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According to this answer, Lorentz Ether Theory (LET) is experimentally indistinguishable from SRT, because the Ether is immobile and practically redundant.

However, on quantum scales, photons are considered to be excitations of the EM field. Is this field "stationary", like ether in LET, then, in order to be in agreement with SRT? I am probably mixing apples and oranges here, but I wonder what's the point of this field then, compared to ether?

Furthermore, I've seen articles (disclaimer: not really scientific) which try to explain the double slit experiment as if the "ether" (or whatever "medium" or "field" you are considering) is being disturbed by the photon into waves, similar to what a tiny rock would do when thrown into water. Observing a photon at one of the slits would absorb (much of) its energy, making it unable to disturb the medium enough to show the pattern.

Mathematically, it would probably boil down to classical interference explanation, which I believe correctly calculates the behavior, but I am presuming it would be a much more down to earth explanation compared to "shut up and calculate".

So obviously, something is wrong about this reasoning?

(Update)

To address the comment from @CuriousOne, I believe that many misconceptions exist about QM ("nobody understands quantum mechanics"), but it's still strange to have this information presented in this way in e.g. Wikipedia, which everyone is free to edit and improve.

For example, if you read about the double slit experiment on Wikipedia:

The modern double-slit experiment (...) displays the fundamentally probabilistic nature of quantum mechanical phenomena.

or

Wheeler's delayed choice experiments demonstrate that extracting "which path" information after a particle passes through the slits can seem to retroactively alter its previous behavior at the slits.

Internet is filled with articles and "interpretations" like this, which seem far away from Occam's razor (I hate comments that photons "don't like to be observed" and "send FTL information to their entangled buddies when someone observes them"), but still the usual answer on Physics SE is "everything you've read online is wrong". You could probably even recommend 10 books on the subject, and each of them would have a different "interpretation".

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  • $\begingroup$ Either it's experimentally indistinguishable and then it's being washed out by Occam's razor or it's not indistinguishable and then the proponents of the hypothesis have to do their homework and propose an experiment to distinguish it. The double slit experiment is fully Lorentz invariant (interference does not go away or change except for the Lorentz factors when it's seen from a moving observer), so it's not clear to me how it would change anything. In all honesty... when people start combining trivial stuff into non-trivial scenarios, it starts smelling of pseudo-science. $\endgroup$ – CuriousOne Mar 19 '16 at 9:05
  • $\begingroup$ @CuriousOne: I think the issue here is that quantum mechanics explains the double slit experiment using probability (especially when single photons/electrons are being fired), and the "path" of this "single electron" is a rather unintuitive notion. This is then further complicated by the fact of "observing" the electron at the slit, which all sounds more voodoo than these simple attempts of explanation. $\endgroup$ – Groo Mar 19 '16 at 9:29
  • $\begingroup$ @Groo: There are no paths in quantum mechanics. That's an old misconception that just doesn't want to die. So is the use of "probability". Quantum mechanics is perfectly causal, it's just uncertain when measurements are involved, but that's not the same as probability. There is nothing complicated about observing electrons. Plenty of companies will gladly sell you the right equipment for that. $\endgroup$ – CuriousOne Mar 19 '16 at 9:31
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    $\begingroup$ the QFT creation and annihilation operator fields are Lorenz invariant by construction, so your thinking of them as a lorenz invariant ether is fine. $\endgroup$ – anna v Mar 19 '16 at 11:36
  • $\begingroup$ @curiousone: the "path" I am mentioning is presumed in experiments which try to detect "which way the photon went", and then remove interference by doing so. I never wrote there is anything complicated in observing them, not sure how you got that impression. $\endgroup$ – Groo Mar 19 '16 at 11:52
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There is first quantization, which is the first course in quantum mechancs, the solutions of Schrodinger's equation and then the relativistically covariant Dirac and Klein Gordon. A basic postulate of quantum mechanics is that the solutions of these equations called wavefunctions, when complex conjugated and squared give the probability for the problem under study, for example the x,y,z of the electron around the hydrogen atom, a probability locus called orbital.

But physics is much more than observations of atomic spectra. Scattering experiments had to be explained, first in cosmic ray experiments, and then in accelerators. This introduced the concept of the scattering matrix and the need to calculate wavefunctions for systems more complicated than just a potential well with analytic solutions. The perturbative expansion of the needed functions gave first order approximations and then corrections that could describe experiments.

This led to quantum field theory: the simplest, ground state solution for the equation appropriate to the interaction, a plane wave, was taken as a ground state on which creation and annihilation operators followed the manifestation of the scattered particles. So the field is a photon field, an electron field, etc, in potentia. The operators operating on this field at (x,y,z,t)would excite a photon, for example, with the appropriate energy.

Calculations were greatly helped because in the QFT formalism Feynman diagrams gave a clear recipe of how to calculate crossections of scatterings and decays of elementary particles, using these annihilation and creation operators on the ground state of the fields under study.

However, on quantum scales, photons are considered to be excitations of the EM field.

In QFT the field exists, and yes , it is like a Lorenz invariant ether, at every (x,y,z,t), but its expectation value is zero, i.e. there are no photons unless a photon creation operator acts on that (x,y,z) at time t.

Is this field "stationary", like ether in LET, then, in order to be in agreement with SRT?

It is Lorenz invariant.

I am probably mixing apples and oranges here, but I wonder what's the point of this field then, compared to ether?

It is part of an elegant mathematical formalism that allows to model elementary particle interactions.

Furthermore, I've seen articles (disclaimer: not really scientific) which try to explain the double slit experiment as if the "ether" (or whatever "medium" or "field" you are considering) is being disturbed by the photon into waves, similar to what a tiny rock would do when thrown into water. Observing a photon at one of the slits would absorb (much of) its energy, making it unable to disturb the medium enough to show the pattern.

This is wrong . There are experiments which show that changing the boundary conditions at the slits is what is changing the solution of the quantum mechanical equation "photon + two slits" . What happens that there is a probability for the photon to interact with the detector at the slit, and change direction.

Mathematically, it would probably boil down to classical interference explanation, which I believe correctly calculates the behavior, but I am presuming it would be a much more down to earth explanation compared to "shut up and calculate".

It is not classical interference, because single photons at a time also show the interference pattern. In classical interference it is the electromagnetic wave that shows interference. The mathematics may be similar but the physics model different. For single photons what is waving is the probability of its manifesting at the (x,y) of the screen.

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  • $\begingroup$ +1 Thanks a lot for taking the time to write such an elaborate explanation. $\endgroup$ – Lou Mar 19 '16 at 15:39
  • $\begingroup$ Just a quick clarification: when you say no photons unless a photon creation operator acts - the operator acts at the moment the photon hits the detector? Because otherwise it would seem creation operator needs to act along all (x,y,z,t) from the moment the photon is sent from the source towards the slits? Unless I misunderstood that sentence. $\endgroup$ – Lou Mar 19 '16 at 16:07
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    $\begingroup$ no, only on the path. Curiousone is correct that a "path" has a fuzzy quantum mechanical definition, but you could think of it as for example, a free electron running straight is represented by consecutive creation and annihilation operators as it moves. wherever the electron is , the creation operator has given a non zero expectation value, where it hits the detector too. $\endgroup$ – anna v Mar 19 '16 at 16:15
  • $\begingroup$ @annav: Nobody has ever observed an electron running on a straight path at the quantum level. The very formalism of QFT says exactly the opposite: in a strict classical particle picture "the particle" would have to take ALL possible paths, which, of course, is just an artifact of mathematics when you convert a wave equation from its differential to its integral form. $\endgroup$ – CuriousOne Mar 20 '16 at 1:32
  • $\begingroup$ @CuriousOne that was what I wanted to convey with "fuzzy" . The centers of fuzziness correspond to the path of the electron we see as a track in a detector. $\endgroup$ – anna v Mar 20 '16 at 4:01

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