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The maximally mixed state maximizes finite quantum entropy

Statement

Let H be a nonzero finite-dimensional complex inner-product space of dimension d, and let ρ be a density operator on H. Then

S(ρ)≤ln⁡d,

with equality if and only if ρ=IH/d.

Facts & Assumptions

Given: The finite-dimensional state ρ and nonzero space H in the Statement.

[F1]

Finite-dimensional von Neumann entropy: S(ρ)=−tr⁡(ρln⁡ρ), with zero-eigenvalue terms omitted.

[F2]

Finite-dimensional quantum relative entropy: on support inclusion, D(A∥B)=tr⁡(Aln⁡A)−tr⁡(Aln⁡B).

[F3]

Nonnegativity and equality for finite quantum relative entropy: for finite-dimensional density operators ρ,σ, D(ρ∥σ)≥0, with equality exactly when ρ=σ.

[F5]

Proof

1.1F2F4F5givenalgebra

The operator τ:=IH/d is a density operator: it is nonnegative and has trace d/d=1 by [F4]–[F5]. Its support is all of H, so supp⁡(ρ)⊆supp⁡(τ) and the finite formula in [F2] applies.

2.1F1F2F3F4F6step 1.1algebra

Spectral calculus gives ln⁡τ=ln⁡(1/d)IH=−(ln⁡d)IH by [F4], [F6]. Hence [F1]–[F2] yield D(ρ∥τ)=tr⁡(ρln⁡ρ)−tr⁡(ρln⁡τ)=−S(ρ)+ln⁡d, using tr⁡(ρ)=1 by [F4]. By [F3], this is nonnegative, so S(ρ)≤ln⁡d.

3.1F1F3step 2.1algebra∎

Equality S(ρ)=ln⁡d is equivalent by step 2.1 to D(ρ∥τ)=0. The equality case of [F3] then gives ρ=τ=IH/d. Conversely, substituting ρ=IH/d into [F1] gives S(ρ)=ln⁡d.

Depends on

Used by

Dependency tree · two levels

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Sources