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is it really so clear t...Dear authors of the blog,<br /><br />is it really so clear that one needs a preferred (albeit dynamical) foliation of spacetime into spacelike hypersurfaces for relativistic BM?<br /><br />To me, the situation seems to be more complicated: the foliation is not necessary to formulate a Lorentz covariant and nonlocal law of motion (this could e.g. be done by forward and backward light cones as well, see http://arxiv.org/abs/quant-ph/0105040 for a similar idea).<br />The crucial argument in favor of a foliation then seems to be that one has no idea how to statistically analyze laws of motion different from ones generated by velocity vector fields.<br />This argument is, however, just a "human" one: one does not know better at the moment. Could it not be that for relativistic physics of N non-independent particles, the way how to do statistics has to be changed in general?<br /><br />An indication that this might be so is Wheeler-Feynman electrodynamics (http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.21.425, http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.17.157): Its laws of motion are very natural from a certain perspective but can also not be formulated using a velocity vector field. Consequently, no way of analyzing the statistics is known.<br />In the expectation that many people might object to the theory because of its unusual view towards "causality", it is in any case an example for how the question of the analyzability of law that are not of a velocity vector field type arises, also in a context different from relativistic BM.<br /><br />I'm sometimes wondering why this line of thought is not discussed more intensely when it comes to relativistic BM...<br /><br />Best, Matthias<br />Anonymoushttps://www.blogger.com/profile/14975076584073019448noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-44279529954160288062014-06-10T05:12:55.285-07:002014-06-10T05:12:55.285-07:00Hi Robert,
" If the q values has any physica...Hi Robert,<br /><br />" If the q values has any physical effect, it must be possible to separate out a subset of the original ensemble"<br /><br />I think this is not correct for BM. Let me illustrate why:<br />"Having a physical effect" could mean e.g. that the initial position influences where the particle ends up in the experiment. This is so in BM.<br />Why should it therefore be possible by the experimental (and statistical means) that you can *control* the position directly? In fact, one can prove that for BM this is not possible in the generic situation of quantum equilibrium. See http://arxiv.org/pdf/quant-ph/0308039v1.pdf, ch. 11 "Absolute Uncertainty" (pp. 46).<br /><br />The simple argument that particle positions are physical in BM is that you need something to represent what is happening in physical space in your theory. Otherwise (strictly speaking and not accepting tacit assumptions or conventions) the theory is an empty mathematical formalism.<br /><br />Best, Matthias<br />Anonymoushttps://www.blogger.com/profile/14975076584073019448noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-46417833267265397362014-06-10T05:05:25.690-07:002014-06-10T05:05:25.690-07:00Hi Florin,
the hydrogen situation in BM is only s...Hi Florin,<br /><br />the hydrogen situation in BM is only so for the ground state. In different states the electron would - according to BM - move e.g. in circles or even quite chaotically (e.g. in superpositions of different eigenstates, see the illustration at http://bohm-c705.uibk.ac.at/ ).<br />So if you want to model a situation in BM in which the electron moves around the core, you should better choose the wave function accordingly. Otherwise you just refer to a different physical situation than you inteded to.<br /><br />The possibility to radiate and fall into the core depends on a mechanicsm that would allow this. The normal Schrödinger equation for the hydrogen atom simply does not include such a term. So an electron that is modeled by this equation also cannot do this.<br />Don't apply the intuition about a different (and classical) theory here!<br /><br />Best, MatthiasAnonymoushttps://www.blogger.com/profile/14975076584073019448noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-69654207569348787512014-06-06T13:52:05.047-07:002014-06-06T13:52:05.047-07:00Hilary Putnam: “This leads me to wonder if the dif...Hilary Putnam: “This leads me to wonder if the difficulty over many decades of finding a compelling scientific realist picture that answers the "collapse" dilemmas is not, at bottom, due to the fact that we are prematurely trying to say what the psi-function is and how it figures in the dynamics before physics has come up with a good account of what its "theater of operation" really is; … As I said (typed) a moment ago, that can't be the true theater of operations, and it may well be premature to try to resolve the "collapse/no collapse" problem until …”<br /><br /><br />Indeed, ‘theater’ before anything else.<br /><br />Let us consider only 3-player-theater (time, space and particles). While all the three have the duality-personality, I would like to ‘simplify’ the issue as below. <br />Space, 99% wave.<br />Time, 99% wave.<br />Particle (such as electron), 99% particle.<br /><br />That is, the wave-function of electron is actually describing a ‘particle’ floating on the space-time-wave. Electron as a ‘particle’ is always as a ‘whole’ while its ‘position and momentum’ are depending upon the space-time-wave (described by the electron-wave function). Thus, there is no ‘collapse’ issue as there is nothing to collapse about. The visualized collapse of electron in our measurement is caused by the space-time wave which is ‘structured’ by the space-time force.<br /><br />F (space-time Force) = K ħ/ (delta S x delta T); the ‘theater’.<br /> K (coupling constant, dimensionless); ħ (Planck constant); S (space); T (time). <br />Then, delta P (linear momentum) = F x delta T = K ħ/ (delta S) <br />So, delta P x delta S = K ħ<br />When, K is near to 1 (but a bit smaller than 1), then delta P x delta S > ħ (the uncertainty principle).<br />When K ħ is near to (0 ħ), the F is ‘gravity’.<br />Tienzenhttps://www.blogger.com/profile/05842156512465678309noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-37892281603610706462014-06-03T06:17:21.083-07:002014-06-03T06:17:21.083-07:00Thanks for your response, Florin, but I think you ...Thanks for your response, Florin, but I think you missed the point. BM asks us to consider an ensemble of systems which all have the same pilot wave but different initial values of the position variable (let's call it q). If the q values has any physical effect, it must be possible to separate out a subset of the original ensemble: let's say all those initial q values that result in the particle going through the top slit. Then we have a new ensemble that violates the predictions of QM (100% probability of top slit vs. 50%).<br /><br />If it is not possible to use the q values to create such a sub-ensemble, then the q values have no physical reality - they have no physical effect. <br /><br />As far as the theorem you mention, it holds under the assumption that the initial distribution of q values corresponds to the square modulus of the pilot wave. If the filtering process is possible, then this assumption is violated and the so the theorem doesn't hold. <br /><br />Von Neumann wrote a paper showing that the filtering process must be possible if the quantity has any physical effect. (He describes the process as creating a membrane that is permeable to systems with some values of the variable (q) and impermeable to others.) I don't have the reference at hand but you can find it in a footnote in his book.Robert Oerterhttps://www.blogger.com/profile/09708981993708509662noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-63977189988850922132014-06-02T12:31:02.684-07:002014-06-02T12:31:02.684-07:00Dear Robert,
I am not a supporter of BM, but I wi...Dear Robert,<br /><br />I am not a supporter of BM, but I will attempt to give an answer in BM spirit (the real Bohmians please jump in and correct me if I am saying something wrong). The way the question is posed is unclear. What does "create a filter that will sweep up a subset of the systems that are represented by a particular pilot wave." actually mean? How is the filter going to be implemented in practice?<br /><br />Let me speculate on the filter implementation in a particular case: in a double slit experiment. Let us block one of the slits (filtering out the paths that go thorough there). Then what happens? The interference pattern vanishes. What is the BM explanation? The quantum potential changes. Therefore the assertion "But sweeping up a subset in this way would lead to an ensemble that has a non-quantum mechanical distribution" is false.<br /><br />Therefore the solution to your objection is to consider contextuality: whenever one changes the experimental setting, the quantum potential in BM changes as well and no predictions contradicting standard QM predictions are possible. In fact it is a mathematical theorem that BM recovers exactly the standard QM predictions.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-57462772548316057532014-06-02T11:55:41.633-07:002014-06-02T11:55:41.633-07:00Thanks for posting this interesting discussion. Ma...Thanks for posting this interesting discussion. Maybe someone here can answer a question that's been bugging me ever since I read von Neumann's Mathematical Foundations of Quantum Mechanics.<br /><br />Von Neumann uses a "filter argument" in his derivation of the quantum mechanical entropy. The filter argument says that if there is any physical difference between two systems, then a filter can be created that will let one pass through but not the other. <br /><br />Applied to Bohmian QM, the filter argument says that if the Bohmian position variables have any physical effect, then we can create a filter that will sweep up a subset of the systems that are represented by a particular pilot wave. But sweeping up a subset in this way would lead to an ensemble that has a non-quantum mechanical distribution, so Bohmian QM would make different predictions from standard QM.<br /><br />On the other hand, if the Bohmian position variables have absolutely no physical effect, then they are unphysical by definition, and can be dropped from the theory.<br /><br />This seems to put an end to Bohmian QM. But none of the discussions I've read of Bohm's approach make any mention of von Neumann's filter argument. (Note this argument is completely different from the one set out in von Neumann's "no hidden variables proof.")<br /><br />Does anyone know of a response to this objection?Robert Oerterhttps://www.blogger.com/profile/09708981993708509662noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-62532318327355771482014-05-29T14:21:36.309-07:002014-05-29T14:21:36.309-07:00My first thought on this question was that Bell...My first thought on this question was that Bell's theorem works on a sufficiently neutral basis (mostly independent of actual QM theory) so as to let us reflect on the measurement problem without bothering too much of issues such as the reconciliation of QM and GR.<br />However after a second thought... Bell's theorem involves at least space time.<br />Moreover recent research seems to establish a conceptual link between quantum entanglement and wormholes in GR. See <br />http://mobile.extremetech.com/#/extreme/575-wormholes-are-just-quantum-entangled-black-holes-says-new-research<br />I do not know what it's worth but it might occur that a deeper connection exists between spacetime geometry and the measurement problem after all, which a new physical theory would uncover?<br />Quentin Ruyanthttps://www.blogger.com/profile/18395553776256376317noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-2266917130752929002014-05-29T14:16:42.646-07:002014-05-29T14:16:42.646-07:00My objection for BM starts with its treatment of t...My objection for BM starts with its treatment of the Hydrogen atom: the electron is stationary at a fixed distance from the proton (otherwise the electron would radiate energy). Now in BM trajectories are surreal, so this is not a problem, but why is this distance distinguished from any other distances? Because God made it so? Something is fundamentally wrong here. <br /><br />I think the problem of quantum mechanics interpretation is ill-posed: what is needed is to reconstruct quantum mechanics from first principles. Then the right QM interpretation will follow naturally. What would those principle be? I have a proposal (which I am in the process of writing up for publication):<br /><br />1. laws of nature are invariant under time evolution <br />2. laws of nature are invariant under tensor composition (if system A is described by QM, and system B is described by QM then the composite system A tensor B is described by quantum mechanics as well)<br />3. positivity (ability to have a state space)<br /><br />4. nature violated Bell's inequalities (technical postulate needed to distinguish between classical and quantum mechanics) <br /><br />The full QM in the C* algebra formalism follows as a mathematical theorem from those 4 postulates, nothing more, nothing less.Florin Moldoveanuhttps://www.blogger.com/profile/01087655914212705768noreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-5710735987537150972014-05-29T12:25:58.360-07:002014-05-29T12:25:58.360-07:00An interesting read. IMHO the compelling scientifi...An interesting read. IMHO the <i>compelling scientific realist picture that answers the collapse dilemma</i> is akin to the optical Fourier transform. A photon is an extended-entity wave, when you detect it with an electron which is another wave, you see a dot on the screen. But photons and electrons are as pointlike as seismic waves and hurricanes. <br /><br />As for what psi-wavefunction is, see weak-measurement work by Aephraim Steinberg et al and Jeff Lundeen et al. Mindful of LIGO think "displacement current", think "spacewarp". <br /><br />Unifying GR and QM is nice, but a red herring. A photon conveys inertia from the emitting body to the absorbing body, and inertial mass equates to active gravitational mass, so the virtual photon is a virtual graviton too. Besides, remember the current in the wire? The electromagnetic forces largely cancel, but not quite. Then when you stop that current, they still don't. <br /><br />Cosmological processes don't take place in spacetime, they take place in space. Spacetime is static. The map is not the territory. And curved spacetime relates to <a href="http://cpl.iphy.ac.cn/EN/Y2008/V25/I5/1571" rel="nofollow">inhomogeneous space</a>. Remember Wheeler's geons? The strong curvature regime isn't down near the event horizon. It's nothing to do with gravity. It's everything to do with electromagnetism. I kid ye not. Check out <a href="https://www.google.co.uk/#q=electromagnetic+geometry" rel="nofollow">Percy Hammond</a>. <br /><br />John Duffield<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8010406940823317481.post-53651516379664766252014-05-29T08:28:11.309-07:002014-05-29T08:28:11.309-07:00Welcome to the blogosphere! Now you just need to a...Welcome to the blogosphere! Now you just need to add some Everettians to your list of correspondents.Sean Carrollhttps://www.blogger.com/profile/07597705630986874497noreply@blogger.com