In my very limited understanding of geometric quantization, we quantize spin by choosing as our phase space $S^2$ with a suitably normalized area form as the symplectic form. Depending on the normalization we get out a finite-dimensional Hilbert space corresponding to some spin $j$ particle.

My question is if/how angular momentum addition may be understood in this picture. In the Hilbert-space picture, we can write things like $1/2\otimes 1/2=0\oplus 1$; i.e., we may decompose tensor products of reps of $SU(2)$ into irreps. If we naively try translating such an equation into phase space, I assume tensor products of Hilbert spaces correspond to Cartesian products of phase spaces, but I have no idea how direct sums would be interpreted (disjoint unions?). Does thinking carefully about quantizing products of spheres give a nice geometric interpretation of angular momentum addition in phase space?


1 Answer 1


I could give you an answer by barking up a very different tree indeed! In phase space QM, and not, repeat not geometric quantization, you may work on flat phase spaces and forfeit spheres altogether, the way you actually do in Hilbert spaces. If you can stomach that, read on, otherwise not, lest you feel your expectations betrayed.

Plain vanilla phase-space QM in a 2d phase space replicates the standard ladder algebraic features of quantized angular momentum in Hilbert space, with very parallel algebraic tricks, cf Exercise 0.8 of p.33 of that book. A long moment's thought should convince you then that Hilbert space addition of angular momentum maps isomorphically to a phase space equivalent, on tensor-product phase spaces, one-to-one corresponding to the Hilbert spaces. The Hilbert space comultiplication representing Kronecker multiplication of SU(2), $$ \Delta(J_a)\equiv J_a \otimes 1\!\!\!1 + 1\!\!\!1 \otimes J_a $$ will satisfy the same Lie Algebra as the left-and-right $J_a$s, and the Casimir $\Delta(J_a)\Delta(J_a)$ will be reducible to a direct sum of the Casimirs of the Clebsched reduced reps, so, then the respective eigenvalues times the respective identities in the block subspaces of each irrep.

The linear algebra is identical in phase space, but, as you might recall, operator multiplication amounts to star-multiplication of plain phase-space functions in each phase space of the tensor product. To eschew spinors as a first step, try Kronecker-multiplying two spin 1s, so, then 33=531, so, then, spin 2, 1, and 0. Half-integral spins can then be introduced in standard ways, cf. Brif & Mann.

  • $\begingroup$ Thank you so much for your answer! If I understand correctly, you are saying that on a 6-d phase space (I wasn't quite sure on how to do it with just one position coordinate) with coordinates $(\vec{x},\vec{p})$, if we define $\vec{L}=\vec{x}\times\vec{p}$, the $L_i$'s will commute under the moyal bracket identically to the way the corresponding operators do in QM, so on the product phase space with coordinates $(\vec{x}_1,\vec{p}_1,\vec{x}_2,\vec{p}_2)$ the usual highest-weight decomposition procedure works the exact same way. (1/3) $\endgroup$ Apr 9, 2016 at 22:17
  • $\begingroup$ I was hoping there could be more geometrical picture. For instance, the adjoint rep for spin-1/2 is $0\oplus 1$, so my understanding is we may visualize hermitian operators over the spin-1/2 hilbert space (including density matrices) as functions over $S^2$, to first order in spherical harmonics, and endowed with the "fuzzy sphere" noncommutative product. Now $1/2\otimes 1/2=0\oplus 1$, and the adjoint rep of $0\oplus 1$ breaks up as a direct sum of operators from 0 to 0, from 0 to 1, from 1 to 0, and from 1 to 1. (2/3) $\endgroup$ Apr 9, 2016 at 22:17
  • $\begingroup$ I wanted a way to visualize the "fuzzy functions" on $S^2\times S^2$ breaking up as a direct sum in a similar way, hopefully with a nice picture that would work for any addition-of-angular-momentum situation, but that may not be doable. In general, I was curious how phase spaces like $S^2\times S^2$ would be quantized. I apologize for confusing the terms "phase space QM" and "geometric QM," which are probably both different than what this is. Also, if I should edit my question with these details (I can also try to write more clearly and give better examples), let me know. (3/3) $\endgroup$ Apr 9, 2016 at 22:17
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    $\begingroup$ In complete agreement with the (1/3) part. I was glib about spheres, your (2&3/3) as there are several schools with strong feelings about these, in phase space... My own favorites are Atakishyev et al. but that's just me...I find them easier to read... $\endgroup$ Apr 9, 2016 at 22:33

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