# PBR no-go theorem: Is ψ epistemic or ontic? An experimental test

1. Nov 5, 2012

### bohm2

This was a recent experiment done to determine the epistemic versus ontic status of ψ:
Can different quantum state vectors correspond to the same physical state? An experimental test
http://lanl.arxiv.org/pdf/1211.0942.pdf

2. Nov 6, 2012

### Demystifier

Do the results of this experiment tell as anything that we didn't already know?

Is any SPECIFIC previously used interpretation of QM ruled out?

I think the answer to both questions is - NO.

3. Nov 6, 2012

### bohm2

It arguably rules out the Rob Spekkens/Leifer ψ-epistemic models.

4. Nov 6, 2012

### Demystifier

But did anybody (except perhaps Spekkens and Leifer) seriously used these models and honestly believed that they might be correct?

5. Nov 6, 2012

### Quantumental

I fail to see that they have been ruled out?

Personally I think that some deeper theory that will unify QM and GR will still be "found" some day. I just don't believe that either Bohm, Everett or any of the other "candidates" are right.

6. Nov 6, 2012

### bohm2

Why? I would imagine that the authors would argue that for all intensive purposes ψ-epistemic models are ruled out even though some unlikely and/or far-fetched loopholes mentioned in the article still exist (like in Bell's). Interestingly, some of the authors (e.g. Rudolph) were ψ-epistemic enthusiasts. Spekkens appears to be a hold-out, though. I'm not sure if this experiment would change his position. Beyond these models suggested by these authors, aren't there other ψ-epistemists, also?

EDIT: Another experiment drawing similar conclusions just published:
Experimentally probing the reality of the quantum state
http://lanl.arxiv.org/pdf/1211.1179.pdf

Last edited: Nov 6, 2012
7. Nov 7, 2012

### Ilja

I would like to point to the fact that epistemic interpretations which correspond to the dBB rule of defining effective wave functions for subsystems by $\psi(q,t)=\Psi(q,q_{env}(t))$, where $q_{env}(t)$ is the trajectory of the configuration, cannot be ruled out by similar arguments if they give the $\Psi$ of the whole universe an epistemic interpretation.

The point is that all the properties of the wave function of any accessible system which make it look like a physical entity it borrows in that formula from $q_{env}(t)$.

In particular, the prescription to make the PBR experiment can, from this point of view, never create a state where it is unclear how it has been prepared. The prepared state is always part of greater system, which contains the devices used for the preparation. This environment can, of course, also include some random throwing of dices which preparation procedure has to be used. But, anyway, $q_{env}(t)$ contains also the result of this dice throwing.

So $\psi_1(q)=\Psi(q,q_{prep\uparrow},\uparrow)$ and $\psi_2(q)=\Psi(q,q_{prep\downarrow},\downarrow)$, thus, even if ψ itself is purely epistemic, the λi of the two states will be different already because their preparation procedure is different.

8. Nov 7, 2012

### bohm2

Last edited: Nov 7, 2012
9. Nov 7, 2012

### DrChinese

Considering that the Bohmian view* essentially says that there IS a specific well-defined value for non-commuting observables, although we cannot know it: I would think it is just a matter of time before theorems in the spirit of PBR clearly conflict with the Bohmian hypothesis (non-local hidden variables).

*As I understand it, that being a limited understanding.

10. Nov 7, 2012

### Ilja

At least at the current moment I think so.

I would not completely exclude that it nonetheless may cause problems. The paleoclassical interpretation is, last but not least, a radical version where the wave function of the whole universe has no own real degrees of freedom. And one has to think about PBR if one constructs a subquantum theory which gives QT in some limit.

But I argue, indeed, that PBR is not an impossibility proof for the paleoclassical interpretation.

No, there is only a well-defined value for the configuration q(t). Momentum is not defined. The result of momentum measurements depends on the configuration of the momentum measurement device, except in the case of a momentum eigenstate.

But why you think that an impossibility theorem may be possible? That dBB in quantum equilibrium gives QT predictions is a simple theorem. That quantum equilibrium will be approached is Valentinis subquantum H-theorem - ok, with caveats similar to the original H-theorem, but I cannot see how this may become a problem. And I see no reason to doubt that dBB in itself is consistent.

The wave function as being real is at least a popular variant, if not the standard dBB interpretation.

11. Nov 7, 2012

### bohm2

I think most Bohmians treat the wave function as ontic, so they would rejoice in the PBR theorem? In fact, Valentini referred to PBR at "the most important general theorem relating to the foundations of quantum mechanics since Bell's theorem". Having said that, I'm not sure if PBR would have any implications for Bohmians who treat the wave function as "less ontic" (more nomological) like the DGZ group/Maudlin? I doubt it, though, since there is no essential place for any notion of probability in Bohmian mechanics, at least, regarding the basic description of particle motion. But, I haven't seen anyone from this group comment on the PBR theorem even though I've been searching.

Last edited: Nov 7, 2012
12. Nov 7, 2012

### DrChinese

1. An element of dBB I usually forget about... I guess no well-defined spin either.

2. Bohm2 said essentially the same thing. I guess I just don't understand that. If there are (nonlocal) hidden variables that determine outcomes, how can a probability amplitude/wave function have reality? I thought outcomes could be predicted if all input values were known. Seems to me that HV implies a form of determinism which in turn implies well-defined values for elements of reality.

13. Nov 7, 2012

### Ilja

The simplest approach is indeed without spin.

There have been different approaches to relativistic QFT for fermion fields, from Bell (beables for quantum field theory) to Colin and Struyve.

But I favour, of course, my own approach. In arXiv:0908.0591 I present a way to obtain two Dirac fermions together with a scalar field from canonical quantization of a scalar field with W-like potential. Once scalar fields in itself present no problem for dBB field theory, this solves the problem even if dBB is not even mentioned there.

ψ is in dBB theory not a probability amplitude, but a field with guides the configuration q(t).

There is only a special state, named quantum equilibrium, where we have a connection between ψ and probabilities for q in form of the Born rule. It is because of a "subquantum H-theorem" that arbitrary probability densities become, after some time, almost indistinguishable from this quantum equilibrium. So there is no fundamental connection at all.
Probabilities play the same role as in classical thermodynamics.