# Geometric quantization: why are the prequantum operators self-adjoint?

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).

• It doesn't even make a lot of sense to claim that an (unbounded) operator is self-adjoint without telling on what domain. – Christian Remling Apr 20 '16 at 2:04
• @ChristianRemling: Isn't it obvious from the text between parantheses? – Alex M. Apr 20 '16 at 7:17

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.
• 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". – Alex M. May 5 '16 at 16:36