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Complement and degradation maps of the erasure channel

Statement

For $E_{d, ϵ}$ , choose the Stinespring isometry in The finite-dimensional erasure map is a quantum channel, with environment $E = A \oplus C e_{E}$ . Its complement is

$E_{d, ϵ}^{c} (X) = ϵ J_{E} X J_{E}^{*} + (1 - ϵ) tr (X) ∣ e_{E} ⟩ ⟨ e_{E} ∣ .$

The channel is degradable for $0 \leq ϵ \leq \frac{1}{2}$ and antidegradable for $\frac{1}{2} \leq ϵ \leq 1$ .

Facts & Assumptions

Given: An input dimension $d \geq 1$ and erasure probability $ϵ \in [0, 1]$ .

[F1]

Finite-dimensional quantum erasure channel: the receiver output space is the orthogonal sum of the data space and its erasure flag, and the data map is the isometric inclusion $J_{B}$ .

[F2]

Degradable and antidegradable channels: degradability means that a channel from $B$ to $E$ simulates the complement, and antidegradability means a channel from $E$ to $B$ simulates the original channel.

[F3]

The finite-dimensional erasure map is a quantum channel: $E_{d, ϵ}$ is a finite-dimensional channel.

[F4]

A finite Kraus sum is completely positive: every finite Kraus sum is completely positive.

[F5]

Trace preservation is Kraus normalization: a finite Kraus sum is trace-preserving when the sum of $K_{j}^{*} K_{j}$ is the identity.

[F6]

Finite-dimensional quantum channel: a channel is complex-linear, completely positive, and trace-preserving.

Proof

1.1F1F3constructalgebra

Use the isometry $V_{ϵ} x = \sqrt{1 - ϵ} J_{B} x \otimes e_{E} + \sqrt{ϵ} e \otimes J_{E} x$ from the proof of The finite-dimensional erasure map is a quantum channel. Tracing the first factor $B$ leaves the data block $ϵ J_{E} X J_{E}^{*}$ and the flag block $(1 - ϵ) tr (X) ∣ e_{E} ⟩ ⟨ e_{E} ∣$ ; the cross terms vanish because the data and erasure subspaces are orthogonal. The data block precedes the flag block because $E = A \oplus C e_{E}$ .

2.1F1F2F3F4F5F6step 1.1algebraconstructchoose

Suppose $0 \leq ϵ \leq \frac{1}{2}$ and set $α = ϵ / (1 - ϵ) \in [0, 1]$ . Choose an orthonormal basis $(u_{j})_{j < d}$ of the data summand of $B$ , and define the finite Kraus family from $B$ to $E$ by $K_{d a t a} = \sqrt{α} J_{E} J_{B}^{*}$ , $K_{j} = \sqrt{1 - α} ∣ e_{E} ⟩ ⟨ u_{j} ∣$ for $j < d$ , and $K_{e} = ∣ e_{E} ⟩ ⟨ e_{B} ∣$ . The sum of adjoint products is $α Π_{A} + (1 - α) Π_{A} + ∣ e_{B} ⟩ ⟨ e_{B} ∣ = I_{B}$ , so the induced map $D_{ϵ} (X) = \sum K X K^{*}$ is a channel by [F4]–[F6]. On an erasure-channel output, its data block has weight $α (1 - ϵ) X = ϵ X$ and its flag block has weight $[(1 - α) (1 - ϵ) + ϵ] tr (X) = (1 - ϵ) tr (X)$ . Thus $D_{ϵ} E_{d, ϵ} = E_{d, ϵ}^{c}$ , proving degradability.

3.1F1F2F3F4F5F6step 1.1algebraconstructchoose∎

Suppose $\frac{1}{2} \leq ϵ \leq 1$ and set $β = (1 - ϵ) / ϵ \in [0, 1]$ . Choose an orthonormal basis $(u_{j}^{E})_{j < d}$ of the data summand of $E$ and define a finite Kraus family from $E$ to $B$ by $L_{d a t a} = \sqrt{β} J_{B} J_{E}^{*}$ , $L_{j} = \sqrt{1 - β} ∣ e_{B} ⟩ ⟨ u_{j}^{E} ∣$ for $j < d$ , and $L_{e} = ∣ e_{B} ⟩ ⟨ e_{E} ∣$ . Its adjoint products sum to $I_{E}$ , so its Kraus map $A_{ϵ}$ is a channel by [F4]–[F6]. Applying it to the complement in step 1.1 gives data weight $β ϵ X = (1 - ϵ) X$ and erasure-flag weight $[(1 - β) ϵ + (1 - ϵ)] tr (X) = ϵ tr (X)$ . Hence $A_{ϵ} E_{d, ϵ}^{c} = E_{d, ϵ}$ , proving antidegradability. At $ϵ = \frac{1}{2}$ both parameters equal one and both relations hold.

Depends on

Used by

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John Watrous, The Theory of Quantum Information, §8.3.2 and §8.3.3 (standard reference, not scraped)
Mark M. Wilde, From Classical to Quantum Shannon Theory, second-edition prepublication v8, §24.7 (standard reference, not scraped)