When Poisson and Moyal Brackets are equal?

TL;DR

This paper proves that if the Poisson and Moyal brackets coincide for a smooth Hamiltonian and all bounded observables, then the Hamiltonian must be at most quadratic, linking to foundational quantization theorems.

Contribution

It establishes a characterization of Hamiltonians for which Poisson and Moyal brackets are identical, showing they are at most quadratic polynomials.

Findings

Poisson and Moyal brackets are equal only for quadratic Hamiltonians.

The result connects to the Groenewold-van Hove Theorem.

Identifies conditions under which classical and quantum brackets coincide.

Abstract

In the phase space $\R^{2d}$, let us denote $\{A,B\}$ the Poisson bracket of two smooth classical observables and $\{A, B\}_\circledast $ their Moyal bracket, defined as the Weyl symbol of $i[ A, B]$, where $ \hat A$ is the Weyl quantization of $A$ and $[ \hat A, \hat B]= \hat A \hat B- \hat B \hat A$ (commutator). In this note we prove that if a smooth Hamiltonian $H$ on the phase space $\R^{2d}$, with derivatives of moderate growth, satisfies $\{A,H\}= \{A, H\}_\circledast$ for any smooth and bounded observable $A$ then $H$ must be a polynomial of degree at most 2. This is related with the Groenewold-van Hove Theorem \cite{Gotay, Groen, vHove} concerning quantization of polynomial observables.