Shared-reference protocols: what label erasure leaves observable

Write the preparation before describing its erasure

Choose computational presentations |0⟩,|1⟩ for two carriers and prepare |Φ+⟩=(|00⟩+|11⟩)/√2. This preparation supplies a particular pairing. Its local density matrices are each I₂/2, so neither endpoint alone reveals that pairing. The joint observables do: ⟨X⊗X⟩=1, ⟨Y⊗Y⟩=−1 and ⟨Z⊗Z⟩=1.

Now add an independent classical register L in a state σ_L. The preparation is σ_L⊗|Φ+⟩⟨Φ+|. Discarding L, resetting L, or applying any trace-preserving channel only to L leaves AB exactly in the original Bell state. Renaming or forgetting that label has not acted on the pairing. The source's stronger result says that channels fixing both full carrier algebras pointwise preserve every joint reference, not only this example. Completed records, v21, Proposition 4.7, p.8.

The displayed Pauli axes are part of the stated measurement calibration. A passive frame change transforms both the state and its observables; it does not physically scramble A while leaving the detector axes fixed. This is why a reference must be inventoried as prepared data rather than described as an unobservable basis name.

Depolarizing A is a different protocol

Let D_s(M)=sM+(1−s)Tr(M)I₂/2 act on A and identity act on B. With s=e^−γt, this is a trace-preserving completely positive channel for every γ,t≥0. Its action on the Bell preparation is

In the order |00⟩,|01⟩,|10⟩,|11⟩, the two outer diagonal entries are a=(1+s)/4, the two inner diagonal entries are b=(1−s)/4, and the only off-diagonal entries are ρ_00,11=ρ_11,00=s/2. This directly determines the entire state without a sampling or continuum approximation.

The source uses this mathematical mixing form on the selector, where it preserves the retained carrier. Applying it here to A deliberately changes the register assignment. It is therefore a comparison protocol, not an implementation that both erases the Bell pairing and satisfies the source's full carrier-preservation premise. V21, Definition 4.2, Theorem 4.3 and Proposition 4.7, pp.7–8.

The channel explorer labels these alternatives explicitly. Its label-erasure branch keeps s=1 throughout; the rate slider does not secretly introduce carrier noise into that branch.

Use joint data to test the channel claim

Taking either partial trace gives I₂/2 for every s. The same unchanged marginals therefore occur for a pure Bell pair, an entangled mixed state, a separable correlated state and the limiting product state. The distinguishing joint measurements are

In the observable picture, D_s(X)=sX. Thus A's full algebra is not fixed when s<1, even though this input's A marginal is unchanged. Checking a universal algebra-preservation assertion on one maximally mixed marginal cannot establish it.

Full joint tomography can provide much stronger information under an additional trusted assumption: only a local channel on A acted after the known Bell preparation. The identity |Φ+⟩⟨Φ+|=½Σ_i,j|i⟩⟨j|⊗|i⟩⟨j| means that the output's B matrix blocks encode the images of all A matrix units. Under that local-channel assumption, an unchanged complete Bell state determines the identity channel. Unchanged marginals do not.

This elementary reconstruction spells out the quantifiers in the source's operator theorem. It also identifies the additional preparation and locality assumptions required before experimental state data can establish a statement about the channel on arbitrary inputs. V21, Proposition 4.7 and discussion of reference preparations, p.8.

Name both comparison states in a distance calculation

The Bell vector is an eigenvector of ρ with eigenvalue (1+3s)/4; every orthogonal Bell direction has eigenvalue (1−s)/4. Squaring and summing these eigenvalues gives purity (1+3s²)/4. The Bell overlap ⟨Φ+|ρ|Φ+⟩ is (1+3s)/4.

Define trace distance as D(ρ,σ)=½‖ρ−σ‖₁. Relative to the initial Bell state, the difference has eigenvalues −3(1−s)/4 and three copies of (1−s)/4. Relative to the maximally mixed product reference I₄/4, its eigenvalues are 3s/4 and three copies of −s/4. Hence

Their sum is 3/4 along this particular straight mixture segment. It is not an information-conservation law for arbitrary channels. One distance increases while the other decreases because their reference states differ.

These are derived diagnostics of the declared qubit example, supplementing the reference-preservation statement in v21, p.8.

Detect entanglement through a matrix calculation

Partial transpose on B moves the corner coherences to the |01⟩,|10⟩ block. That block becomes [[b,s/2],[s/2,b]], while the two outer entries remain a. Its four eigenvalues are (1+s)/4 three times and (1−3s)/4 once. The negativity, defined as the sum of magnitudes of negative partial-transpose eigenvalues, is therefore

A separable mixture Σp_jρ_A,j⊗ρ_B,j has positive partial transpose because each ρ_B,j^T is positive. A negative eigenvalue therefore certifies entanglement. Here it occurs exactly for s>1/3. For γ>0, the boundary time is t_*=ln3/γ. With γ=1, t=ln2 gives negativity 1/8 and t=ln3 gives zero.

A directly measurable witness reaches the same threshold. The Bell correlation combination C=⟨XX−YY+ZZ⟩ is 3s. For a product state with Bloch vectors a,b, its magnitude is at most ‖a‖‖b‖≤1; convex mixing preserves this bound. Thus C>1 also certifies entanglement. The negative sign on YY follows from the chosen |Φ+⟩ reference, not from a negative measurement probability.

Prove what happens beyond the witness threshold

For this family, zero negativity can be supplemented by an explicit separable decomposition. At s=1/3, mix six product states with equal weights. Use the same ±X eigenstates at A and B, opposite ±Y eigenstates, and the same ±Z eigenstates. Averaging a positive and negative pair on each axis cancels all single-qubit Bloch terms and retains the chosen correlation sign.

For 0≤s≤1/3, write ρ(s)=3sρ(1/3)+(1−3s)I₄/4. Both terms are separable and their weights are nonnegative. This proves separability throughout that interval without relying on an inference from a vanishing numerical witness alone.

Separable does not mean uncorrelated. At s=1/3 the three joint correlations remain ±1/3 and the state still differs from I₄/4 by trace distance 1/4. Only the limiting state s=0 is the maximally mixed product. Label erasure never crosses this boundary because it leaves the Bell reference untouched. This separates loss of entanglement, loss of all joint correlations and loss of a classical register into three different statements.

A relational pair is not a standalone phase reference

The Bell preparation names a relational resource. A global invariant pairing with a retained dual reference need not give either subsystem a standalone coherent phase. The source makes this distinction concrete for a determinant line D: a globally invariant superposition of vacuum and a D⊗D* pair can have zero reduced vacuum-to-D coherence after the dual is discarded. A calibrated coherent reservoir instead transfers that coherence through an explicit neutral swap. V21, Propositions 6.7–6.8 and discussion, pp.14–15; see also v20, §20.3, pp.60–61.

For a covariant channel Λ the source bounds a standalone output coherence by |⟨τ|Λ(ρ)|0⟩|≤½‖ρ−Tρ‖₁, where T averages over the full input symmetry. A balanced pure determinant reference has coherence 1/2. This is a phase-asymmetry bound, not the Bell negativity and not an energy estimate.

A careful protocol therefore records which systems are prepared coherently, which remain accessible, which are discarded, and which comparison observables are calibrated. Common origin by itself supplies none of these specifications. The qubit example illustrates that accounting; it does not reconstruct the rank-three/rank-two carrier choice or its determinant tensor.

Reproduce the example and preserve its interpretation

The explorer accepts 0≤γ≤10 and 0≤t≤10 in declared channel units. It returns the full density matrix, partial transpose, both marginals, correlations, spectra, purity, two named trace distances and negativity. At γ=0 or t=0 the depolarization protocol is identity. Its exact zero-retention limit is a mathematical endpoint; at finite positive γt the exponential remains positive.

The independent audit constructs the channel by exponentiating its 16-dimensional Lindblad superoperator, then computes matrix eigenvalues, partial traces and trace norms. This checks the closed formulas against a separate calculation rather than drawing their curves and assuming agreement. Floating-point results near an entanglement boundary should be read at numerical precision; the analytical threshold remains s=1/3.

The model supplies one channel-evolution parameter, not extra physical time axes. The six inner generative levels remain nested roles followed by the seventh completion. A physical interpretation would still have to select the preparation, noise mechanism, units and admitted future tests. Those are precisely the interfaces left explicit in v21, §§5 and 8, pp.9–10 and 16–17.

Continue reading

• Calculate the Bell-reference example
• Exact selector channels and their fixed algebras
• Use a reference to resolve loop transport
• Test a specified finite model