Em03-G Coherence Transfer and Recovery Limits in Physical Scalar Radiation Noncommuting exterior-source interactions, referenced packet channels, and local-access requirements

We solve an effective exterior-source model consisting of two temporally ordered couplings of a degenerate qubit to the massless s-wave scalar field on Schwarzschild. The first coupling uses the source observable $Z$, and the second uses $X$. Both source profiles are smooth, compact, strictly exterior, and have full angular-shell support. The initial scalar field is the independently specified Unruh state.

The exact radiation channel is evaluated on every source matrix unit. Its outgoing packet restrictions are normal channels with explicit interference terms, rather than independently assigned displaced Gaussian outputs. We also calculate the retained-source channel and construct a complete complementary channel in a fixed purified field representation.

A phase-matched one-packet comparison model admits an explicit mathematical decoder. Its worst-case referenced recovery error approaches zero when the first displacement grows and the second decreases with their symplectic pairing fixed. Filtered compact bulk approximations realize this comparison with ordered exterior profiles and a source-input half-diamond error bound. Consequently, finite compact parameters exist for which an identified outgoing scalar packet has a provably entangled reference-output state. This is a sufficient transmission regime, not a classification of every parameter choice.

At every finite first-pulse coefficient, the retained source has nonzero input distinguishability. We derive a corresponding nonzero recovery lower bound, while keeping Bell fidelity, worst-case entanglement fidelity, and half-diamond distance distinct. Complete environmental leakage bounds retain unselected radiation, future-horizon degrees of freedom, and background purification.

We prove a new common second-moment bound for the interference-bearing outputs and use it for local bounded-Weyl probability certificates. Coherent local packet access is not assumed or constructed. Causal support, finite signed excess radiative energy, interference energy, switching work, and resource growth are treated separately.

The interactions are effective exterior source laws, not a microscopic gravitational derivation. No black-hole interior information escape or evaporation theorem is established.

The principal chain is

The operational receiver is a separate chain:

The first chain is constructed here. A coherent local realization of the second chain is not asserted.

Em03-A–C Em03A,Em03B,Em03C provide the selected stationary scalar prescription and finite-packet context. Em03-D Em03D supplies a local bounded-observable architecture. Em03-E Em03E has additional finite radiation payloads. Em03-F Em03F instead identifies source-dependent information in the scalar field itself, but its single-source-observable radiation channel is measure-and-prepare.

Quantum transmission through scalar fields already has explicit nonperturbative protocols in flat and curved spacetimes Simidzija,KasprzakTjoa. We do not claim novelty for its possibility, for controlled displacement, or for the use of noncommuting source observables. Simple-generated qubit–field calculations also have established precedents Tjoa. The present contribution is the stated compact Schwarzschild source realization, its exact interference-bearing packet channel, and its source-input and recovery certificates.

release gates. We distinguish:

1. Information-theoretic scalar transmission: the identified outgoing packet channel and a mathematical CPTP decoder on that packet;

2. Local coherent reception: a specified spacetime receiver whose quantum output has a proved referenced comparison with the packet decoder.

Passing the first gate does not pass the second.

boundary. All constructions and estimates are analytical. Compilation, formal verification, numerical execution, calibration, and physical implementation are not asserted. A theorem about a specified effective interaction is not a derivation of its gravitational origin.

. For normal states and channels,

The channel distance includes arbitrary external references. We use squared pure-target fidelity. In particular,

Worst-case entanglement fidelity is defined separately in Section sec:recovery.

Suppose

For every deterministic radiation-only access channel $\mathcal{A}$ and decoder $\mathcal{D}$,

is entanglement breaking. Its qubit Bell fidelity is at most $1/2$, and its recovery half-diamond error is at least $1/2$.

Proof. The displayed channel remains measure-and-prepare. For a referenced input its output is a sum of positive product operators. Postprocessing on the receiver side preserves separability. A separable two-qubit state has Bell overlap at most $1/2$: the pure-product bound follows by Cauchy–Schwarz and the mixed bound by convexity. The Bell-projector test gives the distance lower bound.

This is the standard entanglement-breaking mechanism EntanglementBreaking, with its numerical consequence proved directly. Better reception cannot recreate source coherence already absent from F's radiation output. The modification below is therefore made at the source.

In physical units the background is the right Schwarzschild exterior,

Killing time is normalized at infinity.

For the estimates we measure lengths in units of $r_{\mathrm{h}}$. Suppressing tildes,

Physical frequencies and energies are ${r_{\mathrm{h}}}^{-1}$ times the displayed values. A dimensionless bulk scalar smearing corresponds to physical smearing with factor ${r_{\mathrm{h}}}^{-3}$. Packet normalization is preserved by

The free real scalar has

Let $Y_0=(4\pi {)}^{-1/2}$. The selected s-wave field is

with reduced operator

For a real compact reduced test $f$, define

Because $\mathrm{d}{\mathrm{vol}}_g=r^2\mathsf{f}(r)\mathrm{d}t\mathrm{d}x\mathrm{d}\Omega$,

All finite compact reduced supports give compact supports strictly outside the horizon.

We use complete stationary massless-scalar scattering on Schwarzschild, the free normally hyperbolic Cauchy and Green-operator theory, and the bulk Hadamard identification of the Unruh boundary prescription. The horizon input is centered quasifree with

the infinity input is vacuum, and the inputs are independent.

The classical scattering and Green inputs are Dimock,BGP. The selected bulk state construction is DMP. They do not supply the new source dynamics or a coherent receiver.

Use horizon-incident and infinity-incident modes $q_{\mathrm{H}}$, $q_{\mathrm{I}}$, normalized as in Em03-D–F. Their spectral transform is

It is unitary from $L^2(\mathbb{R})$ onto $L^2((0,\infty );{\mathbb{C}}^2)$, and takes $\mathsf{L}$ to multiplication by ${\omega }^2$.

The field expansion is

Consequently

Annihilation tests are conjugate-linear:

Set

Then $\varphi (\mathfrak{F}[f])=X(k(f))$.

Define

The ordinary two-channel norm is $\Vert \cdot {\Vert }_0$. We have

For real $f\in {C_c}^{\infty }({\mathbb{R}}^2)$, its coefficient $k(f)$ has finite covariance norm and obeys

The instantaneous coefficient

is smooth in the covariance norm and has compact temporal support.

Proof. Here are the specific s-wave threshold facts needed for the proof. The zero-energy Jost solutions are

with Wronskian $-1$. Also

The Jost Volterra equations are therefore continuous at zero. On each compact spatial interval they give

for suitable finite $C_X$ and ${\omega }_c>0$. At zero use the Volterra kernel bound $|\sin (\omega z)/\omega |\le |z|$ and the nonzero Wronskian; above ${\omega }_c$ use $\exp (\Vert V{\Vert }_1/{\omega }_c)$ and flux conservation.

Thus a compact spacetime test has

At high frequency, one time integration by parts gives $O({\omega }^{-3/2})$. Further time integrations are available. Since $n_{\mathrm{H}}\le 1/(\beta \omega )$, these bounds prove all three stated integrability claims, including the extra ${\mathsf{C}}^2$ in the last one.

For the instantaneous coefficient,

The threshold estimate controls its low-frequency covariance. Time derivatives introduce finite powers of $\omega$; compact spatial smoothness and spectral unitarity control those powers by applying powers of $\mathsf{L}$ to $f(t,\cdot )$. This proves smoothness of every order.

This is the weighted domain relevant to the source. It does not replace the thermal horizon sector by vacuum.

The message is encoded by a fixed identification into the source qubit $S$. Its computational basis is the $Z$ basis:

The initial state is

The source and receiver have no prior message-dependent field preparation.

Choose real compact profiles $f_1,f_2$ with disjoint time slabs $I_1<I_2$, and set $F_j=\mathfrak{F}[f_j]$. The interaction is

The exact measure convention is

where

Thus $\int {\Phi }_F(t)\mathrm{d}t=\varphi (F)$.

Abbreviate

A useful fixed purified realization is the doubled Fock space with real-linear map

It satisfies

Let $\Omega$ be its vacuum. Its physical Weyl subalgebra has exactly the prescribed characteristic function. No global density in the undoubled stationary radiation Fock space is assumed.

Write $N_{\mathrm{d}}$ for doubled Fock particle number. Segal fields are essentially self-adjoint on finite-particle vectors, and $\mathcal{D}((N_{\mathrm{d}}+1{)}^{1/2})$ is their common form domain. Weyl displacements preserve this domain.

The interaction (eq:interaction) has a unitary propagator. After both slabs have ended, its endpoint is

The individual phases are

The two factors preserve the same number form domain. On that domain their chronological composition solves the strong Schrödinger equation into the Hilbert space.

Proof. For one slab and source observable $O\in \{X,Z\}$, $O^2=I$. If $K(t)=g_j{\int }_{I_j\cap \{u\le t\}}{k}_{j,u}\mathrm{d}u$, its propagator is

Chronological Weyl products give exactly the displayed scalar phase. Smooth covariance-norm continuity from Lemma lem:compactregularity gives the strong product limit.

For a differentiable coefficient path $r(t)$, the Weyl relation and the Stone derivative give, on the number form domain,

The scalar derivative cancels ${\partial }_t{\gamma }_j$, proving the Schrödinger equation on the slab. Creation/annihilation estimates and the displacement identity show domain invariance.

The slabs do not overlap. Their endpoint propagators therefore compose in chronological order on the same domain, proving (eq:UG). Differentiating the inner product of two candidate solutions gives uniqueness. No central-commutator claim is made for overlapping $X$- and $Z$-couplings.

Below we omit the irrelevant common scalar phase.

For signs $z,x\in \{+1,-1\}$, put

For $z=+1$ identify $|z⟩=|0⟩$, and for $z=-1$ identify $|z⟩=|1⟩$. Let

Then

These branch phases are part of the dynamics. Dropping them before interference calculations is not valid.

First define the field-output channel as a map into state functionals:

Equivalently, its values on source matrix units $E_{zw}=|z⟩⟨w|$ are

This formulation does not assume a global field density.

For a weighted field coefficient $l$, write

The exact source-to-field channel obeys

The characteristic functional of every input follows by linearity. In particular, off-diagonal source dependence is generally nonzero.

If $a_2=0$, the result is the single-$Z$-observable channel of Em03-F. If $a_1=0$, it is a single-$X$-observable measure-and-prepare channel.

Proof. Let $r_{xz}=-z{a}_1-x{a}_2$. Weyl multiplication gives

The phase from $s(r_{xw},r_{xz})=x(w-z)\theta$ cancels the explicit branch-phase difference. For $z=w$, the coefficient $m_{xz}{m}_{xw}=1/2$. For $z\ne w$, it is $x/2$. Summing over $x$ proves all four formulas. The two reductions follow either from the endpoint unitary or from these characteristic functions.

A nonzero off-diagonal output is coherence dependence, not by itself an entanglement-transmission theorem. For example, the single-$X$ reduction already depends on computational off-diagonals and is nevertheless entanglement breaking.

Stationary scattering gives

For normalized $h$,

The complete outgoing coefficient of pulse $j$ is

where $\mathsf{S}(\omega )$ is the unitary scattering matrix. Its infinity component is

Thus

Every selected mode below is an outgoing mode of this scalar, not an additional radiation payload.

Choose orthonormal compact-band outgoing packets $h_1,\dots ,h_m$. Put

Its dimension is at most $m+2$, and its occupation matrix is finite. Choose an ordinary orthonormal basis whose first $m$ vectors are the selected packets.

The baseline restriction to ${\mathcal{F}}_{\mathrm{s}}(\mathfrak{k})$ has a normal centered gauge-invariant Gaussian density ${\rho }_{\mathfrak{k}}$. It need not factor between the selected packet space and its orthogonal complement in $\mathfrak{k}$.

Proof. In an orthonormal basis $e_i$, its occupation matrix is

It is positive and finite. Diagonalize $A$ by a passive finite-mode unitary. Each diagonal covariance is represented by the geometric density of that mean. Their product, rotated back, has the required Gaussian characteristic function and trace one. Uniqueness of finite-mode Weyl characteristic functions identifies the restricted state. Off-diagonal covariance is retained.

For a branch pair define the trace-class operator

The selected packet channel is the normal CPTP map

Its normalized reference-output state is

Equations (eq:fourchars) determine its selected Weyl characteristic functions exactly.

The diagonal outputs are equal-weight mixtures of the two displaced baseline packet states corresponding to $-z{a}_1-a_2$ and $-z{a}_1+a_2$. The off-diagonal outputs are interference operators and are not positive preparation states.

Proof. Both source endpoint factors act only on $S\otimes {\mathcal{F}}_{\mathrm{s}}(\mathfrak{k})$. Apply them to $\rho \otimes {\rho }_{\mathfrak{k}}$, trace the source and unselected participating modes, and use (eq:branches). This proves the trace-class formula and complete positivity with trace preservation on every source operator and reference extension.

For a diagonal source input there are two orthogonal final source $X$-branches, each of weight $1/2$. Each full-field Weyl conjugation displaces the selected Gaussian baseline. For off-diagonals the two branch coefficients have opposite signs; their trace is zero by (eq:fourchars) at $l=0$. There is no measure-and-prepare interpretation for those terms.

The rest of the participating field has not been prepared in vacuum or assumed independent. Only its actual restricted density was used.

Define

For input Bloch vector $(u_x,u_y,u_z)$, the retained source has Bloch vector

Equivalently,

For every finite $a_1$, $q>0$.

Proof. Write $A=X(a_1)$, $B=X(a_2)$. The second pulse leaves $X$ invariant, while the first gives

Averaging in the centered Gaussian state gives $qX$.

For $Y$,

In a diagonal $Z$-block, the first pulse shifts $B$ by $-z\theta$. The diagonal expectation is therefore $r\sin (2\theta )$, independent of $z$. In an off-diagonal block use

The cosine sandwich contributes $c_{+}Y$.

Similarly,

Its diagonal average is $r\cos (2\theta )Z$, and the sine sandwich gives $c_{-}X$. This proves the adjoint identities and all displayed formulas.

A complement cannot be defined by tracing selected modes from an unpurified mixed packet marginal. We fix the following representation.

In the doubled realization of Section sec:dynamics, the regular selected finite-mode CCR representation has a tensor factorization

The selected packet algebra acts on the first factor. One way to obtain this factorization is to complete its finitely many canonical quadratures to a real symplectic basis in their finite-dimensional doubled one-particle span, implement that finite canonical transformation, and leave the orthogonal doubled Fock space unchanged. This is the finite CCR multiplicity construction.

We fix this factorization once. The vector $\Omega$ is its fixed purification of the selected background. The environment contains:

1. unselected participating radiation;

2. all other outgoing-infinity modes;

3. the future-horizon field algebra;

4. the purification degrees of freedom of the incoming thermal horizon sector.

These physical field subalgebras need not be independent tensor factors in the chosen background. They are not discarded from the environmental representation.

Define the Stinespring isometry

Its accessible output is the selected packet factor, and its environment is $S\otimes {\mathcal{H}}_{\mathrm{env}}$.

A complete complementary channel is

Tracing its environmental factor gives the retained-source channel of Theorem thm:source.

The finite participating calculation of Theorem thm:normal gives the same accessible channel. Its canonical finite-background purification is embedded isometrically into this full environmental interface.

Proof. The isometry follows from unitarity of $U_G$ and normalization of $\Omega$. Taking the two respective partial traces gives the accessible channel and (eq:complement). A further trace gives ${\mathcal{B}}_G$.

For the last assertion, factor the finite participating CCR space from the full regular representation. Its vector restriction is a purification of ${\rho }_{\mathfrak{k}}$. A Schmidt decomposition embeds the canonical purification space into the remaining representation space. Since $U_G$ acts only on the participating physical modes and the source, this embedding intertwines the dilations. No source-dependent information is removed by that replacement of purification coordinates.

We do not identify retained-source leakage with complete leakage. It is a lower bound on the latter because discarding environmental systems is contractive.

For a selected packet channel $\mathcal{N}$, define

For a qubit density $\rho$, choose a purification $|{\psi }_{\rho }⟩$ and put

Its value is independent of the purification convention. This is squared worst-case entanglement fidelity against identity.

The general recovery/environment literature treats metric-specific duality and continuity BenyOreshkov,KSW. The constants below are instead derived directly for the present qubit input.

Let $\mathcal{N}$ be a qubit-input channel with a fixed complete complement. If a decoder has Bell fidelity $F_e$, then the complementary outputs for any orthogonal qubit pair have trace distance at most

There is also a replacement state $\tau$ on the original complete environment with

Proof. Dilate the decoder and apply the complete evolution to a Bell input. The result is a pure vector $|\Psi ⟩$ on reference, decoded qubit, and enlarged environment. Its projection onto the target Bell vector has norm $\sqrt{F_e}$. For $F_e>0$, normalize that projection to an environmental vector $|e⟩$. Then

The assertion for $F_e=0$ is trivial. Tracing the decoded qubit and decoder environment shows that the normalized Choi state of the original complement is within $\sqrt{1-F_e}$ of $I_R/2\otimes \tau$, for the corresponding reduced $\tau$.

For an orthogonal input pair with Bloch direction $L$,

The complete trace norm of the difference bounds this partial contraction. Hence

which is the first assertion.

For a qubit-input Hermiticity-preserving map $\Lambda$,

Indeed every pure referenced qubit input is $(A\otimes I)|\Phi ⟩$, with $\Vert A{\Vert }^2\le 2$; apply the trace-norm product bound and then convexity. Reference dimension two suffices by Schmidt decomposition. The Choi estimate proves the replacement-channel bound.

For every fixed compact two-pulse model and every selected packet family,

Every decoder also obeys

For every replacement channel on the complete complement,

Proof. The retained-source outputs for the two $X$-eigenstate inputs have an $X$-Bloch separation $2q$. Their distance is at least $q$. The complete complementary distance for that pair is therefore at least $q$.

Lemma lem:leakageBell gives $q\le 2\sqrt{1-F_e}$. Thus $1-F_e\ge q^2/4$. The Bell projector gives $d_{\diamond }(\mathcal{DN},\mathrm{id})\ge 1-F_e$. Worst-case fidelity is no larger than Bell fidelity.

For any fixed replacement output, the triangle inequality for the two complementary inputs gives maximal replacement error at least half their distance, hence at least $q/2$.

The theorem excludes exact finite-coupling packet recovery. It does not exclude a family with recovery error tending to zero while $q$ tends to zero and resources diverge.

Fix one normalized compact-band outgoing packet $h$, and put

For $\lambda >0$, choose comparison coefficients

Then

The endpoint unitary $U_0$ acts only on the source and this one outgoing packet. The packet's initial density is the geometric density of mean ${\nu }_h$.

The coefficients in (eq:matched) are not compact bulk tests. They are a comparison model; compact realizations are constructed in Section sec:compact.

Let $|\chi ⟩=(|0⟩+\mathrm{i}|1⟩)/\sqrt{2}$, the $Y=+1$ source state. Define the deterministic packet decoder

Its output source basis is identified with the original message. It uses a fresh independent qubit in $|\chi ⟩$, the inverse two-pulse unitary on that qubit and packet, and packet discard. It is CPTP on every normal packet input.

For a state ${\zeta }_{SB}$, let $P=|\chi ⟩⟨\chi |\otimes I$, and $p=1-\mathrm{tr}(P\zeta )$. Then

The bound retains arbitrary systems in $B$.

Proof. Purify $\zeta$ and apply $P$. The rank-two pure-state calculation, followed by partial trace, gives

Moreover

where ${{\zeta }_B}^{\mathrm{bad}}={\mathrm{Tr}}_S((I-P)\zeta (I-P))⪰0$ has trace $p$. This additional term costs $p/2$.

For the comparison channel ${\mathcal{N}}_0$, define

Then

Consequently

A useful explicit upper estimate is

Proof. Theorem thm:source gives the final source $Y$-mean

For every input and every reference extension, the probability outside $|\chi ⟩$ is therefore at most $p_{\lambda }$.

Let ${\zeta }_{RSh}$ be the actual joint source-packet output. Lemma lem:reset compares it with $|\chi ⟩⟨\chi |\otimes {\zeta }_{Rh}$. Apply ${U_0}^{†}$ on $Sh$, retaining $R$, and trace the packet. The first state returns the input exactly; the second gives $({\mathrm{id}}_R\otimes {\mathcal{D}}_{\lambda }{\mathcal{N}}_0)(\rho )$. Contractivity proves the uniform referenced bound.

Since $1-e^{-x}\le x$,

This proves the explicit estimate and convergence. An effect expectation changes by at most trace distance, giving both fidelity bounds.

No product assumption about this packet and unselected background modes is needed: $U_0$ acts only on $Sh$, so its reduced channel is determined by the packet marginal.

Define

This is a bounded self-adjoint operator on reference and packet, with $\Vert {\mathcal{W}}_{\lambda }\Vert \le 1/2$.

For every separable reference-packet state $\tau$,

For the comparison channel,

Thus $E_{\lambda }<1/2$ proves a non-entanglement-breaking packet channel.

Proof. Local decoding preserves separability, and a separable decoded two-qubit state has Bell overlap at most $1/2$. For $J_0$, its decoded Bell overlap is at least $1-E_{\lambda }$.

The witness is a mathematical reference-packet witness. A local implementation of its decoder is not inferred from its boundedness.

We reproduce the needed bulk approximation rather than assuming that a packet coefficient is already a compact source.

For a compact-band coefficient $u$, choose a time center $t_j$. Let

Set real Cauchy data by

They have coefficient ${\omega }^{-2}u$ at $t_j$ and belong to every integer Sobolev space.

Choose a real $\chi \in {C_c}^{\infty }((-2,2))$, equal to one on $[-1,1]$. For $L\ge 1$, let

Let $s_L$ be their homogeneous Cauchy solution at $t_j$. With ${\eta }_T$ zero before $t_j-T$ and one after $t_j+T$, define

Its support is contained in

The real compact test just defined satisfies

where, with $\delta q=q_L-q,\delta p=p_L-p$,

Proof. The time-slice identity is

Integration by parts proves it using the conserved Cauchy pairing and $\int {\eta }_T^{\prime }=1$. Applying $-{{\partial }_t}^2$ multiplies its coefficient by ${\omega }^2$. Therefore the coefficient error is

Use $|z-\mathrm{i}\omega w{|}^2\le 2|z{|}^2+2{\omega }^2|w{|}^2$ and $1+2n_{\mathrm{H}}\le 1+2/(\beta \omega )$. Spectral unitarity gives the displayed bound.

The cutoffs converge in $H^2$ for $p$ and $H^3$ for $q$. Derivative tails and ${{\partial }_x}^r\chi (x/L)=O(L^{-r})$ prove this. The potential and its derivatives are bounded. The fractional norms are controlled by these integer graph norms, using ${\lambda }^{3/2},\lambda \le 1+{\lambda }^2$ and ${\lambda }^{5/2}\le 1+{\lambda }^3$. Hence ${\Delta }_L\to 0$.