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A superposition of distinct energy eigenrays need not be stationary

Physical scope

A refutation, inside the framework's autonomous-dynamics reading, of the informal claim that every superposition of two or more distinct energy eigenrays of an isolated system is a stationary state with time-independent outcome laws: in the explicit two-level model with two distinct energies, an equal-weight superposition carries the bounded flip observable with an outcome probability oscillating between one and zero. The item refutes that specific universal claim; it asserts no apparatus, no measurement statistics from a laboratory and no empirical result, and it does not deny that individual eigenrays are stationary or that special superpositions can be stationary.

Statement refuted

Adopt the autonomous-dynamics reading of the framework (Assigned energy observable, unitary flow and the two pictures of autonomous evolution, P4). Then: every preparation of an isolated system that is a superposition of two or more energy eigenrays with distinct eigenvalues is a stationary state, in the sense that all of its outcome laws are independent of time; in particular a superposition of two eigenrays of the two-level model is stationary.

Facts & Assumptions

Given: The adopted autonomous-dynamics reading of the framework with the two-level model data of Precession of a two-level preparation: exact phases, probabilities and a Bloch-vector rotation: the Hilbert space $H = C^{2}$ , distinct energies $E_{1} \neq E_{2}$ in joules, the assigned energy observable $H_{o p} = E_{1} T_{u_{1}, u_{1}} + E_{2} T_{u_{2}, u_{2}}$ with flow $U (t)$ and frequency $ω = (E_{2} - E_{1}) / ℏ \neq 0$ , the bounded self-adjoint flip observable $X = T_{u_{1}, u_{2}} + T_{u_{2}, u_{1}}$ , and the unit vector $ψ_{0} = (u_{1} + u_{2}) / \sqrt{2}$ ; the Axiom of Countable Choice and the Axiom of Choice, the choice hypotheses of the cited spectral suppliers (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice).

[F1]

The model data and the law of $X$ . In the two-level model, $ψ (t) = U (t) ψ_{0} = α e^{- i E_{1} t / ℏ} u_{1} + β e^{- i E_{2} t / ℏ} u_{2}$ for unit $ψ_{0} = α u_{1} + β u_{2}$ , $⟨ X ⟩ (t) = 2 Re (α \overline{β} e^{i ω t})$ , and $μ_{ψ (t)}^{X} ({+ 1}) = (1 + ⟨ X ⟩ (t)) / 2$ (Precession of a two-level preparation: exact phases, probabilities and a Bloch-vector rotation clauses (i), (iii) and (v)).

[F2]

Stationary states are the eigenrays. A pure preparation $[ψ]$ is stationary, that is $[ψ (t)] = [ψ (0)]$ for every $t$ , exactly when $ψ$ is an eigenvector of $H_{o p}$ ; for a non-eigenvector the ray is not constant (Energy conservation, conserved sharp observables and the stationary eigenrays clause (c)).

[F3]

Trigonometric values. For real $x$ one has $\cos (x + π) = - \cos x$ and $\cos (- x) = \cos x$ ; in particular $\cos π = - 1$ and $\cos (- π) = - 1$ (Quarter-turn values and shifts by pi/2 and pi, Parity and the Pythagorean identity for sine and cosine).

[F4]

Euler form of the phase. $e^{i θ} = \cos θ + i \sin θ$ for real $θ$ , so $2 Re (e^{i θ}) = 2 \cos θ$ (Euler's formula: $\exp (i θ) = \cos θ + i \sin θ$ for every real $θ$ ). Moreover $\ker (\exp) = 2 π i Z$ , so for real $θ$ one has $e^{i θ} = 1$ exactly for $θ \in 2 π Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

Counterexample

technique · direct; evaluate the outcome law of the flip observable of the two-level model at two times and show that it takes different values

1.1givenF1F4algebra

[The preparation and its frequency] The vector $ψ_{0} = (u_{1} + u_{2}) / \sqrt{2}$ is a unit vector with $α = β = 1 / \sqrt{2}$ , so $α \overline{β} = 1 / 2$ and, by [F1], $⟨ X ⟩ (t) = 2 Re (\frac{1}{2} e^{i ω t}) = \cos ω t$ by [F4], with $ω = (E_{2} - E_{1}) / ℏ \neq 0$ because $E_{1} \neq E_{2}$ .

2.1givenstep 1.1F1F3F4algebra

[Two values of the law] At $t = 0$ one has $e^{i ω \cdot 0} = 1$ since $0 \in \ker (\exp)$ by [F4], so $⟨ X ⟩ (0) = 2 Re (1 / 2) = 1$ by step 1.1 and the law of $X$ in $[ψ_{0}]$ is $μ_{ψ_{0}}^{X} ({+ 1}) = (1 + 1) / 2 = 1$ . At $t_{0} : = π / ∣ ω ∣ > 0$ one has $ω t_{0} = \pm π$ , so $\cos (ω t_{0}) = \cos (\pm π) = - 1$ by [F3], and therefore $μ_{ψ (t_{0})}^{X} ({+ 1}) = (1 - 1) / 2 = 0$ .

3.1givenstep 1.1step 2.1F1F2algebra∎

[The claim fails] The outcome law of the bounded assigned sharp observable $X$ takes the value $1$ at $t = 0$ and the value $0$ at $t = t_{0} > 0$ in the same preparation $[ψ_{0}]$ ; hence this law is not independent of time and the superposition $[ψ_{0}]$ is not stationary. By [F2] this is exactly as it must be, since $ψ_{0}$ is not an eigenvector of $H_{o p}$ : $H_{o p} ψ_{0} = (E_{1} u_{1} + E_{2} u_{2}) / \sqrt{2}$ is a multiple of $ψ_{0} = (u_{1} + u_{2}) / \sqrt{2}$ only if $E_{1} = E_{2}$ . This refutes the displayed universal claim.

Remarks

What is not refuted. Each eigenray is stationary by [F2], and a superposition whose nonzero coefficients all carry the same eigenvalue is stationary too; the refuted assertion is the universal one about superpositions of distinct eigenvalues. The failure is produced by the relative phase $e^{i ω t}$ , not by any external disturbance.
The two-times computation is exact model arithmetic. Both probabilities are consequences of the framework's outcome law for the declared finite model; no detector, efficiency, resolution or recorded data is involved.

A superposition of distinct energy eigenrays need not be stationary

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

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