can be written as: \begin{verbatim} w := new; (Cout,G) := DecoderUnitary(R,w); k := Enter(G); Drop(k); returnQ(Cout); \end{verbatim} All unused work wires belong to \(G\). The consumed qubits are work or syndrome qubits, not the recovered diary output. Applying it locally to the coherent body gives the joint observable channel \[ (\DD\otimes\id_B)\VV:C\to C_{\mathrm{out}}\otimes B. \] The observer's first-factor marginal is \(\DD\NN\). A strict one-register observable implementation is: \begin{verbatim} (R,B) := CoherentBody(C); k := Enter(B); Drop(k); Cout := Decode(R); returnQ(Cout); \end{verbatim} Its channel is again \(\DD\NN\), by Lemma~\ref{lem:erasure-interface}. This added readout is not inferred from an autonomous Hamiltonian. The joint observable return \(C_{\mathrm{out}}B\) does not silently become the complete physical dilation of the decoder. Decoder work, measurement carriers, and omitted apparatus outputs must be added to that complete ledger. For a prescribed terminal observation \(\MM\), contraction gives \begin{equation} \ddia(\MM\DD\NN,\MM) \le\ddia(\DD\NN,\id_C). \label{eq:record-cert} \end{equation} The converse for one chosen observation is false: computational dephasing preserves computational readout but removes off-diagonal matrix units. %===================================================================== \section{Classical records and exact destination examples} \label{sec:examples} \subsection{Record-only reconstruction} \begin{theorem}[Classical-output obstruction, locally restated] \label{thm:classical} Let \(\mathcal E\) be a measure-and-prepare channel from \(C\), and let \(\DD\) be any CPTP decoder to \(C\). Then \begin{equation} \ddia(\DD\mathcal E,\id_C)\ge1-d^{-1}. \label{eq:classical} \end{equation} The bound applies to every finite adaptive process whose only retained output is classical. Computational readout followed by basis-state preparation attains it. \end{theorem} \begin{proof} A measure-and-prepare channel has the form \[ \mathcal E(X)=\sum_x\Tr(F_xX)\sigma_x. \] Its Bell-input output, and any locally decoded version, is separable across reference and output \cite{HorodeckiShorRuskai}. For normalized product vectors, \[ \abs{\langle\Phi_d,u\otimes v\rangle}^2\le d^{-1}. \] Convexity gives Bell overlap at most \(d^{-1}\) for every separable state. Testing the target Bell projector proves \eqref{eq:classical}. A channel to a finite classical register has POVM form. Any decoder from it prepares an output state for each record. Finite adaptive processing before the final record does not change this measure-and-prepare form. For attainability, let \[ Z_d=\sum_{j=0}^{d-1}e^{2\pi\ii j/d}\kb j j. \] Computational dephasing is \[ \Delta_d=\frac1d\sum_{k=0}^{d-1}\Ad_{Z_d^k}. \] These \(d\) unitary channels have orthogonal Bell outputs. Lemma~\ref{lem:unitary-mixture} gives \(\ddia(\Delta_d,\id_C)=1-d^{-1}\). \end{proof} This is Hawk04-B's quantitative record-only obstruction \cite[\texttt{thm:classical}]{CheungHawk04B}. A surviving quantum memory or quantum side information supplied to the receiver changes the hypothesis. \begin{example}[The complete diary is in the residual] \label{ex:bin} For \[ V\ket\psi=\ket0_R\ket\psi_B, \] radiation is constant and has recovery error \(a_d\). The complete residual exactly recovers the diary. Its computational record alone cannot do so: Theorem~\ref{thm:classical} applies. Readout-and-drop leaves radiation unchanged. \end{example} \begin{proof} The channels are a replacement channel and the identity, up to local isometric dictionaries. Equation~\eqref{eq:ad} gives the radiation error. The readout assertions follow from Theorem~\ref{thm:classical} and Lemma~\ref{lem:erasure-interface}. \end{proof} \begin{example}[A diary-independent record can assist correction] \label{ex:assistance} For a qubit, set \[ V\ket\psi = \frac1{\sqrt2} \left( \ket\psi_R\ket0_B+Z\ket\psi_R\ket1_B \right). \] Radiation has optimal recovery error \(1/2\). Computational readout of \(B\) gives a uniform bit \(x\), independent of the logical input. Supplying that bit and applying \(Z^x\) to \(R\) gives exact quantum recovery. The bit alone is not a replacement for the diary. \end{example} \begin{proof} The unnormalized branch is \[ \rho\longmapsto\frac12Z^x\rho Z^x. \] Its probability is \(1/2\), and correction by \(Z^x\) followed by summation returns \(\rho\). Without the record, the channel is \(\Delta_Z=(\id+\Ad_Z)/2\). The two inputs \(\ket+\) and \(\ket-\) have identical dephased outputs, giving recovery error at least \(1/2\) by the triangle inequality. The identity decoder attains \(1/2\) by Lemma~\ref{lem:unitary-mixture}. For a normalized input, the complete residual is \[ \CC(\rho) = \frac12 \begin{pmatrix} 1&\Tr(Z\rho)\\ \Tr(Z\rho)&1 \end{pmatrix}. \] It is not constant. Only its computational record is diary independent. \end{proof} These are local destination tests corresponding to Hawk04-B's residual and assistance examples \cite[\texttt{ex:bin}, \texttt{ex:assistance}]{CheungHawk04B}. They concern quantum access and classical assistance, not thermodynamic branch-reuse advantages. \subsection{A channel-level check of the two-state control} The next example reproduces the information-theoretic structure of Hawk04-B's four-Majorana control without reconstructing its Hamiltonian \cite[\texttt{thm:control}]{CheungHawk04B}. \begin{example}[Complete preservation without either local inverse] \label{ex:control} Let \(0\le c\le1\), \(z=\sqrt{1-c^2}\), and choose \(0\le\theta\le\pi/4\) with \(\cos(2\theta)=c\). Define \[ \ket{r_0}=\cos\theta\ket0+\sin\theta\ket1, \qquad \ket{r_1}=\cos\theta\ket0-\sin\theta\ket1, \] and the isometry \[ V_c\ket i=\ket{r_i}_R\ket i_B. \] Then \begin{align} \NN_c(\rho) &=\rho_{00}\kb{r_0}{r_0} +\rho_{11}\kb{r_1}{r_1}, \label{eq:controlN}\\ \erec(\NN_c) &=\frac{3-z}{4}, \label{eq:controlR}\\ \erec(\CC_c) &=\frac{1-c}{2}, \label{eq:controlB}\\ \dleak(\CC_c) &=\max\left\{\frac12,\frac{1+2c}{4}\right\}. \label{eq:controlleak} \end{align} The complete output is exactly reversible. For \(0