# DBB and non-locality

1. Sep 17, 2010

### LukeD

Where does non-locality come from in dBB?

I've heard that when dealing with multiple particles, dBB is a non-local theory.
The standard knowledge from studying Bell's inequalities is that any hidden-variable theory must be either non-local or non-realist. I'm ok with non-realist theories, but non-local theories weird me out when I'm trying to describe physics that I think should be completely local.

So I'm wondering: What is dBB's description of a situation where non-locality shows up (I've heard that EPR is a good example)? In what sense is dBB "non-local"? Is there any way of interpreting non-locality in dBB as being due to local, but non-realist, effects?
Links to articles would be appreciated in lieu of or in addition to explanations.

--Sorry if this is answered clearly in another thread. I've been searching for the past few hours and haven't found a treatment of this.

Last edited: Sep 17, 2010
2. Sep 18, 2010

### unusualname

dBB type theories are an alternative to "non-realist" explanations. The whole point is that they are non-local by design.

The modern consensus is that some form of non-locality exists in nature at least at planck-scale geometry, and for macroscopic non-local correlations in EPR etc the Holographic Principle is a possible mechanism.

But I don't think there's an accepted explanation for how the non-locality mechanism works in dBB theories, probably best for now to interpret the pilot wave as a descriptive idea rather than a "physical" non-local field.

3. Sep 18, 2010

### LukeD

Ok, I did some more research.
It seems to me that the non-locality shows up when you try to make a 2nd order differential equation for the trajectories. The 1st order equation, however, seems to be completely local.
So the dynamics of dBB are completely local, but concepts like "force" do not seem to be.

4. Sep 18, 2010

### Galap

Yes. Exactly.

And as for nonlocality weirding you out, as humbling as it may seem, we must always remember that the way the universe is isn't based on what human beings do or to not find sensible or intuitive. in fact, it CAN'T be sensible or intuitive to us; our brains are extremely tiny compared to the entire system. If part of the universe could emulate perfectly the whole thing, information constraints would be violated and the rest of the universe would be redundant.

5. Sep 20, 2010

### Demystifier

That is wrong. In the 1st order form, the velocity of one particle depends on the instantaneous positions of all other particles. This is not local. And indeed, it must be nonlocal according to the Bell theorem, because it is a realist theory.

6. Sep 20, 2010

### Demystifier

Re: Where does non-locality come from in dBB?

I never understood people who find non-realism less weird than non-locality.

After all, the good old high-school Newtonian gravity is nonlocal. When you learned about this theory the first time, did it really look so weird to you?

http://xxx.lanl.gov/abs/quant-ph/0607057 [Foundations of Physics, Vol. 37 No. 3, 311-340 (March 2007)]

Last edited: Sep 20, 2010
7. Sep 21, 2010

### LukeD

Edit: Actually, I guess to summarize my question: Doesn't the velocity only depend on the local wavefunction? I thought non-localities only showed up when you tried to take the wave away. The wave propagates so you clearly can't take it away without non-locality. Is there another reason for non-locality though?
Below is a method of... i guess bringing the wave back into dBB? I don't know that it's local though.

-----

Well.. standard Quantum Mechanics is already pretty "non-realistic", and it's local. To me adding more non-realistic elements is just like adding an unobservable field or wave to your theory, and these exist all over the place in Physics, so I have no problem with it.
Non-Locality, on the other hand, just doesn't seem useful. You have many more options for calculating a local law than a non-local law (for instance, it is easier to parallelize the calculations)

Ok.. but now let's add a non-realistic element ala the Quantum Trajectory Method.
In QTM, we do our calculations with the whole set of Bohmian Trajectories at once. We use $$|\psi|^2$$ as a distribution of particle positions (rather than having just 1 point), and we have a velocity field. The velocity field is derived from an action field S that follows the usual differential equation from dBB:
$$\frac{\partial S(\mathbf{x},t)}{\partial t} = -\left[ V + \frac{1}{2m}(\nabla S(\mathbf{x},t))^2 -\frac{\hbar ^2}{2m} \frac{\nabla ^2R(\mathbf{x},t)}{R(\mathbf{x},t)} \right]$$
where R is $$|\psi|$$ (square root of the distribution of positions)

The velocity field is then calculated as $$\nabla S /m$$, and the distribution of particle positions updates via the velocity field.
The wave function is then R*e^(iS/hbar)

--

This seems completely local. Am I wrong that QTM is a local, non-realistic formulation of dBB? I know that the velocity in dBB is supposedly non-local in the multi-particle case, but I just don't see where any non-locality would come into these formulas.

Last edited: Sep 21, 2010
8. Sep 21, 2010

### LukeD

Oh, I just remembered:

Newtonian Gravity is local. Poisson's Equation is local and it contains the full dynamics of gravity. The Newtonian field can even be quantized with spin-0 (Schroedinger) particles! I don't think that could be done if its dynamics weren't local (but the particles are massless and travel at infinite speed, so I might be wrong about that)

Last edited: Sep 22, 2010
9. Sep 22, 2010

### Demystifier

No!
If you know ONLY the wave function and the position of ONE particle, you CANNOT calculate the velocity of that particle. Instead, you must also know the positions of all other particles.

10. Sep 22, 2010

### Demystifier

No!
Poisson equation is local, but acceleration of ONE particle CANNOT be calculated by knowing ONLY the solution of the of the Poisson equation and position of that particle. Instead, you must also know the positions of all other particles.

11. Sep 22, 2010

### Demystifier

LukeD, perhaps you are confused by the difference between 1-particle case and many-particle case. The 1-particle Bohmian mechanics is indeed local. However, the many-particle Bohmian mechanics is not local, provided that the wave function is such that there is entanglement between the particles. Similarly, Newtonian gravity is local for the 1-particle case, but not for the many-particle case.

12. Sep 22, 2010

### unusualname

If Newtonian Gravity was local it wouldn't even predict the stability of planetary orbits. Newtonian Gravity is (in)famous for being precisely non-local.

Gravity in General Relativity however is local, and propagates at c, you need quite difficult calculations to explain why this (hardly) effects the planetary orbits.

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

(QFT is local by design, so a graviton inspired field theory of gravity will also be local)

13. Sep 22, 2010

### Demystifier

Re: Where does non-locality come from in dBB?

The simplest answer to the question above is:
From the fact that the many-particle wave function depends on many particle positions at the same time.

This means that QM is, in a certain formal sense, intrinsically nonlocal even without dBB. See also
http://xxx.lanl.gov/abs/quant-ph/0703071
for an elaboration of that view.

14. Sep 22, 2010

### Demystifier

Here you (like many others) are failing to distinguish field operators from quantum states. The field operators are indeed local in QFT, but nonlocality (more precisely, nonlocal correlations) is a property of quantum states. The nonlocality of many-particle states is related to the fact that QFT contains also NONLOCAL OPERATORS, which are certain PRODUCTS of many local field operators.

Or loosely speaking:
(LOCAL X LOCAL) + (LOCAL X LOCAL) = NONLOCAL
That's how nonlocal entanglement emerges from otherwise local quantum theory, such as QFT.

15. Sep 22, 2010

### unusualname

correlation functions in QFT are purely mathematical devices, they don't imply any physical non-locality. In a QFT of gravity, gravity will propagate at speed c.

16. Sep 22, 2010

### Demystifier

They DO imply (or are related to) nonlocal EPR correlations, which, by the way, are MEASURED. If it is not physical for you, then you have a very unusual definition of the word "physical".

But perhaps it should not be surprising that someone who calls himself unusualname uses unusualmeanings of the words. :-)

17. Sep 23, 2010

### unusualname

Yes they do imply nonlocal correlations, (by "correlation function" I mean propagator which is a mathematical device to calculate probability amplitudes) but the correlations are purely probabilistic results. I don't think many people believe a physical non-local field mechanism is responsible for the correlations, a la dBB.

But obviously, no one really knows the physical basis behind the correlations, so it is possible that the mathematics are describing a physical non-local effect. (Personally I think it's more likely that the reality we are observing is a reconstruction or projection of the simple smooth mathematical space upon which the field theory is constructed, and the non-locality results from the reconstruction or projection)

18. Sep 23, 2010

### Demystifier

That is true. However, the number of people who believe that reality does not exist before we measure it - is also small. Yet, the Bell theorem shows that at least one of these two weird options (neither of which is believed by many people) should be true. So the fact that not many people believe in any of these two options is merely a consequence of their ignorance. If they were aware of the meaning of the Bell theorem, the number of supporters of both of the two options would be much larger.

19. Sep 23, 2010

### inflector

One needs to make a choice in the matter, once one understands the implications of the state of the experiments.

I too, find non-locality to be more to my liking than non-reality. That's why I believe that dBB is the way forward.

20. Sep 24, 2010

### zonde

I wanted to ask if you consider this definition as belonging to option "reality does not exist before we measure it"?
If two observables are non-commuting then at least one of them can not be unambiguously defined as property of single particle.