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Suppose we have the (un)usual Schrödinger representation $\pi'(\cdot)$ of the Heisenberg algebra, along with the extension in $\mathbb{sl}(2,\mathbb{R})$ for the quadratic polynomials. It is assumed that

$$ \pi'(q^3) = -iQ^3;\quad \pi'(p^3) = -iP^3, $$

where $[Q,P] = i\mathbb{I}$ (the usual commutator in quantum mechanics). I want to prove that

$$ \pi'(q^2 p) = -\frac{i}{2}(Q^2P + PQ^2)\quad (1) $$

using the Poisson bracket $\lbrace q^3,p^2\rbrace = 6q^2p$.

Is (1) correct? The most that I can do is, using that $\pi'$ is a linear map

$$ \begin{split} \pi'(\frac{1}{6}\lbrace q^3,p^2 \rbrace) &= \frac{1}{6}[\pi'(q^3),\pi'(p^2)] \\ &= -\frac{1}{6}[Q^3,P^2]\\ &= -\frac{i}{6}(3QP^2 + 3P^2Q + 2QPQ), \end{split} $$

but I cannot get rid of the term $2QPQ$. Or is it that my commutator computation is wrong?

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The last line of your commutator evaluation is definitely wrong.

It appears you are striving to calculate the Groenewold anomaly but I have failed to understand your bizarre "linear" map of "the usual Schrödinger representation 𝜋′(⋅)" you posit to antihermitean operators, so I'll avoid it completely.

I'll just evaluate a few commutators correctly for you: $$ [Q^3,P^2]= 3i ~(PQ^2 + Q^2 P),\\ [Q^2,P^3]= 3i ~(QP^2 + P^2 Q), ~~\leadsto \\ [[Q^3,P^2], [Q^2,P^3]]= 12[Q^3,P^3] +\mathbf{36}~. $$ The last (boldfaced), constant, term is the celebrated Groenewold anomaly (has an $\hbar^2$ factor when you reinstate ℏ in your Heisenberg commutation relation).

Virtually straightforward to compute in deformation quantization.

It corrects the classical limit expression $$ \{\{q^3,p^2\}, \{q^2,p^3\}\}= 12\{q^3,p^3\} , $$ which has lost quantum information, as usual.


Edit post comments

Having found a copy of Woit's book, I at last appreciated the sadistic notation. His conceit (so 1930s) is to define a trial quantization map $i\pi'(q,p)$, Hermitian, for the unambiguous low polynomial orders, and then attempt to ladder his way up to higher orders defining them through the chimeric desideratum $$ \pi'(\{ f,g\})= [\pi'(f),\pi'(g)], $$ producing $$ i\pi'(qp)=i\pi'(\{q^2,p^2\}/4)= -\frac{i}{4}[Q^2,P^2]=\tfrac{1}{2}(PQ+QP), $$(in Weyl ordering), and further producing your (1), which, however, is not in Weyl ordering! (You simply mishandled the commutator).

He then concludes his chapter by pointing out the hapless chimeric desideratum above is inconsistent, as the quartic term $\pi'(q^2p^2)$ is ill-defined, as evaluated in two alternate PB pathways. Mathematicians have these circuitous dandy arguments, but the point is no consistent map can be found. Above, I condensed Groenewold's direct argument in one line, just as in our booklet.

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  • $\begingroup$ Well, the anti-Hermitian part it's Peter Woit's definition that appears in his book "Quantum Theory, Groups and Representations" :). $\endgroup$ Nov 18, 2021 at 9:10
  • $\begingroup$ I think his motivation is between Mathematics and Physics and the differences in the literature. From the group property of $GL(n,\mathbb{R})$ and a member $A=e^{t X}$ of it, the property $A A^\dagger = \mathbb{I}$ forces to have $X^\dagger = -X$. Then he says "Note that physicists often choose to define the Lie algebra in these cases as self-adjoint matrices ... We will not use this definition... we want to think of the Lie algebra as a real vector space..." $\endgroup$ Nov 18, 2021 at 13:22
  • $\begingroup$ But this is only notation... I don't understand the "hapless quantization map" question but I was checking actually yesterday the answer that you linked and Sec. 14 of the book "Quantum Mechanics in Phase Space". $\endgroup$ Nov 18, 2021 at 13:28
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    $\begingroup$ No, no quantization scheme is consistent. The whole dream of quantization is misplaced: it is like trying to determine a series for a function relying on its first term and a God-given recipe. There are several quantum systems, all logically consistent and legitimate, as per nature!, with a common classical limit. It is nice to have canonical guidelines, but that's all they are. At the end of the day, you guess a richer theory, starting from a limited corner/limit of it... $\endgroup$ Nov 18, 2021 at 15:39
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    $\begingroup$ As for his problem 1, Exercise B.8, multiply his last two lines by 9 to compare to our "concise" book. The above expression here has to be rearranged and divided by 12 to compare. All three expressions for the anomaly are identical. $\endgroup$ Nov 18, 2021 at 16:07

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