Pipeline-generated

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Quantum Information Measures — Examples

1 · Prerequisites

2 · Summary

These finite examples compute the entropy of a diagonal qubit and the Holevo information of an orthogonal ensemble, exhibit normalized states with infinite quantum relative entropy when support inclusion fails, and calculate the negative conditional entropy of a Bell pair. They preserve the natural-log convention and make the zero-eigenvalue terms explicit.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Entropy of a diagonal qubit state

Example

For $p \in [0, 1]$ , let $ρ_{p}$ be the qubit state with matrix $diag (p, 1 - p)$ . Its entropy is $S (ρ_{p}) = - p \ln p - (1 - p) \ln (1 - p)$ nats, with zero terms interpreted as $0 \ln 0 = 0$ . In bits it is $h_{2} (p) : = - p \log_{2} p - (1 - p) \log_{2} (1 - p)$ , where $\log_{2} u : = \ln u / \ln 2$ , with the same endpoint convention. Thus pure basis states have entropy $0$ , and the maximally mixed qubit has entropy $\ln 2$ nats, or one bit.

Facts & Assumptions

Given: $p \in [0, 1]$ , the diagonal density operator $ρ_{p}$ , and the entropy definition.

[F1]

Finite-dimensional von Neumann entropy: entropy is the negative sum of $λ \ln λ$ over positive eigenvalues, and entropy in bits divides by $\ln 2$ .

[F2]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: a non-negative diagonal operator has the displayed diagonal entries as its eigenvalues.

[F3]

Density operator: a density operator is non-negative and has trace one.

[F4]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\ln (1 / 2) = \ln 1 - \ln 2 = - \ln 2$ .

Verification

1.1givenF1F2F3algebra

The diagonal entries $p, 1 - p$ are nonnegative and sum to one, so $ρ_{p}$ is a density operator by [F3]. By [F2] its eigenvalues are exactly $p, 1 - p$ . Applying [F1] gives $S (ρ_{p}) = - p \ln p - (1 - p) \ln (1 - p)$ , with a zero-eigenvalue term omitted.

2.1givenF1F4step 1.1algebracases∎

If $p = 0$ or $p = 1$ , the only positive eigenvalue is $1$ , so $S = 0$ . If $p = 1 / 2$ , [F4] gives $S = - 2 (\frac{1}{2} \ln \frac{1}{2}) = \ln 2$ , and [F1] gives $S_{2} = 1$ . For general $p$ , the stated definition of $\log_{2}$ gives $S_{2} = h_{2} (p)$ .

ExampleConstruction: Literature-sourcedVerification: AI-generatedOpen item page →

An orthogonal ensemble attains its Holevo information

Example

Let $X$ have a finite probability vector $(p_{x})$ and let $H_{B} = C^{∣ X ∣}$ have orthonormal basis $(e_{x})_{x \in X}$ . Set $ρ_{x} : = P_{x}$ , where $P_{x} v : = ⟨ v, e_{x} ⟩ e_{x}$ , and measure with the projective POVM $(P_{y})_{y \in X}$ . Define the joint law $p_{x y} : = p_{x} tr (ρ_{x} P_{y})$ and its output marginal $p_{Y} (y) : = \sum_{x} p_{x y}$ . With $H (p) : = - \sum_{x : p_{x} > 0} p_{x} \ln p_{x}$ and $I (X : Y) : = \sum_{x, y : p_{x y} > 0} p_{x y} \ln (p_{x y} / (p_{x} p_{Y} (y)))$ , the Holevo information and the measurement mutual information both equal $H (p)$ .

Facts & Assumptions

Given: The probability vector, orthonormal basis, ensemble, and POVM in the Example.

[F1]

Finite classical-quantum states and Holevo information: $χ = S (\overset{ˉ}{ρ}) - \sum_{p_{x} > 0} p_{x} S (ρ_{x})$ .

[F2]

Finite-dimensional von Neumann entropy: a state with eigenvalue list $(λ_{j})$ has entropy $- \sum_{λ_{j} > 0} λ_{j} \ln λ_{j}$ .

[F3]

Finite POVMs and their outcome law and POVM outcome weights are probabilities: the outcome weight of $P_{y}$ in state $ρ_{x}$ is $tr (ρ_{x} P_{y})$ , and the weights form a probability vector.

[F4]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: for $p_{x} > 0$ , $\ln (1 / p_{x}) = - \ln p_{x}$ .

Verification

1.1givenF1F2algebra

The average state $\overset{ˉ}{ρ} = \sum_{x} p_{x} P_{x}$ is diagonal in $(e_{x})$ with eigenvalues $p_{x}$ , so $S (\overset{ˉ}{ρ}) = H (p)$ by [F2]. Each $P_{x}$ has eigenvalues $1$ and zeros, hence $S (P_{x}) = 0$ ; [F1] gives $χ = H (p)$ .

2.1givenF3F4step 1.1algebra

By [F3], $p_{x y} = p_{x} tr (P_{x} P_{y}) = p_{x} δ_{x y}$ . Thus $p_{Y} (y) = p_{y}$ , and in the sum defining $I (X : Y)$ only $x = y$ with $p_{x} > 0$ contributes, giving $I (X : Y) = \sum_{x : p_{x} > 0} p_{x} \ln (1 / p_{x}) = H (p)$ by [F4].

3.1step 1.1step 2.1algebra∎

Combining steps 1.1 and 2.1 gives $I (X : Y) = χ$ for the displayed orthogonal ensemble. This computation does not assert equality for a non-orthogonal ensemble.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Normalized states can have infinite quantum relative entropy

Statement refuted

Every pair of density operators on the same finite-dimensional space has finite quantum relative entropy.

Facts & Assumptions

Given: A two-dimensional register $H = C^{2}$ with computational orthonormal basis $e_{0}, e_{1}$ .

[F1]

Finite-dimensional quantum relative entropy: $D (ρ ∥ σ) = + \infty$ whenever $supp (ρ) ⊈ supp (σ)$ .

[F2]

Density operator: a density operator is a non-negative endomorphism of trace one.

[F3]

Finite registers and their ordered computational bases: $e_{0}, e_{1}$ are orthonormal.

[F4]

Counterexample

1.1givenF2F3F4construct

Define $P_{j} x : = ⟨ x, e_{j} ⟩ e_{j}$ for $j = 0, 1$ , and set $ρ = P_{0}$ , $σ = P_{1}$ . Each is an orthogonal rank-one projection, hence non-negative by [F4]; in the basis $e_{0}, e_{1}$ each has trace one, so both are density operators by [F2].

2.1givenF1F3step 1.1algebra∎

The range/support of $P_{0}$ is $C e_{0}$ , while the support of $P_{1}$ is $C e_{1}$ ; these are orthogonal by [F3]. Thus $supp (ρ) ⊈ supp (σ)$ , and [F1] gives $D (ρ ∥ σ) = + \infty$ . This contradicts the asserted finiteness.

ExampleConstruction: Literature-sourcedVerification: AI-generatedOpen item page →

A Bell pair has negative quantum conditional entropy

Example

Let $H_{A} = H_{B} = C^{2}$ with computational bases $e_{0}, e_{1}$ , let $ϕ = (e_{0} \otimes e_{0} + e_{1} \otimes e_{1}) / \sqrt{2}$ , and define the pure state $ρ_{A B} x : = ⟨ x, ϕ ⟩ ϕ$ . Then the conditional entropy is $S (A ∣ B)_{ρ} = - \ln 2$ nats.

Facts & Assumptions

Given: The two-qubit vector $ϕ$ and the rank-one state $ρ_{A B}$ in the Example.

[F1]

Finite quantum mutual, conditional, and conditional mutual information: $S (A ∣ B)_{ρ} = S (ρ_{A B}) - S (ρ_{B})$ .

[F2]

Finite-dimensional von Neumann entropy: $S (ρ) = - \sum_{λ > 0} λ \ln λ$ .

[F3]

Density operator: a state is non-negative with trace one.

[F4]

Partial trace over a finite tensor factor: ${Tr}_{B}$ sums the matching second-factor basis indices.

[F5]

Partial trace over the first tensor factor: ${Tr}_{A}$ sums the matching first-factor basis indices.

[F6]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\ln$ is strictly increasing, $\ln 1 = 0$ , and $\ln (1 / 2) = \ln 1 - \ln 2 = - \ln 2$ .

Verification

1.1givenF2F3construct

The vector $ϕ$ is a unit vector by orthonormality of the product basis, so the rank-one projection $ρ_{A B}$ has eigenvalues $1, 0, 0, 0$ and is a state by [F3]. Hence $S (ρ_{A B}) = 0$ by [F2].

2.1givenF4F5step 1.1algebra

Expanding $ρ_{A B}$ gives diagonal terms $\frac{1}{2} P_{e_{0} \otimes e_{0}} + \frac{1}{2} P_{e_{1} \otimes e_{1}}$ and two off-diagonal cross terms. The partial traces [F4]–[F5] remove those cross terms because their traced basis indices differ, and leave $ρ_{A} = ρ_{B} = \frac{1}{2} (P_{e_{0}} + P_{e_{1}}) = I_{2} / 2$ .

3.1givenF1F2F6step 1.1step 2.1algebra∎

The eigenvalues of $I_{2} / 2$ are $1 / 2, 1 / 2$ , so [F2] and [F6] give $S (ρ_{B}) = - 2 (\frac{1}{2} \ln \frac{1}{2}) = \ln 2$ . Since $2 > 1$ , strict increase in [F6] gives $\ln 2 > \ln 1 = 0$ . Substitution into [F1] and step 1.1 yields $S (A ∣ B)_{ρ} = 0 - \ln 2 = - \ln 2 < 0$ .

Quantum Information Measures — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Entropy of a diagonal qubit state

Example

Facts & Assumptions

Verification

An orthogonal ensemble attains its Holevo information

Example

Facts & Assumptions

Verification

Normalized states can have infinite quantum relative entropy

Statement refuted

Facts & Assumptions

Counterexample

A Bell pair has negative quantum conditional entropy

Example

Facts & Assumptions

Verification

Sources