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An orthogonal finite ensemble block sum is a density operator

Statement

Let $X$ be a finite nonempty set, identify its labels with the computational orthonormal basis $(ξ_{x})_{x \in X}$ of a register $H_{X}$ (Finite registers and their ordered computational bases), and define the orthogonal rank-one projection $Π_{x} : H_{X} \to H_{X}$ by $Π_{x} z : = ⟨ z, ξ_{x} ⟩ ξ_{x}$ under the root linear-first convention. Let $H_{B}$ be a nonzero finite-dimensional complex inner-product space. For probabilities $p_{x} \geq 0$ with $\sum_{x \in X} p_{x} = 1$ and density operators $ρ_{x}$ on $H_{B}$ , the operator

$ρ_{X B} : = \sum_{x \in X} p_{x} Π_{x} \otimes ρ_{x}$

is a density operator on $H_{X} \otimes H_{B}$ .

Facts & Assumptions

Given: The finite label set $X$ , the register $H_{X}$ , the space $H_{B}$ , the probability vector $(p_{x})$ , and the density operators $(ρ_{x})$ in the Statement.

[F1]

Finite registers and their ordered computational bases: the computational vectors $(ξ_{x})$ form an orthonormal basis of their finite register.

[F2]

Density operator: each $ρ_{x}$ is non-negative with trace one.

[F3]

Ordered tensor registers, their inner product, and wire permutations: the tensor product carries the product inner product and the displayed factor order.

[F4]

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: a non-negative operator is self-adjoint with nonnegative quadratic form.

[F5]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: trace is the sum of diagonal entries in any ordered basis.

Proof

1.1givenF1F2F3F4algebra

Every $v \in H_{X} \otimes H_{B}$ has a unique form $v = \sum_{x \in X} ξ_{x} \otimes v_{x}$ . Orthonormality in [F1] and the product inner product in [F3] give $⟨ ρ_{X B} v, v ⟩ = \sum_{x} p_{x} ⟨ ρ_{x} v_{x}, v_{x} ⟩ \geq 0$ by [F2]. In the same product basis $ρ_{X B}$ is block diagonal with self-adjoint blocks $p_{x} ρ_{x}$ , so it is self-adjoint and non-negative by [F4].

2.1givenF2F5step 1.1algebra∎

By [F5], $tr (ρ_{X B}) = \sum_{x} p_{x} tr (ρ_{x}) = \sum_{x} p_{x} = 1$ . Thus $ρ_{X B}$ is a density operator by [F2].

Depends on

Used by

Finite classical-quantum states and Holevo information Definition

Dependency tree · two levels

30 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, §5.3.3 (standard reference, not scraped)
Mark M. Wilde, From Classical to Quantum Shannon Theory, arXiv:1106.1445v8, §11.1 (standard reference, not scraped)