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Definition in geometric quantization

We consider the notion of quantum observables in the the context of geometric quantization. See also quantum operator (in geometric quantization).

On a symplectic manifold

Let $\left(X,\omega \right)$ be a (pre-)symplectic manifold, thought of as the phase space of a physical system.

Assume that $\omega$ is prequantizabe (integral) and let $\nabla :X\to BU\left(1{\right)}_{\mathrm{conn}}$ be a prequantum bundle $E\to X$ with connection for $\omega$, hence with curvature ${F}_{\nabla }=\omega$. Write ${\Gamma }_{X}\left(E\right)$ for the space of smooth sections of the associated complex line bundle. This is the prequantum space of states.

Definition

For $f\in {C}^{\infty }\left(X,ℂ\right)$ a function on phase space, the corresponding pre-quantum operator is the linear map on prequantum states

$\stackrel{^}{f}:{\Gamma }_{X}\left(E\right)\to {\Gamma }_{X}\left(E\right)$\hat f : \Gamma_X(E) \to \Gamma_X(E)

given by

$\psi ↦-i{\nabla }_{{v}_{f}}\psi +f\cdot \psi \phantom{\rule{thinmathspace}{0ex}},$\psi \mapsto -i \nabla_{v_f} \psi + f \cdot \psi \,,

where

• ${v}_{f}$ is the Hamiltonian vector field corresponding to $f$;

• ${\nabla }_{{v}_{f}}:{\Gamma }_{X}\left(E\right)\to {\Gamma }_{X}\left(E\right)$ is the covariant derivative of sections along ${v}_{f}$ for the given choice of prequantum connection;

• $f\cdot \left(-\right):{\Gamma }_{X}\left(E\right)\to {\Gamma }_{X}\left(E\right)$ is the operation of degreewise multiplication pf sections.

Remark

In terms of Higher geometric prequantum theory we may, as discussed there, identify the Poisson bracket Lie algebra $\mathrm{𝔓𝔬𝔦𝔰𝔰𝔬𝔫}\left(X,\omega \right)$ with the Lie algebra of the group of automorphism $\mathrm{exp}\left(O\right):\nabla \stackrel{\simeq }{\to }\nabla$ regarded in the slice over $BU\left(1{\right)}_{\mathrm{conn}}$. Moreover, the space of sections is equivalently the space of maps $\Psi :\nabla \to ℂ//U\left(1{\right)}_{\mathrm{conn}}$ in the slice from $\nabla$ into the differential refinement of the smooth universal line bundle $ℂ//U\left(1\right)\to BU\left(1\right)$. In this formulation the action of prequantum operators is just the precomposition action

$\stackrel{^}{\mathrm{exp}\left(O\right)}:\left(\nabla \stackrel{\Psi }{\to }ℂ//U\left(1{\right)}_{\mathrm{conn}}\right)↦\left(\nabla \stackrel{\mathrm{exp}\left(O\right)}{\to }\nabla \stackrel{\Psi }{\to }ℂ//U\left(1{\right)}_{\mathrm{conn}}\right)\phantom{\rule{thinmathspace}{0ex}}.$\widehat{\exp(O)} \colon (\nabla \stackrel{\Psi}{\to} \mathbb{C}//U(1)_{conn}) \mapsto (\nabla \stackrel{exp(O)}{\to} \nabla \stackrel{\Psi}{\to} \mathbb{C}//U(1)_{conn}) \,.

Now after a choice of polarization a quantum state is a prequantum wave function which is covariantly constant along the Lagrangian submanifolds of the foliation. Not all prequantum operators will respect the space of such quantum states inside all quantum states. Those that do become genuine quantum operators.

Definition

Let $𝒫$ be a polarization of the symplectic manifold $\left(X,\omega \right)$. then a quantum state or wavefunction is a prequantum state $\psi$ such that $\nabla \Psi$ vanishes along the leaves of the polarization.

A quantum operator is a prequantum operator which preserves quantum states among all prequantum states.

Proposition

A prequantum operator given by a Hamiltonian function $f$ with Hamiltonian vector field ${v}_{f}$ is a quantum operator, def. 2, with respect to a given polarization $𝒫$ precisely if its flow preserves $𝒫$, hence precisely if

$\left[{v}_{f},𝒫\right]\subset 𝒫\phantom{\rule{thinmathspace}{0ex}}.$[v_f, \mathcal{P}] \subset \mathcal{P} \,.
Example

Over a phase space which is a cotangent bundle and with respect to the corresponding canonical vertical polarization, a Hamiltonian function is a quantum operator precisely if it is at most linear in the canonical momenta.

See for instance (Blau, around p. 35)

(…)

References

See the references at geometric quantization.

Relevant lecture notes include

Revised on April 12, 2013 00:32:43 by Urs Schreiber (131.174.41.18)