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\chapter{Finite-block coding and uncertainty in the loss record}
\label{chap:finite-coding}

The extraction identity separates the physics of the bath from the problem of distributing quantum information across the surviving registers. A complete rate statement nevertheless needs a finite coding theorem. In particular, a small field error per block cannot be declared harmless and then repeated an unlimited number of times. The construction in this chapter encodes directly in one growing physical system. Its finite error bound is combined with whatever approximation budget the field model supplies.

The mathematical ingredient is a two-copy Haar calculation inside an arbitrary protected subspace. No permutation invariance, tensor-product structure of the subspace, or special stabilizer description is required. This generality is useful for feature sectors whose basis states are defined by several rounded spatial charges. The method is an explicit specialization of decoupling-based quantum coding \cite{decoupling}. The new uses in this manuscript concern the protected sectors and their uniform transfer across bath histories, rather than a new claim of priority for random coding itself.

\section{Operational criterion and the role of the flags}
\label{sec:fin-criterion}

Let $P$ be a rank-$D$ projector on $N$ physical qubits, and let $V:\mathbb C^K\rightarrow\operatorname{ran}P$ be an isometric encoding. A reference $F$ of dimension $K$ purifies the maximally mixed logical state. The corresponding encoded state is
\begin{equation}
 \ket{\Psi_V}_{FA}=\frac1{\sqrt K}\sum_{i=1}^K\ket i_F V\ket i,
 \qquad A=\{1,\ldots,N\}.
\end{equation}
The entanglement-transmission error of a recovered logical channel $\mathcal L$ is
\begin{equation}\label{eq:fin-ent-error}
 e(\mathcal L)=1-\bra{\Phi_K}(\id_F\otimes\mathcal L)
       (\ket{\Phi_K}\bra{\Phi_K})\ket{\Phi_K}.
\end{equation}
This criterion tests preservation of unknown coherent amplitudes and their correlations with an outside system. It is stronger than testing a list of logical basis states, but a finite value of \eqref{eq:fin-ent-error} is not itself a dimension-independent diamond-norm guarantee. The distinction is quantified below.

For an erased set $E\subseteq A$, write $R=A\setminus E$ and use a decoder $\mathcal D_E$ acting on $R$. The classical erasure record specifies $E$ exactly. An exogenous loss law is a probability distribution $\mu$ on these subsets that is selected independently of the unknown logical input. The corresponding flagged channel is
\begin{equation}\label{eq:fin-flagged}
 \mathcal M_\mu(\rho)=\sum_{E\subseteq A}\mu(E)\ket E\bra E_C
                  \otimes\Tr_E(V\rho V^\dagger).
\end{equation}
The decoder reads $C$ and applies $\mathcal D_E$. Consequently its entanglement error is the ordinary average of the branch errors. A correlated distribution of loss locations is allowed in this definition. Independence among sites will be used only when deriving particular tail estimates.

The physical extraction theorem gives a stronger kind of uniformity than ordinary random-coding arguments. Once $V$ and the family $\{\mathcal D_E\}$ have been chosen, the same family composed with the prescribed extractor works for every exactly protected bath history. There is no additional union bound over bath Hamiltonians, bath dimensions, or their time dependences. Approximate field protection introduces a channel-distance budget, denoted by $\zeta$ in this chapter.

\section{The exact Haar second moment}
\label{sec:fin-haar}

Put $\Omega=P/D$, with marginals $\Omega_E$ and $\Omega_R$. Choose $V$ uniformly from isometries into $\operatorname{ran}P$. The random encoded state has complementary marginal
\begin{equation}
 \rho_{FE}(V)=\Tr_R\ket{\Psi_V}\bra{\Psi_V}.
\end{equation}
Its fixed decoupling target is $\tau_F\otimes\Omega_E$, where $\tau_F=I_F/K$. Choosing a target that is independent of the sampled encoding makes both the Haar calculation and the later simultaneous-pattern bound transparent.

\begin{lemma}[Sector-restricted erasure decoupling]
\label{lem:subspace}
\label{lem:fin-haar}
Assume $1\leq K\leq D$ with $D>1$. For an erased subsystem of dimension $d_E=2^{|E|}$, define
\begin{equation}
 X_E(V)=\rho_{FE}(V)-\tau_F\otimes\Omega_E.
\end{equation}
Then
\begin{equation}\label{eq:fin-exact-variance}
 \mathbb E_V\Tr X_E(V)^2
 =\frac{DK-1}{K(D^2-1)}
       \left[D\Tr\Omega_R^2-\Tr\Omega_E^2\right].
\end{equation}
In particular,
\begin{align}
 \mathbb E_V\|X_E(V)\|_1^2&\leq\frac{K d_E^2}{D},
       \label{eq:fin-second-trace}\\
 \mathbb E_V\|X_E(V)\|_1&\leq\sqrt{\frac{K d_E^2}{D}}.
       \label{eq:fin-first-trace}
\end{align}
The dimension-one logical case admits a trivial perfect decoder irrespective of these estimates.
\end{lemma}

\begin{proof}
Identify $\operatorname{ran}P$ with a $D$-dimensional space and write the random code vectors as $\ket{u_i}=U\ket i$ for a Haar unitary $U$. The identity and swap span the commutant of $U\otimes U$. If $Y$ is an operator on two copies of this space, its Haar average is
\begin{equation}\label{eq:fin-haar-twirl}
 \mathbb E_U(U\otimes U)Y(U^\dagger\otimes U^\dagger)
       =aI+b\mathsf S,
\end{equation}
where $\mathsf S$ exchanges the two copies and
\begin{equation}
 a=\frac{\Tr Y-\Tr(\mathsf SY)/D}{D^2-1},\qquad
 b=\frac{\Tr(\mathsf SY)-\Tr Y/D}{D^2-1}.
\end{equation}
These coefficients follow by preserving the trace and the trace against $\mathsf S$; thus no normalization convention for the Haar integral remains implicit.

The swap trick for $\Tr\rho_{FE}^2$ involves
\begin{equation}
 Y=\sum_{i,j=1}^K\ket i\bra j\otimes\ket j\bra i.
\end{equation}
Its trace is $K$, while $\Tr(\mathsf SY)=K^2$. Substituting into \eqref{eq:fin-haar-twirl}, dividing by $K^2$, and returning to the physical embedding gives
\begin{align}\label{eq:fin-purity}
 \mathbb E_V\Tr\rho_{FE}(V)^2
 ={}&\frac{D(D-K)}{K(D^2-1)}\Tr\Omega_E^2\\
 &+\frac{D(DK-1)}{K(D^2-1)}\Tr\Omega_R^2.\nonumber
\end{align}
Here the two swap contractions are
\begin{equation}
 \Tr[(P\otimes P)\mathsf S_E]=\Tr(\Tr_RP)^2,
 \quad
 \Tr[(P\otimes P)\mathsf S_R]=\Tr(\Tr_EP)^2.
\end{equation}
Also $\mathbb E\rho_{FE}=\tau_F\otimes\Omega_E$. Therefore subtracting the target purity $K^{-1}\Tr\Omega_E^2$ from \eqref{eq:fin-purity} proves \eqref{eq:fin-exact-variance}.

The coefficient of $\Tr\Omega_R^2$ in \eqref{eq:fin-exact-variance} is at most one, because $K\leq D$, and the other term is nonpositive. Moreover $P\leq I_{RE}$ implies
\begin{equation}\label{eq:fin-marginal-bound}
 \Omega_R\leq\frac{d_E}{D}I_R,
 \qquad
 \Tr\Omega_R^2\leq\frac{d_E}{D}.
\end{equation}
The operator $X_E$ acts on a space of dimension $K d_E$. Thus $\|X_E\|_1^2\leq Kd_E\Tr X_E^2$. Taking expectations and using \eqref{eq:fin-marginal-bound} gives \eqref{eq:fin-second-trace}; Jensen's inequality gives \eqref{eq:fin-first-trace}.
\end{proof}

The exact variance retains information about the geometry of $P$ through its two marginal purities. The dimension-only bound discards that information. A computational study can therefore report both quantities: the dimension-only value supplies a universal certificate, while a directly calculated purity may yield a stronger certificate for a particular erased set. Neither requires guessing a decoder numerically.

\subsection{From complementary decoupling to a decoder}

For a fixed $V$ and $E$, Uhlmann's theorem compares a purification of $\rho_{FE}$ with a purification of $\tau_F\otimes\Omega_E$. It produces a survivor operation whose recovered logical system has entanglement fidelity at least
\begin{equation}
 F_e\geq\left(1-\tfrac12\|X_E(V)\|_1\right)^2
\end{equation}
when the expression in parentheses is nonnegative. Additional local ancillas can be included in the operation. Hence a decoder exists with
\begin{equation}\label{eq:fin-branch-error}
 e_E(V)\leq\min\{1,\|X_E(V)\|_1\}.
\end{equation}
The target is fixed, but the chosen isometry in Uhlmann's theorem can depend on $V$ and $E$. It does not depend on the logical state or on a protected bath history. This dependence is exactly what the classical erasure record makes available.

The statement is an existence result, not a polynomial-time decoding algorithm. Forming the full Choi matrix or finding the optimizing purification can be exponentially expensive in $N$. Reversible ranking of a spatial sector addresses a different computational problem: it describes the inner sector embedding, not the outer near-capacity quantum erasure decoder.

\section{An information-spectrum bound for any external loss law}
\label{sec:fin-spectrum}

Let $T=|E|$ be the total number of erased registers under $\mu$. Lemma~\ref{lem:subspace} and \eqref{eq:fin-branch-error} immediately imply the following statement.

\begin{theorem}[Finite coding under an arbitrary exogenous loss law]
\label{thm:fin-law}
For every rank-$D$ sector and logical dimension $K\leq D$, there is an encoding and a family of flag-dependent decoders such that
\begin{equation}\label{eq:fin-spectrum}
 e_\mu\leq
 \sum_E\mu(E)\min\left\{1,\sqrt{\frac{K4^{|E|}}D}\right\}.
\end{equation}
For every integer cutoff $0\leq u\leq N$, the simpler certificate is
\begin{equation}\label{eq:fin-cutoff}
 e_\mu\leq\Pr_\mu(T>u)+\sqrt{\frac{K4^u}{D}}.
\end{equation}
If extraction and field approximation introduce a uniform half-diamond budget $\zeta$, add $\zeta$ to the right-hand sides.
\end{theorem}

\begin{proof}
For every branch choose the decoder just described. Average \eqref{eq:fin-branch-error} over $V$ and $\mu$, and use concavity of $x\mapsto\min\{1,x\}$ together with \eqref{eq:fin-first-trace}. At least one encoding attains no more than this ensemble average. For \eqref{eq:fin-cutoff}, use the trivial infidelity bound one above the cutoff and $4^{|E|}\leq4^u$ below it. Finally, any fixed decoder is contractive for trace distance; a half-diamond perturbation changes the probability of the entanglement-fidelity test by at most $\zeta$.
\end{proof}

Only the distribution of $T$ appears in the universal bound, although the decoder still uses the complete pattern $E$. In particular, spatial correlations between losses do not invalidate the theorem. For a sequence of exogenous laws, if $T_N/N$ is with high probability at most $p+o(1)$ and $\log_2D_N=N-o(N)$, all rates below $1-2p$ are achievable. Independence is a sufficient condition for that concentration, not a logical requirement of the coding argument. An exact converse for a general correlated loss channel is a separate question.

The mean loss fraction by itself is insufficient. Consider the law that erases every register with probability $p$ and erases none with probability $1-p$. Every individual qubit has the same marginal loss probability $p$ as in an independent model. Nevertheless, the all-erased branch leaves only a flag, so no growing logical system can have vanishing average transmission error at fixed $p>0$. Its loss count does not concentrate around $Np$, and \eqref{eq:fin-cutoff} correctly retains a nonvanishing bad-event term.

\subsection{Selecting a logical dimension at a specified error}

Suppose the total target infidelity is $0<\varepsilon<1$, and an independently proved field and implementation budget is $0\leq\zeta<\varepsilon$. Set $e=\varepsilon-\zeta$. For each cutoff with $\Pr(T>u)<e$, define
\begin{equation}\label{eq:fin-logical-choice}
 L(u)=\left\lfloor\log_2D-2u
       +2\log_2\bigl[e-\Pr(T>u)\bigr]\right\rfloor.
\end{equation}
Any nonnegative integer $\ell\leq L(u)$ is certified with $K=2^\ell$. Maximizing $L(u)$ over integer cutoffs gives a usable finite optimization involving only a loss-count tail. If no positive integer is certified, the theorem has not demonstrated useful transmission at those resources; the zero-qubit task is always trivial. This avoids presenting an asymptotic rate as a one-block guarantee.

\section{Independent but unequal losses}
\label{sec:fin-heterogeneous}

Let qubit $j$ be erased independently with probability $p_j$. Put
\begin{equation}
 m=\sum_{j=1}^Np_j,
 \qquad v=\sum_{j=1}^Np_j(1-p_j).
\end{equation}
The exact generating function of $T$ is
\begin{equation}
 \mathbb E z^T=\prod_{j=1}^N(1-p_j+p_jz).
\end{equation}
Its coefficients can be evaluated by a one-dimensional dynamic program in $O(N^2)$ arithmetic operations. Feeding that tail into \eqref{eq:fin-logical-choice} is often preferable to using a closed-form concentration inequality.

For a simple analytic certificate, Hoeffding's inequality gives
\begin{equation}\label{eq:fin-hoeffding}
 \Pr(T>m+s)\leq e^{-2s^2/N}.
\end{equation}
Choose
\begin{equation}\label{eq:fin-u-hoeffding}
 u=\min\left\{N,\left\lceil
       m+\sqrt{\frac N2\ln\frac2e}\right\rceil\right\}.
\end{equation}
Then $\Pr(T>u)\leq e/2$, including the case $u=N$ where the tail is zero. The choice
\begin{equation}\label{eq:fin-explicit-l}
 \ell\leq\left\lfloor\log_2D-2u-2\log_2\frac2e\right\rfloor
\end{equation}
therefore achieves total error at most $\varepsilon$. Away from saturation at $u=N$, the loss of logical qubits consists of the sector deficit, twice the mean number erased, a fluctuation term of order $\sqrt{N\ln(1/e)}$, and a logarithmic accuracy term. Those terms have distinct physical meanings and should not be merged into one fitted overhead.

When the variance is small, Bernstein's inequality can improve the cutoff:
\begin{equation}
 \Pr(T-m>s)\leq\exp\left[-\frac{s^2}{2(v+s/3)}\right].
\end{equation}
With $L=\ln(2/e)$, it is enough to take
\begin{equation}\label{eq:fin-bernstein}
 s=\sqrt{2vL}+\frac{2L}{3}.
\end{equation}
Indeed this value satisfies $s^2\geq2L(v+s/3)$. Alternatively one can minimize the Chernoff expression
\begin{equation}
 \Pr(T\geq u)\leq\inf_{\theta>0}
       e^{-\theta u}\prod_j(1-p_j+p_je^\theta).
\end{equation}
All three routes enter the same coding bound; choosing a sharper tail does not change the encoding theorem.

For orientation, assume $p_j=0.1$, a sector satisfying
$\log_2D\geq\lfloor N-3(\log_2N)^2\rfloor$, target error $\varepsilon=0.01$, and field budget $\zeta=0.001$. Equations~\eqref{eq:fin-u-hoeffding}--\eqref{eq:fin-explicit-l} give the following conservative certificates. They illustrate the inequalities under an assumed dimension bound, rather than performance of an implemented code.
\begin{center}
\begin{tabular}{rrrr}\toprule
$N$ & Cutoff $u$ & Certified logical qubits $\ell$ & $\ell/N$\\\midrule
$10{,}000$ & $1{,}165$ & $7{,}124$ & $0.712400$\\
$100{,}000$ & $10{,}520$ & $78{,}116$ & $0.781160$\\
$1{,}000{,}000$ & $101{,}644$ & $795{,}504$ & $0.795504$\\\bottomrule
\end{tabular}
\end{center}
The limiting independent-erasure rate is $0.8$. The table also shows why a leading-rate theorem should not be confused with a small-system performance claim.

\subsection{A finite-resource design rule}

The cutoff is a design variable, not an extra physical measurement. Once a loss model has been calibrated, the exact tail probabilities can be tabulated for all $N+1$ possible cutoffs. The largest certified logical dimension follows by evaluating \eqref{eq:fin-logical-choice} at each admissible cutoff. There is no assumption that the resulting integer function is convex or that a normal approximation is accurate in the relevant tail.

The same procedure compares alternative protected-sector constructions. Suppose a feature or cell parameter $b$ supplies a certified dimension $D(b)$ and field budget $\zeta(b)$. For a common target $\varepsilon$, discard values with $\zeta(b)\geq\varepsilon$, insert the remaining budgets into \eqref{eq:fin-logical-choice}, and maximize the resulting logical-qubit certificate over $(b,u)$. This is a finite optimization over proved inequalities. It makes explicit why increasing a cell's logical dimension can be counterproductive if its field-approximation error consumes nearly all the allowable infidelity.

The guarantee remains conservative in several identifiable places. It uses only a largest eigenvalue bound on $\Omega_R$; a marginal-purity computation may be better. It bounds every good pattern by the largest allowed $d_E$; the full spectrum sum may be better. A concentration inequality can overestimate a known loss-count tail. Finally, Haar random coding certifies existence rather than the performance of a chosen efficiently decoded family. These are possible improvements to the certificate, rather than unaccounted physical resources.

For unequal losses, $1-2m/N$ is a generally valid lower rate obtained from this argument when positive. It is not asserted to be the exact capacity of every heterogeneous channel. In particular, some qubits can have $p_j>1/2$, and omitting poor sites can improve on a code that treats the complete array uniformly. When all component erasure channels are degradable, their product admits the usual additive coherent-information analysis; the present dimension-only construction need not realize every finite-resource refinement of that analysis.

\section{Heterogeneous register dimensions and weighted losses}
\label{sec:fin-qudits}

Nothing in the two-copy Haar proof depends on the physical registers being qubits. Let site $j$ have finite dimension $d_j\geq2$, and put
\begin{equation}
 A_N=\sum_{j=1}^N\log_2d_j,
 \qquad W(E)=\sum_{j\in E}\log_2d_j.
\end{equation}
The full input dimension is $2^{A_N}$, while the erased subsystem has dimension $d_E=2^{W(E)}$. The identity $P\leq I$ still gives $\|\Omega_R\|_\infty\leq d_E/D$. Therefore all preceding statements remain valid with $|E|$ replaced by the weighted loss $W(E)$. For example,
\begin{equation}\label{eq:fin-weighted}
 e_\mu\leq\Pr[W(E)>w]+2^{[\log_2K+2w-\log_2D]/2}+\zeta.
\end{equation}
If the protected-sector dimension satisfies $\log_2D\geq A_N-\Delta_N$, then the effective finite dimension budget is $A_N-\Delta_N-2w$. Losing a high-dimensional register consumes more of that budget than losing a qubit. Merely counting the number of missing sites would hide this distinction.

For independent losses, let $Y_j$ be the Bernoulli indicator that site $j$ is erased and write $a_j=\log_2d_j$. The random weighted loss is $W=\sum_ja_jY_j$, with mean $\bar w=\sum_jp_ja_j$. Hoeffding's inequality now gives
\begin{equation}
 \Pr(W-\bar w>s)\leq
       \exp\left[-\frac{2s^2}{\sum_ja_j^2}\right].
\end{equation}
Thus a fixed total target error can be certified by setting
\begin{equation}
 w=\bar w+
       \sqrt{\tfrac12\left(\sum_ja_j^2\right)\ln(2/e)}
\end{equation}
and substituting this value into the weighted version of \eqref{eq:fin-explicit-l}. A noninteger threshold causes no difficulty because $W$ need not be integer. If a discrete distribution is evaluated exactly, the threshold can instead be chosen from its finite support.

For a fixed local dimension $d$ and identical erasure probability $p$, the leading unassisted benchmark is $(1-2p)_+\log_2d$ logical qubits per physical register. The coding argument approaches this value whenever $\Delta_N=o(N)$ and the field approximation error vanishes. This is a change of information units, not an increase in the number of protected physical particles. If local dimensions grow with $N$, a statement normalized by $N$ and one normalized by $A_N$ can have different meanings; the weighted bound keeps the correct normalization visible.

The protected noise model must also generalize appropriately. A character sector can be formed from integer-valued diagonal charges with more than two onsite eigenvalues, but its dimension must be counted for that chosen representation. The qubit counting formula from the spatial chapters cannot simply be reused after replacing each qubit by an arbitrary spin or oscillator. An infinite-dimensional oscillator needs an energy or excitation cutoff before the finite-dimensional lemma applies. Such a cutoff creates another approximation budget if the physical state can leave it.

This weighted formulation is useful for mixed memories containing qubits, finite spin manifolds, or truncated photonic registers. It identifies exactly which part of the coding proof is universal---the rank-$D$ random-subspace argument---and which part is specific to the hardware model---the available sector dimension, its noise algebra, the loss instrument, and the cost of its approximation. The distinction prevents a finite-dimensional capacity theorem from becoming an unsupported claim about unconstrained bosonic communication.

\section{One code for several loss laws or many prescribed patterns}
\label{sec:fin-compound}

The distinction between an unknown bath and an unknown loss distribution is important. Exact extraction removes the former uncertainty without an ensemble penalty. The latter uncertainty still changes which endpoint erasure channels the outer code must handle. A second-moment argument gives an explicit price for that uncertainty.

\begin{proposition}[Finite compound family of exogenous loss laws]
\label{prop:fin-compound}
Let $\mu_1,\ldots,\mu_M$ be loss laws. For each law choose a set of good patterns $G_a$, with bad-event probability at most $\tau_a$. There is a single encoding and a single family of flag decoders satisfying
\begin{equation}\label{eq:fin-compound}
 \max_a e_{\mu_a}\leq\max_a\tau_a+
 \sqrt{\frac KD\sum_{a=1}^M
             \sum_{E\in G_a}\mu_a(E)4^{|E|}}.
\end{equation}
In particular, if every law has $\Pr(T>u)\leq\tau$, then
\begin{equation}\label{eq:fin-compound-cutoff}
 \max_a e_{\mu_a}\leq\tau+\sqrt{\frac{MK4^u}{D}}.
\end{equation}
The uniform approximation budget $\zeta$ can again be added.
\end{proposition}

\begin{proof}
For a sampled $V$, use the same branch decoder from \eqref{eq:fin-branch-error}. Let
\[
 Z_a(V)=\sum_{E\in G_a}\mu_a(E)\|X_E(V)\|_1.
\]
Cauchy--Schwarz with these subnormalized weights gives
\[
 Z_a(V)^2\leq\sum_{E\in G_a}\mu_a(E)\|X_E(V)\|_1^2.
\]
The largest $Z_a$ is at most $(\sum_a Z_a^2)^{1/2}$. Take its expectation, apply Jensen, and then apply \eqref{eq:fin-second-trace}. Some encoding attains the resulting bound. Its bad branches contribute at most $\max_a\tau_a$, proving \eqref{eq:fin-compound}. A common size cutoff gives \eqref{eq:fin-compound-cutoff}.
\end{proof}

At a fixed error target, the factor $M$ costs $\log_2M$ logical qubits in \eqref{eq:fin-compound-cutoff}. Thus subexponentially many calibrated external loss scenarios can be covered without changing a leading-rate result. A continuum of laws can also be treated when it has a finite total-variation net. For a fixed code its branch infidelities lie in $[0,1]$, so replacing one law by another changes average error by at most their total-variation distance. A net of radius $\nu$ adds $\nu$ to \eqref{eq:fin-compound}. The size of that net, rather than the raw number of physical parameters describing a law, is what enters this sufficient bound.

Taking each law to be a point mass gives a simultaneous guarantee for an externally chosen family $\mathcal F$ of erased subsets:
\begin{equation}\label{eq:fin-pattern}
 \max_{E\in\mathcal F}e_E
       \leq\sqrt{\frac KD\sum_{E\in\mathcal F}4^{|E|}}.
\end{equation}
For all subsets of size at most $\lfloor pN\rfloor$, $0<p<1/2$, the sum is bounded by
\begin{equation}
 \sum_{t=0}^{\lfloor pN\rfloor}\binom Nt4^t
       \leq(N+1)2^{N[\hbin(p)+2p]}.
\end{equation}
Consequently a rank-$2^{N-o(N)}$ protected sector admits the conservative simultaneous-pattern rate
\begin{equation}\label{eq:fin-adversarial-rate}
 R<1-2p-\hbin(p)
\end{equation}
whenever the right-hand side is positive. This sufficient rate is weaker than the independent-erasure rate. It is not claimed to be optimal for adversarial erasure correction. It also remains an entanglement-transmission statement for each externally specified pattern; it is not a theorem about a quantum instrument whose choice of pattern measures the logical input.

\section{Approximation budgets and state-dependent loss selection}
\label{sec:fin-instruments}

Let $\Xi_0$ denote an ideal encoded, flagged channel and let $\Xi$ be the actual channel on the same logical input and classical-output space. Suppose
\begin{equation}\label{eq:fin-instrument-close}
 \tfrac12\|\Xi-\Xi_0\|_\diamond\leq\chi.
\end{equation}
For any fixed flag-controlled decoder, its recovered entanglement infidelity changes by at most $\chi$. This follows directly from contractivity and the fact that fidelity with a specified pure Bell state is the probability of a two-outcome measurement. Equation~\eqref{eq:fin-instrument-close} therefore gives a sufficient way to include weak state-dependent flag selection or other unmodeled backaction: establish closeness of the complete instrument, not merely of its distribution of flags on selected test states.

Knowing the flag probabilities alone does not control the instrument. A channel can emit exactly the same constant flag as the ideal no-loss process while applying an unreported logical dephasing measurement. Its flag distribution is identical, although an encoded Bell pair can lose its coherence. Likewise, a probability comparison on logical basis states does not detect a hidden measurement in that basis. These examples explain why input-independent selection is an operational assumption in the main theorem, not a disposable technicality.

Physical removal channels also need to be distinguished from amplitude damping. An ideal qubit erasure removes both basis states with the same probability and reports an orthogonal flag. An excitation-loss process can have a jump probability proportional to $\langle n_j\rangle$. Some constant-excitation codes have special detected-jump properties, but those are separate statements about the relevant Kraus operators. The present erasure bounds do not automatically turn every physical damping process into \eqref{eq:fin-flagged}.

Several certified approximation stages can be combined safely. If state preparation, field approximation, extractor implementation, and readout each have half-diamond errors at most $\zeta_1,\ldots,\zeta_s$ relative to their prescribed reference operations, telescoping the composition and using contractivity gives total budget
\begin{equation}
 \zeta\leq\sum_{i=1}^s\zeta_i.
\end{equation}
Terms that are already included in another stage must not be counted twice. Nor can a state-specific trace-distance test be substituted for a half-diamond estimate without an additional argument. The master theorem for spatial features uses a field perturbation bound of the latter, stronger kind.

\section{Converse checks and what the coding theorem does not provide}
\label{sec:fin-converse}

The independent qubit erasure channel has unassisted quantum capacity $(1-2p)_+$ \cite{bennett,barnum}. For $p\geq1/2$, its complementary receiver can simulate the actual receiver by erasing a further fraction of the systems it obtains. The channel is therefore antidegradable, and input restriction to a protected sector cannot turn it into a positive-rate unassisted quantum channel. For $p<1/2$, the standard erasure-capacity converse applies whenever the compound bath family includes the zero-bath history. Field protection can recover that rate, but cannot raise the worst-case rate above the noiseless-bath erasure benchmark.

For comparison, a nontrivial exact code ($K\geq2$) correcting every erased set of size $t$, with $2t\leq N$, obeys the quantum Singleton bound
\begin{equation}\label{eq:fin-singleton}
 \log_2K\leq N-2t.
\end{equation}
A short entropy proof exposes the different quantifiers. Partition the physical system as $ABC$, with $|A|=|B|=t$ and $|C|=N-2t$, and encode a maximally entangled logical-reference state. Exact correction of $A$ and $B$ gives $I(F:A)=I(F:B)=0$. Purity and subadditivity imply
\begin{align}
 S(F)+S(A)=S(FA)=S(BC)&\leq S(B)+S(C),\\
 S(F)+S(B)=S(FB)=S(AC)&\leq S(A)+S(C).
\end{align}
Adding them gives $S(F)\leq S(C)\leq N-2t$. If $2t>N$, two disjoint survivor sets of size $N-t$ could each reconstruct the logical state, contradicting no-cloning for $K\geq2$. The dimension-one logical task is trivial and is excluded from this nontrivial-code converse. This converse concerns exact correction of all such patterns. The sufficient bound \eqref{eq:fin-adversarial-rate} is an approximate, random-subspace construction and need not saturate it.

The resource model also matters. Two-way classical assistance permits protocols such as distributing Bell-pair halves and retaining the successfully received ones, with a different erasure benchmark. This manuscript's rate $1-2p$ uses unassisted quantum transmission and known receiver-side flags; it does not assume preshared entanglement or unrestricted two-way purification during the storage interval.

Finally, finite entanglement fidelity and diamond error should not be conflated. For a recovered logical channel on $K$ dimensions, the normalized Choi-state relation and the fidelity--trace-distance inequality give the conservative implication
\begin{equation}
 \tfrac12\|\mathcal L-\id\|_\diamond
       \leq K\sqrt{e(\mathcal L)}.
\end{equation}
The factor $K$ can be exponential in $N$. Thus the finite estimates above do not silently certify a useful worst-case diamond error for an exponentially large logical space. The field-extraction error is controlled in half-diamond norm independently of $K$, while the outer code is evaluated by entanglement transmission. Those two strengths coexist consistently. They specify precisely what is proved, what is transferred uniformly across bath histories, and what remains to be supplied by an efficient architecture.