 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-50/5/d/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 50 / 5 / d / Solution by

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

Let the quantum channel send

B

to

B^{'}

. From its Kraus representation, choose a Stinespring representation

V : B \to B^{'} E

and form

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

The channel output is

σ_{A B^{'}} = Tr_{E} σ_{A B^{'} E}

. Isometric invariance of Von Neumann entropy gives

S (B^{'} E)_{σ} = S (B)_{ρ}, S (A B^{'} E)_{σ} = S (A B)_{ρ}, S (A)_{σ} = S (A)_{ρ} .

Therefore the quantum mutual information satisfies

I (A : B)_{ρ} = I (A : B^{'} E)_{σ}

. Expanding its loss after discarding the environment gives

I (A : B)_{ρ} - I (A : B^{'})_{σ} = I (A : B^{'} E)_{σ} - I (A : B^{'})_{σ} = S (A B^{'})_{σ} + S (B^{'} E)_{σ} - S (B^{'})_{σ} - S (A B^{'} E)_{σ} .

The final expression is the quantum conditional mutual information

I (A : E ∣ B^{'})

. The Strong subadditivity of Von Neumann entropy states exactly that it is nonnegative. Hence

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

This proves data processing for quantum mutual information for any mixed input and any local quantum channel, not just the atomic channel in earlier parts. The correlations lost are exactly the conditional correlations with the discarded environment.

 Read the full article

 New to topics? Read the docs here!