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An orthogonal ensemble attains its Holevo information

Example

Let X have a finite probability vector (px) and let HB=C∣X∣ have orthonormal basis (ex)x∈X. Set ρx:=Px, where Pxv:=⟨v,ex⟩ex, and measure with the projective POVM (Py)y∈X. Define the joint law pxy:=pxtr⁡(ρxPy) and its output marginal pY(y):=∑xpxy. With H(p):=−∑x:px>0pxln⁡px and I(X:Y):=∑x,y:pxy>0pxyln⁡(pxy/(pxpY(y))), the Holevo information and the measurement mutual information both equal H(p).

Facts & Assumptions

Given: The probability vector, orthonormal basis, ensemble, and POVM in the Example.

[F1]

Finite classical-quantum states and Holevo information: χ=S(ρˉ)−∑px>0pxS(ρx).

[F2]

Finite-dimensional von Neumann entropy: a state with eigenvalue list (λj) has entropy −∑λj>0λjln⁡λj.

[F3]

Finite POVMs and their outcome law and POVM outcome weights are probabilities: the outcome weight of Py in state ρx is tr⁡(ρxPy), and the weights form a probability vector.

[F4]

Verification

1.1givenF1F2algebra

The average state ρˉ=∑xpxPx is diagonal in (ex) with eigenvalues px, so S(ρˉ)=H(p) by [F2]. Each Px has eigenvalues 1 and zeros, hence S(Px)=0; [F1] gives χ=H(p).

2.1givenF3F4step 1.1algebra

By [F3], pxy=pxtr⁡(PxPy)=pxδxy. Thus pY(y)=py, and in the sum defining I(X:Y) only x=y with px>0 contributes, giving I(X:Y)=∑x:px>0pxln⁡(1/px)=H(p) by [F4].

3.1step 1.1step 2.1algebra∎

Combining steps 1.1 and 2.1 gives I(X:Y)=χ for the displayed orthogonal ensemble. This computation does not assert equality for a non-orthogonal ensemble.

Depends on

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