How Many People Actually Understand QM?

  • #51
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How would you show that Born is not right?
What to do if Born is wrong?
“Well, don't worry too much” (Michel).

Consider single electron/photon set up. According to the Born’s statistical interpretation, after beam splitter one obtains the statistical distribution which represents the potential reality carried by the single particle due to interaction with itself. Alternatively, you may treat it as the same particle propagating in the different waveform after interaction with the beam splitter. The difference is that according to the laws of the statistical mechanics, it is impossible to assemble back the initial wave packet. According to deterministic “interpretation” it is enough to include into the measurement assembly a device that will perform the inverse transformation in order to obtain the initial wave packet. If it is correct, then it is clear that the Mach-Zehnder interferometer will do a job. We (I and my son) discussed that as a proposal and turned to the literature to see what A.Tonomura et al. (in particular) are doing last time. We were not surprised to find that they are doing something very similar (“Double-biprism electron interferometry”, App. Phys. Lett., 84(17), 3229 (2004); “Triple-biprism electron interferometry”, JAP, 99, 113502 (2006)). Moreover, it seems that the presented results confirm our expectation. We still do not understand all necessary details of the experiment to be sure that that it (Triple-biprism is expected to present the pattern obtained after the first biprism as in the original 1989 experiment).

From the theoretical side, it was pointed out by Y. Aharonov and L.Vaidman, Phys.Rev.A, 41,11,(1990) that in case of two component wave packet the expansion coefficients are the eigenvalues of two self-adjoint mutually commuting operators (observables) which provide the necessary information about the system and have nothing to do with statistics. Now I work to extend that statement for the n-component case. In principle, I have no problems, but the description is not elegant enough so far and also I would like to see how it works when I consider n-level system. We intend to submit the paper with the detailed discussion for publication when it will be cooked enough.

What to do if Born is wrong? Nothing. You may continue if you wish to consider the deterministic evolution of the probability amplitudes (which describe the potential reality) and the interference effects between them. It is just English or psychology and we both agree that the physics is not there.

To be more serious, I expect to obtain the criterions how the “reading” devices should be constructed in order to extract the information stored in the quantum system without spoiling it (indeed using the collapse phenomenon which Michel consider useless).

With all my sincere respect, Dany.
 
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  • #52
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If there is another view, that has just as much formal "predictive power" than the usual "spacetime manifold" view of relativity, yes, I guess so. Are you hinting at Ashtekar variables (I am only vaguely aware of what it is about, honestly) ?
Well, not really.
In
http://arxiv.org/abs/hep-th/0407228
http://arxiv.org/abs/hep-th/0601027
I object that standard canonical quantization of fields is not covariant with respect to general coordinate transformations. Instead, canonical method of quantization requires a special choice of time. To fix this problem I propose a new method of quantization which, at the same time, is both canonical and covariant. In this method a "special time", or more precisely a "special" foliation of spacetime, emerges dynamically.
More surprisingly, in order to this new covariant method of quantization be consistent with the standard one, I must assume one additional covariant equation. This additional equation turnes out to contain the Bohmian equation of motion.
In this way, at the same time, from the requirement of covariance of quantum theory I obtain a dynamical preferred time and derive (not simply postulate) the Bohmian equation of motion. All this, however, refers to BM of fields, not of particles.

In
http://arxiv.org/abs/hep-th/0702060
I do something quite different, even more surprising. From the assumption of boson-fermion unification, which is suggested by superstring theory, I conclude that boson and fermion particle currents must have more similar forms. However, in the fermionic case, such a current does not transform as a vector, unless a preferred foliation of spacetime exists. In this approach I do not really derive the Bohmian equation of motion, but I argue that a Bohmian equation of motion is the most natural interpretation of these (otherwise physically obscure) currents.

To conclude, in the first approach I have derived the existence of a preferred foliation of spacetime from the assumption of general-relativistic covariance, while in the second approach I have derived the same from the assumption of boson-fermion unification. In these derivations the Bohmian equation of motion has not been assumed, but turned out to fit well with the proposed theory.
 
  • #53
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Just a very quick note (this needs much more time to be read seriously): aren't you taking psi as some kind of scalar, frame-independent quantity here ?
Normally, psi is transforming when looked upon from a different frame, and only psi psi^* is conserved under transformations. This is probably a very naive remark, after just a 10 minute skim through the articles...
 
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Just a very quick note (this needs much more time to be read seriously): aren't you taking psi as some kind of scalar, frame-independent quantity here ?
Normally, psi is transforming when looked upon from a different frame, and only psi psi^* is conserved under transformations. This is probably a very naive remark, after just a 10 minute skim through the articles...
Psi is a scalar.
For a comparison, a wave function of a spinless particle is also a scalar.
When I say scalar, I mean scalar with respect to spacetime coordinate transformations. With respect to rotations in the Hilbert space, it is, of course, a vector.
 
  • #55
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I barely understand the basics :)
 
  • #56
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Psi is a scalar.
For a comparison, a wave function of a spinless particle is also a scalar.
Under a galilean boost, in non-relativistic QM, psi is not a scalar...
 
  • #57
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Under a galilean boost, in non-relativistic QM, psi is not a scalar...
In relativistic QM, psi of a spinless particle is a scalar under arbitrary coordinate transformations, including a galilean boost.
 
  • #58
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In relativistic QM, psi of a spinless particle is a scalar under arbitrary coordinate transformations, including a galilean boost.
You mean "psi" as in KG equation ? But that's not a wavefunction, but a classical field...
If you consider Fock space, then, to a boost corresponds a unitary transformation of the Fock space, such that the momenta after transformation are the transformed momenta before transformation. For me, psi is an element of Fock space in (free) QFT.
I would intuitively understand that if you want to limit yourself to trivial transformations in Fock space (those that leave the projective space invariant) that you've then indeed fixed a foliation of spacetime, which is probably what you do in your paper, no ?
 
  • #59
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1. You mean "psi" as in KG equation ? But that's not a wavefunction, but a classical field...

2. If you consider Fock space, then, to a boost corresponds a unitary transformation of the Fock space, such that the momenta after transformation are the transformed momenta before transformation. For me, psi is an element of Fock space in (free) QFT.
I would intuitively understand that if you want to limit yourself to trivial transformations in Fock space (those that leave the projective space invariant) that you've then indeed fixed a foliation of spacetime, which is probably what you do in your paper, no ?
1. And what is the nonrelativistic limit of psi in KG equation? It is psi in the Schrodinger equation. So, are you claiming that psi in the Schrodinger equation is also a classical field and not a wave function?

2. No! Think about QFT Fock-space states in the functional Schrodinger picture. That psi is a scalar.

In fact, you will understand much better what I am talking about if you take a look at Sec. 8 of my quant-ph/0609163.
 
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