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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.

One two-level state from two different preparation ensembles

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources