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In the context of geometric quantization, we usually look for a map from the Poisson algebra of classical observables to the algebra of quantum observables (or rather, a sub-algebra of the classical observables, depending on polarization).

We also have the following requirements for the quantization map $Q$:

  1. $Q$ is $\mathbb{R}$-linear;
  2. for $f(x) = 1$ the constant map, we get $Q(f) = (-i\hbar)\mathbb{I}$ the identity operator;
  3. brackets map to brackets, i.e. $\{f_1,f_2\} = f_3 \implies [Q(f_1),Q(f_2)] = (-i\hbar)Q(f_3)$; and
  4. for $\{f_j\}$ a complete set of classical observables, the quantized Hilbert space $\mathcal{H}$ must be irreducible under the action of $\{Q(f_j)\}$.

Coming from a math background with a weaker physics background, I'm not sure how to motivate this last point. The first three requirements are easily motivated; the map $Q$ should preserve the Lie algebra structure.

Certainly if our set of observables is complete (i.e. there are no nontrivial $g$ such that $\{f_j,g\} = 0$ for all $j$), it separates points on the symplectic manifold $M$ that we are trying to quantize. So is the condition related to our ability to recover $M$ from $\mathcal{H}$? Part of me doubts this because, from what I can tell, this was a requirement even in Dirac's early publications, which leads me to think that it has physical significance; I'm also not sure if the ability to recover geometric information from the quantized Hilbert space was a major priority in the early-to-mid 20th century.

Any responses or references are highly appreciated.

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This answer is a bit late give how long ago the question was asked! However: The Hilbert space has an inner product that is preserved by the operators. Therefore reducibility (existance of an invariant subspace) implies complete reducibility: a reducible space is the orthogonal direct sum of two invariant component subspaces. We can't get from one subspace to the other by using the complete set of operators, and no measurments can (using only the given operators) detect interference between states in the irreducible components. So we may as well keep only one of them. Thus the basic answer is ``convenience''.

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  • $\begingroup$ Awesome! Thanks for taking the time to respond! I highly appreciate it, even with the gap in time :-) $\endgroup$
    – ptsw
    Feb 15 '17 at 1:44

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