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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Entropy inequalities, recovery boundaries and thermal entanglement — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Identical marginals and unequal quantum information

Example

Let HA=HB=C2 with the fixed orthonormal basis ∣0⟩,∣1⟩ on each factor. Define the Bell density

ρΦ=∣Φ⟩⟨Φ∣,∣Φ⟩=(∣00⟩+∣11⟩)/2,

and the computational-basis-dephased density

ρdiag=12∣00⟩⟨00∣+12∣11⟩⟨11∣.

Both have marginals I2/2, but their dimensionless mutual informations differ: I(A:B)ρΦ=2log⁡2 and I(A:B)ρdiag=log⁡2. The displayed formula defines the dephased state; no measurement or experimental outcome is asserted.

Facts & Assumptions

Given: HA=HB=C2 with the ordered basis ∣0⟩,∣1⟩ and the two densities in the Example. Use natural logarithms.

[F1]

Dimensionless entropy, mutual and conditional information: I(A:B)ρ=S(ρA)+S(ρB)−S(ρAB) and S(ρ)=−Tr⁡(ρlog⁡ρ) with 0log⁡0:=0.

[F2]

Partial trace over a finite tensor factor: tracing the second tensor factor sums matrix entries whose row and column agree in that factor's basis index.

[F3]

Partial traces of a finite bipartite state are states: the partial traces of a finite bipartite density are density operators.

[F4]

Partial trace over the first tensor factor: in the ordered product basis, (Tr⁡AR)jj′=∑iR(i,j),(i,j′).

Verification

1.1givenF2F3F4algebra

The vector ∣Φ⟩ has norm one because ∣00⟩ and ∣11⟩ are orthonormal, so ρΦ is a rank-one projection with trace one. The two orthogonal basis projections defining ρdiag are positive and each has trace one, so their half-weighted sum is also positive with trace one. Expanding ρΦ gives the four terms 12∣00⟩⟨00∣, 12∣00⟩⟨11∣, 12∣11⟩⟨00∣, and 12∣11⟩⟨11∣. In the partial trace over the second factor, the two off-diagonal terms vanish because the traced basis indices differ, by [F2]; in the trace over the first factor, the same is true by [F4]. In both cases the diagonal terms each contribute one half of the corresponding basis projection, so both marginals are I2/2. Applying the same two formulas to ρdiag gives these marginals directly; they are density operators by [F3].

2.1F1step 1.1algebra∎

The eigenvalues of ρΦ are 1,0,0,0, those of ρdiag are 1/2,1/2,0,0, and those of each marginal are 1/2,1/2. Substitution into [F1] yields S(ρΦ)=0, S(ρdiag)=log⁡2, and S(I2/2)=log⁡2. Therefore the mutual informations are 2log⁡2 and log⁡2, respectively, as stated.

Sources