CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Slow variation alone is insufficient

Statement refuted

Assuming Countable Choice, slow scaling by $T$ alone guarantees a uniform $O (1 / T)$ leakage bound for the pointwise ordered upper eigenspace of every finite-dimensional $C^{2}$ Hermitian family, even when its spectral gap may close.

Statement references

Physical adiabatic phase with explicit scale

Facts & Assumptions

Given: Countable Choice, $ℏ > 0$ in joule-seconds, an energy $E_{0} > 0$ in joules, and a duration $T > 0$ in seconds.

[A1]

Countable Choice is The Axiom of Countable Choice ( ${A C}_{ω}$ ), the hypothesis carried by Physical adiabatic phase with explicit scale.

[F1]

Physical adiabatic phase with explicit scale: under Countable Choice, a $C^{2}$ finite Hermitian family with a spectral projection followed continuously in operator norm and separated from its complementary spectrum by a uniform positive gap satisfies the stated $O (1 / T)$ leakage bound.

Counterexample

1.1givenA1algebra

On $C^{2}$ , let $e_{↑} = (1, 0)^{T}$ , $e_{↓} = (0, 1)^{T}$ , and $H (s) = E_{0} (2 s - 1) diag (1, - 1)$ for $s \in [0, 1]$ . For $s < 1 / 2$ the ordered upper eigenvalue is $E_{0} (1 - 2 s)$ and its projection is $P_{u p} (s) = e_{↓} e_{↓}^{*}$ ; for $s > 1 / 2$ it is $E_{0} (2 s - 1)$ and its projection is $P_{u p} (s) = e_{↑} e_{↑}^{*}$ . At $s = 1 / 2$ , $H (s) = 0$ , so the degenerate upper eigenspace is all of $C^{2}$ and its spectral projection is $I_{2}$ ; the one-sided rank-one projections therefore have no continuous rank-one extension. The gap between the two eigenvalues is $2 E_{0} ∣ 2 s - 1 ∣$ , which closes at $s = 1 / 2$ .

2.1givenstep 1.1constructalgebra

The propagator from $s = 0$ solves $i ℏ U_{T}^{'} (s) = T H (s) U_{T} (s)$ and is $U_{T} (s) = \exp ⁣ (- \frac{i T E_{0}}{ℏ} (s^{2} - s) diag (1, - 1)),$ as direct differentiation verifies. In particular $U_{T} (1) = I_{2}$ .

3.1givenF1step 1.1step 2.1algebra

At the endpoints, $P_{u p} (0) = e_{↓} e_{↓}^{*}$ and $P_{u p} (1) = e_{↑} e_{↑}^{*}$ . Therefore $∥ (I - P_{u p} (1)) U_{T} (1) P_{u p} (0) ∥ = ∥ e_{↓} e_{↓}^{*} ∥ = 1$ for every $T > 0$ . No finite constant $C$ independent of $T$ can bound this leakage by $C / T$ . The ordered projection fails the continuity and uniform-gap hypotheses in [F1].

4.1givenstep 1.1step 2.1algebra∎

The constant projection $P_{↓} = e_{↓} e_{↓}^{*}$ commutes with every $H (s)$ and has zero leakage for all $T$ . At the crossing it is not the spectral projection onto the degenerate eigenspace, which is all of $C^{2}$ ; it is a continuous invariant rank-one subspace, not a coherently followed gapped spectral cluster. Thus the example refutes a gap-free uniform tracking claim for the ordered branch; it does not assert that every choice of state at a crossing causes transitions.

Slow variation alone is insufficient

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Counterexample references

Depends on

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