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I'm reading a bit about geometric quantization and, among the axioms of this construction, is one requiring that the operator $\hat f = -\textrm i \hbar \nabla _{X_f} + f$ associated to the classical observable $f$ be self-adjoint whenever $f$ takes real values (where $X_f$ is the Hamiltonian field of $f$ and the second term acts by multiplication). It is not difficult to see that $\hat f$ is symmetric, but why is it also self-adjoint? ($\hat f$ acts on the completion in the $L^2$ norm of the space of sections in some complex Hermitian line bundle $L$ over $M$.) Frustratingly, none of the texts that I have been reading bothers showing this, they just state it (at best).

As a side-note, does anyone know of a serious, solid, trustworthy text on the subject? The ones that I have been reading skip a lot of details and rely on plenty of hand-waving (including Woodhouse's book from '91), what a disappointment).

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    $\begingroup$ It doesn't even make a lot of sense to claim that an (unbounded) operator is self-adjoint without telling on what domain. $\endgroup$
    – Christian Remling
    Commented Apr 20, 2016 at 2:04

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When the function $f$ has a complete Hamiltonian vector field $X_f$, the prequantization operator is the infinitesimal generator of the unique connection preserving lift of the flow of $X_f$ to the the prequantization line bundle.

It is essentailly self adjoint by the Stone theorem. Please see, Śniatycki: Geometric Quantization and Quantum Mechanics, section 3.3.

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    $\begingroup$ I know this, but some authors claim that $\hat f$ is self-adjoint in general, without any condition on $X_f$. Since asking the question, I have convinced myself (without proof!) that they misuse "self-adjoint" for "symmetric". $\endgroup$
    – Alex M.
    Commented May 5, 2016 at 16:36

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