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In quantum mechanics we have quantization map $Q$ that maps classical observables to quantum observables. If symplectic manifold $(M, \omega)$ is the phase space of a classical system then classical observables $\mathcal{F}(M)$ are complex valid functions on $M$ and if $\mathcal{H}$ is the Hilbert space of the quantized system, space $\operatorname{Op}(\mathcal{H})$ of linear operators on $\mathcal{H}$ is the space of quantum observables, so quantization map $Q$ is a linear map

$$ Q : \mathcal{F}(M) \to \operatorname{Op}(\mathcal{H}). $$

One can require that map $Q$ satisfy some condition e.g. Dirac condition, von Neumann condition etc. But I'm interested in the following special case: suppose $M=\mathbb{R}^{2n}$ with "standard" symplectic form and thus Poisson bracket, coordintates $q_i$ and $p_i$ are quantized to operators $Q_i$ and $P_i$ satisfying Heisenberg algebra relations $$ [P_i,Q_j]=\sqrt{-1} \delta_{ij} \operatorname{Id}. $$

Using this I can quantize any polynomial classical observable $f(q,p)$ as $f(Q,P)$ if a choose an ordering rule e.g. Weyl rule, Born-Jordan rule or normal ordering if I use creation and annihilation operators.

So, when I work only with polynomial functions map $Q$ can be defined by quantization of $q_i$ and $p_i$ (Heisenberg algebra) and by choosing an ordering rule.

Suppose $g \subset \mathbb{C}[q_1, \ldots, q_n, p_1, \ldots, p_n]$ is a finite dimensional Lie subalgebra of the Lie algebra of polynomials wrt to Poisson bracket, in other words $g = \operatorname{span}(f_1, \ldots, f_m)$ and $$ \{f_i.f_j\}=\sum_{k=1}^m C^k_{ij} f_k. $$

Set $F_i = Q(f_i)$, is it true that $\operatorname{span}(F_1, \ldots, F_m)$ is a Lie algebra (now bracket is the commutator of operators) isomorphic to the "classical algebra"?

I've heard that this claim "there is no anomaly for finitely many degrees of freedom" from physicist, but what is a proof? Is is written somewhere?

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  • $\begingroup$ No it is not, by the Groenewold-van Hove theorem. The extra constant in his counterexample of a P.B. identity not mapping to the corresponding commutator identity is called the Groenewold anomaly. It is of a very different ilk than the QFT anomalies you have heard statements about. $\endgroup$ – Cosmas Zachos Dec 10 '18 at 22:52
  • $\begingroup$ What is the difference between this anomaly and QFT anomaly? I thought both come from the ordering problem. $\endgroup$ – Alex Dec 14 '18 at 9:00
  • $\begingroup$ Endless question like and here.... QFT anomalies rely on an infinite Dirac sea, but.... $\endgroup$ – Cosmas Zachos Dec 14 '18 at 12:04

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