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Isn't the basic operational reason that the hidden variables are "hidden" the assumption that they can't be measured?

Well in case, given that measurement is realistically speaking, nothing but interaction, by such assumption, are we postulating that the hidden variables do not interact with the physical elements of our reality?

Nonetheless, it seems that one can study the possible features of such an unphysical entity via experiments like Bell's experiment or general theorems like the Kochen-Specker theorem.

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    $\begingroup$ "Hidden" in this context is generally taken to mean "yet undiscovered". $\endgroup$
    – WillO
    Commented May 22, 2022 at 17:05
  • $\begingroup$ @WillO So you mean it's an observable quantity? Take the case of Bohmian Mechanics, the hidden field is the module of the $\psi$ function. How do you measure such field? $\endgroup$
    – Bastam Tajik
    Commented May 22, 2022 at 17:12
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    $\begingroup$ English words often mean different things in different contexts. $\endgroup$
    – WillO
    Commented May 22, 2022 at 17:20
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    $\begingroup$ @BastamTajik Surely the hidden variable is $\psi$ itself, which is a complex valued function ? Although you can measure $|\psi|$, that does not tell you the value of $\psi$. $\endgroup$
    – gandalf61
    Commented May 22, 2022 at 17:29

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Answer to question in 1st paragraph is "no".

The term "hidden" here is not being used to signify "impossible to detect"; it is being used to signify "a physical property whose value influences or determines an outcome which is thought (on some other theory) to come about some other way." The typical situation is that the other theory is quantum physics without further variables, and the hidden variable theory is something like a pilot wave theory. The hidden variable here acts as a causal influence on outcomes which, according to the theory without the hidden variable, are merely random. Another example is when we are concerned with non-local behaviour of the type exhibited by entangled systems. The idea here is that if you want to bring in a model (a hypothesis) in which there are further parameters, i.e. physical properties of the quantum systems, then whatever parameters you use will have to be capable of leading to the observed non-locality. The term "hidden variable" is useful simply to suggest that we are trying to talk about such further parameters in a general way.

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  • $\begingroup$ Sorry, but how do you "measure" the value of the hidden field in Bohmian mechanics at every point of space? $\endgroup$
    – Bastam Tajik
    Commented May 22, 2022 at 17:16
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    $\begingroup$ One accepts any given physical model (e.g. Bohmian model) on the usual basis of internal consistency, parsimony, and agreement with observations. The model itself will have statements about how the physical parameters relate to observations. This does not necessarily require that every parameter value at every event can be deduced exactly. Each model proposes what the physical nature is; if their empirical predictions agree then one accepts one model over another based on ones judgement. Such acceptance is always provisional and pragmatic. $\endgroup$
    – Andrew Steane
    Commented May 22, 2022 at 17:42
  • $\begingroup$ I just understood the answer precisely now. I hope you can change your down vote. Thanks for the answer. $\endgroup$
    – Bastam Tajik
    Commented May 23, 2022 at 9:55
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    $\begingroup$ @BastamTajik (the downvote on the question was not from me; if I thought it was a bad question I would not have answered!) $\endgroup$
    – Andrew Steane
    Commented May 23, 2022 at 11:13
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A particle stream streaks a corner and the particles are deflected in such a way that a periodic intensity distribution can be observed on an observation screen. Any attempt to observe the particles directly behind the edges fails because our (not perfect) observation instrument acts on the particles with an energy that deflects its motion so much that the intensity pattern on the screen is destroyed. Then it is necessary to introduce a hidden variable. In this case the interference.

But the explanation that interference between particles is the cause of particle deflection becomes obsolete for single-particle experiments. The explanation that two edges forming a slit are responsible for the interference is also obsolete, since the intensity distribution also occurs behind a single edge. What variable could be introduced that would not be a hidden one - but then just a known one?

A wide field of physical research are phononic excitations, in which periodic states are excited that propagate in a material (Phonons can be thought of as quantised sound waves, similar to photons as quantised light waves.).

What if one assumes and verifies by experiments that the streaking particles interact with the surface electrons of the edge with their electric and their magnetic field? This would be an example of turning a hidden variable into a measurable variable.

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    $\begingroup$ @HolgerFieldler I up-voted for your 2nd sentence of paragraph one. The term "Hidden Variable" makes things more mysterious than they needs to be. Particles already have at least (6 to 8) variables that are NOT hidden. Mass, Speed, Direction, Polarization, Rotation/Frequency, Coherency/Timing. Considering these already known variables a pattern can be derived, even one particle at a time. $\endgroup$
    – Bill Alsept
    Commented Jun 30, 2022 at 17:13
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    $\begingroup$ @HolgerFieldler you could assume streaking photons interact with edges in several ways. (1) Direct impact to electrons on the surface, which could then be re-emitted in some random direction. (2) Electromagnetically by rotating the photons polarization as it passes by. (3) Scattering off and away from the edge at all angles, based on frequency and proximity to the edge. (4) Diffracting around and behind the edge at all angles based on proximity to the edge (similar to gravity). Photons would scatter and diffract at all angles left and right to cover the screen along the plane. $\endgroup$
    – Bill Alsept
    Commented Jul 1, 2022 at 4:43
  • $\begingroup$ @Bill Alsept Very agree with you. $\endgroup$
    – HolgerFiedler
    Commented Jul 2, 2022 at 2:45

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