$\psi(\tau) = \exp(H\tau)\psi(0)$ solves $\dot\psi = H\psi$

Proved

UnQuantumMechanics.hamiltonian_flow_solution

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

hamiltonian-mechanicslinear-algebramathematical-physics

Let $H$ be a Hamiltonian $2n\times 2n$ matrix and $\psi_0 \in \mathbb R^{2n}$ . Then $\psi(\tau) = \exp(H\tau)\,\psi_0$ solves the linear Hamiltonian equation of motion (35):

\frac{d}{d\tau}\Big(\exp(H\tau)\,\psi_0\Big) = H\,\exp(H\tau)\,\psi_0 \qquad\text{for all }\tau\in\mathbb R .

Formalization Note $\exp$ is the matrix exponential (power series). Only the "is a solution" direction of Eq. (49) is stated; $\psi(0) = \psi_0$ holds since $\exp(0) = \mathbf 1$ .

Preamble

import Mathlib
import Definitions.Def_UnQM_phase_space

open Matrix

Formal statement

namespace UnQuantumMechanics

theorem hamiltonian_flow_solution (n : ℕ) (H : Matrix (PhaseIdx n) (PhaseIdx n) ℝ) (hH : IsHamiltonianMatrix n H)
    (ψ₀ : PhaseIdx n → ℝ) (τ : ℝ) :
    HasDerivAt (fun t : ℝ => NormedSpace.exp (t • H) *ᵥ ψ₀)
      (H *ᵥ (NormedSpace.exp (τ • H) *ᵥ ψ₀)) τ := by sorry

end UnQuantumMechanics

Source

C. Baumgarten, How to (Un-) Quantum Mechanics, arXiv:1810.06981v2 [physics.gen-ph] (v2 dated June 10, 2025), https://arxiv.org/abs/1810.06981, Sec. VI, p. 17, Eq. (49) (solution of Eq. (35))

Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) — same agent that drafted the statements; non-blind

Non-blind read-back — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement, at the explicit direction of the proposal owner. The author knew the intended meaning and the source text while writing it, so it must not be mistaken for a blind, independent audit. An independent blind read-back is still recommended before submission.

For every natural number $n$ , every real $2n\times 2n$ matrix $H$ for which there is a symmetric $A$ with $H = \gamma_0 A$ , every vector $\psi_0\in\mathbb R^{2n}$ and every $\tau\in\mathbb R$ : the vector-valued function $t \mapsto \exp(tH)\,\psi_0$ (matrix exponential defined by the power series, applied to $\psi_0$ ) is differentiable at $\tau$ with derivative $H\,(\exp(\tau H)\,\psi_0)$ .

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).

View graph

ψ(τ)=exp⁡(Hτ)ψ(0)\psi(\tau) = \exp(H\tau)\psi(0)ψ(τ)=exp(Hτ)ψ(0) solves ψ˙=Hψ\dot\psi = H\psiψ˙​=Hψ

$\psi(\tau) = \exp(H\tau)\psi(0)$ solves $\dot\psi = H\psi$