6
$\begingroup$

I'm reviving and expanding this question, because of the new paper today, by Aaronson et al. The more general question is:

How does quantum-potential Bohmian mechanics relate to no-go theorems for psi-epistemic theories?

According to Bohmian mechanics, there is a wavefunction as in conventional quantum mechanics, which evolves according to a Schrödinger equation, and then a set of classical degrees of freedom, which evolve according to the gradient of the phase of this wavefunction. A Born-rule probability distribution for the classical degrees of freedom is preserved by this law of motion, so Bohmian mechanics provides a deterministic hidden-variables theory, albeit a nonlocal one.

However, the equations of motion for the classical variables can be rewritten in terms of two potentials, the local potential appearing in the Schrödinger equation, and a nonlocal "quantum potential" which depends on the form of the wavefunction. This means that Bohmian mechanics for any given wavefunction can be replaced by a nonlocal deterministic theory in which there is no wavefunction. (Note that the quantum potential will be time-dependent unless the wavefunction was an energy eigenstate; thanks to Tatiana Seletskaia for emphasizing this point.)

Meanwhile, there has recently been theoretical interest in ruling out so-called "psi-epistemic" hidden-variables theories, in which the quantum wavefunction is not part of the ontology. Obviously, "quantum-potential Bohmian mechanics" is a psi-epistemic theory. It ought to be relevant to this enterprise. But it may require some mental gymnastics to bring it into contact with the PBR-like theorems, because the mapping from quantum mechanics to these quantum-potential theories is one-to-many: scenarios with the same quantum equations of motion, but different initial conditions for the wavefunction, correspond to different equations of motion in a quantum-potential theory.

NOTE 1: Empirically we only have one world to account for, so a real-life quantum-potential theorist in principle shouldn't even need their formalism to match the whole of some quantum theory. Quantum cosmology only needs one wavefunction of the universe, and so such a theorist would only need to consider one "cosmic quantum potential" in order to define their theory.

NOTE 2: Here is the original version of this question, that I asked in January:

The "PBR theorem" (Pusey-Barrett-Rudolph) purports to show that you can't reproduce the predictions of quantum mechanics without supposing that wavefunctions are real. But it always seemed obvious to me that this was wrong, because you can rewrite Bohmian mechanics so that there's no "pilot wave" - just rewrite the equation of motion for the Bohmian particles, so that the influence coming from the pilot wave is reproduced by a nonlocal potential. Can someone explain how it is that PBR's deduction overlooks this possibility?

$\endgroup$
  • 1
    $\begingroup$ I still believe that my answer - that Bohmian mechanics is psi-ontic - is the correct one, although I haven't read the Aaronson et al. paper yet. I've updated my answer to address your updated question. (The new stuff is in the last paragraph.) $\endgroup$ – Nathaniel Mar 13 '13 at 5:08
  • $\begingroup$ I think a more general way to express this question would be "can Bohmian mechanics be expressed as a psi-epistemic theory?" I suspect the answer to this would be no, but I don't know for sure. (But I don't think there are many users left on this site whose interests are in this area.) $\endgroup$ – Nathaniel Mar 13 '13 at 5:09
  • $\begingroup$ Scott Aaronson has told me 1) that another such representation exists and was discussed here physics.stackexchange.com/questions/38064/… ... $\endgroup$ – Mitchell Porter Mar 13 '13 at 7:16
  • $\begingroup$ ... 2) that PBR theorems are about theories where there's an overlap in the ontic base of the epistemic states, i.e. if psi1 and psi2 are nonorthogonal quantum states, then the corresponding sets of possible ontic states overlap. Hopefully in a few days we can put all these pieces together and figure out a comprehensive answer. $\endgroup$ – Mitchell Porter Mar 13 '13 at 7:18
2
$\begingroup$

I once watched an online seminar by Robert Spekkens, who said something along the lines that no-go theorems are interesting, because they put constraints on what an epistemic interpretation of quantum mechanics can look like. Any no-go theorem makes a certain set of assumptions, and if the theorem is correct, we know that we must avoid at least one of those assumptions if we're to make a successful theory.

The Pusey-Barrett-Rudolph paper spells out some of its assumptions. (They do it most explicitly in the concluding paragraphs.) There may well be additional unmentioned assumptions (e.g. causality), but the ones they specifically mention are:

  • There is an objective physical state $\lambda$ for any quantum system

  • There exists some $\lambda'$ that can be shared between some pair of distinct quantum states $\psi_1$ and $\psi_2$. That is, $p(\lambda=\lambda' | \psi=\psi_1)$ and $p(\lambda=\lambda' | \psi=\psi_2)$ are both non-zero. (This is what it means for an interpretation to be epistemic, according to Spekkens' definition.)

  • The outcomes of measurements depend only on $\lambda$ and the settings of the measurement apparatus (though there can be stochasticity as well)

  • Spatially separated systems prepared independently have separate and independent $\lambda$'s.

It is from these that they derive a contradiction. Any theory that fails to make all of these same assumptions is unaffected by their result.

I'm fairly sure that under the standard formulation of Bohmian mechanics violates the second one, i.e. Bohmian mechanics is not an epistemic theory in the sense of Spekkens. This is because in Bohmian mechanics the physical state $\lambda$ consists of both the particle's real position and the "quantum potential", and the latter is in a one-to-one relationship with the quantum state.

If I understand correctly, you're suggesting that you could instead think of the physical state, $\lambda$, as consisting only of the positions and velocities of the particles, with the non-local potential being considered part of the equations of motion instead. But in this case, the third assumption above is violated, because you still need to know the potential, in addition to $\lambda$, in order to predict the outcomes of measurements. Since this formation would violate the assumptions of Pusey, Barrett and Rudolph's argument, their result would not apply to it.

In your updated question, you clarify that your suggestion is to fix the wavefunction to a particular value. In that case it's surely true that Bohmian mechanics reduces to particles moving according to deterministic but non-local equations of motion. But then you have only a partial model of quantum mechanics, because you can no longer say anything about what happens if you change the wavefunction. My strong suspicion is that if you take this approach, you will end up with an epistemic model, but it will be a model of only a restricted subset of quantum mechanics, and this will result in Pusey et al.'s result not being applicable.

$\endgroup$
  • $\begingroup$ In the usual formulation of Bohmian mechanics, the equations of motion for the particles have some dependence on a wavefunction, which acts as a "pilot wave". But I'm saying you can just substitute a given wavefunction into those equations of motion, so you're left with nothing but particles that interact according to a nonlocal potential. So in this altered formulation, the quantum potential is not part of the physical state, it's part of the equations of motion. $\endgroup$ – Mitchell Porter Jan 6 '13 at 7:00
  • $\begingroup$ Well, ok, but in that case, if I understand you correct, to predict the outcomes of measurements you still need to know the wavefunction, otherwise you'd just have some partially unknown equations of motion. So if you choose not to count the wavefunction of part of $\lambda$, the theory violates the third assumption in my list, so Pusey-Barrett-Rudolph result doesn't apply to it. $\endgroup$ – Nathaniel Jan 6 '13 at 7:46
  • $\begingroup$ I've edited my previous comment into the answer. $\endgroup$ – Nathaniel Jan 6 '13 at 7:59
1
$\begingroup$

In mathematical terms, the PBR theorem shows that the relation between the states of a more fundamental theory and the states of quantum mechanics (wave functions) is functional (i.e., to every state of the more fundamental theory, there belongs at most one wave function; the relation does of course not have to be onto or one-to-one).

If this is enough for you to say "wavefunctions are real" (your quote), then so be it. In general, however, this conclusion cannot "be obtained" (by arguments) from the result mentioned in the first paragraph. This is why one uses a language (psi-epistemic, etc.) which codifies exactly the functional dependency as mentioned above. In my view it is a mistake to take the exclusion of a non-functional relation between the states of a more fundamental theory and wave functions to be a sign that "wavefunctions are real".

Finally, let me just note that a state in Bohmian mechanics consists of the pair (wave-function, positions of particles). Therefore, the relation between the states in this theory and the states in quantum mechanics is certainly functional (projection on the first component if you wish). - As expected no problem and no implications here.

Best, J.K.

Ps: Please point out if there is an error here! Pps: Put in the above way, the PBR theorem is almost "trivially true".

$\endgroup$
0
$\begingroup$

Unfortunately I don't have the points to leave comments on other answers, as Nathaniel is correct, but since no answer has been accepted, here is the answer to your question with a citation to back it up:

You've made an incorrect assumption.

Meanwhile, there has recently been theoretical interest in ruling out so-called "psi-epistemic" hidden-variables theories, in which the quantum wavefunction is not part of the ontology. Obviously, "quantum-potential Bohmian mechanics" is a psi-epistemic theory. It ought to be relevant to this enterprise.

Bohmian mechanics is not ψ-epistemic, it is ψ-supplemented (a subset of ψ-ontic). See http://dx.doi.org/10.1007/s10701-009-9347-0

$\endgroup$
  • $\begingroup$ I am talking about a rewrite of Bohmian mechanics in which the wavefunction is absorbed into the equation of motion for the configurational beables. In such a theory, ψ no longer exists. $\endgroup$ – Mitchell Porter Jun 27 '17 at 5:43
  • $\begingroup$ So like Poirier's ( dx.doi.org/10.1016/j.chemphys.2009.12.024 ) formulation? Since the trajectories are still real and non-overlapping, every physical state is still consistent with only one pure quantum state. Therefore by definition can't be considered ψ-epistemic (as defined by Harrigan and Spekkens above). $\endgroup$ – Toby Hawkins Jun 27 '17 at 11:38
  • $\begingroup$ That seems to be a version of "many interacting worlds", in which all the Bohmian trajectories for a given ψ are real at once. I'm just talking about one trajectory, with an equation of motion in which the role of ψ is entirely replaced by a potential term. $\endgroup$ – Mitchell Porter Jul 3 '17 at 6:00
  • $\begingroup$ This is a common misunderstanding. In BM the guidance field is considered real over 3N-space (in the case of spinless nonrelativistic particles). From Bell (1989), 'No one can understand this theory until he is willing to think of $\Psi$ as a real objective field rather than just a "probability amplitude". Even though it propagates not in 3-space but in 3N-space'. You seem to want to create a unique interpretation as suggested by Norsen (2016). This wouldn't be BM though. $\endgroup$ – Toby Hawkins Jul 3 '17 at 10:42
  • $\begingroup$ I am not talking about a one-particle trajectory. You can have your N particles, the point is that whatever ψ over 3N-space that you start with, you can substitute it into the Bohmian equation of motion for the N particles, and obtain a new global equation of motion with a specific nonlocal potential (Norsen's Q). $\endgroup$ – Mitchell Porter Jul 3 '17 at 11:41

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.