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I have been told that $$[\hat x^2,\hat p^2]=2i\hbar (\hat x\hat p+\hat p\hat x)$$ illustrates operator ordering ambiguity.

What does that mean? I tried googling but to no avail.

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You may be interested in the so called Groenewold van Hoove theorem, this is the famous result in the context of your question. – student Dec 19 '12 at 12:37
Related: – Qmechanic Oct 7 '14 at 16:30
up vote 11 down vote accepted

The ordering ambiguity is the statement – or the "problem" – that for a classical function $f(x,p)$, or a function of analogous phase space variables, there may exist multiple operators $\hat f(\hat x,\hat p)$ that represent it. In particular, the quantum Hamiltonian isn't uniquely determined by the classical limit.

This ambiguity appears even if we require the quantum operator corresponding to a real function to be Hermitian and $x^2 p^2$ is the simplest demonstration of this "more serious" problem. On one hand, the Hermitian part of $\hat x^2 \hat p^2$ is $$ \hat x^2 \hat p^2 - [\hat x^2,\hat p^2]/2 = \hat x^2\hat p^2 -i\hbar (\hat x\hat p+\hat p\hat x)$$ where I used your commutator.

On the other hand, we may also classically write the product and add the hats as $\hat x \hat p^2\hat x$ which is already Hermitian. But $$ \hat x \hat p^2\hat x = \hat x^2 \hat p^2+\hat x[\hat p^2,\hat x] = \hat x^2\hat p^2-2i\hbar\hat x\hat p $$ where you see that the correction is different because $\hat x\hat p+\hat p\hat x$ isn't quite equal to $2\hat x\hat p$ (there's another, $c$-valued commutator by which they differ). So even when you consider the Hermitian parts of the operators "corresponding" to classical functions, there will be several possible operators that may be the answer. The $x^2p^2$ is the simplest example and the two answers we got differed by a $c$-number. For higher powers or more general functions, the possible quantum operators may differ by $q$-numbers, nontrivial operators, too.

This is viewed as a deep problem (perhaps too excessive a description) by the physicists who study various effective quantum mechanical models such as those with a position-dependent mass – where we need $p^2/2m(x)$ in the kinetic energy and by an expansion of $m(x)$ around a minimum or a maximum, we may get the $x^2p^2$ problem suggested above.

But the ambiguity shouldn't really be surprising because it's the quantum mechanics, and not the classical physics, that is fundamental. The quantum Hamiltonian contains all the information, including all the behavior in the classical limit. On the other hand, one can't "reconstruct" the full quantum answer out of its classical limit. If you know the limit $\lim_{\hbar\to 0} g(\hbar)$ of one variable $g(\hbar)$, it clearly doesn't mean that you know the whole function $g(\hbar)$ for any $\hbar$.

Many people don't get this fundamental point because they think of classical physics as the fundamental theory and they consider quantum mechanics just a confusing cherry on a pie that may nevertheless obtained by quantization, a procedure they consider canonical and unique (just hat addition). It's the other way around, quantum mechanics is fundamental, classical physics is just a derivable approximation valid in a limit, and the process of quantization isn't producing unique results for a sufficiently general classical limit.

The ordering ambiguity also arises in field theory. In that case, all the ambiguous corrections are actually divergent, due to short-distance singularities, and the proper definition of the quantum theory requires one to understand renormalization. At the end, what we should really be interested in is the space of relevant/consistent quantum theories, not "the right quantum counterpart" of a classical theory (the latter isn't fundamental so it shouldn't stand at the beginning or base of our derivations).

In the path-integral approach, one effectively deals with classical fields and their classical functions so the ordering ambiguities seem to be absent; in reality, all the consequences of these ambiguities reappear anyway due to the UV divergences that must be regularized and renormalized. The process of regularization and renormalization depends on the subtraction of various divergent counterterms, to get the finite answer, which isn't quite unique, either (the finite leftover coupling may be anything).

That's why the renormalization ambiguities are just the ordering ambiguities in a different language. Whether we study those things as ordering ambiguities or renormalization ambiguities, the lesson is clear: the space of possible classical theories isn't the same thing as the space of possible quantum theories and we shouldn't think about the classical answers when we actually want to do something else – to solve the problems in quantum mechanics.

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Thank you very much, Lubos! – Greta Dec 16 '12 at 19:15

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