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Quantum Channel Capacities and Entanglement Assistance — Examples

1 · Prerequisites

2 · Summary

The noiseless qubit endpoint is checked by the dense-coding protocol, while the qudit erasure examples evaluate Q, P, and C_E at endpoints and in both degradation regimes. These computations illustrate the A-page formulas and do not assert single-letter additivity for general channels.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Entanglement-assisted capacity of a noiseless qubit

Example

For the identity channel on a qubit, the entanglement-assisted classical capacity is $2 \ln 2$ nats per use, or two classical bits per transmitted qubit. A shared Bell pair and the four-message dense-coding protocol attain the value.

Facts & Assumptions

Given: The noiseless identity channel ${id}_{C^{2}}$ and prior shared entanglement.

[F1]

Additive entanglement-assisted classical capacity formula: $C_{E} (N) = \max_{ρ} [S (ρ) + I_{c} (ρ, N)]$ .

[F2]

Complementary channels and coherent information: $I_{c} (ρ, N) = S (B) - S (R B)$ for a purification.

[F3]

The maximally mixed state maximizes finite quantum entropy: for a qubit, $S (ρ) \leq \ln 2$ , with equality at $I / 2$ .

[F4]

The Bell vectors: the four Bell vectors form an orthonormal basis of $C^{2} \otimes C^{2}$ .

[F5]

Dense-coding correctness and resources: the four Pauli encodings on one half of a shared Bell pair produce the four orthogonal Bell states, which Bob distinguishes with zero error using one transmitted qubit.

Verification

1.1F1F2F3givenalgebra

For any qubit input $ρ$ , the identity channel leaves its purification $ψ_{R A}$ pure. Hence $S (R B) = 0$ and $S (B) = S (ρ)$ , so [F2] gives $I_{c} (ρ, id) = S (ρ)$ . By [F1] and [F3], $C_{E} (id) = \max_{ρ} 2 S (ρ) = 2 \ln 2$ .

2.1F4F5step 1.1algebra∎

For a maximally entangled Bell pair, Alice's four Pauli choices produce the four orthogonal Bell states by [F4]–[F5]. Bob's Bell-basis measurement identifies the message without error, so four messages are sent per channel use and the rate is $\ln 4 = 2 \ln 2$ . This attains the capacity computed in step 1.1.

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

Capacity regimes of the qubit erasure channel

Example

For $d = 2$ , the erasure formulas give the following capacities. Entries are shown in bits per channel use; multiply by $\ln 2$ to obtain nats.

$ϵ$	$Q = P$	$C_{E}$	Regime
$0$	$1$	$2$	degradable
$1 / 4$	$1 / 2$	$3 / 2$	degradable
$1 / 2$	$0$	$1$	degradable and antidegradable; self-complementary
$3 / 4$	$0$	$1 / 2$	antidegradable
$1$	$0$	$0$	antidegradable

Facts & Assumptions

Given: A qubit erasure channel, so $d = 2$ , at the five probabilities in the table.

[F1]

Exact quantum, private, and entanglement-assisted erasure capacities: $Q = P = \max {1 - 2 ϵ, 0} \ln d$ and $C_{E} = 2 (1 - ϵ) \ln d$ .

[F2]

Complement and degradation maps of the erasure channel: the receiver and complement outputs exchange the data and erasure probabilities; the channel is degradable for $ϵ \leq 1 / 2$ and antidegradable for $ϵ \geq 1 / 2$ .

[F3]

Finite-dimensional quantum erasure channel: the two outputs use the same data space with a one-dimensional orthogonal erasure flag.

[F4]

Information units and logarithm-base convention: capacities in bits are obtained by dividing nats by $\ln 2$ .

Verification

1.1F1F4givenalgebra

Substituting $d = 2$ into [F1] and dividing by $\ln 2$ gives $Q = P = \max {1 - 2 ϵ, 0}$ bits and $C_{E} = 2 (1 - ϵ)$ bits. At $ϵ = 0, 1 / 4, 1 / 2, 3 / 4, 1$ , these pairs are respectively $(1, 2), (1 / 2, 3 / 2), (0, 1), (0, 1 / 2), (0, 0)$ , which are the table entries.

2.1F2F3step 1.1constructalgebra∎

By [F2], the first two probabilities lie in the degradable region, the last two lie in the antidegradable region, and the half-erasure point lies in both. At $ϵ = 1 / 2$ , [F2] gives $E^{c} (X) = \frac{1}{2} J_{E} X J_{E}^{*} + \frac{1}{2} tr (X) ∣ e_{E} ⟩ ⟨ e_{E} ∣$ , while [F3] gives $E (X) = \frac{1}{2} J_{B} X J_{B}^{*} + \frac{1}{2} tr (X) ∣ e_{B} ⟩ ⟨ e_{B} ∣$ . The unitary identification $J_{B} x \mapsto J_{E} x$ and $e_{B} \mapsto e_{E}$ carries one map to the other, proving self-complementarity up to flag relabeling.

Quantum Channel Capacities and Entanglement Assistance — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Entanglement-assisted capacity of a noiseless qubit

Example

Facts & Assumptions

Verification

Capacity regimes of the qubit erasure channel

Example

Facts & Assumptions

Verification

Sources