Proof of Theorem 1 — $[\sigma,X]=0\iff\mathcal P_X(\sigma)=\sigma$

Proved

ProjectiveMeasurementEquilibration.commute_iff_pinching_eq

by Lucas · Oct 2, 2026 · Mathlib 0df444a (Lean v4.33.1)

pinchingquantum-mechanics

Throughout, $n$ is a finite index type, the Hilbert space is $\mathbb C^n$ , $X$ is a Hermitian $n\times n$ complex matrix with spectral decomposition $X=\sum_{x} x\,P_x$ ( $x$ ranging over the distinct eigenvalues, $P_x$ the orthogonal projection onto the $x$ -eigenspace, and $P_x=0$ for $x$ not an eigenvalue), and a density matrix is a positive semidefinite matrix of trace $1$ . For every $n\times n$ complex matrix $\sigma$ : $\sigma$ commutes with $X$ if and only if $\sigma$ is block diagonal with respect to the eigenspaces of $X$ , i.e.

[\sigma,X]=0\iff \mathcal P_X(\sigma)=\sum_xP_x\sigma P_x=\sigma.

Preamble

import Mathlib
import Definitions.Def_pme_quantum_basics
import Definitions.Def_pme_measurement_conditions

open Matrix Filter Topology

Formal statement

namespace ProjectiveMeasurementEquilibration

theorem commute_iff_pinching_eq {n : Type} [Fintype n] [DecidableEq n] {X : Matrix n n ℂ} (hX : X.IsHermitian)
    (σ : Matrix n n ℂ) :
    X * σ = σ * X ↔ pinching hX σ = σ := by sorry

end ProjectiveMeasurementEquilibration

Source

G. Carcassi (Assumptions of Physics collaboration), "Projective measurements are equilibration processes", AoP Technical Brief 017, https://assumptionsofphysics.org/resources/briefs, proof of Theorem 1 (EQ-HS ⇔ EQ-MEAS, first paragraph), p. 3

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For every finite index type $n$ , Hermitian $X\in M_n(\mathbb C)$ and every matrix $\sigma\in M_n(\mathbb C)$ (arbitrary):

X\sigma=\sigma X\iff\sum_{x\in\Lambda}P_x\sigma P_x=\sigma,

$\Lambda$ the distinct eigenvalues of $X$ and $P_x$ its eigenprojections.

Human review

Endorsed by Shuze Chen · Oct 8, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Oct 8, 2026
Confirmed by the mission captain (proposal self-audit).

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Proof of Theorem 1 — [σ,X]=0 ⟺ PX(σ)=σ[\sigma,X]=0\iff\mathcal P_X(\sigma)=\sigma[σ,X]=0⟺PX​(σ)=σ

Proof of Theorem 1 — $[\sigma,X]=0\iff\mathcal P_X(\sigma)=\sigma$