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  4. What happens when both slits are observed by in the double slit experiment?
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What happens when both slits are observed by in the double slit experiment?

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Offline Colin2B

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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #20 on: 08/11/2019 12:28:03 »
Quote from: Lloyd on 06/11/2019 11:01:39
I think it's fair to say that I could not get into any meaningful discussion with you guys because of our differing levels of scientific/mathematical understanding on the whole subject.
I hope we can still have a meaningful discussion, and I’ll try not to be too technical. The maths is only a description of what happens, but sometimes it’s hard to put into words which is why the popsci press opt out and give oversimplified explanations.

Quote from: Lloyd on 06/11/2019 11:01:39
seemingly a particle before the double slit and a wave after the double slit ... and reverting back to a particle when 'observed' after the double slit.
The wave or particle is only a description of how the light behaves, not really a description of what it is.
Ok, to keep it simple and stick with light. When light is emitted by an atom it is as a very short burst of electromagnetic oscillation - a short wave. So that burst, which we call a photon travels as a wave, but the way we detect it requires it to interact with another atom. That atom could be in a photographic film or in a camera ccd, but that interaction shows up as a dot, just as if a particle had hit the detector. In other words, when the wave hits the detector it can be modelled as a particle. This is why you often hear physicists say the photon travels as a wave but is detected as a particle.
It is fair to say that most modern physicists consider the wave/particle duality thing a bit ‘old ways thinking’, they are more likely to ask how you are detecting or measuring the result of your experiment.

Quote from: Lloyd on 06/11/2019 11:01:39
The point is that the photons travelling through the slits should produce an impact pattern similar to the double slits.
‘Should’, but only if they are travelling and behaving as particles. Electrons are interesting because we tend to think of them as particles - little hard bullets - but when they are moving they clearly behave like waves, if fact when they were first discovered they were called cathode rays (which is where we get the name cathode ray tube).
One thing I should mention here is that the prime objective of physics is not to provide an explanation of why something happens, the objective is to observe, model and predict behaviour; that’s what Newton did, he didn’t explain gravitation he just described what it did - with very useful accuracy. His model was of a force acting at a distance, today we model it as a field through Einstein’s field equations and that’s a little more accurate.
Sometimes we can see clearly how a thing works, eg we start a car and the wheels push against the ground providing a force that moves the car forward. Some other thing are less easy to predict eg we flip a coin, we can’t predict exactly which way up it will land, but we do know that in a large number of flips we will get approximately 50:50 and we can do a lot of predicting with the uncertainty of which way it will land. This same uncertainty works at the quantum level and allows us to use probability to make some very accurate predictions - we might not know the exact position of an electron in an atom, but we can make very accurate predictions about where it will spend the majority of its time.
Probability is worth learning about because it really helps us understand a lot about the world around us. So, I’m going to go a bit technical and hope you will follow us by learning more about probability.
When we fire bullets (particles) at a slit they will, as you said, go through one slit or the other and form the 2 slit shadow we expect. This is because in probability we say the events of a bullet passing through one slit or the other are mutually exclusive (it can’t do both) and more importantly neither action can influence or have any effect on  the other (we say they are uncorrelated). In probability mutually exclusive, uncorrelated evens follow the sum rule, their probabilities add which is what we see with bullets, but with quantum objects which are coherent, it is the square of the sum which gives us the probability of what we see on the screen. There is growing evidence that this is the way the world really works, but at our big world level we lose the coherence between quantum objects and we start to see the probabilities we are used to. However, when we come to modelling the slit experiment we can look at it and say that the electrons going through both slits is one way of modelling it because it works. So when physicists say it goes through both slits and interferes with itself, they are really using a shorthand for ‘that’s the way we can model it’. At the moment we don’t have enough information to know exactly how this works, but we can predict what happens to a very high degree of accuracy.

@jeffreyH  has put up a video of polarisation, this raises another problem which occurs when we view the quantum world. We view polarisation as horizontal/vertical, right/left and electron spin as up/down, just as we would say a coin is heads/tails. But it’s not that simple and the polarisation and spin are best represented as complex numbers (which can also be represented as vectors).

PS Looking back I notice @alancalverd has given a very similar explanation to mine, please read it carefully it’s important. I hope mine has some extra detail you find useful. Remember, 62 is not too old to learn about probability, vectors etc, keeps the grey cells on their toes.
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Offline Lloyd (OP)

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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #21 on: 08/11/2019 17:44:21 »
Time to do some printing I think. Can't beat paper in the hand!
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #22 on: 08/11/2019 19:10:08 »
Quote from: Colin2B on 08/11/2019 12:28:03
When we fire bullets (particles) at a slit they will, as you said, go through one slit or the other and form the 2 slit shadow we expect.
No! There is a calculable probability that a bullet will, in effect,  "go through both slits". As Eddington said, "If a student of physics should fall through the floor and find himself in the room below, he would not be surprised but mildly elated at having observed an extremely rare phenomenon."

That's what I meant by quantum mechanics scaling up to continuum physics at a mesoscopic level. So far, it's only been demonstrated with large molecules like buckyballs, but the maths is sound.

You can work the model two ways: either map an enormous number of parallel universes onto one timeline to get "now", or wait a very long time in one universe for everything to happen. Problem is that once you get beyond 60 carbon atoms, the numbers become ridiculous.
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Offline Halc

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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #23 on: 08/11/2019 19:31:53 »
Quote from: alancalverd on 08/11/2019 19:10:08
Problem is that once you get beyond 60 carbon atoms, the numbers become ridiculous.
They've done it (put in superposition, not fire through slits) to a large enough object to see with the naked eye, so they've gone considerably beyond 60 carbon atoms.  The problem at larger scales is that it is almost impossible to not measure them, in an effort to maintain coherence.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #24 on: 08/11/2019 21:08:45 »
@Halc This is worth a read.

https://physicsworld.com/a/does-general-relativity-violate-determinism-inside-charged-black-holes/
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #25 on: 08/11/2019 22:51:58 »
Quote from: jeffreyH on 08/11/2019 21:08:45
@Halc This is worth a read.
physicsworld :does-general-relativity-violate-determinism-inside-charged-black-holes
Interesting link. Yes, it says that Newton envisioned a classic billiard-ball sort of physics, and it was that way all the way down, and relativity didn't really challenge that (except maybe it does, as the article points out). 
Quantum mechanics definitely threw a wrench into the works for the determinists, but it as well doesn't disprove it since there are very much deterministic interpretations of QM.

I think we'd need a unified field theory to begin to say whether relativity has any concrete stance on the issue.

From the article:
Quote from: Cartlidge
Newton’s mechanics allow us in principle to calculate the exact state of a physical system at any point in the future, provided that we know its initial state perfectly. So too with general relativity: a precise knowledge of space’s geometry and its rate of change in the present enables us in theory to predict exactly how space-time will evolve.
This is clearly false, and known to Einstein at the time.  QM theory says no amount of measuring of a system will allow you to predict it. Einstein was definitely a determinist, as evidenced by his "God doesn't throw dice" quip, but I don't think any of the deterministic interpretations where developed or well known at the time.  There was the 'hidden variables' postulate which said there are variables which cannot be known, but if they were, the future would be perfectly predictable.

Quote
As such, Einstein’s theory is considered by most physicists to be entirely deterministic.
... as it doesn't contradict Newton's 'classic all the way down' assumption.  But the theory also doesn't posit this assumption. It merely declines to challenge it. QM very much challenges it.

Quote
Charged black holes, however, challenge this deterministic picture. The “Reissner-Nordström” solution of general relativity describes a black hole created when a star that is electrically charged and spherical collapses in on itself under the force of gravity. Hidden from view inside such a black hole’s event horizon lies a second boundary known as the Cauchy horizon, beyond which space-time is smooth but indeterminate. In other words, the future can no longer be predicted.
It has also yet to be demonstrated that it is meaningful to speak of the physics inside the event horizon.  It is mathematically infinitely far into our future, and since the black hole will evaporate in finite time, it is questionable if anything 'gets in' so to speak.  Again, a unified theory would help.  The subject has proponents on both sides, and this would best be discussed in a separate thread. I'm quite opinionated on it myself, having the luxury of not completely knowing what I'm talking about.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #26 on: 09/11/2019 07:37:47 »
Quote from: Halc on 08/11/2019 22:51:58
QM theory says no amount of measuring of a system will allow you to predict it.
Be careful how you state this. The test of quantum mechanics is not merely that it explains obvious quantum phenomena like line spectra, but that it scales to classical mechanics for large assemblies. And it does.

The wave function of a billiard ball is negligible in comparison with its apparent diameter, so for all practical purposes (e.g. potting the black) it is adequately predictable. But its structure, mass and elasticity are all calculable from quantum mechanics.

Indeed the "measuring" test isn't fundamental to QM. The classical model of bouncing a photon off an electron sort of illustrates the problem of measurement influencing the measurand, but masks the inherent indeterminacy of Heisenberg's statement that Δp.Δx ≥ h.
« Last Edit: 09/11/2019 12:14:11 by alancalverd »
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Offline Halc

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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #27 on: 09/11/2019 17:09:21 »
Quote from: alancalverd on 09/11/2019 07:37:47
Quote from: Halc on 08/11/2019 22:51:58
QM theory says no amount of measuring of a system will allow you to predict it.
Be careful how you state this.
I thought I was quite careful. It's actually quite a weak statement since it references the limits of measuring something, and not magically having full knowledge of the state of a system.

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The wave function of a billiard ball is negligible in comparison with its apparent diameter, so for all practical purposes (e.g. potting the black) it is adequately predictable.
I would disagree with that. You're thinking of a classic description of the ball, which isn't a wave function.  A wave function might give probabilities of particles being measured here or there or not, and that function is unimaginably complex for something like a billard ball.

So to illustrate the problem, I find the ball in a completely different location 10 seconds from the first measurement.  It's in the corner pocket now (I know there are no pockets in billiards) and it wasn't before. If your 'negligible' wave function didn't predict that, then it wasn't a very accurate wave function for it, was it?

Given a full deterministic interpretation of QM (like what Bohm suggests) and a complete description of the wave function of the ball (which includes the state of moon and anything else within 10 light seconds) plus all the immeasurable hidden variables involved, only then can an accurate prediction of it being in the corner pocket be made.
Other interpretations (mostly ones without hidden variables) might or might not make that prediction. 10 seconds doesn't give much time for quantum indeterminacies to manifest themselves in classic ways, but it can happen explicitly with a quantum amplifier.

Quote
But its structure, mass and elasticity are all calculable from quantum mechanics.
I agree that classic behavior and properties emerge from QM.

Quote
Indeed the "measuring" test isn't fundamental to QM
I agree with this as well, but there are certain interpretations that give fundamental importance to measurements, and they don't all define measurement the same way.
« Last Edit: 09/11/2019 17:11:35 by Halc »
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Offline alancalverd

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Re: What happens when both slits are observed by in the double slit experiment?(
« Reply #28 on: 10/11/2019 00:13:28 »
Quote
In English Billiards, scoring is as follows:
........
A pot: This is when the red ball is struck by the player's cue ball so that the red ends up entering a pocket. This scores three points. If the player’s cue ball strikes the other cue ball resulting it going down the pocket, then this scores two points...…..

My apologies, there are pockets, but the target ball is red, not black.

Anyway, the fact remains that for a billiard ball p and x are so large compared with h that the indeterminacy of its position is negligible even when p is zero, which makes the game playable and double-slit diffraction (or spontaneous appearance in another pocket)  of a billiard ball very difficult to observe. Or, if you like wave functions, the wave function barely extends beyond its classical surface.
« Last Edit: 10/11/2019 00:18:38 by alancalverd »
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #29 on: 10/11/2019 00:40:00 »
Macroscopic objects follow deterministic paths which is why combining relativity and quantum mechanics is so hard. You need a very small scale and high energies. These conditions are found when particles are close to an event horizon. So it is going to be difficult to observe. It may be possible to recreate these condition in future particle accelerators.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #30 on: 10/11/2019 02:11:14 »
Quote from: alancalverd on 10/11/2019 00:13:28
Quote
In English Billiards ...
My apologies, there are pockets.
I guess what I've played isn't English Billiards. Thanks for the education.

Quote
Anyway, the fact remains that for a billiard ball p and x are so large compared with h that the indeterminacy of its position is negligible even when p is zero
Totally agree, but you're talking about measuring it, and not about predicting where it will be in the future. Yes, I can measure the position and momentum of a billiard ball to a lot of zeros of accuracy. No argument there.

Kindly interpret my post as talking about predictions (from a wave function) of the future, and not of measurement of its current state.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #31 on: 10/11/2019 08:26:09 »
Yes, I misinterpreted your scenario. We are in fact in perfect agreement. Because the wave function of a large object is negligible outside its classical radius (indeed you can use that as a definition of "mesoscopic") we can predict where a billiard ball will go. Because the wave function of an electron is much bigger than its radius, atoms and molecules do not collapse.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #32 on: 10/11/2019 14:08:43 »
Quote from: alancalverd on 10/11/2019 08:26:09
Because the wave function of a large object is negligible outside its classical radius (indeed you can use that as a definition of "mesoscopic") we can predict where a billiard ball will go.
If this were true, they could publish tonight's billiards scores in today's paper. Only the one interpretation I mentioned actually asserts the game is fixed (completely determined) like that, and even then there is neither a way to 1) take anywhere near the necessary measurements, nor 2) perform the calculation, so no computer, even hypothetical, can actually perform such a calculation.

This is what I was talking about when discussing predictions of macroscopic objects.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #33 on: 10/11/2019 15:03:34 »
 "   enough charge, damping wins out over amplification and the oscillations die away quickly. As Cardoso explains, the charge and cosmological constant essentially provide repulsive forces that counteract the pull of gravity and so diminish its amplifying effects. "

" In the context of cosmology the cosmological constant is a homogeneous energy density that causes the expansion of the universe to accelerate. Originally proposed early in the development of general relativity in order to allow a static universe solution it was subsequently abandoned when the universe was found to be expanding. Now the cosmological constant is invoked to explain the observed acceleration of the expansion of the universe. The cosmological constant is the simplest realization of dark energy, which is the more generic name given to the unknown cause of the acceleration of the universe. Its existence is also predicted by quantum physics, where it enters as a form of vacuum energy, although the magnitude predicted by quantum theory does not match that observed in cosmology. "

Maybe?
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #34 on: 10/11/2019 18:00:28 »
Quote from: Halc on 10/11/2019 14:08:43
If this were true, they could publish tonight's billiards scores in today's paper.
Indeed, if you knew in advance what shot each player was going to make. Problem is, you don't. Billiards is a competitive game of skill, not random chance. At least that's what good players tell me! You always  have a choice of shots and, like snooker, you can choose a shot that gives you a score or one that puts your opponent in a position from which he cannot score but is likely to foul. The fact that good players rarely foul suggests that for all practical purposes, the behaviour of billiard balls is predictable.

If we took your assertions at face value, nobody would dare to drive a car because it is impossible to predict where it will go.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #35 on: 11/11/2019 02:55:14 »
Quote from: Lloyd on 04/11/2019 09:16:36
I watched as single photons passed through a single slit producing a single bar pattern.
I believe this only happens in simulation, never really happens in actual experiment. The images below show what you'd get from single slit and double slit experiment.

With a normal single slit apperture, you will still get interference pattern due to diffraction effect.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #36 on: 11/11/2019 08:39:06 »
That's interesting hamandi ... a lot of information I am coming across is based on simulation and not the results of experimentation. Simulation is fine up to a point I suspect. But there comes a point when the simulation has to be replicated by experiment in order for it to carry any credence.
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #37 on: 11/11/2019 10:07:39 »
It has been argued elsewhere in this forum that interference and diffraction are not the same, and the difference between Hamandi's excellent images shows this. It's a bit academic and a lot of the maths is common to both phenomena but logically you need two sources or particles to interfere, and only one edge to diffract.

The quantum two-slit problem is that a single particle seems to interfere with itself, thus showing that classical continuum wave equations don't adequately model very sparse (single photon) systems. 
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Re: What happens when both slits are observed by in the double slit experiment?
« Reply #38 on: 11/11/2019 14:45:39 »
Quote from: Halc on 09/11/2019 17:09:21
Given a full deterministic interpretation of QM (like what Bohm suggests) and a complete description of the wave function of the [system] (which includes the state of moon and anything else within 10 light seconds) plus all the immeasurable hidden variables involved, only then can an accurate prediction of [its state a few seconds from now] be made.
This statement is wrong.  One would need to include the full wave function and hidden variables of the universe to make a perfect determination of state 10 seconds from now since any interpretation of the nature I describe above (where the universe has a state) has causes that can come from outside the past light cone of the system being predicted, and from the future as well.

This is sort of true of local interpretations as well.  A given particle has a wave function that says there is a finite probability that it will be measured anywhere, including outside its future light cone. No information can be conveyed that way, so it doesn't seem to violate the information-faster-than-light rule. Anyway, I know of no local interpretation that is hard deterministic: The future state of any system cannot be determined even if all information could be known, even in principle.  Something like MWI is a deterministic interpretation, but only by saying that all the possible states exist.
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