 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-48/2/a/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 48 / 2 / a / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

The channel acts only on the input system

Q

; the reference

R

is untouched. Write

ω_{RB} = (id_{R} \otimes Φ) (∣ Ψ^{ρ} ⟩ ⟨ Ψ^{ρ} ∣_{RQ}) .

The coherent information is

I_{c} (Φ, ρ) = S (B)_{ω} - S (RB)_{ω} = - S (R ∣ B)_{ω} .

The reference marginal remains

ρ_{R}

, whose nonzero eigenvalues agree with those of

ρ

. Unlike a classical information quantity, coherent information can be negative.

Figure 1.
Quantum channel acting on one half of a purification, with untouched reference and discarded environment
. Details.

The lower diagram uses a Stinespring dilation: append an environment in a fixed pure state, apply a joint unitary operator, and discard the environment by partial trace. Before discarding it, the global

RBE

state is pure.

 Read the full article

 New to topics? Read the docs here!