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Pure preparations are well defined by their rays, and phase factors do not change their states

Physical scope

A model-internal well-definedness statement about the state assignment of the non-relativistic quantum framework: with the state postulate adopted, the pure preparation of a nonzero state vector is independent of the chosen representative of its ray, phase-equivalent unit representatives yield the same pure state, distinct rays yield distinct pure states, and the zero vector is assigned no preparation or state. The item makes no empirical claim, asserts no dynamics or probability law, and does not assert that any particular laboratory procedure prepares a ray.

Statement

Adopt the state and composition postulate P1 of the framework (State and composition postulate P1: Hilbert spaces, preparations and composite systems) and use the state-space vocabulary of The state space of the framework: preparations, pure rays and density operators. Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), which is the choice hypothesis of the rank-one suppliers cited below. Let $H \neq {0}$ be the Hilbert space assigned to a declared system, let $ψ, φ \in H^{\times}$ be nonzero state vectors, and let $[ψ], [φ]$ be their rays. Then:

(a) Representative independence. If $φ = z ψ$ for some nonzero $z \in C$ , then the normalized rank-one operators agree, $\frac{T_{φ, φ}}{∥ φ ∥^{2}} = \frac{T_{ψ, ψ}}{∥ ψ ∥^{2}} = T_{ν (ψ), ν (ψ)},$ and this operator is a pure state. In particular the density operator associated with a pure preparation depends only on the ray $[ψ]$ and not on the representative used to name it.

(b) Phase invariance and injectivity. If $u, w \in S (H)$ are unit representatives of the same ray, then $w = ζ u$ for a unique $ζ \in U (1)$ and $T_{w, w} = T_{u, u}$ ; conversely, if two pure preparations $[ψ] \neq [φ]$ are distinct, then their assigned pure states are distinct. Hence the assignment $[ψ] \mapsto T_{ν (ψ), ν (ψ)}$ is a bijection from the set of pure preparations onto the set of pure states.

(c) Exclusion of the zero vector. The zero vector is not a pure preparation of the model: it has no unit representative, the normalization map is undefined at it, the operator $T_{0, 0} = 0$ has trace $0$ and is not a density operator, and the state postulate assigns pure preparations only to unit rays of vectors of norm one. Consequently no pure preparation and no state is assigned to $0$ .

Facts & Assumptions

Given: The adopted state postulate P1 with its Hilbert-space assignment $H$ to the declared system; nonzero state vectors $ψ, φ$ and unit representatives $u, w$ ; the ray $[ψ]$ and normalization $ν$ of State vectors, unit vectors, rays, normalization and phases of a Hilbert space.

[F1]

The state postulate (State and composition postulate P1: Hilbert spaces, preparations and composite systems clauses (i)-(iii)): the model assigns to the system a nonzero separable complex Hilbert space $H$ ; a pure preparation is a unit ray $[ψ] = {z ψ : z \neq 0}$ with $∥ ψ ∥ = 1$ , two nonzero vectors describing the same pure preparation exactly when they differ by a nonzero complex factor; a general preparation is a density operator; the zero vector is not a state and no preparation is assigned to it; and the rank-one operator $T_{ψ, ψ}$ is the density operator associated with the ray of the unit vector $ψ$ .

[F2]

State-space vocabulary (The state space of the framework: preparations, pure rays and density operators, The state space: density operators, pure and mixed states, and purity): the state assigned to a nonzero vector is the normalized rank-one operator $T_{ψ, ψ} / ∥ ψ ∥^{2} = T_{ν (ψ), ν (ψ)}$ , the pure states are the operators $T_{u, u}$ with $u \in S (H)$ , and $D (H)$ consists of the density operators of trace one.

[F3]

Rank-one scaling and equality (Rays, unit vectors and rank-one Hilbert projections, Rank-one operator calculus clauses (a)-(c)): $T_{z ψ, z ψ} = ∣ z ∣^{2} T_{ψ, ψ}$ for $z \neq 0$ ; $T_{ψ, ψ} = ∥ ψ ∥^{2} T_{ν (ψ), ν (ψ)}$ ; $T_{ψ, ψ} = T_{φ, φ}$ holds exactly when $∥ ψ ∥ = ∥ φ ∥$ and $[ψ] = [φ]$ ; and for unit $u, w$ , $T_{u, u} = T_{w, w}$ holds exactly when $w = ζ u$ for some $ζ \in U (1)$ .

[F4]

Normalization and phases (Normalization, ray equivalence and the phase orbits of the unit sphere): $ν (ψ) \in [ψ] \cap S (H)$ , $[ψ] = [φ]$ holds exactly when $ν (φ) = ζ ν (ψ)$ for a unique $ζ \in U (1)$ , and for unit $u, w$ one has $[u] = [w]$ exactly when $w = ζ u$ with unique $ζ \in U (1)$ .

[F5]

Norms (The induced length is a norm): $∥ z ψ ∥ = ∣ z ∣ ∥ ψ ∥$ , $∥ ψ ∥ = 0$ exactly for $ψ = 0$ , and $∥ ψ ∥ \geq 0$ . In particular $∥ 0 ∥ = 0$ , so the scalar $∥ ψ ∥^{- 1}$ used in the normalization is defined only for nonzero $ψ$ .

[F6]

Rank-one trace, including zero (Rank-one operator calculus clauses (b)-(c)): for all $x, y \in H$ , including zero, $T_{x, y}$ is trace class and $tr (T_{x, y}) = ⟨ x, y ⟩$ ; its defining formula gives $T_{0, 0} = 0$ .

Proof

1.1givenF1F2F3F5algebra

[(a) representative independence] Let $φ = z ψ$ with $z \neq 0$ . By [F3] one has $T_{φ, φ} = T_{z ψ, z ψ} = ∣ z ∣^{2} T_{ψ, ψ}$ , and by [F5] the squared norms satisfy $∥ φ ∥^{2} = ∣ z ∣^{2} ∥ ψ ∥^{2}$ with $∣ z ∣^{2} \neq 0$ ; dividing the first identity by the second gives $T_{φ, φ} / ∥ φ ∥^{2} = T_{ψ, ψ} / ∥ ψ ∥^{2}$ , and the second identity of [F3] identifies the common value as $T_{ν (ψ), ν (ψ)}$ , which is a pure state by [F2]. Thus the operator depends only on the ray.

1.2givenF1F3F4algebra

[(b) phase representatives give the same state] Let $u, w \in S (H)$ lie in the same ray. By [F4] one has $w = ζ u$ for a unique $ζ \in U (1)$ ; then $T_{w, w} = T_{ζ u, ζ u} = ∣ ζ ∣^{2} T_{u, u} = T_{u, u}$ by [F3]. Hence multiplying a unit representative by a phase factor does not change the assigned pure state, in agreement with clause (ii) of [F1], which identifies the pure preparation only up to a nonzero complex factor.

2.1givenstep 1.2F2F3F4algebra

[(b) injectivity of the assignment] Suppose two pure preparations $[ψ], [φ]$ have the same assigned pure state, that is $T_{ν (ψ), ν (ψ)} = T_{ν (φ), ν (φ)}$ . Since $ν (ψ), ν (φ) \in S (H)$ , [F3] gives $ν (φ) = ζ ν (ψ)$ for some $ζ \in U (1)$ , and [F4] then gives $[φ] = [ψ]$ . Hence the assignment is injective, and it is surjective onto the set of pure states by [F2]; so it is a bijection from the pure preparations onto the pure states.

3.1givenstep 1.1F1F2F5F6algebra∎

[(c) the zero vector] By [F5], $∥ 0 ∥ = 0$ , so the normalization map $ν (ψ) = ∥ ψ ∥^{- 1} ψ$ is undefined at $ψ = 0$ because the scalar $∥ ψ ∥^{- 1}$ does not exist; the ray of $0$ would be ${z \cdot 0 : z \neq 0} = {0}$ , which contains no unit vector, so clause (ii) of [F1] assigns no pure preparation to it. Moreover $T_{0, 0} = 0$ has $tr (T_{0, 0}) = ⟨ 0, 0 ⟩ = 0$ by [F6], so it is not a density operator of trace one and hence not an element of $D (H)$ by [F2]; clause (iii) of [F1] likewise assigns no general preparation to $0$ . This proves (c) and completes (a)-(c).

Pure preparations are well defined by their rays, and phase factors do not change their states

Statement

Facts & Assumptions

Proof

Depends on

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