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Helstrom Binary Discrimination

Statement

Let $ρ, σ$ be density operators on a finite-dimensional complex Hilbert space. If they are presented with equal prior probability, the greatest success probability of any binary measurement for identifying the state is $\frac{1}{2} (1 + D (ρ, σ)) .$ The binary projective measurement whose first effect is the spectral projection of $Δ = ρ - σ$ onto its strictly positive eigenspaces attains this value.

Facts & Assumptions

Given: Density operators $ρ, σ$ and equal prior probabilities $1 / 2$ .

[F1]

$D (ρ, σ) = \frac{1}{2} ∥ ρ - σ ∥_{1}$ (Trace Distance).

[L1]

A finite POVM has nonnegative effects summing to $I$ and outcome probabilities $p_{a} = tr (ρ E_{a})$ (Finite POVMs and their outcome law). The current QC04 definition body is present as a draft.

[L2]

For Hermitian trace-zero $Δ$ , its positive and negative spectral parts satisfy $tr Δ_{+} = tr Δ_{-} = \frac{1}{2} ∥ Δ ∥_{1}$ (Hermitian Trace Norm Positive Negative Parts). The current QC01 theorem body is present as a draft.

[L3]

If $A, B$ are nonnegative, then $tr (A B) = tr (B^{1 / 2} A B^{1 / 2}) \geq 0$ by cyclicity of trace and nonnegativity (For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ , A non-negative operator has a unique non-negative square root, A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S).

Proof

1.1givenF1L1algebra

Write a binary POVM as $(E, I - E)$ with $0 \leq E \leq I$ . Its equal-prior success probability is $\frac{1}{2} tr (E ρ) + \frac{1}{2} tr ((I - E) σ) = \frac{1}{2} + \frac{1}{2} tr (E Δ), Δ = ρ - σ,$ using $tr σ = 1$ .

1.2givenL2L3algebra

For any such effect, [L2] gives $tr (E Δ) = tr (E Δ_{+}) - tr (E Δ_{-}) \leq tr Δ_{+}$ : the second trace is nonnegative by [L3], and $tr ((I - E) Δ_{+}) \geq 0$ by [L3] bounds the first by $tr Δ_{+}$ . Thus [L2] implies $tr (E Δ) \leq D (ρ, σ)$ .

2.1step 1.2L2

Let $E_{+}$ be the orthogonal projection onto the positive eigenspaces of $Δ$ . Then $E_{+} Δ = Δ_{+}$ , so $tr (E_{+} Δ) = tr Δ_{+} = D (ρ, σ)$ by [L2]. The binary projective measurement $(E_{+}, I - E_{+})$ therefore attains the upper bound from step 1.2.

3.1step 1.1step 1.2step 2.1algebra∎

Substituting the maximum $tr (E Δ) = D$ into the success formula of step 1.1 gives $\frac{1}{2} + \frac{1}{2} D = \frac{1}{2} (1 + D)$ , as claimed.

Depends on

Used by

Measurement Variation At Most Trace Distance Theorem
Trace Distance Channel Contractivity Theorem

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John Watrous, The Theory of Quantum Information (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 2: Foundations I: States and Ensembles (updated July 2015) (standard reference, not scraped)