What are the objections to stochastic quantum mechanics?

I recently discovered the stochastic interpretation of quantum mechanics, which is different from the De Broglie-Bohm theory The best article I found on it was very much a comparison of the two by Bohm himself: here (which is probably not accessible to everyone).

Therefore, a short summary:

• quantum particles perform random walks (Wiener process)
• the theory is non-local, the particles random walk is guided by the drift velocity $$\nabla\Psi/\Psi$$ in contrast to the quantum potential $$\nabla^2\Psi/\Psi$$ of De Broglie-Bohm theory.
• in contrast to De Broglie-Bohm theory, the stochastic trajectories can cross
• in contrast to path integral formulation, there is a definite trajectory, which we can however not know.
• in contrast to the Kopenhagen interpretation, particles are already at a certain position/in a certain state before measurement, we just can't know in which one (due to the stochastic nature). There is thus no collapse of wave function. (i.e. Schrödingers cat is either dead or alive, even before the measurement, we just can't know.)

I think, this interpretation is quite elegant as it avoids two concepts, that seem quite absurd: the wave function collapse of the Kopenhagen interpretation as well as the bizarre non-crossing paths of the De Broglie-Bohm theory. Furthermore, the correspondence principle is quite elegantly resolved: the stochastic fluctuations of the random walk become negligible for large systems. I see that some objections to the De Broglie-Bohm theory apply as well (the need for a guiding wave), but I don't really see them as a problem.

Alors: Why is this interpretation not more popular/well known?

Edit: There has already been this question here with some instructive answers.

• I'm really curious about how does it explain the double slit experiments? – RobertSzili Oct 15 '20 at 9:58
• there is a wave function $\Psi$, which guides the particles. This wave function takes into account both slits and interferes with itself. A measurement destroys the interference pattern, since it interacts with the particle, creating another wave function, which lacks this pattern. With weakly interacting measurements, the pattern can still be observed: journals.aps.org/prd/abstract/10.1103/PhysRevD.19.473 – Libavius Oct 15 '20 at 10:13
• @RobertSzili The pilot wave encapsulates all the non-locality. It does the self-interference bit and then tells the particle where to go. Less clear is how it can tell the particle to turn a corner as it passes through its designated slit, thereby changing its momentum and velocity vectors in apparent violation of the associated conservation laws. – Guy Inchbald Oct 15 '20 at 10:25
• @Libavius Conservation laws apply to individual quantum interactions, which is the one and only reason they must also apply in aggregate or "classical" situations. – Guy Inchbald Oct 15 '20 at 11:54
• I have seen various suggestions. Broadly, they seem built around the idea that the pilot wave can interact with other waves to make other particles move over and compensate. But I have not dug into any in detail - most seem to conceptually muddle their maths and its interpretation to at least some degree - and I do not know where the stochastic model fits in that picture. – Guy Inchbald Oct 15 '20 at 18:00

According to this text (https://web.math.princeton.edu/~nelson/papers/talk.pdf) gives different correlation values. See on the last pages.

Edit: It's about a two HO system. The theory gives different correlation values between the states of the HOs than QM's predictions.

• Answers should not rely on a link to a page, but should have enough information to be standalone if the link should ever go dead. You need to provide some summary of the linked material (or it's key points) to achieve this. – StephenG Oct 15 '20 at 12:06
• Maybe Nelsons stochastic mechanics is different from the one described in the Bohm-Hiley article I cited. The talk you linked to states, that for the two HOs' Markov processes : "This is a diffusion process and it eventually loses all memory of where it started." I think, this is a misconception: since the HOs are entangled, the system cannot be described by two individual Markov processes. Instead, there should be one 2D Markov process. This way, it would never "lose the memory". Or am I mistaken? – Libavius Oct 15 '20 at 12:43