 Solution

Inverting the detector rotation at this time givesFor a representation with one system factor and one environment factor per displayed detector state, each parenthesized system–environment vector would have to be a product state. But the first has coefficient matrix $\mathrm{diag}(c,s)$ in the indicated orthonormal bases; the second has matrix $\mathrm{diag}(s,-c)$. Their matrix ranks are two whenever $cs\ne 0$. Equivalently, their two nonzero Schmidt coefficients are $|c|,|s|$, which no independent basis changes can remove. Thus the proposed factorization is impossible when $\sin \theta \ne 0$.

The exception matters: at $\theta =m\pi$, the detector basis is the original basis up to phases or interchange, and a factorization does exist. In particular, the literal condition $\theta \ne 0$ is insufficient if it includes $\theta =\pi$. The intended obstruction concerns genuinely mixed detector bases.

The environment has selected robust detector records: its interaction commutes with the original projectors, and orthogonal environmental records suppress their interference in the reduced density matrix. A rotated detector branch instead leaves the system and environment entangled. This is environment-induced pointer-basis selection, the issue analyzed by W. H. Zurek in his apparatus-environment model.^[ref]

This addresses the basis ambiguity but does not by itself select one realized outcome from the global superposition. Moreover, a two-state environment has recurrences: $\cos 2A$ returns to unit modulus, so the toy model does not establish irreversible macroscopic collapse. A large environment can make records effectively persistent over experimental times, but that requires further dynamical assumptions. The distinction is part of the measurement problem, not an algebraic error cured by a partial trace.