Em04-A Exact Reduction, Two-Coordinate Scaling, and Transient Entanglement in a Locally Dissipative XY Chain Local Instruments and Validated Approximation Interfaces

We specify a finite open XY chain with uniform zero-temperature relaxation and local dephasing. Its evolution from a single excitation admits an exact vacuum-plus-single-excitation reduction to an $N\times N$ density matrix and a scalar survival probability. We derive explicit concurrence, logarithmic-negativity, purity, and computational-basis-coherence formulas, including the complete partial-transpose spectrum. At fixed coherent time and fixed dimensionless detunings, two dissipative coordinates $(\chi ,r)$ are required: excitation survival and a short-time concurrence expansion disprove single-$\chi$ universality. We prove entanglement generation, dissipative decay, conditional mixing, and controlled finite-window limits. Separate results establish finite local instrument consistency and typed deferred readout. For an independently specified oscillator, we prove compressed-position POVM identities, state-preparation and tail bounds, a composite-trapezoidal error estimate, and an arithmetic-aware completion to a valid POVM. A cellwise residual theorem provides a validated state-error interface without assuming positivity of the approximation. Mathematical proofs are distinguished from source alignment, numerical regression, validated numerical certificates, and document construction. No execution or completed verification package is asserted.

This is an independent mathematical construction under specified quantum semantics. Density operators, completely positive channels, Lindblad dynamics, and Born probabilities are assumptions, not conclusions derived from a measurement-strategy characterization. No earlier landmark manuscript is a formal premise of the proofs.

The spin-chain and oscillator constructions are distinct models. Neither is identified with a gravitational mediator, quantum field, microscopic evaporation channel, or characteristic-product coefficient. Documentary programme restrictions are recorded separately in Appendix app:scope.

We distinguish:

Analytical proof.

A theorem established by an explicit argument under stated hypotheses.

Source alignment.

A documented comparison with an identified implementation, including source locations and conventions.

Numerical regression.

Reproducible floating-point comparisons with predeclared tolerances.

Validated numerical certificate.

Outward-rounded enclosures and a finite error budget for a specified calculation.

Document construction.

Successful compilation, resolved references, and inspection of the PDF.

The theorem and proof environments below belong to the first category. A proved error-bound theorem is an interface for a numerical certificate; it is not evidence that its hypotheses have been numerically enclosed for a particular run.

The supplied verification plan reports that an originating simulator export was supplied as documentary material and describes its conventions. These reported correspondences are recorded in Appendix app:alignment. This revision does not include an archived, line-numbered copy of that export or a completed source-inspection record. Source alignment therefore remains to be independently established; it is not replaced by the assertion that code was unavailable, nor by an unsupported assertion that inspection passed.

No execution archive, compiler log, independent symbolic-check output, validated numerical certificate, or machine-checked proof accompanies this revision. The execution-dependent release gates in Appendix app:verification remain unverified.

For matrices,

For density operators,

All entanglement logarithms are base two. A density operator is positive semidefinite and has trace one.

Let $N\ge 2$, $J>0$, ${\gamma }_{\mathrm{r}},{\gamma }_{\mathrm{d}}\ge 0$, and ${ϵ}_k\in \mathbb{R}$, $0\le k<N$. On $\mathcal{H}=({\mathbb{C}}^2{)}^{\otimes N}$, use

and

Subscripts denote tensor-local operators. Set

For any operator $L$, define

The master equation is

There are no bath-raising operators. The default initial state is

where $|e_k⟩$ has its unique excitation at site $k$.

The one-excitation hopping amplitude is $J$, because of the factor $1/2$ in (eq:H). With ${L_k}^{\mathrm{d}}=\sqrt{{\gamma }_{\mathrm{d}}/2}{Z}_k$, an isolated qubit's dephasing rate is ${\gamma }_{\mathrm{d}}$, while an intersite one-excitation coherence receives dephasing rate $2{\gamma }_{\mathrm{d}}$. Replacing this collapse operator by $\sqrt{{\gamma }_{\mathrm{d}}}{Z}_k$ doubles these dephasing contributions.

The relaxation vacuum is $|0⟩$, since $a=|0⟩⟨1|$. Software operator names do not substitute for checking this action.

The finite-dimensional equation (eq:master) has a unique solution for all $t\ge 0$. Its propagator is completely positive and trace preserving (CPTP). Every CPTP map $\Phi$ satisfies

Proof. Existence and uniqueness follow from $\rho (t)=e^{t\mathcal{L}}\rho (0)$. Enumerate the collapse operators as $L_{\ell }$, and put

Then $\mathcal{L}=\mathcal{A}+\mathcal{J}$. The maps $e^{t\mathcal{A}}(M)=e^{tK}M{e}^{t{K}^{†}}$ and $\mathcal{J}$ are completely positive. The norm-convergent Dyson expansion for $e^{t\mathcal{L}}$ consists of integrals of compositions of these maps. Complete positivity is preserved by these integrals, sums, and finite-dimensional limits. Cyclicity gives $\mathrm{Tr}\mathcal{L}(M)=0$, so the propagator preserves trace. This proves the needed sufficiency statement of the GKSL framework GKS,Lindblad,Kuramochi directly.

For contraction, write the Hermitian Jordan decomposition $M=M_{+}-M_{-}$, $M_{\pm }\ge 0$. Positivity and trace preservation give

Let $V:{\mathbb{C}}^N\to \mathcal{H}$ be the isometry $V|k⟩=|e_k⟩$, and let $\Delta (M)=\mathrm{diag}M$ in the site basis.

The Hamiltonian commutes with

The linear class

is invariant under the generator, with equations

No vacuum–single-excitation coherences are generated.

Proof. The identity

shows excitation conservation by hopping. The detuning terms also commute with ${\hat{N}}_{\mathrm{exc}}$. The vacuum energy is $E_{\Omega }=\frac{1}{2}{\sum }_k{ϵ}_k$, and $V^{†}(H-E_{\Omega }I)V=h$, proving the Hamiltonian term.

Since $a_k|e_i⟩={\delta }_{ki}|\Omega ⟩$, the relaxation jump term is

The anticommutator term on the one-excitation space is $-{\gamma }_{\mathrm{r}}R$, because ${\sum }_k{n}_k$ restricts to its identity.

Finally,

For $i\ne j$, conjugation of $|e_i⟩⟨e_j|$ changes sign at exactly sites $i,j$. Each contributes $-{\gamma }_{\mathrm{d}}$. Diagonal excitation matrix units and the vacuum projector are unchanged. These calculations prove all stated block equations.

For the specified model, let $S_0$ be any density matrix on ${\mathbb{C}}^N$, and take $\rho (0)=V{S}_0{V}^{†}$. There is a unique density matrix $S(t)$ satisfying

Writing $q(t)=e^{-{\gamma }_{\mathrm{r}}t}$, the full state is exactly

In particular,

The default preparation corresponds to $S_0=|0⟩⟨0|$.

Proof. Tracing (eq:R) gives $(\mathrm{Tr}R{)}^{\prime }=-{\gamma }_{\mathrm{r}}\mathrm{Tr}R$, with initial trace one, hence $\mathrm{Tr}R=q$. Equation (eq:p) gives $p=1-q$. For finite $t$, $q>0$, and substituting $R=qS$ gives (eq:S). This conditional equation is itself GKSL:

Thus it has a unique physical solution. The reconstructed state satisfies the full equation and initial condition, so uniqueness proves the identity.

The physical tensor factorization remains that of $N$ sites. The $N$-level matrix $S$ is a computational reduction, not a new factorization for evaluating entanglement.

Uniform relaxation is a hypothesis of this scalar-survival theorem. Nonuniform relaxation still preserves the vacuum-plus-one-excitation block class, but generally replaces $-{\gamma }_{\mathrm{r}}R$ by $-\{D,R\}/2$, with $D$ the diagonal rate matrix; the universal scalar factorization above then need not hold. Bath raising, higher-excitation preparations, and vacuum–excitation coherences are not covered by the theorem.

For a two-qubit density matrix $\omega$, let

The concurrence is

where the decreasing ${\lambda }_{\ell }$ are the nonnegative square roots of the eigenvalues of $\omega \tilde{\omega }$ Wootters.

For distinct sites $i,j$, in the basis $(|00⟩,|01⟩,|10⟩,|11⟩)$,

Consequently

and

Proof. An excitation outside the pair reduces to $|00⟩$. Only coherences supported entirely in the pair survive the partial trace, giving the displayed matrix. Set $a=R_{jj}$, $b=R_{ii}$, and $z=R_{ji}$. Direct multiplication with the spin-flipped matrix gives square-root eigenvalues

Positivity implies $|z|\le \sqrt{ab}$. Substitution yields $C=2|z|$, and the same positivity inequality followed by the arithmetic–geometric mean inequality proves the bounds.

For a partition $A\bigsqcup B=\{0,\dots ,N-1\}$ with both parts nonempty, let $R_{AB}$ have rows in $A$, columns in $B$. Use

with computational-basis partial transpose VidalWerner.

Let $p_0=1-q$ and $s={\Vert R_{AB}\Vert }_{\mathrm{F}}$. As a multiset, the complete spectrum is

where

Zeros already present in the first three multisets are retained. The formula holds for $s=0$ as well.

Therefore

Within the state class (eq:reduction),

Also,

Proof. A cross matrix unit transforms according to

The vacuum and the states with one excitation on each side therefore carry the block

The cross-sector basis is orthonormal. For $s>0$, restriction to the vacuum and $v/s$ gives ${\lambda }_{\pm }$; the remaining directions give zeros. For $s=0$, choose any one cross-sector basis vector instead, giving eigenvalues $p_0,0$. Such a vector exists because both parts are nonempty.

The other nonzero blocks are $R_{AA}$ and ${R_{BB}}^{\mathsf{T}}$, on mutually orthogonal local single-excitation spaces. All remaining sectors are zero. Their dimensions give exactly $2^N-N-2$ additional zeros in (eq:PTspec). The positive local blocks have combined trace $q$, giving the trace norm and the negativity formulas.

If $R_{AB}\ne 0$, then ${\lambda }_{-}<0$. Every separable state has positive partial transpose, by transposing the positive second factor in each product term of a separable decomposition. Hence this state is entangled. If $R_{AB}=0$, the vacuum term and the two local excitation blocks are positive product terms. Normalizing the nonzero terms by their traces gives a separable decomposition.

Finally,

Insert this into (eq:EN). The inequality $\sqrt{(1-q{)}^2+q^2}\le 1$ proves the last bound.

Use the specified normalized diagnostic

Its numerator is the usual basis-dependent ${\ell }_1$ coherence BCP; the denominator is an additional convention.

For (eq:reduction),

For ${\gamma }_{\mathrm{r}}>0$,

Proof. The vacuum and excitation blocks are orthogonal, proving (eq:purity). All full-state off-diagonal entries belong to $qS$. Positivity and Cauchy–Schwarz give

Differentiating conditional purity eliminates the commutator, and $\mathrm{Tr}(S(S-\Delta S))={\Vert S-\Delta S\Vert }_{\mathrm{F}}^2$. The difference from the vacuum has blocks $-q$ and $qS$, so its trace norm is $2q$. The limiting purity follows from (eq:purity).

For the default pure preparation with ${\gamma }_{\mathrm{d}}=0$, $S$ remains pure and $\mathrm{Tr}{\rho }^2=(1-q{)}^2+q^2$. At positive relaxation this decreases to $1/2$ at $t=(\log 2)/{\gamma }_{\mathrm{r}}$ and subsequently increases to one. At fixed $t>0$, it is also nonmonotone as a function of ${\gamma }_{\mathrm{r}}$. Vacuum convergence does not prove monotonicity in relaxation rate. The normalization in (eq:coherence) must likewise be separated from any claimed chain-length dynamical effect.

Protocol A fixes physical observation times $t$. Protocol B fixes coherent observation times $\tau =Jt$. Let

When ${\Gamma }_2>0$, define

Equivalently,

Here ${\Gamma }_2$ is an isolated qubit's transverse decay rate under the specified convention, not the intersite dephasing rate.

With $\tilde{h}=h/J$, Protocol-B dynamics satisfy

After fixing $N,S_0,{ϵ}_i/J$, the pair $(\chi ,r)$ determines these trajectories. A common detuning shift has no effect.

Multiplying $J,{ϵ}_i,{\gamma }_{\mathrm{r}},{\gamma }_{\mathrm{d}}$ by a common $c>0$ replaces the physical-time trajectory $\rho (t)$ by $\rho (ct)$ and leaves the corresponding Protocol-B trajectory unchanged. Nevertheless two models with the same finite $\chi$ and different $r$ cannot have identical full Protocol-B trajectories.

Proof. Substitution proves the scaled equations. A common detuning shift changes $\tilde{h}$ by a scalar identity, which disappears from the commutator. Simultaneous scaling changes the full generator to $c\mathcal{L}$, because collapse amplitudes scale by $\sqrt{c}$. Finally, the excitation expectation is $e^{-2r\tau /\chi }$, which differs for different $r$ at every $\tau >0$.

If ${\Gamma }_2=0$, both rates vanish. The model is unitary, and $(\chi ,r)$ is not a finite coordinate chart. Limits $\chi \to \infty$ at fixed $r$ are meaningful, but the exactly nondissipative model has no distinguished $r$.

For the default localized preparation and ${\Gamma }_2>0$,

The formula holds for arbitrary fixed real detunings and includes $r=0,1$. In physical time, for all nonnegative rates, it reads

Proof. Set $P=|0⟩⟨0|$, $d_i=-{ϵ}_i/J$, and $\kappa =2(1-r)/\chi$. For $\mathcal{A}(M)=-\mathrm{i}[\tilde{h},M]-\kappa (M-\Delta M)$, analyticity gives

Direct matrix multiplication on the open chain yields

Thus

The detuning term affects the modulus only at third order. Multiplying by $2q$, with $q=1-(2r/\chi )\tau +O({\tau }^2)$, proves (eq:shortC). The same calculation with physical coefficients proves the last formula, including zero dissipation. All remainders concern a fixed model, not unbounded detuning families.

The survival law and (eq:shortC) independently establish the failure of general single-$\chi$ universality.

For the default localized preparation, finite rates, and finite detunings, $C_{01}>0$ on a sufficiently short positive-time interval. Every cut separating sites $0,1$ is entangled there.

If ${\gamma }_{\mathrm{r}}>0$, every pairwise concurrence and bipartite logarithmic negativity tends to zero. The concurrence $C_{01}$ attains a strictly positive global maximum at finite positive time.

Proof. The leading coefficient in the physical-time expansion is $2J>0$. A cut separating the two sites contains the nonzero coherence in $R_{AB}$, so Theorem thm:negativity applies. For ${\gamma }_{\mathrm{r}}>0$, $q\to 0$, and (eq:Cbound), (eq:ENbound) prove decay. A continuous function vanishing at zero and infinity but positive somewhere attains a positive maximum on a sufficiently large compact time interval, which is also its global maximum.

If $J>0$ and ${\gamma }_{\mathrm{d}}>0$, then

For the fixed model there are $M,g>0$ such that

Consequently when ${\gamma }_{\mathrm{r}}=0,{\gamma }_{\mathrm{d}}>0$, all the entanglement diagnostics above tend to zero, and the full limiting state is the separable state

Proof. On all complex matrices, the conditional generator $\mathcal{A}(M)=-\mathrm{i}[h,M]-2{\gamma }_{\mathrm{d}}(M-\Delta M)$ obeys

If $\mathcal{A}(M)=\lambda M$, $M\ne 0$, and $\mathrm{Re}\lambda =0$, then $M$ is diagonal. The diagonal equation gives $\lambda M_{ii}=0$, hence $\lambda =0$. The neighboring off-diagonal equations give $J(M_{i+1,i+1}-M_{ii})=0$. Connectedness implies $M=c{I}_N$.

The invariant trace-zero subspace therefore has its finite spectrum strictly in the left half-plane. Jordan decomposition supplies an exponential bound, with $g$ chosen smaller than the spectral decay rate to absorb possible polynomial factors. Apply this bound to $S_0-I_N/N$. The limiting matrix has no off-diagonal entries; the entanglement formulas vanish and the displayed mixture is separable.

A general mixed $S_0$ need not generate entanglement: $S_0=I_N/N$ is already stationary conditionally. The generation theorem is therefore not asserted for all one-excitation density matrices. With no dissipation, decay also need not occur; the disorder-free two-site chain has $C_{01}(\tau )=|\sin (2\tau )|$.

Fix $N,S_0,\tilde{h},r$, and let

For finite $\chi >0$,

All diagnostics above converge uniformly to their unitary counterparts on every fixed $[0,T]$. In particular,

Proof. Variation of constants with the unitary propagator $\mathcal{U}$ gives

Both matrices in the integrand difference are density matrices, so its trace norm is at most two. This proves (eq:Sunit). Reconstruction gives

For Hermitian $M$ and distinct orthonormal $u,v$, $2|⟨u|M|v⟩|\le {\Vert M\Vert }_1$, by testing against a phase-aligned Hermitian off-diagonal operator of norm one. Together with (eq:C), this proves the concurrence bound. The remaining diagnostics are continuous functions of $q,S$; uniform continuity on their compact finite-dimensional state domain gives the stated uniform convergence.

For two models with the same $J,{\gamma }_{\mathrm{r}},{\gamma }_{\mathrm{d}},S_0$, and Hamiltonians $\tilde{h},{\tilde{h}}^{\prime }$,

Proof. Duhamel's formula has Hermitian source $-\mathrm{i}[\tilde{h}-{\tilde{h}}^{\prime },S^{\prime }(s)]$, of trace norm at most $2{\Vert \tilde{h}-{\tilde{h}}^{\prime }-cI\Vert }_{\infty }$. Contractivity from Lemma lem:physical proves the conditional distance bound. The vacuum weights agree, so reconstruction multiplies the conditional distance by $q$.

The finite-window estimate is not a fitted saturation law, global monotonicity theorem, or uniform large-time limit. The detuning bound does not establish a sharp disorder threshold. Ensemble crossover statements require a distribution and statistic. Protocols A and B answer different observation questions.

For the localized preparation, $N=2$, and equal detunings, let $\kappa =2(1-r)/\chi$ and define

Then

For $r=1$, this is $e^{-2\tau /\chi }|\sin (2\tau )|$.

Proof. Write $S_{01}=\mathrm{i}y$, $z=S_{00}-S_{11}$. The equations are

Hence $y^{\prime \prime }+\kappa y^{\prime }+4y=0$, $y^{\prime }(0)=1$, whose solution is $y=e^{-\kappa \tau /2}{F}_{\kappa }(\tau )$. Use $C=2q|y|$.

For stable overdamped evaluation one can instead use the identical expression

This is an analytical identity, not an executed numerical result.

The ordinary finite-channel framework is described in Watrous. The identities needed here are proved explicitly.

A finite instrument from ${\mathcal{H}}_{\mathrm{in}}$ to ${\mathcal{H}}_{\mathrm{out}}$ has Kraus operators $M_{a\mu }$ with

Set

The record includes every outcome, including failure labels.

For a density matrix $\rho$:

1. $p_a=\mathrm{Tr}{\Phi }_a(\rho )=\mathrm{Tr}(E_a\rho )\ge 0$, and ${\sum }_a{p}_a=1$.

2. If $p_a>0$, the conditional state is ${\Phi }_a(\rho )/p_a$. If $p_a=0$, then ${\Phi }_a(\rho )=0$; no normalized branch is assigned.

3. $\hat{\Phi }$ is CPTP.

4. Actual tensor-local instruments on disjoint factors commute as selective superoperators after identifying output types.

5. A remote trace-preserving channel leaves local outcome probabilities unchanged, including for entangled inputs.

Proof. Kraus positivity, cyclicity, and completeness prove (i). A positive matrix of trace zero is zero, proving (ii). The record Kraus operators $|a{⟩}_R\otimes M_{a\mu }$ are complete, proving (iii). For separate factors, composition in either order gives the same Kraus operator $M_{a\mu }\otimes K_{b\nu }$, proving (iv).

For a remote channel with ${\sum }_b{K}_b^{†}{K}_b=I_B$,

For a supplied single update operator, require ${M_a}^{†}{M}_a=E_a$; for several operators use their sum. Declared footprints do not establish tensor locality. Remote postselection is not a trace-preserving unconditioned channel. Finally, equality of effects does not imply equality of instruments.

For a measured factor $S$, complement $B$, and computational basis $|z{⟩}_S$, define

Reading $A$ computationally and retaining $S$ gives

Reading $A$ and discarding $S$ gives instead

These identities hold with any external reference and with the same fixed continuation channel on $B$ in both experiments.

Proof. Expand $M={\sum }_{z,w}|z⟩⟨w{|}_S\otimes M_{zw}$. Coherent recording gives ${\sum }_{z,w}|z⟩⟨w{|}_S\otimes |z⟩⟨w{|}_A\otimes M_{zw}$. Reading $A$ removes the $z\ne w$ terms and produces ${\sum }_z|z⟩⟨z{|}_R\otimes |z⟩⟨z{|}_S\otimes M_{zz}$. Tracing out $S$ gives the destructive formula. A reference simply enlarges the untouched matrix coefficients. A fixed channel on $B$ replaces each $M_{zw}$ by its image.

Coherent recording itself is not a classical-record channel. The output types above differ. Arbitrary later interactions with the retained measured factor, carrier, or outcome-dependent feedback require additional hypotheses.

Let ${\mathcal{H}}_{\mathrm{osc}}=L^2(\mathbb{R})$ have oscillator number basis $|n⟩$, $n\ge 0$. In this section $a$ denotes the oscillator annihilation operator, not the two-level matrix of Definition def:model. Let

be multiplication by $x$, with PVM

For an integer $d\ge 1$, define

The compressed effects form a countably additive normalized POVM:

Their sum over a countable measurable partition is $I_d$. They are generally not projections and are not generally the spectral effects of $P_d{X}_{\mathrm{osc}}{P}_d$.

Proof. Positivity follows by testing on vectors in ${\mathcal{H}}_d$. Compression preserves countable additivity of matrix elements, and $P_{d}E(\mathbb{R})P_d=P_d$. For $d=1$, the vacuum position density is ${\pi }^{-1/2}{e}^{-x^2}$, giving $E_1((-\infty ,0])=I_1/2$, not a projection. The compressed position operator is zero, whose spectral effect for this set is instead $I_1$.

For any oscillator state $\rho$, suppose

Embedding ${\rho }_d$ into the full space,

For any finite state ${\sigma }_d$,

Every effect-probability difference is bounded by the corresponding trace distance.

Proof. A purification $|\Psi ⟩$ of $\rho$ and its normalized projection have overlap $\sqrt{1-{\delta }_d}$. Their pure-state trace distance is $\sqrt{{\delta }_d}$, by diagonalizing their difference on their two-dimensional span. Partial trace is contractive, proving the first assertion. The triangle inequality proves (eq:stateerror). For an effect $0\le F\le I$, the Jordan decomposition of a trace-zero state difference gives $|\mathrm{Tr}(F(\rho -\sigma ))|\le d_{\mathrm{tr}}(\rho ,\sigma )$.

Normalized projection is a preparation comparison, not a claim that postselected normalization defines a linear deterministic channel.

For finite $\alpha \in \mathbb{C}$, let $\mu =|\alpha {|}^2$ and

Its omitted probability is

For $\mu >0$ and $d>\mu$,

for $\mu =0$, ${\delta }_d=0$.

Define

These preparations need not agree, but

where

Proof. The squared coefficients are Poisson probabilities. For a Poisson variable $K$, $\mathrm{Pr}(K\ge d)\le \exp (\mu (e^s-1)-sd)$. Choose $s=\log (d/\mu )>0$.

For displacement, write $A=\alpha a^{†}-\overline{\alpha a}$, $A_d=P_{d}A{P}_d{|}_{{\mathcal{H}}_d}$, and $u(s)=P_d|s\alpha ⟩$. Differentiation gives