Bohmian trajectories vs. Feynman path integrals, continuous paths?

After reading some of the other posts online, I'm clear on the fact that Bohmian trajectories (of the de Broglie Bohm formulation) and the paths of the Feynman path integral formulation are very different things.

I'm wondering (and it's a naive question, no doubt), when talking about Bohmian paths - if you have a particle at position A, and then later observe the particle at position B, are you assured that the path the particle took from A to B was continuous? That is, there is no uncertainty and the particle is assumed to 'exist' along the whole trajectory? (wondering in part how the de Broglie Bohm formulation deals with particle creation / annihilation along a trajectory)

• Neither Bohmian trajectories nor paths in the path integral formalism are physical (i.e. experimentally measurable) and one shouldn't concern oneself with either except as intermediate calculations steps in "shut up and calculate" formalisms. In QM there simply is nothing that can be righteously called "particle A and particle B at positions $x_A$ and $x_B$". It would take infinite energy to localize a particle state at exactly one coordinate, which is clearly impossible, hence unphysical. – CuriousOne Oct 26 '15 at 0:05
• The Bohmian mechanics quarrels with the relativity. The very idea of continuous trajectories quarrels with the relativity (see Hardy's so-called "paradox"). There is a generalization of the Bohmian mechanics based on the idea of full/empty waves. This idea also implies continuous trajectories s.t. quarrels with the relativity. However, denying this idea we remain with extremely difficult problems. Fortunately, recent experiments open hopes about elucidating these questions. You can ask me more by writing me mail, I don't post answers on this site - the moderators know my email. – Sofia Oct 26 '15 at 0:27