The following was meant to be a comment rather than an answer. However, since it was a bit long for a comment so I am writing it in the answer box.

In the case of a field theory, states can be thought of as functions on the space of boundary conditions on a spatial slice. This is so because the space of boundary conditions on a spatial slice is the configuration space and (by the definition of canonical quantization) the quantum states are functions on the configuration space.

Now, in the case of a field theory in the complex plane with the radial direction taken as the time direction, spatial slices are of the form of cirlces. Therefore, the quantum states are now functions on the space of boundary conditions on a circle of fixed radius. We can chose any circle of nonzero radius to define our quantum space.

If, we denote by $H$ our space of states then the path integral on an annulus $A$ with fixed boundary conditions on its inner and outer boundary circles, defines a map

$$T_A : H\to H$$

This is the statement of the first integral in your question. Given a state on the inner boundary circle, we can get a state on the outer boundary circle by doing the path integral.

Now, instead of an annulus, consider a disc. In this case we have only one boundary. If we insert a local functional $\mathcal{O}(\phi(0),\partial_{z}\mathcal{O}(0))$ at the origin and do path integral on the whole disc then we'll of course get a quantum state in $H$ (fixing a boundary condition and then doing the path integral will give us a number. Thus, path integral will give a function on the space of boundary conditions on the boundary of the disc which is, by definition, a quantum state). Thus, path integral on the disc $D$ (of say unit radius) with a local functional inserted at the origin will define a map

$$T_{D}:\{\text{space of local functionals at the origin}\}\to H$$

This holds in any two dimensional field theory. However, in case of a conformal field theory, the above map's dependence on the geometry of the disc is much simpler than in a theory without conformal symmetry.

This post imported from StackExchange Physics at 2015-03-30 13:51 (UTC), posted by SE-user user10001