Measurement of entangled particles causes dechorence at a distance?

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  • #1
craigi
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If we entangle particles and separate them by a large distance, can the action of measuring one cause decoherence at the other's location?

If yes then does this violate relativisitic causality? Could we not use this process to transmit information instantaneously?

If no then why not? Is there a fundamental difference between the process of direct measurement of the local particle and the indirect measurement of the distant particle? Is decoherence any observer relative phenomena?
 
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  • #2
Nugatory
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You don't even need to appeal to decoherence to find an apparent faster-than-light influence. We could just as easily say that when either particle is measured the wave function of both collapses, and indeed that's how it was often described before decoherence was understood.

However, there is no violation of relativistic causality - because there's no causality involved.

I measure my particle and get spin-up, you measure yours and get spin-down. Is that because you went first, lucked into a spin-down measurement and collapsed my particle's wave function into the spin-up state, or because I went first, lucked into a spin-up measurement and collapsed your particle's wave function into the down state? That question is completely meaningless if the two measurement events are separated by a space-like interval because there's no ordering between them - neither one can be said to have happened first. All we know is that when we get together afterwards and compare notes, I measured spin-up and you measured spin-down, and that result is independent of which measurement "happened first".
 
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  • #3
craigi
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You don't even need to appeal to decoherence to find an apparent faster-than-light influence. We could just as easily say that when either particle is measured the wave function of both collapses, and indeed that's how it was often described before decoherence was understood.

However, there is no violation of relativistic causality - because there's no causality involved.

I measure my particle and get spin-up, you measure yours and get spin-down. Is that because you went first, lucked into a spin-down measurement and collapsed my particle's wave function into the spin-up state, or because I went first, lucked into a spin-up measurement and collapsed your particle's wave function into the down state? That question is completely meaningless if the two measurement events are separated by a space-like interval because there's no ordering between them - neither one can be said to have happened first. All we know is that when we get together afterwards and compare notes, I measured spin-up and you measured spin-down, and that result is independent of which measurement "happened first".

The difference is that in the standard EPR paradox that you describe, you don't know when or indeed if, I have measured my particle. We can confer later using sub-luminal information transfer to work it out, but you don't know instantenously.
 
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  • #4
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The difference is that in the standard EPR paradox that you describe, you don't know when or indeed if, I have measured my particle. We can confer later using sub-luminal information transfer to work it out, but you don't know instantaneously.
Ah - I'm sorry, I did misunderstand the question you were asking.
 
  • #5
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No, there is no way to know instantaneously whether the other particle was measured. That would convey information FTL (faster than light).
 
  • #6
craigi
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No, there is no way to know instantaneously whether the other particle was measured. That would convey information FTL (faster than light).

Are you saying that we can indirectly measure distant particles without causing decoherence? If so, why is the local measurement different to the distant, indirect measurement?
 
  • #7
DrChinese
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Are you saying that we can indirectly measure distant particles without causing decoherence? If so, why is the local measurement different to the distant, indirect measurement?

Measuring one places the other in a state in which it is now separate from the combined pair. So the decoherence is immediate and non-local. But there is no effective method to determine that change has occurred.
 
  • #8
craigi
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Measuring one places the other in a state in which it is now separate from the combined pair. So the decoherence is immediate and non-local. But there is no effective method to determine that change has occurred.

If this is true then why is non-local decoherence impossible to observe when local decoherence is observable?
 
  • #9
dauto
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because it is non-local. Rule # 1: no information can be transmitted FTL period.
 
  • #10
DrChinese
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If this is true then why is non-local decoherence impossible to observe when local decoherence is observable?

Regardless of distance, you cannot detect decoherence by looking at a single particle (or system).
 
  • #11
craigi
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because it is non-local. Rule # 1: no information can be transmitted FTL period.

I don't doubt it.

The question is designed to establish why that is true in the case of decoherence with entanglement.
 
  • #12
craigi
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because it is non-local. Rule # 1: no information can be transmitted FTL period.

I don't doubt it.

The question is designed to establish why that is true in the case of decoherence with entanglement.

Regardless of distance, you cannot detect decoherence by looking at a single particle (or system).

Are you sure about this?

A single particle or system can interfere with itself and can be detected by checking for this interference, right?
 
  • #13
DrChinese
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A single particle or system can interfere with itself and can be detected by checking for this interference, right?

Generally, entangled particles do not self-interfere. After they decohere, they still don't particularly look any different unless you do something to make them such. For example, you could focus and therefore collect an entangled particle into a very small point and then send it through a double slit. You would see interference. But that would be true of any photon, entangled or unentangled.
 
  • #14
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A single particle or system can interfere with itself and can be detected by checking for this interference, right?

yes the single particle can (and always will) interfere. however the interference pattern is not clear. it's a mixture of various patterns.

a co-incidence counter is required to filter/extract the interference pattern.


to make the interference pattern clearer we need to make the photon more coherent.
as we start making the photon more coherent, the entanglement starts weakening.

Uncertainty principle (complementarity to be more precise) says - we cannot have your cake (neat interference pattern) and eat it too (strong entanglement)

for more details you can see post 52 in the below link

https://www.physicsforums.com/showthread.php?t=320334&page=3
 
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  • #15
craigi
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Generally, entangled particles do not self-interfere. After they decohere, they still don't particularly look any different unless you do something to make them such. For example, you could focus and therefore collect an entangled particle into a very small point and then send it through a double slit. You would see interference. But that would be true of any photon, entangled or

Suppose the distant particle is in a superposition of spin states and entangled with the local particle. A distant observer can check at any time if this superposition is still present using an interference based observation.

The local observer can, at any time measure the spin of the local particle ending the distant superposition. The distant observer now knows that the local particle has already been observed.

This situation can't be consistent with relativistic causality.

How do we resolve this?
 
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  • #16
DrChinese
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Suppose the distant particle is in a superposition of spin states and entangled with the local particle. A distant observer can check at any time if this superposition is still present using an interference based observation.

The local observer can, at any time measure the spin of the local particle ending the distant superposition. The distant observer now knows that the local particle has already been observed.

This situation can't be consistent with relativistic causality.

How do we resolve this?

There are a couple of specific elements you need to be aware of. Suppose we have distantly separated entangled particles called Alice and Bob:

a. Coherence (and lack thereof) is a function of position and momentum distribution. Which means that Alice and Bob start out with sufficiently wide a spread that there is no interference to be had. This is just a restatement of what we have already said. There is no measurement uou can perform on Alice or Bob at this point that will detect if either is still in a superposition. They are still entangled, and there is no interference. This too is just a restatement of what we have already said.

b. So what you are thinking is that there is some manipulation which can be performed on Alice which changes the above and you will get interference for Bob. That is not correct. The only way to get interference from Bob is to first change Bob to be coherent. And there is no way for Alice to do that! It is true that measuring Alice may cast Bob into a known state, but none of those are coherent for Bob.

c. To be more specific: suppose you measure Alice's spin. You now know Bob's spin. But Bob is still not coherent, nothing has changed in the position or momentum bases.

d. A rule you should be aware of: after measuring Alice's spin per c. above, Alice and Bob are still entangled! That is because collapse has only occurred on the polarization basis. Commuting bases are not affected. You can entangle particles on all or just some bases. For example, a single Type I PDC crystal produces partially entangled photon pairs. Because they are NOT polarization entangled. Vice versa, measuring the polarization of of fully entangled photons does not necessarily eliminate all entanglement on them. So they can retain their superposition.

e. So what happens if we measure Alice as to position? We cast Bob into a known position too. Which is tantamount to knowing which slit Bob will traverse. So there is no interference for Bob. If we don't measure Alice, Bob is not coherent and there is no interference. Either way, no interference.

Nature is quite amazing!
 
  • #17
billschnieder
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For example, you could focus and therefore collect an entangled particle into a very small point and then send it through a double slit. You would see interference.

No you won't, a single particle can not create an interference pattern, ever!
 
  • #18
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If we entangle particles and separate them by a large distance, can the action of measuring one cause decoherence at the other's location?
No.

If no then why not?
One should distinguish wave-function collapse from decoherence. Unlike the former, the latter can be derived from the Schrodinger equation, which cannot involve non-local change of the state.
 
  • #19
billschnieder
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If we entangle particles and separate them by a large distance, can the action of measuring one cause decoherence at the other's location?

If yes then does this violate relativisitic causality? Could we not use this process to transmit information instantaneously?

Coherence is not a property of a single particle. You need at least two particles to talk of coherence. It is a global property of at least 2 particles. Therefore anything which disturbs the relevant aspect of the one of the particles immediately destroys the coherence of the system of particles globally. But you won't know that until you have informasyion from both particles.

This does not violate causality because you are dealing with non-local global information to start with. Its like you had a long string attached to the two particles. Then you can say they are LINKED, which is a global property of both particles. Cutting the string locally at one end does not cut the string at the other instantaneously but it immediately destroys their LINK no matter how long the string is. At the other end, you won't know that the link was destroyed until that information travels to you.
 
  • #20
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A single particle or system can interfere with itself and can be detected by checking for this interference, right?

in non linear quantum mechanics yes.
unlike of standard quantum mechanics (a linear one).


.
 
  • #21
craigi
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No you won't, a single particle can not create an interference pattern, ever!

This is only really true on a technicality. A single particle does interfere with itself. We can demonstrate this by performing the double slit experiment with single photons. The "pattern" only emerges after we've accumulated enough to be able to see a pattern created by them. So it's true in the respect that a single particle can never create a pattern, but that's completely independent of interference.

A single particle or system can interfere with itself and can be detected by checking for this interference, right?

in non linear quantum mechanics yes.
unlike of standard quantum mechanics (a linear one).

If this is true, then how do you use linear quantum mechanics to explain the results of Young's double slit experiment, performed with single photons, without self-inteference?

Coherence is not a property of a single particle. You need at least two particles to talk of coherence. It is a global property of at least 2 particles. Therefore anything which disturbs the relevant aspect of the one of the particles immediately destroys the coherence of the system of particles globally. But you won't know that until you have informasyion from both particles.

Single particle coherence is deeper that what I'd envisaged for this question. The question was really referring to a distant macroscopic quantum system undergoing decohorence due the measurement of a local particle that is in some way entangled with it.

This does not violate causality because you are dealing with non-local global information to start with. Its like you had a long string attached to the two particles. Then you can say they are LINKED, which is a global property of both particles. Cutting the string locally at one end does not cut the string at the other instantaneously but it immediately destroys their LINK no matter how long the string is. At the other end, you won't know that the link was destroyed until that information travels to you.


No.

One should distinguish wave-function collapse from decoherence. Unlike the former, the latter can be derived from the Schrodinger equation, which cannot involve non-local change of the state.

Same question to both of you:

If measurement of a local particle causes a distant particle to take a distinct state, then this state can no longer be part of a superposition, right? Why would this distinct state not be capable of causing decoherence at the distant particle's location?

...

Thanks for your response. I have more questions for you too. It'll just take a little more time to formulate them.
 
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  • #22
craigi
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There are a couple of specific elements you need to be aware of. Suppose we have distantly separated entangled particles called Alice and Bob:

a. Coherence (and lack thereof) is a function of position and momentum distribution. Which means that Alice and Bob start out with sufficiently wide a spread that there is no interference to be had. This is just a restatement of what we have already said. There is no measurement uou can perform on Alice or Bob at this point that will detect if either is still in a superposition. They are still entangled, and there is no interference. This too is just a restatement of what we have already said.

b. So what you are thinking is that there is some manipulation which can be performed on Alice which changes the above and you will get interference for Bob. That is not correct. The only way to get interference from Bob is to first change Bob to be coherent. And there is no way for Alice to do that! It is true that measuring Alice may cast Bob into a known state, but none of those are coherent for Bob.

This isn't where I was going with this.

What was really talking about was, suppose we create 2 entangled particles. We keep one locally, in an un-measured state. The other, we put into a macroscopic system that is capable of detecting whether it is still in a superposition.

So here, the difference with what you suggest is that Alice isn't manipulating Bob to check for superposition, rather Bob is being manipulated so that at some future time, it can be known whether Bob is still in a superposition.

So now we send off Bob far away. And then use the measurement process on Alice to cast Bob into a distinct state in order to send a signal to that location.

c. To be more specific: suppose you measure Alice's spin. You now know Bob's spin. But Bob is still not coherent, nothing has changed in the position or momentum bases.

d. A rule you should be aware of: after measuring Alice's spin per c. above, Alice and Bob are still entangled! That is because collapse has only occurred on the polarization basis. Commuting bases are not affected. You can entangle particles on all or just some bases. For example, a single Type I PDC crystal produces partially entangled photon pairs. Because they are NOT polarization entangled. Vice versa, measuring the polarization of of fully entangled photons does not necessarily eliminate all entanglement on them. So they can retain their superposition.

e. So what happens if we measure Alice as to position? We cast Bob into a known position too. Which is tantamount to knowing which slit Bob will traverse. So there is no interference for Bob. If we don't measure Alice, Bob is not coherent and there is no interference. Either way, no interference.

I'd originally discounted a double slit interference as being too simple to illustrate my question, but you've given me an idea.

Suppose we create 2 entangled photons. We keep Alice, locally and send Bob off into the distance towards a double slit arrangement far away, then after another large distance we have a detector.

Now there is a time, where Bob is between the slits and the detector where we can choose to either measure Alice's position or not. That decision will determine whether Bob can contribute to an interference pattern.

Now suppose we do this with a million Alices and million Bobs, we have enough to form an interference pattern, or not, at a distance, depending upon whether we measure Alice's position.

Is there a process or a quirk of quantum mechanics than prevents this information being transmitted faster than the speed of light? If so, then it's probably the same thing that answers my original question.
 
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  • #23
DrChinese
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Coherence is not a property of a single particle. You need at least two particles to talk of coherence. It is a global property of at least 2 particles. Therefore anything which disturbs the relevant aspect of the one of the particles immediately destroys the coherence of the system of particles globally. But you won't know that until you have informasyion from both particles.

This does not violate causality because you are dealing with non-local global information to start with. Its like you had a long string attached to the two particles. Then you can say they are LINKED, which is a global property of both particles. Cutting the string locally at one end does not cut the string at the other instantaneously but it immediately destroys their LINK no matter how long the string is. At the other end, you won't know that the link was destroyed until that information travels to you.

craigi,

As you mention, bill's point is a technicality. I think everyone here understood that we were discussing groups of entangled photons. It is equally true that entanglement cannot be determined with an experiment on a single pair of photons, interference likewise, coherence likewise.

From past experience, I will tell you that bill is a man with a very specific agenda because he denies entanglement is an actual state. He pushes local realism, which was largely discredited decades ago. He and I have sparred for years. His presence usually signals that the thread is getting ready to change focus.

I mostly agree with his comments above, and he is knowledgeable, but he mixes his agenda in quickly. Hopefully this time he will be more helpful, as I know he can be. I am saying all this to save you time and to try to salvage further discussion on this topic.

And to be technical back to him: we don't know if causality is violated or not (as evidenced by entanglement experiments). There are generally accepted interpretations both ways. There are several retro-causal and time symmetric interpretations, for example. And many consider standard QM to lack causality anyway. There are also non-local deterministic theories.
 
  • #24
DrChinese
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We create 2 entangled particles. We keep one locally, in an un-measured stated. The other we put into a macroscopic system that is capable of detecting whether it is still in a superposition.

There is no such macroscopic system. Any experiment on Bob will have the same outcome (as far as can be known) regardless of whether Bob is still entangled or not.

Further, there is no known point in time (spacetime) when entanglement of separated particles can be said to absolutely cease. It is purely out of convenience that it is usually assigned to the earlier measurement of the pair. But all statements are equally valid if you assert it is the later of the two which causes entanglement to cease.

Keep in mind that you can entangle a pair of photons that no longer exist! That is done in some entanglement swapping experiments. I can give references if that will help, but it is a subject for another thread.
 
  • #25
DrChinese
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I'd originally discounted a double slit interference as being too simple to illustrate my question, but you've given me an idea.

Suppose we create 2 entangled photons. We keep Alice, locally and send Bob off into the distance towards a double slit arrangement far away, then after another large distance we have a detector.

Now there is a time, where Bob is between the slits and the detector where we can choose to either measure Alice's position or not. That decision will determine whether Bob can contribute to an interference pattern.

Now suppose we do this with a million Alices and million Bobs, we have enough to form an interference pattern, or not, at a distance, depending upon whether we measure Alice's position.

I once had the same idea (for an FTL device), and that is when I first discovered that entangled photons do not produce interference patterns. :smile:

http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

See page S290, figure 2
 
  • #26
craigi
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craigi,

As you mention, bill's point is a technicality. I think everyone here understood that we were discussing groups of entangled photons. It is equally true that entanglement cannot be determined with an experiment on a single pair of photons, interference likewise, coherence likewise.

From past experience, I will tell you that bill is a man with a very specific agenda because he denies entanglement is an actual state. He pushes local realism, which was largely discredited decades ago. He and I have sparred for years. His presence usually signals that the thread is getting ready to change focus.

I mostly agree with his comments above, and he is knowledgeable, but he mixes his agenda in quickly. Hopefully this time he will be more helpful, as I know he can be. I am saying all this to save you time and to try to salvage further discussion on this topic.

And to be technical back to him: we don't know if causality is violated or not (as evidenced by entanglement experiments). There are generally accepted interpretations both ways. There are several retro-causal and time symmetric interpretations, for example. And many consider standard QM to lack causality anyway. There are also non-local deterministic theories.

I'm interested in interpretations of QM too and I'm more than happy to entertain unpopular ideas, but as you suggest, let's keep discussions on that in their own threads.

I once had the same idea (for an FTL device), and that is when I first discovered that entangled photons do not produce interference patterns. :smile:

http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

See page S290, figure 2


Brilliant.

That page answers my question perfectly. I'm more than happy to extrapolate that explanation to my original question.

As ever, it's distrubing how QM always seems to be one step ahead. It's yet another way that our intuition from the classical world is broken.
 
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  • #27
DrChinese
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1. I'm interested in interpretations of QM too and I'm more than happy to entertain unpopular ideas, but as you suggest, let's keep discussions on that in their own threads.

2. As ever, it's distrubing how QM always seems to be one step ahead. It's yet another way that our intuition from the classical world is broken.

1. The interpretations I mentioned are all generally accepted as viable ones.

2. Which is why I mentioned: Nature is quite amazing! :smile: Always outsmarting our attempts to make something of quantum non-locality.
 
  • #28
billschnieder
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A single particle does interfere with itself. We can demonstrate this by performing the double slit experiment with single photons.
...
If this is true, then how do you use linear quantum mechanics to explain the results of Young's double slit experiment, performed with single photons, without self-inteference?
But haven't you misinterpreted the results of this experiment? The experiment shows that you only need a single particle in a slit at a given time to obtain interference, it does not demonstrate that the photon is interfering with itself. Besides, without the slit, you have no interference, so you could say the photon is interfering with the slits not itself. But this is probably off-topic anyway.

Single particle coherence is deeper that what I'd envisaged for this question. The question was really referring to a distant macroscopic quantum system undergoing decohorence due the measurement of a local particle that is in some way entangled with it.
It doesn't matter, my answer still applies to this. Coherence is the property of a system of more than one particle. In your example, you have a particle which is entangled with a macroscopic entity, the the particle moves far away and you are asking if measuring the the distant particle causes decoherence of the local macroscopic entity. What I was explaining is that you should realize that the "System" in your question is the "macroscopic entity" + "particle". it is this combined system that has the coherence. If you disturb the particle, you have distroyed the coherence in the system between the particle and the macroscopic entity. This coherence by definition is a global (read non-local) property of the system, since the system is non-local.

Same question to both of you:

If measurement of a local particle causes a distant particle to take a distinct state, then this state can no longer be part of a superposition, right? Why would this distinct state not be capable of causing decoherence at the distant particle's location?
Measurement of a local particle does not cause a distant particle to take a distinct state (though some may disagree). It changes the information content of the "non-local" wavefunction describing the system which includes both particles. Some will call this "wave-function collapse". As to the second part of your question. Indeed measuring the local particle disturbs it and as such causes decoherence of the system which includes both particles. This decoherence does not violate causality because it refers to destroying the non-local correlation between the particles. In case you wonder, non-local coherence is not mysterious. Any two clocks will have coherence irrespective of how far apart they are. It is non-local just because we have chosen to contemplate two distant particles together as a single system and once we do that, we have to be consistent in our descriptions.
 
  • #29
craigi
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But haven't you misinterpreted the results of this experiment? The experiment shows that you only need a single particle in a slit at a given time to obtain interference, it does not demonstrate that the photon is interfering with itself. Besides, without the slit, you have no interference, so you could say the photon is interfering with the slits not itself. But this is probably off-topic anyway.

The reason that we say that the particle interferes with itself pertains to the wave properties of particles. A classical wave interferes with itself in the same way. The mathematical description of the self-interference is as we'd expect, from the geometry in both cases.

I can't envisage a mathematical formulation that could reproduce the results of the experiment so neatly, where the particle doesn't interfere with itself, only with the slits. Do you have a mathematical description to support your idea? Are there others that do?

Whatever it does, it would need to feel out the geometry of the entire arrangement and we'd end up with is something that is mathematically equivelant to doing this via wave propogation anyway.

It doesn't matter, my answer still applies to this. Coherence is the property of a system of more than one particle. In your example, you have a particle which is entangled with a macroscopic entity, the the particle moves far away and you are asking if measuring the the distant particle causes decoherence of the local macroscopic entity. What I was explaining is that you should realize that the "System" in your question is the "macroscopic entity" + "particle". it is this combined system that has the coherence. If you disturb the particle, you have distroyed the coherence in the system between the particle and the macroscopic entity. This coherence by definition is a global (read non-local) property of the system, since the system is non-local.

I have no problem in considering the particle and a distant macroscopic entity as a single system. It is the decoherence in exactly such a system, that the question was about.

Measurement of a local particle does not cause a distant particle to take a distinct state (though some may disagree). It changes the information content of the "non-local" wavefunction describing the system which includes both particles. Some will call this "wave-function collapse". As to the second part of your question. Indeed measuring the local particle disturbs it and as such causes decoherence of the system which includes both particles. This decoherence does not violate causality because it refers to destroying the non-local correlation between the particles. In case you wonder, non-local coherence is not mysterious. Any two clocks will have coherence irrespective of how far apart they are. It is non-local just because we have chosen to contemplate two distant particles together as a single system and once we do that, we have to be consistent in our descriptions.

I certainly wouldn't use the word coherence, particularly in the quantum sense, to describe the correlation between the readings on two clocks.
 
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  • #30
DrChinese
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... In case you wonder, non-local coherence is not mysterious. Any two clocks will have coherence irrespective of how far apart they are. It is non-local just because we have chosen to contemplate two distant particles together as a single system and once we do that, we have to be consistent in our descriptions.

There you go again, making a purely classical assertion regarding a quantum process. Stop this nonsense now, or I will report you. You know the rules here, bill.

Entanglement is not classical, is nothing like two synchronized clocks, and Bell's Theorem demonstrates that clearly. Further, entangled systems *may* violate causality, that is not clear at this time.
 
  • #31
billschnieder
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The reason that we say that the particle interferes with itself pertains to the wave properties of particles.
I understand that it is an inference, but certainly not "demonstrated" by experiment as suggested in your previous post. I was just pointing out to you that if particles were interfering with themselves as you say, we would have interference everywhere without any need for slits but we don't. Even the popular idea that photons interfere with each other is not correct. Photons are bosons they can interact with fermions but not other bosons. The slits have lots of fermions. This is not my idea, this is standard physics.

A classical wave interferes with itself in the same way. The mathematical description of the self-interference is as we'd expect, from the geometry in both cases.
No question about that. Epicycles also explained the motion of the planets pretty well.

I can't envisage a mathematical formulation that could reproduce the results of the experiment so neatly. Do you have a mathematical description to support your idea? Are there others that do?
The originators of epicycles couldn't envisage anything better either. Every generation does the best it can. I'm just prodding your imagination to think carefully about some of the assumptions you have taken for granted which may be the source of some misunderstandings evidenced in your original question.Like I said, this is not my idea. But I'll give you not just one but several mathematical descriptions which do not use waves but are based on quantized momentum transfer from discrete particles.


Whatever it does, it would need to feel out the geometry of the entire arrangement and we'd end up with is something that is mathematically equivelant to doing this via wave propogation anyway.
No question that the wave propagation description provides a workable "mathematical" solution to the puzzle. But as demonstrated by Duane and Compton many years ago, it is not the only one.

I have no problem in considering the particle and a distant macroscopic entity as a single system. It is the decoherence in exactly such a system, that the question was about.
And my answer was that once you start talking of coherence in such a system, you can not later separate it and talk of coherence only in one part without reference to the other as implied in your question. It is not a question of classical vs quantum. It is a question of consistency vs inconsistency and being clear of what we are talking about.
 
  • #32
naima
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I once had the same idea (for an FTL device), and that is when I first discovered that entangled photons do not produce interference patterns. :smile:

http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf

See page S290, figure 2

Hello Dr Chinese
I ignored that.
Zeilinger says that photons registered in the focal plane of the Heisenberg lens are associated with photons which make an interference pattern. i guess that they must be chosen in a pattern with no interference.
Why cannot we detect all the photons in the focal plane? the interference pattern would be seen in the double slit screen (and that is impossible).
 
  • #33
meBigGuy
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Maybe I am missing the whole point, but when you measure the first particle you no longer have an entangled system. The second particle is just an electron with a state. You can find out things about it or do tricky experiments or play weird games by knowing the results of the first measurements, but that is all.

Of course single electrons interfere with themselves when they "combine" after taking different paths. Otherwise there could be no interference pattern when we look at a lot of them. Or, am I again missing the whole point?
 
  • #34
naima
Gold Member
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Look at fig 3 in zeilinger paper
It is the paper cited by Dr Chinese.
Zeilinger says that all the points on the screen behind the two slits are not registered by coincidence by photons in the focal plane. I wonder why Alice cannot get all the photons in the focal plane.
 
  • #35
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Maybe I am missing the whole point, but when you measure the first particle you no longer have an entangled system.

My first thought was the particles remain entangled until one or the other is observed. That particle then becomes entangled with whatever observes it and is no longer entangled with the other particle.

However it seems a detailed analysis shows its a bit more complicated than that when considered in the light of the ensemble interpretation I hold to:
http://arxiv.org/ftp/quant-ph/papers/0404/0404011.pdf

Thanks
Bill
 
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