arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 2207.08798math-phmath.MP
收藏

当Poisson和Moyal括号相等时？

When Poisson and Moyal Brackets are equal?

Didier Robert

阅读 PDF
arXiv 页面 ↗HTML 原文 ↗

摘要：

在相空间$\R^{2d}$中，我们将两个光滑经典观测量的泊松括号表示为$\{A,B\}$，将Moyal括号表示为$\{A, B\}_\circledast $的Weyl符号，其中$ \hat A$是$A$和$[ \hat A, \hat B]= \hat A \hat B- \hat B \hat A$（交换子）的Weyl量子化。本文证明了：如果相空间$\R^{2d}$上的光滑Hamilton $H$，具有中等增长的导数，对任何光滑有界的可观测$A$，满足$\{A,H\}= \{A, H\}_\circledast$，则$H$必是次数至多为2的多项式.这与关于多项式观测量的量子化的Groenewold-van Hove定理\cite{Gotay, Groen, vHove}有关。

英文摘要：

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.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

↑