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The selective Lueders update by an orthogonal projection

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a nonzero complex Hilbert space (Hilbert space), let $P$ be an orthogonal projection on $H$ (The Hilbert orthogonal projection onto a closed subspace) and let $ρ$ be a density operator with $p : = tr (ρ P) > 0$ (Density operators and scalar spectral probability clause (a)). Define the selective update $ρ_{P} : = \frac{P ρ P}{p} .$ Then:

(a) $P ρ P$ is a positive trace-class operator with $tr (P ρ P) = p$ , so $ρ_{P}$ is a density operator;

(b) $tr (ρ_{P} P) = 1$ and $P ρ_{P} P = ρ_{P} = P ρ_{P} = ρ_{P} P;$ in particular $ρ_{P}$ commutes with $P$ and is fixed by compression with $P$ ;

(d) if $ρ = T_{u, u}$ for a unit vector $u$ with $P u \neq 0$ , where $T_{u, u} x = ⟨ x, u ⟩ u$ (Rank-one operator calculus), then $ρ_{P} = T_{P u, P u} / ∥ P u ∥^{2}$ , the pure state of the normalized vector $P u$ ; the value depends only on the ray of $P u$ , meaning its equivalence class of nonzero vectors under $x \sim y$ if $y = z x$ for some $z \in C ∖ {0}$ .

No physical interpretation of the update is asserted.

Facts & Assumptions

Given: A nonzero complex Hilbert space $H$ ; an orthogonal projection $P$ on $H$ ; a density operator $ρ$ with $p = tr (ρ P) > 0$ ; an orthogonal projection $Q$ with $Q P = 0$ ; a unit vector $u$ with $P u \neq 0$ .

[F1]

Orthogonal projections. An orthogonal projection $P$ on $H$ is a bounded linear operator with $P^{2} = P = P^{*}$ , $∥ P ∥ \leq 1$ and $⟨ P x, y ⟩ = ⟨ x, P y ⟩$ for all $x, y \in H$ (The Hilbert orthogonal projection onto a closed subspace).

[F2]

Density operators. A density operator $ρ$ is self-adjoint, positive and trace class with $tr ρ = 1$ , and it admits an expansion $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ with $(u_{n})$ a finite or countable orthonormal family and weights $w_{n} \geq 0$ , $\sum_{n} w_{n} = 1$ (Density operators and scalar spectral probability clause (a)).

[F3]

Rank-one calculus. $T_{u, v} x = ⟨ x, v ⟩ u$ is trace class with $tr (T_{u, v}) = ⟨ u, v ⟩$ and $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ ; for bounded $B$ one has $B T_{u, v} = T_{B u, v}$ and $T_{u, v} B = T_{u, B^{*} v}$ (Rank-one operator calculus clauses (b)-(d)). For any $x, y \in H$ , $T_{x, x}^{*} = T_{x, x}$ and $⟨ T_{x, x} y, y ⟩ = ∣ ⟨ y, x ⟩ ∣^{2} \geq 0$ , so $T_{x, x}$ is positive under the general Hilbert-space definition (Self-adjoint, positive, unitary and normal operators).

[F4]

Trace-class ideal and cyclicity. If $T$ is trace class and $B$ bounded, then $B T$ and $T B$ are trace class with $tr (B T) = tr (T B)$ and $tr$ linear and trace-norm continuous (Trace class is a two sided Banach operator ideal, Nuclear bounded cyclicity for trace-class operators clause (b)).

[F5]

Positivity of quadratic forms. If $S$ is positive on the Hilbert space $H$ , then $⟨ S x, x ⟩ \geq 0$ for every $x$ ; sums of positive operators are positive, and $ρ = T_{u, u}$ with $∥ u ∥ = 1$ is a density operator (Self-adjoint, positive, unitary and normal operators, [F2], [F3]).

Proof

1.1givenF1F2F4F5algebra

[The unnormalized update] Since $P = P^{*}$ and $ρ = ρ^{*}$ by [F1] and [F2], $(P ρ P)^{*} = P^{*} ρ^{*} P^{*} = P ρ P$ , so $P ρ P$ is self-adjoint. For $x \in H$ , $⟨ P ρ P x, x ⟩ = ⟨ ρ P x, P x ⟩ \geq 0$ by [F1] and [F5], so $P ρ P \geq 0$ . With $P$ bounded and $ρ$ trace class, $ρ P$ and then $P (ρ P) = P ρ P$ are trace class by [F4], and $tr (P ρ P) = tr (ρ P P) = tr (ρ P) = p$ by cyclicity [F4] and $P^{2} = P$ [F1].

2.1step 1.1F2algebra

[Clause (a)] Since $p > 0$ , the operator $ρ_{P} = P ρ P / p$ is self-adjoint and positive with trace $p / p = 1$ , hence a density operator by [F2]; this is (a).

2.2step 1.1F1F4algebra

[Clause (b)] By $P^{2} = P$ and cyclicity [F4], $tr (ρ_{P} P) = tr (P ρ P P) / p = tr (ρ P P P) / p = tr (ρ P) / p = 1$ . Moreover $P ρ_{P} P = P (P ρ P) P / p = P ρ P / p = ρ_{P}$ and $P ρ_{P} = P P ρ P / p = ρ_{P} = ρ_{P} P P = ρ_{P} P$ ; thus $ρ_{P}$ commutes with $P$ and is fixed by compression with $P$ . This is (b).

2.3step 1.1F4algebra

[Clause (c)] If $Q P = 0$ then $Q ρ_{P} Q = Q P ρ P Q / p = 0$ as an operator identity, and $tr (ρ_{P} Q) = tr (P ρ P Q) / p = tr (Q P ρ P) / p = 0$ by cyclicity [F4]. This is (c).

3.1step 1.1F3F4algebra∎

[Clause (d)] Let $ρ = T_{u, u}$ with $∥ u ∥ = 1$ . By [F3], $P ρ P = P T_{u, u} P = T_{P u, u} P = T_{P u, P u}$ , whose trace is $∥ P u ∥^{2} > 0$ , and $p = tr (T_{u, u} P) = tr (P T_{u, u}) = ⟨ P u, u ⟩ = ∥ P u ∥^{2}$ by [F3] and cyclicity [F4]; hence $ρ_{P} = T_{P u, P u} / ∥ P u ∥^{2}$ . For $z \in C ∖ {0}$ the identity $T_{z P u, z P u} = ∣ z ∣^{2} T_{P u, P u}$ from the conjugate-linearity of $(u, v) \mapsto T_{u, v}$ [F3] shows that the state depends only on the ray of $P u$ . This is (d).

The selective Lueders update by an orthogonal projection

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources