Quantization of the Nambu bracket

The most simple quantum mechanical system consists of a canonical pair of operators $$\{P, Q\}$$ satisfying

$$P Q - Q P = i \hbar.$$

It is well known that there is a unique (modulo unitary maps) irreducible representation of this algebra (actually, of the Weyl algebra generated by $$P, Q$$), given by

$$Q f(x) = x f(x),$$ $$P f(x) = i \hbar f'(x).$$

The quantization of Nambu mechanics is generated by a canonical triple $$\{P, Q, R\}$$ and takes the following form:

$$P Q R - P R Q - Q P R + Q R P + R P Q - R Q P = i \hbar.$$

What are the irreducible representations of this algebra? Is there a unique representation just like in the case of ordinary quantum mechanics?

• Yes and no... See this 2003 paper. The answer is "yes", despite unfounded myths to the contrary, for an even number of variables, unlike your 3 variables here. Basically, you are lacking the canonical momentum analog of R here. For an even number of variables, e.g. P,Q,R,S this is feasible and elegant (see above) and nature has already done it for superintegrable systems. But for an odd one, you must somehow do it for the larger even case, and then somehow marginalize, factor out, or disentangle the 4th variable S. It is a long story. Nov 3 '19 at 15:24
• @CosmasZachos I don’t mind long stories, especially if you could outline the story in an answer posted on this site. Nov 3 '19 at 15:29
• Well, all the information is in the 2003 paper and the one precursor and 4 followups thereof. They cover the waterfront, replete with examples. I'm not sure if you are asking for a condensed tutorial. The elegant path goes through phase-space quantization, since these systems really "want" to be described in phase space... narrow questions are easiest to answer, but "show me something like that" ones not. Nov 3 '19 at 15:33
• @CosmasZachos thanks for the reference, I’m looking into it as we speak. I would love to see a condensed tutorial (especially posted by one of the authors), and I’m sure that other users who upvoted this question would like that, too; but that’s assuming that you have enough free time to spare, of course. Nov 3 '19 at 15:39
• ...as for the Stone-von-Neumann theorem analog you are asking, I have no idea, but the indisputable "owner" of ternary algebras is M Bremmer in Saskatchewan and I wouldn't be shocked if he has some analog... Intriguingly, the algebra enters in infinite dimensional generalizations. Nov 3 '19 at 15:50

for example, an explicit realization of $$P,Q,R$$ would already be a very helpful answer (even with the details of how to obtain this realization missing).
You just consider the Casimir-normalized generators of so(3), $$P=\left(\frac{\hbar}{L^2}\right )^{1/3} {L_1}, \qquad Q=\left(\frac{\hbar}{L^2}\right )^{1/3} L_2 , \qquad R= \left(\frac{\hbar}{L^2}\right )^{1/3} L_3 ,$$ so that the Nambu bracket you wrote reduces to $$P [Q, R] - Q [P, R] + R [P, Q] = i \frac{L^2}{L^2} \hbar = i\hbar.$$ So consider the irrep of your choice.
PS in response to comment. To appreciate how radically nontrivial this ternary QNB structure is, consider that a trivial $$R=1\!\!1$$ does not yield a vanishing QNB, $$[P,Q,R]= R [P,Q] \neq 0 ~~~~~!$$ This subverts the classical solenoidal flow Nambu noticed, and makes it impossible to pair up the trivial constant $$R$$ with a meaningful canonical momentum.
• Interesting, so the representation theory mimics that of $SO(3)$? Nov 3 '19 at 17:19