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The zero vector determines no pure preparation and no state

Physical scope

A refutation, inside the framework's state semantics, of the claim that every vector of the assigned Hilbert space determines a pure preparation with the zero vector corresponding to the zero operator: the zero vector has no unit representative, the normalization map is undefined at it, and the zero operator has trace zero and is not a density operator, so under the state postulate no preparation and no state is assigned to it. The item refutes that specific informal claim; it asserts no empirical result and no claim about nature.

Statement refuted

Adopt the state postulate of the framework, and let $H \neq {0}$ be the Hilbert space assigned to a declared system. Then: every vector of $H$ determines a pure preparation of the system; the zero vector determines the zero operator $0$ , which is the state of the system when no preparation is made, and the normalization convention $0 / ∥ 0 ∥ = 0$ supplies the unit representative of its ray.

Facts & Assumptions

Given: The adopted state postulate P1 with its assigned Hilbert space $H \neq {0}$ ; the ray, normalization and unit-sphere notions of State vectors, unit vectors, rays, normalization and phases of a Hilbert space; the rank-one operators $T_{u, v} x = ⟨ x, v ⟩ u$ , whose calculus assumes the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )); the zero vector $0 \in H$ .

[L1]

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

[L2]

Rays and normalization (State vectors, unit vectors, rays, normalization and phases of a Hilbert space, The induced length is a norm): rays are defined only for nonzero vectors, $S (H) = {ψ : ∥ ψ ∥ = 1}$ , $ν (ψ) = ∥ ψ ∥^{- 1} ψ$ requires the nonzero scalar $∥ ψ ∥^{- 1}$ , and $∥ 0 ∥ = 0$ with $∥ z v ∥ = ∣ z ∣ ∥ v ∥$ .

[L3]

Rank-one operators at zero (Rank-one operator calculus, Rays, unit vectors and rank-one Hilbert projections): $T_{0, v} x = ⟨ x, v ⟩ 0 = 0$ for every $x$ , so $T_{0, 0}$ is the zero operator, and $tr (T_{0, 0}) = ⟨ 0, 0 ⟩ = 0$ ; a state has trace one by [L1].

[L4]

The scalar field (Field, The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ): $C$ is a field containing $R$ ; in $C$ only nonzero elements have multiplicative inverses; the expression $0^{- 1}$ and the quotient $0 / 0$ are not defined, and the formula for $ν$ is a scalar multiplication by the inverse of the norm.

Counterexample

technique · direct; examine the zero vector against each clause of the refuted statement

1.1givenL1L2algebra

[No unit representative] By [L2], $∥ 0 ∥ = 0$ and $∥ z \cdot 0 ∥ = ∣ z ∣ ∥ 0 ∥ = 0$ for every scalar $z$ , so no scalar multiple of $0$ has norm $1$ ; hence the set ${z \cdot 0 : z \neq 0} = {0}$ contains no unit vector, and it is not a ray in the sense of [L2], which is defined only for nonzero generating vectors.

2.1givenstep 1.1L2L4algebra

[Normalization is undefined] The normalization formula $ν (ψ) = ∥ ψ ∥^{- 1} ψ$ uses the inverse of the scalar $∥ ψ ∥$ ; at $ψ = 0$ this inverse would be $0^{- 1}$ , which does not exist in the field $C$ of [L4]: for every scalar $a$ , distributivity gives $a 0 = a (0 + 0) = a 0 + a 0$ , hence $a 0 = 0 \neq 1$ , so $a$ cannot be an inverse of $0$ . Thus the displayed "convention" $0 / ∥ 0 ∥ = 0$ is therefore not an instance of the definition but a separate stipulation. Even granting the stipulation its value, $∥ 0 ∥ = 0 \neq 1$ by [L2], so it does not produce a unit representative; the stipulation is self-defeating as a normalization rule.

2.2givenstep 1.1L1L3algebra

[The zero operator is not a state] By [L3], $T_{0, 0}$ is the zero operator with $tr (T_{0, 0}) = 0$ ; a density operator has trace one by [L1], so $0 \notin D (H)$ and the zero operator is not a state of the model. Consequently the zero vector does not determine a state through the rank-one operator either.

3.1givenstep 1.1step 2.1step 2.2L1L2algebra∎

[Failure at every step] The refuted statement fails on the zero vector in all of its assertions: there is no unit representative of $0$ by step 1.1, the normalization is undefined rather than $0$ by step 2.1, and the zero operator is not a density operator by step 2.2; and clause (ii) of the state postulate in [L1] expressly assigns no preparation to the zero vector. Hence the claim that every vector of the assigned Hilbert space determines a pure preparation, with the zero vector as the state of an empty preparation, is refuted.

Remarks

The refutation is a statement about the model's state semantics: the zero vector and the zero operator are excluded by the definitions and by the state postulate. It is not an empirical claim, and it does not say that the framework's assignment of states is complete or exact.
The exclusion of the zero vector is one reason the unit ray, rather than the raw vector, is the carrier of a pure preparation: rescaling a nonzero vector by any nonzero complex factor keeps its ray, whereas no rescaling can make the zero vector a unit vector.

The zero vector determines no pure preparation and no state

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

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