Discontinuity of paths in phase space path integrals

Question

Berezin's famous paper "Feynman path integrals in a phase space" discusses the space of paths on which the phase space path integral is concentrated. In particular, these paths are known to be discontinuous, though not necessarily in all coordinates because a trade-off can be made achieving continuity in a position coordinate, say, at the expense of its momentum coordinate becoming less continuous, meaning possessing larger jump contributions to its behavior.

However, the Klauder and Daubechies papers on the coherent state path integral, i.e. the phase space path integral using the Glauber-Sudarshan P symbol (sometimes called the contravariant or lower symbol) of the Hamiltonian operator, expresses matrix elements of the time evolution operator as a limit of Feynman-Kac type expressions. They regularize the paths using a finite diffusion constant, allowing them to use a genuine Wiener measure on path space, and recover the coherent state path integral as the limit that the diffusion constant diverges. The Wiener integral is concentrated on continuous paths, and so there is no longer any need to trade off continuity in one coordinate for "worse" discontinuity in another; they're all continuous.

How can the limit taken in the Klauder & Daubechies paper recover the discontinuity of the paths in the phase space path integral?

Answer 1

This is an interesting question. I'm rather fond of coherent state phase-space path integrals, but their rigorous aspects are quite tricky (particularly issues of operator ordering). I'm not an expert on the proper measure theory, but it's interesting that the semi-classical analysis also has continuous-> discontinuous trajectory feature. Take as the action density $$ L_m=\frac{m}{2} (\dot x_1^2 +\dot x_2^2) + \frac{1}{2}(\dot x_1 x_2-\dot x_2 x_1)+H(x_1,x_2) $$ so that the $m\to 0$ leads to the phase space action with $x_1\to q$ and $x_2\to p$, but $m>0$ looks like a conventional Lagrangian for a Feynman or Wiener integral. Ask for a classical solution of the equation of motion from $(x_1^{(1)},x_2^{(1)})$ to $(x_1^{(2)},x_2^{(2)})$ then for all $m>0$ there will be a smooth solution, but for $m=0$ there will generically be no such path. The limit of the $m>0$ smooth path will exist, but will be discontinuous.