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Isham in his Lectures on Quantum Theory, Chapter 5, General Formalism of Quantum Theory, Section 5.2.1 makes states the following two assumptions for quantising a classical system.

Q1 The quantisation procedure is linearity preserving, i.e., the quantisation map $f\mapsto\hat{f}$ is linear, so that $$r_1f_1+r_2f_2\mapsto r_1\hat{f_1}+r_2\hat {f_2}.$$ Q2 The quantisation procedure is function-preserving in the following sense. If $\mathcal{F}:\mathbb{R}\to\mathbb{R}$ is any real function, the real-valued function $\mathcal{F}(f)$ on $\mathcal{S}$ is defined as $$\mathcal{F}(f)(s):=\mathcal{F}(f(s))\text{ for all s} \in\mathcal{S}.$$ Then if the operator that represents $f$ is $\hat{f}$, the operator that represents $\mathcal{F}(f)$ is $\mathcal{F}(\hat{f})$, i.e., $$\widehat{\mathcal{F}(f)}=\mathcal{F}(\hat{f}).$$

Now, in the following comments (comment 2, page 92-93), Isham makes it clear that these two rules give contradictions. However in the final comment (comment 3, page 96), he states that the function-preserving assumption "is normally assumed". Even in deriving a key theorem (on page 99), he uses this.

Question: What is the actual status of this assumption, Q2? Is it accepted even though it leads to contradictions?

BioPhysicist
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Atom
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    I think this is a reference to the van Hove problem, excellently discussed in Chernoff, P. R. "Mathematical obstructions to quantization." Hadronic Journal 4, no. 3 (1981): 879-898. In particular, it fails for observables like $x^2p^3$ but we have no setup to measure this kind of observable. The ordering problem is NOT solved. See also https://physics.stackexchange.com/q/550612/36194 and https://physics.stackexchange.com/q/345859/36194 – ZeroTheHero May 15 '20 at 15:03
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    After Todorov's friendly review linked in the questions/answers linked above, you could try Gotay et al 1996."Quantization is not a functor". – Cosmas Zachos May 15 '20 at 16:09

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Q2 was never assumed, and the practitioners know it is by far the weakest and notoriously misinterpreted paragraph in von Neumann's basic book on foundational QM, ritually, and virtually comically.

His acolyte H Groenewold resolved it in his monumental 1946 thesis, but the message has evidently not fully sunk in yet. The different Lie algebras underlying classical versus QM fare spectacularly differently under the Q2 map. People draw guidance/inspiration from it (Q2) when extending classical physics (e.g. the Boltzmann factor, or angular momentum) to the quantum domain, which is what "assume" means here: a background shadow; but they are very careful to avoid proving the earth is thereby flat, and delight in focussing on the novel, genuine, quantum behaviors involved.

Here is a stark example. Absorbing superfluous pesky constants in the variables, the classical oscillator hamiltonian is $H= \tfrac{1}{2} (p^2+ x^2)$. You have quantized it in introductory QM as $$ \hat H= \tfrac{1}{2} (\hat p^2+ \hat x^2), $$ but note neither of the summands commutes with the sum, and hence eigenvectors and eigenvalues are not shared.

Now, in QM, you have studied the Boltzmann operator, $$ \exp (-\beta\hat H)= \exp (-\beta (\hat p^2+ \hat x^2)/2 ), $$ and you might have inferred it as a slam-dunk application of Q2.

However, the classical ancestor of this is $$ \exp (- \beta H )= \exp (- \beta( p^2+ x^2)/2 )\\ = \exp (- \beta p^2/2 )\exp (- \beta(x^2/2 )= ... , $$ or any algebraic rearrangement of the canonical variables.

You then appreciate Q2 will give you an infinity of differing, inequivalent answers, depending on which rearrangement of classical canonical variables you start with.

It turns out that in the maximally symmetric arrangement of $\hat x$ and $\hat p$ operators, dubbed "Weyl ordering", the classical variables expression ("Wigner image") yielding the above simple QM one is a famously magnificent mess, $$ \exp_\star \left ( -\beta H \right )= \left ( \cosh ( \hbar \beta / 2 )\right ) ^{-1} \exp\left ( \frac{-2}{\hbar} H\tanh( \hbar \beta /2)\right ) , $$ as it is not even classical-looking: even though the canonical variables are classical, there is nontrivial $\hbar$ dependence due to the $\star$ operator which entails $\hbar$ and was invented by Groenewold to ensure this Q2-substitute is consistent.

So Q2 "assumed" is a shared metaphor: one tastefully chooses what to extend, by inspired guesswork, and develops, by experience, a sixth-sense of avoiding the problematic and the inconsistent. Trying to quantize problems in messy recondite geometry manifolds is still a challenge, problems to be solved. Q2 will never get you there by itself.

Now, to be sure, if only one observable (here H) is involved, there is no non-commutativity structure to snag, and the sole observable is the collection of its eigenvalues, so, in a trivial way, Q2 works and is assumed. The mayhem starts when the observable is a function of others, (x,p), and ordering problems arise, as illustrated.

Cosmas Zachos
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  • Why do you emphasise “that neither of the summands commutes with the sum”? – Atom May 16 '20 at 16:41
  • Overall, a great answer! But I don’t get the one paragraph with Weyl ordering. (I’m yet to come across it.) :) – Atom May 16 '20 at 16:48
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    Isn't it obvious? The two classically equivalent ancestors will yield two wildly inequivalent Q2 expressions. – Cosmas Zachos May 16 '20 at 16:48
  • Oh! Yeah, I see it now! Thanks! But still not sure of the Weyl ordering. Can you point me to some readable resource. – Atom May 16 '20 at 16:53
  • Here is an update of a booklet of ours on phase space quantum mechanics. The above freaky star exponential is an analytic continuation of equation (80). Phase space QM basically expresses standard Hilbert space QM in classical c-number x and p variables. – Cosmas Zachos May 16 '20 at 17:20