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 Incoming links: Relative-entropy barycenter of a quantum ensemble

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 5 / iv / Solution Created 2026-10-03 Updated 2026-10-05

By Donald's identity, the objective equals

\sum_{j} p_{j} D (ω_{j} ∥ ρ) = \sum_{j} p_{j} D (ω_{j} ∥ \overline{ω}) + D (\overline{ω} ∥ ρ) .

The nonnegativity of quantum relative entropy makes the last term at least zero, with equality exactly when

ρ = \overline{ω}

. Hence the unique minimizing density operator is the ensemble average and

min_{ρ} \sum_{j} p_{j} D (ω_{j} ∥ ρ) = \sum_{j} p_{j} D (ω_{j} ∥ \overline{ω}) = S (\overline{ω}) - \sum_{j} p_{j} S (ω_{j}) .

For the classical-quantum state

σ_{XB}

, its marginal on

X

H (p)

, its marginal on

B

is

\overline{ω}

, and its joint entropy is

H (p) + \sum_{j} p_{j} S (ω_{j})

. Substituting in the definition of quantum mutual information gives

ρ min j \sum p_{j} D (ω_{j} ∥ ρ) = χ ({p_{j}, ω_{j}}) = I (X : B)_{σ}, ρ_{m i n} = \overline{ω} .

This is the relative-entropy barycenter of a quantum ensemble: the average state minimizes mean distinguishability from the ensemble. Its minimal value is exactly the Holevo quantity.

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