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Quantum Source Compression and Classical Communication over Quantum Channels — Examples

1 · Prerequisites

2 · Summary

These calculations turn the two A-page rates into qubit and binary-ensemble examples. They also show directly that a selected measurement can leave information unused: the Holevo quantity bounds every measurement, but a fixed POVM need not attain it.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Schumacher compression of a diagonal qubit source

Example

For $0 \leq p \leq 1$ , let $ρ_{p} = diag (p, 1 - p)$ be a qubit source state. Its asymptotic quantum compression threshold is $h (p) : = - p \ln p - (1 - p) \ln (1 - p) nats per source qubit,$ with $0 \ln 0 = 0$ . Equivalently the threshold is $h_{2} (p) = h (p) / \ln 2$ transmitted qubits per source qubit.

Facts & Assumptions

Given: The diagonal qubit density operator $ρ_{p}$ , $0 \leq p \leq 1$ .

[F1]

Finite-dimensional von Neumann entropy: $S (ρ) = - \sum_{λ_{j} > 0} λ_{j} \ln λ_{j}$ , with $0 \ln 0 : = 0$ ; the bit entropy is $S_{2} = S / \ln 2$ .

[F2]

Finite-dimensional Schumacher quantum source coding theorem: the optimal reliable compression rate of a finite-dimensional IID source is $S (ρ)$ nats per source letter.

Verification

1.1givenF1casesalgebra

The eigenvalues of $ρ_{p}$ are $p$ and $1 - p$ . Substitution into the entropy definition [F1] gives $S (ρ_{p}) = - p \ln p - (1 - p) \ln (1 - p)$ when $0 < p < 1$ ; at $p = 0$ or $1$ , the zero-term convention gives $S (ρ_{p}) = 0$ .

2.1F1F2step 1.1algebracases∎

Apply the Schumacher theorem [F2] to $ρ_{p}$ . Its threshold is $S (ρ_{p}) = h (p)$ in nats; dividing by $\ln 2$ converts the logarithmic code-dimension rate to bits, hence to qubits sent per source qubit. The endpoint rates are zero, while $p = 1 / 2$ gives $\ln 2$ nats, or one qubit per source qubit.

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Holevo information of an equiprobable binary pure-state channel

Example

Let a binary cq channel send equiprobable pure qubit states $∣ ψ_{0} ⟩, ∣ ψ_{1} ⟩$ with overlap magnitude $c = ∣ ⟨ ψ_{0}, ψ_{1} ⟩ ∣ \in [0, 1]$ . This ensemble has Holevo information $χ = h ⁣ (\frac{1 + c}{2}) nats per channel use,$ where $h (t) = - t \ln t - (1 - t) \ln (1 - t)$ with $0 \ln 0 : = 0$ . If $χ > 0$ , every rate $R$ with $0 \leq R < χ$ is achievable for this memoryless channel; if $c = 1$ , the states coincide and no positive rate follows from this ensemble.

Facts & Assumptions

Given: Two unit vectors in a two-dimensional complex inner-product space and the equiprobable ensemble of their pure-state density operators.

[F1]

Finite-input memoryless cq channel: each input symbol is sent to its assigned output density operator and block outputs are tensor products.

[F2]

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

[F3]

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely: finite-dimensional Hermitian operators have orthonormal eigenbases.

[F4]

Density operator: a pure-state projector has trace one and is a density operator.

[F5]

Holevo–Schumacher–Westmoreland capacity theorem for finite cq channels: the finite cq channel capacity is the maximum ensemble Holevo information, and every rate below that capacity is achievable by reliable memoryless block codes.

Verification

1.1F3F4givenalgebraconstruct

Put $\overset{ˉ}{ρ} = \frac{1}{2} ∣ ψ_{0} ⟩ ⟨ ψ_{0} ∣ + \frac{1}{2} ∣ ψ_{1} ⟩ ⟨ ψ_{1} ∣$ . Its trace is one. After changing the global phase of $∣ ψ_{1} ⟩$ and choosing an orthonormal basis, write $∣ ψ_{0} ⟩ = (1, 0)$ and $∣ ψ_{1} ⟩ = (c, \sqrt{1 - c^{2}})$ . Then $\overset{ˉ}{ρ} = \frac{1}{2} (\begin{array}{cc} 1 + c^{2} & c \sqrt{1 - c^{2}} \\ c \sqrt{1 - c^{2}} & 1 - c^{2} \end{array}),$ whose trace is $1$ and determinant is $\frac{1}{4} ((1 + c^{2}) (1 - c^{2}) - c^{2} (1 - c^{2})) = (1 - c^{2}) / 4$ . Thus its characteristic polynomial is $t^{2} - t + (1 - c^{2}) / 4$ , and [F3] gives eigenvalues $(1 + c) / 2$ and $(1 - c) / 2$ .

2.1F1F2F5step 1.1algebracases∎

Each pure output state has eigenvalues $1, 0$ , hence entropy zero by [F2]. By the symmetry $h (t) = h (1 - t)$ in its displayed definition, the average entropy from step 1.1 is $h ((1 + c) / 2)$ , so [F2] gives the displayed $χ$ . This ensemble is among those maximized in [F5], so $C_{c q} \geq χ$ ; the rate conclusion follows from [F5] for every $0 \leq R < χ$ when $χ > 0$ . If $c = 0$ , the value is $\ln 2$ ; if $c = 1$ , both outputs coincide up to phase and the value is zero.

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

A fixed POVM need not attain ensemble Holevo information

Statement refuted

Every fixed POVM applied to an ensemble extracts its full Holevo information.

Facts & Assumptions

Given: The equiprobable ensemble of the computational basis states $∣ 0 ⟩, ∣ 1 ⟩$ on a qubit, and the fixed $X$ -basis POVM with effects $∣ + ⟩ ⟨ + ∣, ∣ - ⟩ ⟨ - ∣$ , where $∣ \pm ⟩ : = (∣ 0 ⟩ \pm ∣ 1 ⟩) / \sqrt{2}$ .

[F1]

Finite classical-quantum states and Holevo information: the ensemble Holevo information is the entropy of the average state minus the average output entropy.

[F2]

Finite mutual information: for finite classical variables, mutual information measures dependence and equals zero when the joint law is the product of its marginals.

[F3]

Finite POVMs and their outcome law: the outcome probability on state $ρ$ for POVM effect $E_{y}$ is $tr (ρ E_{y})$ .

[F4]

The finite Holevo measurement bound: every finite POVM has classical mutual information at most the ensemble Holevo information.

[F5]

Density operator and Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely: the pure qubit basis projectors and their average $I / 2$ are density operators with the displayed spectra.

Counterexample

technique · direct

1.1F1F5givenalgebra

The average state is $I / 2$ , while each individual state is pure. By [F1] and [F5], its Holevo information is $S (I / 2) - 0 = \ln 2$ nats.

2.1F2F3F4step 1.1algebra∎

For either input $∣ 0 ⟩$ or $∣ 1 ⟩$ , the squared inner products with $∣ + ⟩$ and $∣ - ⟩$ are both $1 / 2$ by [F3]. Hence every joint input-output probability is $1 / 4$ , and each marginal probability is $1 / 2$ . The finite mutual-information sum in [F2] is therefore $\sum_{x, y} (1 / 4) \ln ((1 / 4) / ((1 / 2) (1 / 2))) = 0$ . This is strictly below $\ln 2$ from step 1.1, while still satisfying the Holevo upper bound [F4].

Quantum Source Compression and Classical Communication over Quantum Channels — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Schumacher compression of a diagonal qubit source

Example

Facts & Assumptions

Verification

Holevo information of an equiprobable binary pure-state channel

Example

Facts & Assumptions

Verification

A fixed POVM need not attain ensemble Holevo information

Statement refuted

Facts & Assumptions

Counterexample

Sources