This is a thought-experiment, see "Quantum Mechanics, Local Realistic Theories, and Lorentz-Invariant Realistic Theories", Phys. Rev. Lett., Vol. 68, No. 20, page 2981, year 1992, that rules out local hidden variables. But does it also rule out nonlocal influence of the measurement result of one particle, on the wave-function of another particle, before it was measured? Is this paradox a proof of impossibility of Bohm's interpretation of the quantum mechanics?

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The experiment goes as follows: an electron $e^-$ and a positron $e^+$ land, each one, on a beam-splitter, $BS1^-$ respectively $BS1^+$. One gets the following joint wave-function of two independent particles:

$ (1) \ e^+ \to \frac {|v^+> + i|u^+>}{\sqrt (2)} $

$ (2) \ e^- \to \frac {|v^-> + i|u^->}{\sqrt (2)} $

However, the couple $|u^+>|u^->$ annihilates at the point P, with production ofgamma rays, s.t. an entanglement appears

$ (3) \ |\psi> = \frac {1}{2} (|v^+>|v^-> + i|u^+>|v^-> + i|v^+>|u^-> + |2\gamma>). $

The "remainder" of the beams of the positron and electron are gathered by the beam-splitters $BS2^-$ and $BS2^+$, at which the following transformations take place.

$ (4) \ |v^±> → \frac {i|c^± > + |d^± >}{\sqrt (2)}, $

$ (5) \ |u^±> → \frac {|c^± > + i|d^± >}{\sqrt (2)} . $

So, in all, we get after $BS2^-$ and $BS2^+$ one gets the following wave-function:

$ (6) \ |\psi> → \frac {1}{4} (−3|c^+>|c^−> + i|c^+>|d^−> + i|d^+>|c^−> − |d^+>|d^−> + |2\gamma>) $.

There is nothing special until now, unless the experiment is judged by people traveling in opposite frames of coordinates.

In what follows we are interested in the detections in the detectors $D^+$ and $D^-$. Let's accompany an analyst that travels in a frame $I^+$ in which the positron is at rest. From his point of view, the positron reaches $BS2^+$ before the electron reaches $BS2^-$, s.t. after the positron reaches $BS2^+$ the wave-function is

$ (7) \ |\psi> = \frac {1}{4}(−|c^+>|u^−> + 2i|c^+>|v^−> + i|d^+>|u^−> + |2\gamma>) .$

According to the before last term, the detection in $D^+$ leaves the electron on the path $u^-$.

But an analyst that travels in the frame $I^-$ in which the electron is at rest, would hold the opposite, i.e. that the electron reaches $BS2^-$ before that positron reaches $BS2^+$, and the following wave-function is bound to appear

$ (8) \ |\psi> = \frac {1}{4}(−|u^+>|c^-> + 2i|v^+>|c^-> + i|u^+>|d^-> + |2\gamma>) .$

So, he holds that after the detection in $D^-$, the positron should have been left on the path $u^+$.

Here is the problem: the combination $|u^+>|u^->$ does not exist, it was destroyed into gamma rays. And if the gamma rays appeared, the detections in the detectors $D^±$ and $C^±$ wouldn't be obtained.

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    $\begingroup$ Bohm's interpretation cannot be falsified in an experiment that does not falsify quantum mechanics itself. $\endgroup$ Nov 8 '14 at 17:30
  • $\begingroup$ Sofia's reaction: this comment is not rigorously correct. Bohm's interpretation is NOT fully equivalent with the quantum mechanics (QT), it is based on MORE assumptions than QT. $\endgroup$
    – Sofia
    Nov 8 '14 at 17:36
  • $\begingroup$ Yes, but these extra assumptions like the assumption of "quantum equilibrium" cannot be falsified in experiments that are consistent with QM, because you can always assume that these conditions are met. Only the opposite is possible, e.g. where you could see a violation of QM consistent with a Bohmian quantum non-equilibrium situation. $\endgroup$ Nov 8 '14 at 17:39
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    $\begingroup$ You didn't discuss the difference in emission time (lower left corner) ... I suppose both are coming from a common point. But what about length and time contraction of the different paths? You get something like an inferometer. $\endgroup$
    – JDługosz
    Dec 21 '14 at 12:19
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    $\begingroup$ Sofia, please take care not to edit your posts too much. Making 3 or 4 edits to a post is fine, but any time it seems like you might have more edits than that, save up your changes and only make an edit when you have a substantial amount of stuff to change. Editing a single post 8 or 10 or more times is usually too much. $\endgroup$
    – David Z
    Dec 21 '14 at 16:38

The answer I give here is the result of many discussions with experts, and I heard many opinions in all the directions. So, I write here what is known to me by the time that I give this answer.

1. There are three interpretations known as the most elaborated, most investigated:

  • a) The standard (or Copenhagen) interpretation + the von Neumann's reduction postulate, which can act non-locally;
  • b) Bohm's interpretation, that rules out the collapse, and recognizes that entanglements work on a non-local base;
  • c) GRW (Ghirardi, Rimini and Weber) interpretation that tries to simulate the collapse by modifying the Schrodinger equation, i.e. introducing a stochastic term and non-local potential.

2. As detailed above, the idea that the entanglements resort to some sort of non-local transmission of information between the measured particles, could not be avoided. On top of this problem comes the relativity theory and tells us that space-separated measurements on particles belonging to an entanglement, don't have an absolute time-order. According to some frame of coordinates the particle A is measured first, while according to another frame, in relative movement, the particle B is measured first. This fact opens a question: $$The \ measurement \ of \ one \ of \ particles \ collapses \ the \ entanglement?$$

How can it be possible? There exists frames of coordinates according to which, by the time the above particle is measured, the other particle didn't reach the detector.

$$Then, it \ is \ not \ at \ the \ contact \ with \ a \ macroscopic \ detector \ that \ the \ collapse \ occurs?$$

$$If \ so, \ could \ it \ be \ that \ Bohm's \ interpretation, \ that \ eliminates \ the \ collapse, \ is \ correct?$$

3. Hardy's thought experiment (actually performed by Aephraim Steinberg's group), shows that Bohm's interpretation has difficulties with the relativity, as explained in the question. This thought-experiment belongs to the class of contextual experiments and applies counterfactual reasoning, but Bohm's interpretation is compatible with counter-factual reasoning.

As a consequence, some researchers are inclined to think that the quantum world admits a preferred frame, despite the theory of relativity which refutes such things.

  • $\begingroup$ There is no problem between Bohm's formulation and relativity, except when people applies the non-relativistic version to relativistic domains, but well the same difficulties arise if one was to apply non-relativistic wavefunction or the non-relativistic density matrix formulations to relativistic situations. $\endgroup$
    – juanrga
    Oct 9 '16 at 13:37
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    $\begingroup$ Don't hurry to take celebrated physicists for fools. The fact that the Bohmian mechanics clashes with relativity was widely acknowledged, and first of all by dedicated Bohm's supporters. Nobody said that the wave-function in Hardy's article is a solution of the Schrodinger equation. It may be as well a solution of Dirac's equation. Best regards, Sofia $\endgroup$
    – Sofia
    Oct 11 '16 at 1:27
  • $\begingroup$ Bohmian mechanics has been misunderstood by some of his supporters, but the same happens with other formulations. And being a "celebrated physicist" doesn't avoid one from making mistakes. I recall all the claims about how collapse in Copenhagen formulation violated relativity, until relativistic collapse models showed non-local quantum measurements in agreement with relativity. I don't know any clash of Bohmian mechanics with relativity wasn't a misunderstanding or an attempt to apply the non-relativistic formulation to relativistic domain. Best. $\endgroup$
    – juanrga
    Oct 11 '16 at 18:38
  • $\begingroup$ @juanrga You see, such things as you say, "Bohmian mechanics has been misunderstood by some of his supporters", are just words, in absence of specific examples. Let me tell you the Bohm's formalism was rigorously proved to clash with the experiment, even without using moving frames, by Partha Ghose, 16 years ago, and recently by me. But, such a talk of just exchanging general comments leads nowhere. Unfortunately, I am no more an active user of this site. If you are interested in rigorous talk, I can give you another address. Best regards, Sofia $\endgroup$
    – Sofia
    Oct 13 '16 at 11:30
  • $\begingroup$ Really? The same Partha Ghose that now claims that quantum mechanics is an approximation to classical mechanics? About his "rigorous" work on Bohm, I find a glaring mistake already in his equation (3). He doesn't understand ensembles and about trajectories he just reveals to be another critic that misunderstand the theory. He applies fermionic trajectory theory to bosons and claims to find a conflict with experiment. He is wrong. $\endgroup$
    – juanrga
    Oct 15 '16 at 18:25

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