How much information about a quantum operator is determined by its Poisson bracket Lie algebra?

Question

Hamiltonian quantum mechanics is often built using many ideas from Hamiltonian classical mechanics like the Poisson bracket to determine the commutator between quantum operators, which is appropriate given they are both Lie algebras. It also seems that often the specification of the Lie algebra alone is enough like, for example, the position and momentum operator. We can define many momenta, all of which are equivalent in a measurable sense given it is still conjugate to position. However, in other cases, like spin and angular momentum, it seems the Lie algebra alone misses vital information about the spectrums of each. How much information does this Lie algebra provide and how much is determined by its definition alone?

Answer 1

It's all in the Lie algebra, provided one understands it properly. There is a language of QM more amenable to comparing and contrasting with CM in phase space, summarized in a concice treatise by Curtright, Fairlie, and minder. But, here, I'll not use it directly—merely through its lessons.

The key point is the 3-dim PB Lie algebra $\{x,p\}=1$ is not the whole story, by far; the elements of it are mere "generators" of the rest of its elements ("presentation"), in the mathematics, not physics, sense. The PB algebra is the infinite-dimensional one $\{f(x,p),g(x,p)\}=...$, where the crucial (omitted but understood) step is that the bracket is evaluated by the chain rule for commutative xs and ps.

By contrast, the quantum commutator algebra $[\hat f (\hat x, \hat p), \hat g (\hat x, \hat p), ]=...$ is very different, since one now forfeits commutativity for $\hat x, \hat p$ in evaluating this infinite slew of commutators.

Because of that, the two algebras are very, very different: the PBs comprise a Wigner-İnönü contraction of the quantum commutator algebra, preserving its infinite dimension. This goes under the rubric of "Groenewold's theorem", and it dramatizes and circumscribes the differences between PBs and quantum commutators.

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When, however, you focus on subalgebras of the respective infinite dimensional ones, e.g. , the so(3) subalgebra of the three classical $\vec L$s (whose PBs close among themselves), and the quantum operators $\hat{\vec {L}}$ (which close under commutation into the very same algebra), then the identical Lie algebras dictate identical representation theories, eigenvectors, eigenvalues, ladder operators, vector operators, etc... $\leadsto$ The algebra dictates all.

• You already know this, since you use the same spherical harmonics eigenfunctions in the QM of the spherically symmetric Coulomb potential in hydrogen atom; but also in classical electromagnetism, fluid mechanics, and solutions of the Laplace and Helmholtz equations, as well, for example!

If you were confused about how classical angular momenta have eigenfunctions, lowest weight states, etc, you might have to ask about the commutator form of PBs, $$ [\tilde f, \tilde g]= k ~ \widetilde{\{f,g\} } $$ for some convenience relative normalization constant k and $$ \tilde f \equiv k \bigl ( (\partial_{x^i}f) ~\partial_{p^i} - (\partial_{p^i}f) ~\partial_{x^i} \bigr ). $$ These operators acting on phase space, of course, can have discrete eigenvalue spectra, and phase-space (the relevant vector space) eigenfunctions.

The classical angular momenta then, $\vec L\equiv \vec r\times \vec p$ obey the PB algebra $$ \{L^a,L^b \}= \epsilon^{abc} L^c, $$ and, consequently, (you may confirm), $$ [ \tilde L^a,\tilde L^b]=k \epsilon^{abc} \tilde L^c, $$ whose representation theory is the same as that of the quantum deformation of L, since it obeys the same so(3) Lie algebra, $$ [\hat L^a, \hat L^b]= i\hbar \epsilon ^{abc} \hat L^c ~. $$ For convenience, take $k=i\hbar$, the bridge between PBs and commutators.

Compare the eigenvalues of $\hat L_z=\hat x \hat p_y -\hat y \hat p_x $, whose action on coordinate functions is $-i\hbar (x\partial_y-y\partial_x)=-i\hbar \partial _\phi $, so its eigenfunctions are $\exp(im\phi)$ with eigenvalues $\hbar m$.

The above defined classical operators, then, are $$\tilde L_z=-k(x\partial_{y} -y\partial_x +p_x\partial_{p_y}-p_y\partial_{p_x}) .$$ Acting on $\exp (im\phi)= \exp (im \arctan (y/x))$, it yields an eigenvalue $-ikm\mapsto \hbar m$, etc. You may check the rest of the ladder representation construction.

The takeaway is that the Lie algebra rules: if you are only talking about a unique finite-dimensional Lie algebra in isolation, the PB or MB particularities of the operators does not matter much. However, if you are really talking about CM versus QM, underlain by dramatically different infinite-dimensional Lie algebras, then, of course, the corresponding operators are very different and display dramatically differing behaviors.

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• Spin is a delicate special case aside, and quantifying it and limiting it in phase space is a tricky cottage industry I'm sure you would not wish to get into until first appreciating the above trail-map.