Path integral formulation of Bohmian mechanics

[eljose]

If Bohmain mechanics is true then the path integral:

$$\int{d[\phi]}e^{(i/\hbar)\int_{a}^{b}Ldt$$ where the Lagrangian is:

$$L=(1/2)m(dx/dt)^{2}-V(x)+(\hbar^{2}/2m)\nabla^{2}\rho$$

should be equal to its semiclassical expansion...(as in both cases are trajectories) my question is how would one reformalize Bohmian mechanics by means of path integrals?..thanks.

Anohter question if a path integral calculated exactly gives the Schroedinguer equation..then its semiclassical expansion wouldn,t give us the Hamilton-Jacobi equation?..thanx.

 

[vanesch]

If Bohmain mechanics is true then the path integral:
$$\int{d[\phi]}e^{(i/\hbar)\int_{a}^{b}Ldt$$ where the Lagrangian is:
$$L=(1/2)m(dx/dt)^{2}-V(x)+(\hbar^{2}/2m)\nabla^{2}\rho$$
should be equal to its semiclassical expansion...

Maybe I misunderstand you, but I thought that the "classical path" in Bohmian mechanics would be the classical treatment (only the extremum of the action, and not the path integral) WITH the "quantum potential" which is to be equivalent to the quantum treatment (the path integral with the phase contribution of all paths) of the Lagrangian WITHOUT quantum potential, no ?
I mean: Bohmian mechanics reduces to the running in parallel of:
"standard quantum mechanics state vectors" using the standard Lagrangian
and
"Newtonian mechanics of particles according to the standard Lagrangian PLUS (or rather, minus, for the Lagrangian) the "quantum potential" which is calculated on the basis of the standard QM state vector.
So we have two different lagrangians here: one WITH and one WITHOUT the quantum potential (derived from the state vector description of the standard quantum problem).
...
Or maybe I'm totally missing your point...
cheers,
Patrick.