Why don't the Slits collapse the wave function?

[selfAdjoint]

No, we don't know that. You forget that such a theory does not allow "free will", there is no statistical independence between detector orientations and the particle source.

Aren't you confusing Local and Global realism? "Local", in local realistic theories, means on the scale of the particle interactions, not on the scale of the lab; that last requires an additional assumption, as was pointed out years ago, at the time the original Aspect experiment was discussed.

 

[Hans de Vries]

Never made such a suggestion

Yes, you did. You said literally "MUCH smaller than your number" with the
word "MUCH" in capitals.

- and you are still completely missing the point, as if you didn't understand what or how the HUP works,

No I'm absolutely not missing the point. There is nothing here which
is in conflict with HUP. There is no limit in how exact you can measure
the momentum in QM as long as you do it over a large enough traject.

But have your way, start a new thread when you or one of yours actually runs such an experiment some day in the future, I’ll unsubscribe here, this thread is long enough as it is.

It should not be a matter of prevailing or conceding here. Much of this
has also to do with your discussion style. As long as you throw things
at me like "you are completely missing the point", then expect me to
respond.


Regards, Hans.

 

[RandallB]

Yes, you did. You said literally "MUCH smaller than your number" with the word "MUCH" in capitals.

How could that have anything to do with deflection angle as you never gave that detail.

It should not be a matter of prevailing or conceding here. Much of this has also to do with your discussion style. As long as you throw things at me like "you are completely missing the point", then expect me to respond.

Sorry fair point,
But don't expect me to continue, as you don't work through a complete detailed example your plan to do such a test, and others a lot smarter than me have already said it is an impossible test.
IMO there is no point in arguing over it; and it’s fine to imagine someone will be able to do it some day if you like.
I’m well convinced that they are correct that it cannot be done and no one will, and don’t need my imagination stimulated on the point thank you.

 

[ueit]

Aren't you confusing Local and Global realism? "Local", in local realistic theories, means on the scale of the particle interactions, not on the scale of the lab; that last requires an additional assumption, as was pointed out years ago, at the time the original Aspect experiment was discussed.

AFAIK "local" means that every interaction propagates with a limited speed (c). The range of EM interaction is infinite so I don't see how the scale makes any difference.
Why is this even relevant to my point (that is, in a deterministic theory the detector orientation is a result of the deterministic law and therefore cannot be assumed to be random)?
Bell's theorem assumes randomness therefore is irrelevant when discussing deterministic theories (fallacy of circular reasoning).
I agree however that a deterministic theory has to account for EPR experiments but this is not a problem as no theory to date explains them.

 

[DrChinese]

Bell's theorem assumes randomness therefore is irrelevant when discussing deterministic theories (fallacy of circular reasoning).
I agree however that a deterministic theory has to account for EPR experiments but this is not a problem as no theory to date explains them.

Bell's Theorem does not assume randomness. It states that a result at one location is not affected by the choice of measurement at another. There is absolutely no requirement that the selection of spin measurement angles be random either, another common myth.

As to your second point: QM is a theory which explains the observed results perfectly, as they are exactly as predicted: cos^2.

 

[ueit]

Bell's Theorem does not assume randomness. It states that a result at one location is not affected by the choice of measurement at another. There is absolutely no requirement that the selection of spin measurement angles be random either, another common myth.

In a deterministic universe there is no choice. If you want to verify a LHV theory you have to study how the "choice" is determined at each location by the deterministic law. The two measurement "choices" are the result of this law applied to their common past. You cannot assume that the two detectors are independent as long as they share a common past.

As an analogy, think about the orbit of Hyperion (or a stray comet, or whatever). It is chaotic, for all practical purposes it looks random. If you are staying on its surface with an accelerometer you would be able to determine the "instantaneous" position of the Sun, according to Newton's law. How do you explain that? Did Hyperion's wave function collapsed when you took your measurement, or it's just the deterministic local GR acting on this complex system?

As to your second point: QM is a theory which explains the observed results perfectly, as they are exactly as predicted: cos^2.

QM doesn't explain anything, just computes the probability of the final result. It doesn't state the reason we have the correlations, just that they occur.

 

[DrChinese]

QM doesn't explain anything, just computes the probability of the final result. It doesn't state the reason we have the correlations, just that they occur.

I don't think you understand what a fundamental theory is. QM is a fundamental theory, so naturally there is no explanation of why h has the value it has. Nor why the other physical elements work as they do; nor would any suchg explanation be expected. This is why QM is an actual theory, where as your "strict determinism" is not a theory at all.

If you postulate strict determinism, it is incumbent on you to offer a complete and consistent theory which can be competitive to QM (i.e. at least the same scope and predicability). For example: if there is strict determinism, then why cannot all attributes of a particle be measured to unlimited accuracy (in violation of the HUP) ? This glaring inconsistency undermines your approach. Bell's whole point was that:

QM + (assumptions of locality & realism) was inconsistent with QM alone.

So how can you say:

QM + (assumptions of locality & strict determinism) is consistent with QM alone.

...without first demonstrating this to be true?

My point is - if your strict determinism were true - that every single particle in the universe must contain "DNA" which allows it to know how to act for every single interaction it will ever have in the future. And it must have enough of this DNA so that in experiments on different particles - but just some, those we call entangled - that the results of separate experiments yield results according to a statistical distribution as predicted by QM. Whew! By my estimate, that would be essentially an infinite amount of information to carry around. And yet there is not the slightest evidence of this DNA to date, as no known internal structure exists for any particles. When you produce some evidence (or even a testable element) to support this hypothesis, we can discuss. Meanwhile, your theory is purely "ad hoc" and of no use.

 

[ueit]

I don't think you understand what a fundamental theory is. QM is a fundamental theory, so naturally there is no explanation of why h has the value it has. Nor why the other physical elements work as they do; nor would any suchg explanation be expected. This is why QM is an actual theory, where as your "strict determinism" is not a theory at all.

A theory can only be considered fundamental if it can explain any phenomenon in its range. QM cannot predict single events (the value of the spin measured on an arbitrary axis, the time when a decay takes place, etc.) therefore it is only a statistical approximation to a fundamental theory.
The so-called "pure randomness" of QM is nothing but another example of bad logic. It is known that by squaring the amplitude of the wave function, you get a probability but this does not imply that a non-probabilistic description cannot be found. Assuming so, is both irrational and unscientific because it puts an arbitrary boundary to our knowledge. If every theory used the same line of reasoning (there is no explanation, it just is the way it is) we would still be throwing virgins into volcanoes to have good crops or whatever.

If you postulate strict determinism, it is incumbent on you to offer a complete and consistent theory which can be competitive to QM (i.e. at least the same scope and predictability). For example: if there is strict determinism, then why cannot all attributes of a particle be measured to unlimited accuracy (in violation of the HUP) ? This glaring inconsistency undermines your approach.

1. If all that exists are billiard balls, in order to see one of them you have to hit it with another one. This will disturb the system so your knowledge will not allow you to make a prediction.

2. Contrary to your belief we can measure with unlimited accuracy both momentum and position of a particle by simply detecting it at an arbitrary large distance from the source. You cannot use the information to make a prediction but non-predictability is not the same thing as non-determinism.

I see no glaring inconsistency here.

Bell's whole point was that:

QM + (assumptions of locality & realism) was inconsistent with QM alone.

No, his point was:

QM + (assumptions of locality & realism and free choice) was inconsistent with QM alone.

Please read "Speakable and Unspeakable in Quantum Mechanics" by Bell to see him saying that, with his own words.
If you look carefully at the statement above you can see that, in the case of deterministic lhv's QM is superfluous, because "determinism & free choice" is false as required by the law of non-contradiction (determinism != free choice).
The theorem can only apply to non-deterministic lhv theories but I don't care much about them anyway.

So how can you say:

QM + (assumptions of locality & strict determinism) is consistent with QM alone.

...without first demonstrating this to be true?

I have nothing to demonstrate. Take out the assumption of free choice and see what remains from Bell's theorem.

I'll not make a fallacy myself though. Just because Bell's theorem does not apply to determinism doesn't mean that determinism is compatible with QM. However, this is a subject to investigate, not to dismiss as impossible.

My point is - if your strict determinism were true - that every single particle in the universe must contain "DNA" which allows it to know how to act for every single interaction it will ever have in the future. And it must have enough of this DNA so that in experiments on different particles - but just some, those we call entangled - that the results of separate experiments yield results according to a statistical distribution as predicted by QM. Whew! By my estimate, that would be essentially an infinite amount of information to carry around. And yet there is not the slightest evidence of this DNA to date, as no known internal structure exists for any particles. When you produce some evidence (or even a testable element) to support this hypothesis, we can discuss. Meanwhile, your theory is purely "ad hoc" and of no use.

Where did I speak about DNA? It's the most absurd caricature of determinism I've ever heard of, and I don't see how my analogies with gravity could be interpreted that way.
So, if you want a model of deterministic theory look at GR. The planets do not need to carry a "DNA" telling them how to move in the universe. Each particle in an EPR experiment (including the ones in the detectors, source, experimenters) follows a path in accordance with the forces acting on it.

You can make the first steps to extending Bell's theorem to deterministic theories by proving that the spins of an electron and positron produced by the decay of a positronium "atom" are not correlated to the orientation of a distant magnet (Stern-Gerlach detector).

 

[Dense]

One thing I don't understand is how the screen with the slits do not collapse the wave function until there is a measurement. Wouldn't bouncing off the sides of the slit constitute some sort of interaction?

What is it about the "observer" or measurement device that would interact differently with the photon than how the photon interacts with the slit?

The difference is that the slits merely limit the potential paths of the photon, while an observation captures the path actually taken.

Think of it this way. A photon is created. It is going to travel at velocity c. It could go in a bunch of different directions. Imagine that -- before the photon even begins to travel -- all the possible paths it could take are carved out in space.

Now think of all these potential paths in this frozen instant in time. Think of them acting outside of time, if you will.

You've got one slit. The potential paths through that slit go right through pretty straight.

You've got two slits. The potential paths go through both slits, and interfere with each other on the other side.

Now start time up again, and release the photon. It's going to travel along only one of all the potential paths.

If it follows one of the potential paths that goes through one of the two slits, then that photon is going to follow a path that looks as if it "interfered with itself." In fact, it's not interfering with anything. It's merely following a potential path that already interfered with another, before the photon traveled down that path.

Does that help?

 

[kliide]

Thanks all.

jpr0 rephrased my question perfectly and DrChinese's answer registered on my brain. Dense answered my question with this: "The difference is that the slits merely limit the potential paths of the photon, while an observation captures the path actually taken." Wherein I can see that you're saying that limiting the path of the photon means terminating their paths at the point where they impact the slit rather than pass through.

In retrospect, I don't know why I was asking about the electrons that had impacted the sides of the slits as it had nothing to do with the experiment. I think I was visualizing photons that contained paths that would deflect off the side of the slit and still impact the measurement device. I originally believed that the intent of the experiment was to say that even these photons maintained wave characteristics once impacting the slit and would only lose their wave characteristic when "measured". I couldn't figure how a measurement would differ from coming into contact with any other type of subatomic particle. Poorly formed question. Great answers, anyhow.

DrChinese's explanation that the slits simply terminate the paths of the photons that impact them makes sense to me.

Anyhow, thanks all.

 

[Anonym]

DrChinese's explanation that the slits simply terminate the paths of the photons that impact them makes sense to me.

In the path integral view, you consider all possible ways that the photon could arrive there. Those possibilities give rise to interference, and a pattern results.

In sum: The interference pattern represents the subset of photons that went through the slits. In effect, the detection mechanism is designed to be a fertile target for the source photons and the rest of the apparatus is not. That is the only distinction..

DrChinese’s explanation makes no sense to me. He starts with the single photon and end up with subset of some set of the incident photons. We consider here a single particle phenomenon. The incident photon/electron etc. are described by the coherent wave packet. And his answer to your original question is bla,bla,bla. In addition, his answer violates unitarity which is maintained in the discussed experiments.

The "slit" in the idealized case is simply an illustration of separate paths for the photon, electron, neutron, buckyball, etc. As I've said, that isn't the real issue. Now, if the slit happens to be a metallic device that can somehow detect things like electric field of an electron or a photon, THEN it is now a detector that can tell you if a photon, electron, or whatever, passed through it. This is now a different set up.

I consider that the only correct answer presented to your question (Why don't the Slits collapse the wave function?), however I do not agree that the reflected and transmitted wavefunctions describe separate paths for the photon, electron, neutron, buckyball, etc. As mentioned by Zz, that isn’t the real issue.

Regards, Dany.

 

[DrChinese]

DrChinese’s explanation makes no sense to me. He starts with the single photon and end up with subset of some set of the incident photons. We consider here a single particle phenomenon. The incident photon/electron etc. are described by the coherent wave packet.

Well, that is the path integral explanation. As ZapperZ points out, the general version is one in which there are a number of ways to reduce the set of paths.

When there is one slit, there are still multiple paths being traversed and there is still interference. However, such interference will NOT create the classic interference pattern in that case.

 

[Anonym]

Well, that is the path integral explanation. As ZapperZ points out, the general version is one in which there are a number of ways to reduce the set of paths.

When there is one slit, there are still multiple paths being traversed and there is still interference. However, such interference will NOT create the classic interference pattern in that case.

My criticism was that you did not explain the difference in the set up. And I don’t see how you maintain unitarity. In addition, in case of one slit we usually are talking about diffraction.

In “Light, particle or wave?” I asked Zz to comment A. Tonomura et al. “Double-biprism electron interferometry”, App. Phys. Lett., 84(17), 3229 (2004) paper. I think, it is obvious that I am very interesting to know your opinion too.

Regards, Dany.

 

[sokrates]

exactly.

There are no particles.

There are localized waves

 

[Invinoveritas]

As for the requirement of 4D spacetime, it is a bit premature to require that there be no additional dimensions. Maybe there are more.

This is a good point.

 

[Invinoveritas]

exactly.

There are no particles.

There are localized waves

Stupid question.. how does a localized wave travel in a vaccum?

 

[Count Iblis]

Why don't the Slits collapse the wave function?

Because the wavefunction of the slits is itself collapsed by the environment. If we assume that there are very weak interactions causing the state of the slits to change a bit then the particle moves through them, then this can only be used to detect the "which path information", if the state the slit is in when a aprtcile moves throgh it is orthogonal to the state it would be in if the particle does not move through it. The visibility of the fringes is proportional to the overlap of the two wavefunctions.

If you have a setup with floating slits or floating mirrors and consider the change in the state of the slits/ mirror due to the change in momentum of the particle that moves through it, then one has to take into account that you wouldn't be able to get an interference pattern in the first place if the center of mass position of the slits/mirrors were not determined to within less than a wavelength. By the position-mometum uncertainty relation, this means that the width of the wavefunction in momentum space is much lager than the absorbed momentum.

The center of mass state of a floating mirror of mass M at temperature T is described by a density matrix that has is almost diagonal in the position representation. The off diagonal components are approximately Gaussian with a width of approximately the thermal de Broglie wavelength. This means that we can think of the environment havong measured the center of mass position of the mirror but with an uncertainty of the thermal de Broglie wavelength and that the center of mass of the mirror could be regarded to be in an unknown pure state described by a Gaussian wave function.


In the momentum representation the off diagonal components of the density matrix are then Gaussians with a width of the order of
sqrt[M k T]. So if M is macroscopic, this width is huge compared to the momenta of particles in two slit experiments.


So, we see that explaining interference of particles in two slit experiments requires one to consider the quantum state of the rest of the universe. Decoherence of the many particle wavefunction of objects with which the particles interact explains why these objects are located to well within the wavelength of the particles. The wavefunction of the particle at a fixed position at the screen then has a well determined phase. And this is also related to the fact that the momentum that the slit or mirror may have absorbed is not suficient to gain which path information as the wavefunction of the center of mass of the slits/mirrors is necessarily wider than the absorbed momentum.

 

[alexepascual]

Going back to the original question:
Lets consider sending one photon at a time. There is a probability that the photon will be absorbed or reflected back by the plate. In either case, the photon does not reach the screen and we will not affect the pattern on it one way or another. If the photon is not absorbed or reflected by the plate, then we can assume that the wave goes through both slits. We couuld make the plate thin enough so that any reflection from the edges of the slit can be ignored. But even if we have a plate of a considerable thickness so that part of the wave is reflected by the edge of the slit, when the photon hits the screen, we will not know if the photon "bounced" against the edge of the slit or went thrugh the middle. We could think of the wave as made up o a bunch of possible trajectories for the photon. Depending on the "Interpretation" we could think that if we didn't make a measurement as to which was the exact path the photon took, then it took all paths. It kind of split itself into a bunch of "gost" photons which went in every possible way (not necessarily straight line). The overlap of the trajectories of all these ghost photons is what makes up the wave. Summing up the contributions of each ghost photon on the screen (considering the phases) gives us the probability that the real photon will be found at a particular spot on it. If the screen is made of photografic material, we can say that when some molecules change and leave a permanent record of the position where the photon landed, we have measured the position and the wave function has collapsed.
But let's go back and look at what happens if we consider a ghost photon that "bounced" against the edge of the slit and eventually contributed to the interference pattern on the screen. In this case, there is no permanent record made on the edge of the slit. If the photon was a little macroscopic ball, it could have left an indentation on the edge of the slit. But a photon (for our purposes) can only do one of two things: "bounce" o get absorbed. I think we should consider that bouncing is not able to leave a permanent record. If the photon was absorbed, then again it will not reach the screen and can be ignored because it does not contribute to the distribution of dots on the screen.
In the case of photons I don't think we can put a detector that will tell you if a photon went throught a slit or not without drastically affecting it. What I mean by drastic is that if you put a detector there are only two possible outcomes. 1) the detector absorves the photon or it doesn't. If it does, then this photon is eliminated from consideration as it does not reach the screen. If it doesn't and you do see a photon hitting the screen, then it means that it went through the other slit. In this case the mere presence of the detector eliminates all the possible paths through that slit (This is a detector that does not let any photons go through). So in this case if we send many photons, the interference pattern does not show up. This would be similar to just covering one slit instead of using the detector. So this example using photons is not very good if we want to use a detector. We could perhaps think of a more engineous setup where we convert the photon into two photons of longer wavelength and absorve one but that would change the experiment into something much more complicated than a simple double-slit. If we used finite rest-mass particles, we could do some kind of detection where the particle does go through the slit. Perhaps some of the smallest particles we could use would be electrons, but in this case we can't use slits because the wavelength of an electron is too small and the way to obtain an interference pattern is using a crystal.
So any double-slit experiment using electrons or other particles and putting a detector behind the slit which let's the particle go through is just a thought experiment that may help understanding some of the issues involved but can't be carried out just as presented.
But let's consider one of these thought experiments. We could think of a particle such as an electron still exhibiting behavior similar to the photon. So most of what I said above would apply, except that we could put a detector and still have the particle go through. If the detector clicks, then we know that the particle went through this slit and we now have a permanent record of it which colapses the wave function in the sense that it eliminates the part of the wave that would have gone thrugh the other slit. (no interference)
If the detector does not click and we do see a particle hitting the screen, then we can assume it went through the other slit. In this case the part of the wave that would have gone through the slit that contains the detector dissapears (no interference). So in both cases just the presence of the detector destroys the interference pattern.
If we consider those paths in which the particle "bounces" against the edge of the slit, we should consider that it just bounced, without leaving a permanent record of it. In this case the interference pattern is not destroyed. If it does "smash" something when hitting the wall (edge of the slit) then this would represent a measurement. It would leave a permanent record. We could use some powerful microscope and detect the change on the edge of the slit. Using other mechanism of detection such a looking for a change in momentum which would be imparted on the wall by the bouncing electron, would be kind of tricky and it would in the end give the same result.
I don't think my explanation is complete but it gives an intuitive idea of why the slit may in most cases not act as a detector. (And I think I didn't use big words). I think it happens very often that people with more years of study try to explain simple things using complex concepts when simpler concepts would do. This is very discouraging to the person who is trying to learn as it seems that in order to understand the most simple things you would need to understand the more complex. But you can't understand the more complex if you don't first understand the more simple. Well, that could be the topic for a different thread...

 

[DrChinese]

1) the detector absorves the photon or it doesn't. If it does, then this photon is eliminated from consideration as it does not reach the screen. If it doesn't and you do see a photon hitting the screen, then it means that it went through the other slit. In this case the mere presence of the detector eliminates all the possible paths through that slit (This is a detector that does not let any photons go through). So in this case if we send many photons, the interference pattern does not show up...

There is a way to learn which slit the photon pass through without absorbing it. If you put a polarizer in front (or behind) each slit, and the polarizers are crossed (at 90 degrees apart as to their relative orientation), then the photon passes through. Because of the polarizer, you know which slit it passed through (or could learn later); therefore there is no interference pattern.

 

[alexepascual]

There is a way to learn which slit the photon pass through without absorbing it. If you put a polarizer in front (or behind) each slit, and the polarizers are crossed (at 90 degrees apart as to their relative orientation), then the photon passes through. Because of the polarizer, you know which slit it passed through (or could learn later); therefore there is no interference pattern.

I can see two situations here. If we send toward the slits a circular polarized photon, then each component of linear polarization would pass through each slit and then re-combine. The interference pattern would still be there.
If we sent a linearly polarized photon with either horizontal or vertical polarization, then we would know which slit it went through. Every photon that lands on the screen would be coming from the slit that corresponds to its initial polarization. In this case I guess the interference pattern would be destroyed. But I also think that 50% of the time the photon would be absorbed by the polarizer film in the slit with crossed polarization. If you use a polarizer that works by reflection, in that case you could absorb the photon after being reflected. So I don't know if this would qualify as a non-absorption detection experiment. It would also be equivalent to putting any obstacle in front of one of the slits. What you are detecting is the photon that doesn't make it through and not the one that does.
Well Dr. Chinese you got me thinking for a while. This is an interesting modification of the double-slit experiment and I had fun thinking about it.

 

[alexepascual]

There is a way to learn which slit the photon pass through without absorbing it. If you put a polarizer in front (or behind) each slit, and the polarizers are crossed (at 90 degrees apart as to their relative orientation), then the photon passes through. Because of the polarizer, you know which slit it passed through (or could learn later); therefore there is no interference pattern.

I can see two situations here. If we send toward the slits a circular polarized photon, then each component of linear polarization would pass through each slit and then re-combine. The interference pattern would still be there.
If we sent a linearly polarized photon with either horizontal or vertical polarization, then we would know which slit it went through. Every photon that lands on the screen would be coming from the slit that corresponds to its initial polarization. In this case I guess the interference pattern would be destroyed. But I also think that 50% of the time the photon would be absorbed by the polarizer film in the slit with crossed polarization. If you use a polarizer that works by reflection, in that case you could absorb the photon after being reflected. So I don't know if this would qualify as a non-absorption detection experiment. It would also be equivalent to putting any obstacle in front of one of the slits. What you are detecting is the photon that doesn't make it through and not the one that does.
Well Dr. Chinese you got me thinking for a while. This is an interesting modification of the double-slit experiment and I had fun thinking about it.