# Bohmian trajectories vs. Feynman paths, always continuous?

1. Oct 25, 2015

### asimov42

After reading some of the other posts on the Forum, 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)

Thanks all.

2. Oct 25, 2015

### bhobba

Yes that's true.

Particle creation and annihilation is part of Quantum Field Theory where the fundamental thing is the field - not a particle. Here is how it works. You take a field. Model it as a large number of blobs and apply quantum rules to those blobs. You then take the blob size to zero and you get a QFT. Particles then emerge from the field - but it's the field that is quantised. The field blobs are what you apply the BM interpretation to.

The particles arise in an abstract way similar to the quantum harmonic oscillator:
https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

Thanks
Bill

3. Oct 26, 2015

### atyy

An argument for Bohmian trajectories to be smooth is given in http://arxiv.org/abs/1112.2034 "Furthermore, as (10) is a smooth function of its arguments, one expects that particle trajectories are smooth too."

I am not sure whether the "expects" is the physicists intuition or whether it can be stated as a mathematical requirement.

4. Oct 26, 2015

### bhobba

It's a particle under the influence of the quantum potential. That its continuously differentiable is the usual assumption in mechanical systems where particles are subject to a potential.

Thanks
Bill

5. Oct 27, 2015

### muscaria

Although the phase function is multivalued, its gradient is a single valued function and we have a velocity field picture where trajectories (integral curves) cannot cross or touch. The effective potential (Q + V) always acts so as to preserve this single-valuedness , which follows from the properties of the phase/Hamilton's principle function(action).

6. Oct 27, 2015

### atyy

But could a more generalized form admit non-differentiable trajectories? Before the comment on smooth trajectories, Nikolic says "In principle, the trajectories could be stochastic." So could one have something like a Wiener process in which the trajectories are not differentiable?

Edit: The OP asked about continuous, and the Wiener process trajectories are continuous even if they are not differentiable. So maybe asking whether the trajectory could be like white noise, in which the trajectories are not continuous, would be a better question.

Last edited: Oct 27, 2015
7. Oct 27, 2015

### bhobba

It obeys a differential equation which means at least its first derivative must exist. Normally, in mechanics, it is assumed to be continuously differentiable.

See its guiding equation
http://plato.stanford.edu/entries/qm-bohm/#eqs

I suspect, similar to rigged Hilbert spaces, it would be fruitful to consider the physically realizable solutions as continuously differentiable and rapidly decreasing to all orders so as to apply Fourier transforms easily.

Thanks
Bill

Last edited: Oct 27, 2015
8. Oct 27, 2015

### glaucousNoise

What's the actual upshot of a Wiener process being non-differentiable? Usually you have finite data and are computing finite differences to approximate derivatives, and the time series of a Wiener process is no exception, so it's not obvious to me there is any penalty for this fact.

9. Oct 27, 2015

### Demystifier

Yes. See e.g. the recent paper
http://lanl.arxiv.org/abs/1510.06391
and references therein.