 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-60/4/iv/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 60 / 4 / iv / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

Take a Stinespring dilation

V : B ⟶ B^{'} E

of the local quantum channel and define

τ_{A B^{'} E} = (I_{A} \otimes V) ρ_{A B} (I_{A} \otimes V^{†})

. Tracing out

E

gives the prescribed output, with

A^{'} = A

. An isometry preserves the nonzero eigenvalues of a density operator; therefore

S (B^{'} E)_{τ} = S (B)_{ρ}

,

S (A B^{'} E)_{τ} = S (A B)_{ρ}

, and

S (A)_{τ} = S (A)_{ρ}

.

Consequently

I (A : B^{'} E)_{τ} = I (A : B)_{ρ}

. The loss of quantum mutual information is

I (A : B)_{ρ} - I (A^{'} : B^{'})_{σ} = I (A : B^{'} E)_{τ} - I (A : B^{'})_{τ} = S (A B^{'})_{τ} + S (B^{'} E)_{τ} - S (B^{'})_{τ} - S (A B^{'} E)_{τ} = I (A : E ∣ B^{'})_{τ} \geq 0.

The last inequality is Strong subadditivity of Von Neumann entropy, or nonnegativity of quantum conditional mutual information. Hence

I (A^{'} : B^{'})_{σ} \leq I (A : B)_{ρ} .

This mutual-information loss as conditional mutual information shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.

 Read the full article

 New to topics? Read the docs here!