What is an obstruction in quantization?

I've found that obstructions object of the study of a mathematical theory, previously concerned with homotopy. The problem is that to explain what an obstruction is they recur to topology, of which I have no knowledge.

I just need a qualitative explanation, sufficient to understand why is Weyl ordering obstruction free and what implications it has. For example, I guess that Weyl ordering being obstruction free doesn't imply that we get the correct result of a quantum Hamiltonian using its classical correspondent, and why or why not?

To summarize:

  • What is an obstruction in quantization?

  • Why is Weyl ordering obstruction free? And why is that relevant?

Where I originally saw the affirmation that Weyl ordering is obstruction free, was in Rakic & Prvanovic's 2003 paper Weyl ordering rule and new Lie bracket of quantum mechanics

  • 1
    $\begingroup$ Is this from a reference? $\endgroup$
    – Qmechanic
    Commented Jun 27, 2019 at 16:36
  • 1
    $\begingroup$ Are you trying to understand Gotay et al 1996? $\endgroup$ Commented Jun 27, 2019 at 16:39
  • $\begingroup$ @Qmechanic where I saw this was on the reference I've put in the post now. The original reason I started searching was to find why Weyl ordering is useful. $\endgroup$ Commented Jun 27, 2019 at 16:53
  • 1
    $\begingroup$ OP's link refers to Gotay et al 1996. $\endgroup$
    – Qmechanic
    Commented Jun 27, 2019 at 17:37
  • 1
    $\begingroup$ See Chernoff, P. R. "Mathematical obstructions to quantization." Hadronic Journal 4, no. 3 (1981): 879-898 which predates Gotay by 15 years. $\endgroup$ Commented Jun 28, 2019 at 12:22

1 Answer 1


OK, I might try a qualitative story as a placeholder until I appreciate your question. So, a soft rant, as the comment format cannot accommodate it.

The basic facts are probably covered in my answer to this question and the paradigmatic obstruction to quantization is the "Groenewold anomaly" therein. To my understanding, it has nothing to do with homotopy. It is simply that, expressed in phase-space, the classical Lie bracket, the Poisson B, specifies a different Lie algebra than that of quantum commutators, the Moyal B in phase space. And consequently various identities in the PB algebra do not hold in the MB algebra (there are remainders ~ "obstructions" like the above mentioned one).

Such mismatches are well appreciated in elementary QM when it comes to the operator angular momentum quadratic Casimir $\mathbf L\cdot \mathbf L$ versus its classical limit, subject to algebraic manipulations of MBs and PBs respectively.

In short, the quantum objects are fundamentally different than their classical limits ($\hbar \to 0$) and obey different relations: there is no quantization functor. This is what one learns in QM courses.

Classical phase space functions can be mapped invertibly to symmetric operators in Hilbert space through the Weyl map. However, many operators we use in QM (like the square of the angular momentum above), when re-expressed in such a Weyl-symmetrized form, normally contain residual $\hbar$-pieces, so they present in phase space as nonclassical, $\hbar$-dependent functions. More importantly, their multiplications map to nonlocal "star products in phase space".

These are quite nonclassical and $\hbar$-dependent, precisely because the plain operator product of two Weyl-ordered operators is not automatically Weyl-ordered, and generates $\hbar$-dependent anomalies/obstructions when reordered. You might learn about it in WikiPedia.

However, there is a pernicious residual myth we all grew up with, that if something is a function in phase space it must "therefore" be classical (you saw this isn't so above); and, conversely, if it is operators in Hilbert space, it must be quantum. (Not so!). Your rather alarming odd reference appears to rely on such an unfounded understanding, and introduces an operator commutator isomorphic to the PB, now acting on Hilbert space, imagining that this is "quantum" and not classical; they know it is not our conventional QM, but they might not care. (Weyl-symmetrizing operators has been used by Koopman & non Neumann, and much better, more recently, by Bracken, to express classical mechanics in Hilbert space. Unsurprisingly, you don't get discreet spectra here, the real hallmark of quantization.) I could not muster the patience to convince myself R&P cannot get a Schroedinger equation with discreet spectra. For no such spectra, they haven't really quantized.

The Gotay et al article link is far more careful, and makes no wild claims, by contrast. You might think of it as a trail map at the mathematical level of special recondite spaces and situations where obstruction might be invisible or irrelevant. But not our QM.

A footnote: The Weyl ordering utilized here is particularly simple technically, but, as explored and proven by L Cohen, all such orderings are basically equivalent. Given sufficient misplaced gumption, one might well be able to re-express R&P in alternate orderings.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.