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Qm State Geometry — Examples

1 · Prerequisites

2 · Summary

The companion page to qm-state-geometry records finite worked checks of the state-space objects and one refutation. The normalization and phase check A two-dimensional normalization and phase computation computes norms, normalized representatives and normalized rank-one operators on the explicit vectors $(1, i)$ and $(1 + i) (1, i)$ , verifying that phase-equivalent or rescaled representatives of one ray give the same pure state. The state space of the model's two-level finite-spin system is computed in The state space of a two-level finite-spin system is the Bloch ball: the states are exactly the operators $\frac{1}{2} (I + r \cdot σ)$ with $∣ r ∣ \leq 1$ , the pure states are the boundary points $∣ r ∣ = 1$ , and each is assigned to the corresponding spin-up ray, defined there as the ray of a unit eigenvector of $r \cdot σ$ with eigenvalue $+ 1$ . Each boundary state has one-dimensional range. Pure and mixed states of a fixed two-level internal space performs the corresponding computation for an abstract two-level internal space with a pair of polarization labels, whose photon reading is deliberately deferred to the radiation assumptions of the category.

Two comparisons illustrate the state-versus-ensemble structure of the framework. A coherent two-level preparation and the equal statistical mixture are different states shows that the relative-phase superposition $(h + i v) / \sqrt{2}$ and the equal statistical mixture of the rays $[h]$ and $[v]$ are assigned different states — one pure with one-dimensional range and one mixed with two-dimensional range. Relative to $(h, v)$ , their off-diagonal matrix entries are $(- i / 2, i / 2)$ for the superposition and $(0, 0)$ for the mixture; this is the state-level distinction defined in the linked example. One two-level state from two different preparation ensembles exhibits one mixed two-level state that arises both as an orthonormal ensemble with unequal weights and as the uniform mixture of two non-orthogonal pure states, so the state does not determine the ensemble. The counterexample The zero vector determines no pure preparation and no state refutes the claim that every vector determines a pure preparation with the zero vector as an empty preparation: the zero vector has no unit representative, the normalization map is undefined at it, and the zero operator is not a density operator. A closing audit (What the worked checks of the companion page establish) records what each check establishes and that none of them is an experiment, a measurement prediction or a comparison with data.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passOpen item page →

A two-dimensional normalization and phase computation

Example

Let $H = C^{2}$ carry the standard inner product $⟨ z, w ⟩ = z_{1} \overline{w_{1}} + z_{2} \overline{w_{2}}$ (Real and complex inner product spaces, with the inner product linear in the first argument, Hilbert space; completeness follows from Every finite-dimensional normed space is Banach), and let $e_{1} = (1, 0)$ , $e_{2} = (0, 1)$ . Take $ψ = (1, i), z = 1 + i, φ : = z ψ,$ and verify on this explicit data the normalization and phase statements of Normalization, ray equivalence and the phase orbits of the unit sphere: the norm of $φ$ is $∣ z ∣$ times the norm of $ψ$ , the normalized representative of $φ$ is the phase multiple $(z / ∣ z ∣)$ of the normalized representative of $ψ$ , the rays of $ψ$ and $φ$ coincide, and the normalized rank-one operators of $ψ$ , $φ$ and $i ψ$ are the same pure state The state space: density operators, pure and mixed states, and purity.

Facts & Assumptions

Given: The Hilbert space $H = C^{2}$ with the standard inner product and basis $(e_{1}, e_{2})$ ; the vectors $ψ = (1, i)$ , $φ = z ψ$ with $z = 1 + i$ , and $i ψ$ ; the ray, normalization and phase notions of State vectors, unit vectors, rays, normalization and phases of a Hilbert space; and the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the choice hypothesis of the cited rank-one and state-space suppliers.

[L1]

Standard computations in $C$ : $i^{2} = - 1$ , $(1 + i)^{2} = 2 i$ , $(1 + i) (1 + i) - 1$ expansions below are termwise multiplications with $\overline{1 + i} = 1 - i$ , $∣ 1 + i ∣^{2} = 2$ , $∣ i ∣ = 1$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L2]

Norm and pairing: $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ with $⟨ z, w ⟩ = z_{1} \overline{w_{1}} + z_{2} \overline{w_{2}}$ , so $∥ (a, b) ∥^{2} = ∣ a ∣^{2} + ∣ b ∣^{2}$ , and $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ (The induced length is a norm, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L3]

Normalization and phases (Normalization, ray equivalence and the phase orbits of the unit sphere): $ν (v) = ∥ v ∥^{- 1} v$ , $[v] = {w : w = t v, t \neq 0}$ , and $ν (t v) = (t / ∣ t ∣) ν (v)$ for $t \neq 0$ ; $[v] = [w]$ if and only if $ν (w) = ζ ν (v)$ for some $ζ \in U (1)$ .

[L4]

Rank-one operators (Rays, unit vectors and rank-one Hilbert projections, The state space: density operators, pure and mixed states, and purity): $T_{v, v} x = ⟨ x, v ⟩ v$ , $T_{t v, t v} = ∣ t ∣^{2} T_{v, v}$ , $T_{v, v} / ∥ v ∥^{2} = T_{ν (v), ν (v)}$ is a pure state, and for unit $u, w$ one has $T_{u, u} = T_{w, w}$ exactly when $w = ζ u$ with $∣ ζ ∣ = 1$ .

Verification

technique · direct; compute the displayed vectors, their normalizations and the associated rank-one operators

1.1givenL1L2algebra

Compute $φ = z ψ = (1 + i) (1, i) = ((1 + i) \cdot 1, (1 + i) \cdot i) = (1 + i, i - 1)$ . By [L2] and [L1], $∥ ψ ∥^{2} = ∣ 1 ∣^{2} + ∣ i ∣^{2} = 2$ and $∥ φ ∥^{2} = ∣ 1 + i ∣^{2} + ∣ i - 1 ∣^{2} = 2 + 2 = 4$ , so $∥ ψ ∥ = \sqrt{2}$ , $∥ φ ∥ = 2$ , and indeed $∥ φ ∥ = ∣ z ∣ ∥ ψ ∥ = \sqrt{2} \cdot \sqrt{2} = 2$ , verifying the homogeneity used in the normalization lemma.

2.1givenstep 1.1L1L2L3algebra

The normalizations are $ν (ψ) = (1 / \sqrt{2}, i / \sqrt{2})$ and $ν (φ) = ((1 + i) / 2, (i - 1) / 2)$ . With $ζ : = (1 + i) / \sqrt{2}$ one has $∣ ζ ∣^{2} = ∣ 1 + i ∣^{2} / 2 = 1$ , so $ζ \in U (1)$ , and $ζ ν (ψ) = \frac{1 + i}{\sqrt{2}} (\frac{1}{\sqrt{2}}, \frac{i}{\sqrt{2}}) = (\frac{1 + i}{2}, \frac{i (1 + i)}{2}) = (\frac{1 + i}{2}, \frac{i - 1}{2}) = ν (φ),$ which is the identity $ν (z ψ) = (z / ∣ z ∣) ν (ψ)$ ; by [L3] this also identifies $[φ] = [ψ]$ , so $ψ$ , $φ$ and $i ψ$ lie in one ray.

3.1givenstep 2.1L1L4algebra

The rank-one operator of the unit vector $u = ν (ψ) = (1 / \sqrt{2}, i / \sqrt{2})$ has matrix in the basis $(e_{1}, e_{2})$ with columns $T_{u, u} e_{1} = ⟨ e_{1}, u ⟩ u = \overline{u_{1}} u = \frac{1}{\sqrt{2}} u = (\frac{1}{2}, \frac{i}{2})$ and $T_{u, u} e_{2} = ⟨ e_{2}, u ⟩ u = \overline{u_{2}} u = \frac{- i}{\sqrt{2}} u = (- \frac{i}{2}, \frac{1}{2})$ , that is $T_{u, u} = \frac{1}{2} (\begin{array}{cc} 1 & - i \\ i & 1 \end{array}) .$ This matrix is self-adjoint, has trace $1$ , and satisfies $T_{u, u}^{2} = T_{u, u}$ because ${(\begin{array}{cc} 1 & - i \\ i & 1 \end{array})}^{2} = 2 (\begin{array}{cc} 1 & - i \\ i & 1 \end{array})$ by [L1]; hence $T_{u, u}$ is a pure state by [L4].

4.1givenstep 1.1step 3.1L1L2L4algebra∎

For the second representative, $T_{φ, φ} = ∣ z ∣^{2} T_{ψ, ψ} = 2 T_{ψ, ψ}$ by [L4], and dividing by $∥ φ ∥^{2} = 4$ from step 1.1 gives $T_{φ, φ} / ∥ φ ∥^{2} = T_{ψ, ψ} / 2 = T_{ψ, ψ} / ∥ ψ ∥^{2} = T_{u, u}$ ; likewise $T_{i ψ, i ψ} = ∣ i ∣^{2} T_{ψ, ψ} = T_{ψ, ψ}$ , so the phase multiple $i ψ$ determines the same pure state. This verifies on explicit data the representative and phase independence of the normalized rank-one operator.

Remarks

The computation checks the definitions and the normalization lemma on a fixed two-dimensional example. It asserts nothing about any physical system, preparation or apparatus, and it records no measured value.
The phase multiple $i ψ$ and the rescaled representative $φ$ give the same pure state, while the relative phase between the two components of a vector is not removed by any of these operations; the effect of relative phase belongs to the framework's probability law on the observable pages and is not examined here.

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The state space of a two-level finite-spin system is the Bloch ball

Physical scope

A worked computation in the model's two-level finite-spin state space: for the spin-one-half assignment of the framework, the density operators on the two-dimensional internal space are exactly the operators (I + r dot sigma)/2 with |r| <= 1, and the pure states are exactly the boundary points |r| = 1. Rank means the number of vectors in a finite basis of the operator range; each boundary state has one-dimensional range and is the state of the spin-up ray, the ray of a unit eigenvector of r dot sigma with eigenvalue +1. The Bloch ball is a parametrization of the model's states; the item asserts no measurement outcome probability, no experimental result and no interpretation of r beyond the displayed formula.

Example

Consider the model's two-level finite-spin assignment: the internal space $H = C^{2}$ of a spin-one-half system (Identical-particle and spin assignments P6: permutation sectors and finite-spin representations), with the standard inner product and the Pauli matrices $σ_{x} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}), σ_{y} = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array}), σ_{z} = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ of Finite spin matrices clause (e). For $r = (r_{1}, r_{2}, r_{3}) \in R^{3}$ write $r \cdot σ = r_{1} σ_{x} + r_{2} σ_{y} + r_{3} σ_{z}$ and $∣ r ∣^{2} = r_{1}^{2} + r_{2}^{2} + r_{3}^{2}$ . Then, using the state space $D (H)$ of The state space of the framework: preparations, pure rays and density operators:

For an operator $A : H \to H$ , write $ran A : = {A x : x \in H}$ and define its rank by $rank A : = \dim_{C} (ran A)$ , using the finite-dimensional definition of dimension in Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis. When $∣ r ∣ = 1$ , the ray of a unit eigenvector of $r \cdot σ$ with eigenvalue $+ 1$ is called the spin-up ray in direction $r$ in this calculation.

(i) A linear operator $ρ$ on $H$ is a state if and only if there is $r \in R^{3}$ with $ρ = \frac{1}{2} (I + r \cdot σ) and ∣ r ∣ \leq 1;$ that is, the state space is the unit ball of $R^{3}$ in this parametrization.

(ii) Such a state is pure if and only if $∣ r ∣ = 1$ ; and if $∣ r ∣ = 1$ and $ψ$ is a unit eigenvector of $r \cdot σ$ with eigenvalue $1$ , then $ρ = T_{ψ, ψ}$ .

(iii) The value $r = 0$ gives the mixed state $\frac{1}{2} I$ of purity $\frac{1}{2}$ , and $r = e_{3} = (0, 0, 1)$ gives the pure state $\frac{1}{2} (I + σ_{z}) = T_{e_{1}, e_{1}}$ with $e_{1} = (1, 0)$ .

Facts & Assumptions

Given: The two-dimensional complex Hilbert space $H = C^{2}$ with its standard inner product and basis $(e_{1}, e_{2})$ ; the Pauli matrices $σ_{x}, σ_{y}, σ_{z}$ displayed above; the state space $D (H)$ of The state space of the framework: preparations, pure rays and density operators.

[L1]

Pauli algebra (Finite spin matrices clause (e)): $σ_{i}^{2} = I$ for $i = x, y, z$ , $σ_{i} σ_{j} + σ_{j} σ_{i} = 2 δ_{i j} I$ , and the displayed matrices satisfy $σ_{i}^{*} = σ_{i}$ and $tr σ_{i} = 0$ for $i = x, y, z$ ; direct calculation gives $tr I = 2$ .

[L2]

States and purity (The state space of the framework: preparations, pure rays and density operators, Purity, spectral weights and the criterion for a pure state): a state is a self-adjoint positive trace-class operator of trace one; a state is pure exactly when its purity $tr (ρ^{2})$ equals $1$ and exactly when it equals $T_{ψ, ψ}$ for a unit vector $ψ$ ; a mixed state has at least two positive eigenvalues.

[L3]

Rank-one states (Rays, unit vectors and rank-one Hilbert projections): for a unit vector $u$ , $T_{u, u} u = u$ , $T_{u, u}$ vanishes on the orthogonal complement of $u$ , and $T_{u, u}$ is a pure state.

[L4]

Finite-dimensional linear algebra (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , Real and complex inner product spaces, with the inner product linear in the first argument, Hilbert space, Every finite-dimensional normed space is Banach, Eigenspaces of a self adjoint operator are orthogonal, Trace class operator): matrices act on $C^{2}$ by matrix--vector multiplication; a linear map is determined by its values on a basis; $C^{2}$ with the standard inner product is a Hilbert space; and eigenspaces of a self-adjoint operator belonging to distinct eigenvalues are orthogonal. Every endomorphism of this two-dimensional space has finite rank and is trace class by the finite-rank consequence in Trace class operator.

[L5]

Trace, quadratic forms and the choice hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )): the cited state and purity suppliers assume the Axiom of Countable Choice, which is therefore carried as a hypothesis of this computation; for the displayed $2 \times 2$ matrices, traces and quadratic forms are computed by direct matrix algebra. Here an orthogonal projection means a self-adjoint idempotent $P$ ; then $P (I - P) = 0$ and $P^{*} = P$ , so $P x$ is orthogonal to $(I - P) x$ and $⟨ (λ_{1} P + λ_{2} (I - P)) x, x ⟩ = λ_{1} ∥ P x ∥^{2} + λ_{2} ∥ (I - P) x ∥^{2}$ for real $λ_{1}, λ_{2}$ . A complex matrix $A$ is positive exactly when its quadratic form $⟨ A x, x ⟩$ is nonnegative for every $x \in C^{2}$ .

Verification

technique · direct; parametrize the Hermitian trace-one matrices, test positivity via an explicit projection decomposition, and compute the purity

1.1givenL1L4algebra

[(i) parametrization] A general Hermitian $2 \times 2$ matrix with trace one has the form $A = (\begin{array}{cc} a & b - i c \\ b + i c & 1 - a \end{array})$ with $a, b, c \in R$ ; expanding the three Pauli matrices shows $A = \frac{1}{2} I + b σ_{x} + c σ_{y} + (a - \frac{1}{2}) σ_{z} = \frac{1}{2} (I + r \cdot σ), r = (2 b, 2 c, 2 a - 1),$ so every Hermitian trace-one matrix is of the displayed form with a unique real vector $r$ , and conversely every $r \in R^{3}$ gives a Hermitian matrix of trace one.

2.1givenstep 1.1L1L4L5algebra

[(i) positivity exactly for $∣ r ∣ \leq 1$ ] If $r = 0$ , then $ρ = \frac{1}{2} I$ and $⟨ ρ x, x ⟩ = \frac{1}{2} ∥ x ∥^{2} \geq 0$ for every $x$ , while $∣ r ∣ = 0 \leq 1$ . Now suppose $r \neq 0$ and set $n = r / ∣ r ∣$ . By [L1], $(n \cdot σ)^{2} = \sum_{i, j} n_{i} n_{j} σ_{i} σ_{j} = \frac{1}{2} \sum_{i, j} n_{i} n_{j} (σ_{i} σ_{j} + σ_{j} σ_{i}) = ∣ n ∣^{2} I = I$ . Put $P : = \frac{1}{2} (I + n \cdot σ)$ ; then $P^{2} = \frac{1}{4} (I + 2 n \cdot σ + (n \cdot σ)^{2}) = \frac{1}{4} (2 I + 2 n \cdot σ) = P$ and $P^{*} = P$ , so $P$ and $I - P$ are orthogonal projections, and $I + r \cdot σ = (1 + ∣ r ∣) P + (1 - ∣ r ∣) (I - P)$ . For every $x$ the quadratic form of $I + r \cdot σ$ is $(1 + ∣ r ∣) ∥ P x ∥^{2} + (1 - ∣ r ∣) ∥ (I - P) x ∥^{2}$ by [L5], which is nonnegative if $∣ r ∣ \leq 1$ . If $∣ r ∣ > 1$ , then $tr P = \frac{1}{2} (tr I + tr (n \cdot σ)) = 1$ , whereas $tr I = 2$ , so $P \neq I$ . If $\ker P = {0}$ , the two columns of the matrix of $P$ would be linearly independent and hence form a basis of $C^{2}$ ; thus $P$ would be invertible. The identity $P^{2} = P$ would then imply $P = I$ , a contradiction. Therefore there is a nonzero $x \in \ker P$ , and for this vector $⟨ ρ x, x ⟩ = \frac{1}{2} (1 - ∣ r ∣) ∥ x ∥^{2} < 0$ . Hence $ρ = \frac{1}{2} (I + r \cdot σ)$ is positive exactly when $∣ r ∣ \leq 1$ ; together with step 1.1 and the finite-rank trace-class fact in [L4], this proves (i).

2.2givenstep 1.1L1L2algebra

[(ii) purity is $∣ r ∣ = 1$ ] For $ρ = \frac{1}{2} (I + r \cdot σ)$ one computes, using $(r \cdot σ)^{2} = ∣ r ∣^{2} I$ from [L1], $ρ^{2} = \frac{1}{4} (I + 2 r \cdot σ + (r \cdot σ)^{2}) = \frac{1}{4} ((1 + ∣ r ∣^{2}) I + 2 r \cdot σ),$ so $tr (ρ^{2}) = \frac{1}{4} (1 + ∣ r ∣^{2}) tr I = \frac{1}{2} (1 + ∣ r ∣^{2})$ because $tr (r \cdot σ) = 0$ by [L1]. By [L2] the state is pure exactly when this number equals $1$ , that is exactly when $∣ r ∣^{2} = 1$ , which is the first part of (ii).

3.1givenstep 2.1step 2.2L1L2L3L4L5algebra

[(ii) every spin-up unit vector gives its rank-one state] Assume $∣ r ∣ = 1$ and put $A : = r \cdot σ$ , so $A^{2} = I$ , $A^{*} = A$ , and $ρ = \frac{1}{2} (I + A)$ is a self-adjoint idempotent, hence an orthogonal projection by [L5]. Since $tr ρ = 1$ , $ρ \neq 0$ ; choose $y$ with $ρ y \neq 0$ and set $ψ_{0} : = ρ y / ∥ ρ y ∥$ . Then $ψ_{0}$ is a unit vector and $ρ ψ_{0} = ψ_{0}$ , so $A ψ_{0} = ψ_{0}$ . This proves that the spin-up ray in direction $r$ exists. Now let $ψ$ be any unit eigenvector of $A$ with eigenvalue $+ 1$ , as in (ii). There is a nonzero vector $χ \in \ker (A + I)$ : if $A + I$ were injective, its two columns in $C^{2}$ would be linearly independent and hence a basis, making $A + I$ invertible; but $(A - I) (A + I) = A^{2} - I = 0$ would then imply $A = I$ , contradicting $tr A = 0 \neq 2 = tr I$ by [L1]. Normalize $χ$ to have norm one. The vectors $ψ$ and $χ$ belong to distinct eigenspaces of the self-adjoint operator $A$ , so they are orthogonal by [L4]; as two nonzero orthogonal vectors in $C^{2}$ , they form an orthonormal basis. On this basis, $ρ$ and $T_{ψ, ψ}$ both act as the identity on $ψ$ and as zero on $χ$ by [L3]. Since a linear map is determined by its values on a basis by [L4], $ρ = T_{ψ, ψ}$ . Thus every boundary point is the pure state of every unit vector in its spin-up ray, and with step 2.2 every state with $∣ r ∣ < 1$ is mixed.

4.1givenstep 2.2step 3.1L1L2L3algebra∎

[(iii) the two displayed values] For $r = 0$ the operator $ρ = \frac{1}{2} I$ is a state by (i) with purity $\frac{1}{2} (1 + 0) = \frac{1}{2}$ by step 2.2, and it is mixed because $\frac{1}{2} \neq 1$ by [L2]; for $r = e_{3}$ one has $∣ r ∣ = 1$ and $\frac{1}{2} (I + σ_{z}) = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}) = T_{e_{1}, e_{1}}$ by the definition of $T_{u, u}$ , so it is a pure state by [L3]. This verifies (iii) and completes the computation.

Remarks

The Bloch ball is a parametrization of the finite-spin state space of the model, computed on the explicit two-dimensional data of the spin-one-half assignment. The vector $r$ here is a coordinate of the state, not an empirical measurement result, and no outcome probability is computed in this item.
The same computation applies verbatim to any two-level internal space carrying the same operator algebra; the identification of such a two-level space with a specific physical system is made by the model assumption that assigns the space, and no further interpretation is supplied here.

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Pure and mixed states of a fixed two-level internal space

Physical scope

A worked computation in a fixed two-dimensional internal space carrying an orthonormal pair of 'polarization' labels: the state of the diagonal vector is the pure state with matrix (1/2)[[1,1],[1,1]], and the equal statistical mixture of the two basis labels is the mixed state (1/2)I. The two-level space is treated abstractly; the identification of its vectors with photon polarization states, and any statement about photon number, energy or momentum, is deferred to the radiation assumptions of the category and is not made in this item. The item asserts no measurement probability and no empirical result.

Example

Let $H$ be a two-dimensional complex Hilbert space with an orthonormal basis written $(h, v)$ , and use the state space $D (H)$ of The state space of the framework: preparations, pure rays and density operators. Put $u : = \frac{h + v}{\sqrt{2}} \in S (H), ρ_{+} : = \frac{1}{2} T_{h, h} + \frac{1}{2} T_{v, v},$ Here $u$ is the unit vector of the equal-amplitude superposition, with ray $[u]$ , and $ρ_{+}$ is the density operator of the equal statistical mixture of the two basis rays. For an operator $A : H \to H$ , write $ran A : = {A x : x \in H}$ and define its rank by $rank A : = \dim_{C} (ran A)$ , using Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis. Then:

(i) The state assigned to the ray $[u]$ is the pure state $T_{u, u}$ , whose matrix relative to the basis $(h, v)$ is $T_{u, u} = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array});$ its purity is $1$ and its rank is one.

(ii) The mixture is the state $ρ_{+} = \frac{1}{2} I$ , whose matrix relative to $(h, v)$ is diagonal with entries $\frac{1}{2}$ ; its purity is $\frac{1}{2}$ , it is mixed, and it is not a pure state.

(iii) The two states are distinct: $T_{u, u} \neq \frac{1}{2} I$ , and they differ in rank and in purity.

The basis labels $h$ and $v$ are abstract here. The identification of such a two-level space with the polarization states of a fixed propagation mode, and every statement about photon number, energy, momentum or detection, is deferred to the radiation assumptions adopted on the radiation pages of the category; no such identification or statement is made in this item.

Facts & Assumptions

Given: A two-dimensional complex Hilbert space $H$ with orthonormal basis $(h, v)$ and the standard inner product; the unit vector $u = (h + v) / \sqrt{2}$ ; the rank-one operators $T_{x, y} z = ⟨ z, y ⟩ x$ ; the state space $D (H)$ of The state space of the framework: preparations, pure rays and density operators; and the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the choice hypothesis of the cited rank-one and state-space suppliers.

[L1]

Orthonormality and coordinates (Real and complex inner product spaces, with the inner product linear in the first argument, Hilbert space): $⟨ h, h ⟩ = ⟨ v, v ⟩ = 1$ and $⟨ h, v ⟩ = ⟨ v, h ⟩ = 0$ , so $∥ h ∥ = ∥ v ∥ = 1$ and every vector is a unique linear combination of $h$ and $v$ ; in particular $∥ u ∥^{2} = \frac{1}{2} ∥ h + v ∥^{2} = 1$ .

[L2]

Rank-one states (Rays, unit vectors and rank-one Hilbert projections): $T_{x, x} x = ∥ x ∥^{2} x$ ; for a unit vector $x$ the operator $T_{x, x}$ is a pure state and the Hilbert orthogonal projection onto the one-dimensional subspace $C x$ ; its matrix relative to a basis has columns $T_{x, x} e_{j} = ⟨ e_{j}, x ⟩ x$ .

[L3]

Mixtures, purity and mixedness (The state space is convex, closed under countable mixtures, and complete in the trace norm, Purity, spectral weights and the criterion for a pure state, The state space: density operators, pure and mixed states, and purity): a finite convex combination of states with weights summing to one is a state; the purity of a state with positive eigenvalues $λ_{j}$ , counted with multiplicity, is $tr (ρ^{2}) = \sum_{j} λ_{j}^{2}$ ; a state is pure exactly when this equals $1$ and mixed exactly when it has at least two positive eigenvalues.

[L4]

Matrix multiplication and identity (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes): for the displayed $2 \times 2$ matrices, $(A B)_{i j} = \sum_{k} A_{i k} B_{k j}$ , and the identity matrix $I$ has diagonal entries one and off-diagonal entries zero.

Verification

technique · direct; compute the two states in the basis $(h,v)$ and compare their purities and ranks

1.1givenL1L2L3L4algebra

[(i) the pure state of the diagonal ray] For the unit vector $u = (h + v) / \sqrt{2}$ of [L1] one has $⟨ h, u ⟩ = 1 / \sqrt{2}$ and $⟨ v, u ⟩ = 1 / \sqrt{2}$ , so by [L2] the columns of $T_{u, u}$ in the basis $(h, v)$ are $T_{u, u} h = \frac{1}{\sqrt{2}} u = \frac{1}{2} (h + v)$ and $T_{u, u} v = \frac{1}{\sqrt{2}} u = \frac{1}{2} (h + v)$ , that is, the displayed matrix $\frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ . Its square is $\frac{1}{4} (\begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array}) = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}) = T_{u, u}$ by [L4]; its range is the one-dimensional subspace $C u$ , so its rank is one, and its purity is $1$ . By [L3] it is a pure state.

1.2givenL1L2L3L4algebra

[(ii) the mixture is one half times the identity] By [L2] the matrices of the two pure states are $T_{h, h} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $T_{v, v} = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ , so $ρ_{+} = \frac{1}{2} T_{h, h} + \frac{1}{2} T_{v, v} = \frac{1}{2} I$ ; this is a state by [L3], since the weights are nonnegative and sum to one. Its range is $H$ , so its rank is two. Its square is $\frac{1}{4} I$ by [L4]. Since $ρ_{+} h = h / 2$ and $ρ_{+} v = v / 2$ and $(h, v)$ is a basis, its two eigenvalues counted with multiplicity are $1 / 2, 1 / 2$ , so its purity is $(1 / 2)^{2} + (1 / 2)^{2} = 1 / 2$ by [L3]; since $\frac{1}{2} \neq 1$ , the state is mixed by [L3].

2.1givenstep 1.1step 1.2L4algebra∎

[(iii) the states are distinct] By steps 1.1 and 1.2 the difference $T_{u, u} - \frac{1}{2} I = \frac{1}{2} (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ is nonzero, so the two states are distinct elements of $D (H)$ ; their purities $1$ and $\frac{1}{2}$ and their ranks $1$ and $2$ also differ, so the pure superposition and the equal mixture are different states of the model.

Remarks

The computation is a state-level comparison only. Whether the two states differ in the outcome probabilities of some measurement, and with which apparatus, is a question for the framework's probability law and measurement assumptions, which are not used here.
The labels $h$ and $v$ are carried only to fix an orthonormal basis. The physical reading of such a two-level space as photon polarization is an assumption of the radiation pages, not a consequence of this example.

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A coherent two-level preparation and the equal statistical mixture are different states

Physical scope

A worked comparison, inside the framework's state space, of two preparations of a two-level system: the pure preparation of the relative-phase superposition (h + i v)/sqrt2 and the equal statistical mixture of the h and v rays. The two assigned states are computed explicitly and shown to be distinct, with different rank and purity: the superposition is pure and the mixture is mixed. Here rank means the number of vectors in a finite basis of the operator's range. In this item, coherent preparation denotes only the pure ray of (h + i v)/sqrt2, whose density matrix in the ordered basis (h,v) has off-diagonal entries (-i/2, i/2); the equal mixture has zero off-diagonal entries. These are descriptions of assigned states only. The item asserts no outcome probability, no measurement model, no apparatus realization and no empirical result; whether any measurement distinguishes the states belongs to the framework's probability law and is not decided here.

Example

Let $H$ be a two-dimensional complex Hilbert space with orthonormal basis $(h, v)$ , and use the state space $D (H)$ and the preparation semantics of The state space of the framework: preparations, pure rays and density operators. Consider the two preparations $superposition: w = \frac{h + i v}{\sqrt{2}}, equal mixture: ρ_{+} = \frac{1}{2} T_{h, h} + \frac{1}{2} T_{v, v},$ where $w$ is a unit vector and $ρ_{+}$ is the statistical mixture of the two basis rays with weights $\frac{1}{2}, \frac{1}{2}$ . The assigned state matrices, in the given order, are: $T_{w, w} = \frac{1}{2} (\begin{array}{cc} 1 & - i \\ i & 1 \end{array}), ρ_{+} = \frac{1}{2} I$ For an operator $A : H \to H$ , write $ran A : = {A x : x \in H}$ and define its rank by $rank A : = \dim_{C} (ran A)$ , using Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis. Relative to the ordered basis $(h, v)$ , call exactly the ray $[w]$ of this displayed superposition the coherent preparation in this example. Its density matrix has off-diagonal entries $(T_{w, w})_{12} = - i / 2$ and $(T_{w, w})_{21} = i / 2$ ; these are the coherence entries in this basis. The mixture has zero off-diagonal entries. These terms refer only to the displayed state data and assert no outcome-statistical or apparatus property.

The two assigned operators are distinct states of the model: $T_{w, w}$ is pure with purity $1$ and rank one, while $ρ_{+}$ is mixed with purity $\frac{1}{2}$ and rank two. In particular the relative phase of the superposition is retained in the assigned state, and the superposition and the mixture are not the same preparation.

Facts & Assumptions

Given: A two-dimensional complex Hilbert space $H$ with orthonormal basis $(h, v)$ ; the unit vector $w = (h + i v) / \sqrt{2}$ ; the mixture $ρ_{+} = \frac{1}{2} T_{h, h} + \frac{1}{2} T_{v, v}$ ; the state space $D (H)$ of The state space of the framework: preparations, pure rays and density operators; and the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the choice hypothesis of the cited state, mixture and purity suppliers.

[L1]

Inner product and modulus (Real and complex inner product spaces, with the inner product linear in the first argument, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive): $⟨ h, h ⟩ = ⟨ v, v ⟩ = 1$ , $⟨ h, v ⟩ = ⟨ v, h ⟩ = 0$ , $∣ i ∣ = 1$ , and $∥ w ∥^{2} = \frac{1}{2} ∥ h + i v ∥^{2} = 1$ because $∣ 1 ∣^{2} = ∣ i ∣^{2} = 1$ and the two components are orthogonal.

[L2]

Rank-one states and matrices (Rays, unit vectors and rank-one Hilbert projections, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes): for a unit vector $x$ the operator $T_{x, x}$ is the pure state of the ray $[x]$ and the Hilbert orthogonal projection onto the one-dimensional subspace $C x$ ; its matrix in the basis $(h, v)$ has columns $⟨ h, x ⟩ x$ and $⟨ v, x ⟩ x$ . For $h$ and $v$ , these operators are the projections onto $C h$ and $C v$ , with matrices $T_{h, h} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $T_{v, v} = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ .

[L3]

Mixtures and purity (The state space is convex, closed under countable mixtures, and complete in the trace norm, Purity, spectral weights and the criterion for a pure state, The state space: density operators, pure and mixed states, and purity, Statistical mixtures, spectral decompositions and the extremal role of pure preparations): finite mixtures with weights summing to one are states; a state is pure exactly when its purity is $1$ and mixed exactly when it has at least two positive eigenvalues.

Verification

technique · direct; compute both matrices in the basis $(h,v)$ and compare purity and rank

1.1givenL1L2L3algebra

[The superposition state] With $w = (h + i v) / \sqrt{2}$ one has $⟨ h, w ⟩ = 1 / \sqrt{2}$ and $⟨ v, w ⟩ = \overline{i} / \sqrt{2} = - i / \sqrt{2}$ by [L1], so the columns of $T_{w, w}$ in the basis $(h, v)$ are $T_{w, w} h = \frac{1}{\sqrt{2}} w = \frac{1}{2} (h + i v)$ and $T_{w, w} v = - \frac{i}{\sqrt{2}} w = - \frac{i}{2} (h + i v) = \frac{1}{2} (v - i h)$ , giving the displayed matrix $\frac{1}{2} (\begin{array}{cc} 1 & - i \\ i & 1 \end{array})$ ; its square is $\frac{1}{4} (\begin{array}{cc} 2 & - 2 i \\ 2 i & 2 \end{array}) = \frac{1}{2} (\begin{array}{cc} 1 & - i \\ i & 1 \end{array}) = T_{w, w}$ , so its purity is $1$ ; by [L2] its range is the one-dimensional subspace $C w$ , hence it has rank one, and by [L3] it is pure.

1.2givenL1L2L3algebra

[The mixture state] By [L2] one has $ρ_{+} = \frac{1}{2} (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}) + \frac{1}{2} (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}) = \frac{1}{2} I$ , a state by [L3]; its range is $H$ , so it has rank two, while its square is $\frac{1}{4} I$ and its purity is $tr (\frac{1}{4} I) = \frac{1}{2}$ ; since $\frac{1}{2} \neq 1$ , it is mixed by [L3].

2.1givenstep 1.1step 1.2algebra∎

[The states are distinct] The difference $T_{w, w} - ρ_{+} = \frac{1}{2} (\begin{array}{cc} 0 & - i \\ i & 0 \end{array})$ is nonzero, so the two preparations are assigned different states; the purities $1$ and $\frac{1}{2}$ , and the ranks one and two, confirm that one is pure and the other mixed. Hence the relative-phase superposition and the equal-weight statistical mixture are distinct states of the model, and the identity of their diagonal entries alone does not make them equal.

Remarks

The comparison is made at the level of the assigned states only. Which measurements distinguish the two states, and by how much, is a statement about the framework's probability law and measurement assumptions, and is not claimed here.
The diagonal entries of the two matrices agree and only the off-diagonal coherence entries defined above differ; this is the state-level expression of the relative phase of the superposition. No claim about interference or detection is made in this item.

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One two-level state from two different preparation ensembles

Physical scope

A worked computation in the framework's two-level state space showing that two distinct statistical mixtures of pure preparations, with different members and different weights, determine the same mixed preparation: the diagonal state with eigenvalues 3/4 and 1/4 arises both as the orthonormal spectral ensemble with weights 3/4, 1/4 and as the uniform mixture of two non-orthogonal pure states. The item shows that the state does not determine the ensemble; it asserts no measurement probability, no apparatus realization and no empirical result, and it makes no claim of physical incompleteness.

Example

Let $H$ be a two-dimensional complex Hilbert space with orthonormal basis $(e_{1}, e_{2})$ and use the state space and preparation semantics of The state space of the framework: preparations, pure rays and density operators. Let $ρ$ be the diagonal state $ρ = (\begin{array}{cc} 3 / 4 & 0 \\ 0 & 1 / 4 \end{array})$ relative to $(e_{1}, e_{2})$ , and put $ψ = (\cos \frac{π}{6}, \sin \frac{π}{6}) = (\frac{\sqrt{3}}{2}, \frac{1}{2}), φ = (\cos \frac{π}{6}, - \sin \frac{π}{6}) = (\frac{\sqrt{3}}{2}, - \frac{1}{2}) .$ Then:

(i) $ρ = \frac{3}{4} T_{e_{1}, e_{1}} + \frac{1}{4} T_{e_{2}, e_{2}}$ : the orthonormal ensemble with weights $\frac{3}{4}, \frac{1}{4}$ .

(ii) $ρ = \frac{1}{2} T_{ψ, ψ} + \frac{1}{2} T_{φ, φ}$ : the uniform ensemble of the two pure preparations $[ψ]$ and $[φ]$ , which are not orthogonal, since $⟨ ψ, φ ⟩ = \cos^{2} \frac{π}{6} - \sin^{2} \frac{π}{6} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} \neq 0.$

(iii) The two ensembles differ in their members and in their weights, yet they determine the same preparation, whose purity is ${\frac{3}{4}}^{2} + {\frac{1}{4}}^{2} = \frac{5}{8} < 1$ . Hence the assigned state does not determine the ensemble, and two distinct statistical preparations may determine the same state.

Facts & Assumptions

Given: A two-dimensional complex Hilbert space $H$ with orthonormal basis $(e_{1}, e_{2})$ ; the vectors $ψ = (\sqrt{3} / 2, 1 / 2)$ and $φ = (\sqrt{3} / 2, - 1 / 2)$ ; the displayed state $ρ$ ; the state space $D (H)$ of The state space of the framework: preparations, pure rays and density operators; and the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the choice hypothesis of the cited state, mixture and purity suppliers.

[L1]

Inner product and moduli (Real and complex inner product spaces, with the inner product linear in the first argument, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive): $⟨ u, v ⟩ = u_{1} \overline{v_{1}} + u_{2} \overline{v_{2}}$ , so $⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ , $∥ ψ ∥^{2} = \frac{3}{4} + \frac{1}{4} = 1$ , $∥ φ ∥^{2} = 1$ , and $⟨ ψ, φ ⟩ = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$ .

[L2]

Rank-one states (Rays, unit vectors and rank-one Hilbert projections, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes): for a unit vector $x$ the operator $T_{x, x}$ is the pure state of the ray $[x]$ , with matrix in the basis $(e_{1}, e_{2})$ having entries $(T_{x, x})_{j k} = ⟨ e_{k}, x ⟩ ⟨ x, e_{j} ⟩$ ; in particular $T_{e_{1}, e_{1}} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $T_{e_{2}, e_{2}} = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ .

[L3]

Mixtures, purity and non-uniqueness of ensembles (The state space is convex, closed under countable mixtures, and complete in the trace norm, Purity, spectral weights and the criterion for a pure state, Statistical mixtures, spectral decompositions and the extremal role of pure preparations): finite mixtures of states with nonnegative weights summing to one are states; the purity of a state is $tr (ρ^{2})$ and it is mixed exactly when its purity is less than $1$ ; and a preparation is the trace-norm limit of any of its statistical ensembles, distinct ensembles of the same state being allowed.

Verification

technique · direct; compute both ensembles in the basis $(e_1,e_2)$ and compare their data and the resulting state

1.1givenL2L3algebra

[The orthonormal ensemble] By [L2], $\frac{3}{4} T_{e_{1}, e_{1}} + \frac{1}{4} T_{e_{2}, e_{2}} = (\begin{array}{cc} 3 / 4 & 0 \\ 0 & 1 / 4 \end{array}) = ρ$ , a state by [L3]; its positive eigenvalues are $\frac{3}{4}$ and $\frac{1}{4}$ , so its purity is $\frac{9}{16} + \frac{1}{16} = \frac{5}{8} < 1$ and it is mixed.

1.2givenL1L2L3algebra

[The uniform ensemble of non-orthogonal states] With $c : = \cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $s : = \sin \frac{π}{6} = \frac{1}{2}$ one computes $T_{ψ, ψ} = (\begin{array}{cc} c^{2} & c s \\ c s & s^{2} \end{array}) = (\begin{array}{cc} 3 / 4 & \sqrt{3} / 4 \\ \sqrt{3} / 4 & 1 / 4 \end{array})$ and $T_{φ, φ} = (\begin{array}{cc} c^{2} & - c s \\ - c s & s^{2} \end{array}) = (\begin{array}{cc} 3 / 4 & - \sqrt{3} / 4 \\ - \sqrt{3} / 4 & 1 / 4 \end{array})$ by [L2] and [L1]; their average is $\frac{1}{2} (T_{ψ, ψ} + T_{φ, φ}) = (\begin{array}{cc} 3 / 4 & 0 \\ 0 & 1 / 4 \end{array}) = ρ,$ so the uniform mixture of the two non-orthogonal pure preparations also determines $ρ$ .

2.1givenstep 1.1step 1.2L1L3algebra∎

[The ensembles differ] The two ensembles have different member preparations (the orthogonal pair $(e_{1}, e_{2})$ versus the non-orthogonal pair $(ψ, φ)$ with $⟨ ψ, φ ⟩ = \frac{1}{2} \neq 0$ by [L1]) and different weights ( $\frac{3}{4}, \frac{1}{4}$ versus $\frac{1}{2}, \frac{1}{2}$ ); yet by steps 1.1 and 1.2 both determine the same state $ρ$ . Hence two distinct statistical preparations of the model determine the same preparation, and the state does not encode the ensemble.

Remarks

The non-uniqueness exhibited here is a property of the adopted state assignment: the two displayed ensembles receive the same density operator. This computation does not establish that the framework is physically incomplete or determine whether any further preparation data enter predictions. The maximally mixed two-level state gives another instance, treated on the mathematics composites page.
Neither ensemble in the computation is asserted to be realized by any apparatus; whether a laboratory procedure realizes one of them is an identification with uncertainty under the preparation and apparatus assumptions of the framework.

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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.

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What the worked checks of the companion page establish

Physical scope

A scope audit of the companion page of worked checks: it records what each computation verifies, that all of them are finite state-space computations or one refutation of an informal claim about the model's state semantics, and that none is an experiment, a measurement prediction or a comparison with data.

Scope of the worked checks

This remark audits the companion page. It records what each item computes, and what none of them claims. It proves nothing and reports no observation.

Elementary normalization and phase. A two-dimensional normalization and phase computation computes, on the explicit two-dimensional data $ψ = (1, i)$ and $φ = (1 + i) ψ$ , the norms, the normalized representatives, the phase relating them and the normalized rank-one operators; it verifies the normalization and phase class lemma of the main page and exhibits the representative independence of the pure state on a worked vector pair.

Finite spin and the two-level state space. The state space of a two-level finite-spin system is the Bloch ball proves that the states of the model's two-level finite-spin system are exactly the operators $\frac{1}{2} (I + r \cdot σ)$ with $∣ r ∣ \leq 1$ , that the pure states are the boundary points $∣ r ∣ = 1$ , and that each boundary state has one-dimensional range and is the state of the spin-up ray defined there by the eigenvalue $+ 1$ of $r \cdot σ$ . Pure and mixed states of a fixed two-level internal space repeats the corresponding elementary computation for an abstract two-level internal space carrying a pair of polarization labels, showing the pure state of the diagonal vector and the equal mixture of the basis rays. Neither item asserts the identification of the abstract two-level space with photon polarization; that reading belongs to the radiation assumptions of the category, and no statement about photon number, energy or momentum is made here.

Superposition versus mixture, and non-unique ensembles. A coherent two-level preparation and the equal statistical mixture are different states shows that the relative-phase superposition $(h + i v) / \sqrt{2}$ and the equal statistical mixture of the rays $[h]$ and $[v]$ are assigned different states: the first has one-dimensional range and purity $1$ , and the second has two-dimensional range and purity $1 / 2$ . One two-level state from two different preparation ensembles shows the converse direction of freedom: one and the same mixed state is determined by two genuinely different ensembles, an orthonormal pair with unequal weights and a uniform pair of non-orthogonal states. Both computations stop at the level of the assigned states; which outcome probabilities distinguish states, and with which apparatus, is the subject of the framework's probability law and measurement assumptions.

Zero-vector exclusion. The zero vector determines no pure preparation and no state refutes the informal claim that every vector determines a pure preparation with the zero vector as an empty preparation: 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 state. The refutation concerns the model's definitions and assignment; it is not an experiment and it makes no claim about nature.

Empirical neutrality. Every item of this page is a finite computation or the refutation just described. None of them uses measured data, none adopts a law of nature, none predicts the outcome of any measurement, and none compares a model prediction with an observation. Agreement or disagreement with data would test the combined framework and apparatus assumptions, and is not established or claimed here.

Qm State Geometry — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

A two-dimensional normalization and phase computation

Example

Facts & Assumptions

Verification

Remarks

The state space of a two-level finite-spin system is the Bloch ball

Example

Facts & Assumptions

Verification

Remarks

Pure and mixed states of a fixed two-level internal space

Example

Facts & Assumptions

Verification

Remarks

A coherent two-level preparation and the equal statistical mixture are different states

Example

Facts & Assumptions

Verification

Remarks

One two-level state from two different preparation ensembles

Example

Facts & Assumptions

Verification

Remarks

The zero vector determines no pure preparation and no state

Statement refuted

Facts & Assumptions

Counterexample

Remarks

What the worked checks of the companion page establish

Scope of the worked checks

Sources