# Spin states and path integral formulation

I'm trying to understand the path integral formulation of spin. Such a thing, supposedly, can be done. In doing so I've discovered a conceptual problem with my understanding of path integrals in general. This is despite being pretty good at doing actual QFT and QM path integral problems.

My question is how to map the space of classical trajectories to the quantum state space.

Using the Lagrangian formulation, a particle's trajectory lives on its tangent bundle. So, for example, with a particle in $\mathbb{R}$, a particle's trajectory is a path on $T\mathbb{R} \cong \mathbb{R}^2$. These are the paths that go into the path integral formulation, and which yield quantum mechanical states given by square-integrable wavefunctions $\psi: \mathbb{R} \to \mathbb{C}$.

So there is evidently a map that takes $T\mathbb{R} \cong \mathbb{R}^2$ to the Hilbert space of square-integrable function on $\mathbb{R}$. And if I'm understanding the path integral formulation of spin correctly, the same map should take $TS^2$ (the classical spin space) to $\mathbb{C}^2$/{norm = 1} $\cong S^3$, which are the spin states. Is this right? Is there an intuitive or simple way to understand this map?