When does entanglement end?

In summary, the conversation discusses the concept of entanglement and when it ends in the context of polarizing beam splitters and filters. There is a disagreement on whether entanglement ends when particles collide with other particles or when there is an irreversible measurement made. The possibility of reversing a measurement and the role of decoherence in entanglement are also mentioned. The idea of entanglement being a manifestation of the measurement process is brought up, and the concept of polarization by reflection is briefly discussed.
  • #71
Epicurus3 said:
Bell's is NOTHING MUCH, and follows from superposition directly.
This makes absolutely no sense at all, and does little to explain your position.

Other than having an irrational bias against the Irish exactly what is your scientific position.
1) Local Realism is correct – and the Bell proofs against Local Realism are simplistic and flawed.
or
2) The Non-Local QM Copenhagen view is the most complete, but Bell and EPR-Bell experiments do nothing to refute Local Realism (but offering nothing to say why not).
Just state your position clearly:

To say “Bell's follows from superposition directly” only says that Bell agrees with QM’s ability to make predictions. It totally misses the point that Bell only addresses the viability of Local Realism; not the preference of one QM interpretation over another.

If by chance you think you are a Local Realist, you are representing the position irrationally – please read the sickly threads on the top of the forums and abide by the agreements you made on joining PF.
 
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  • #72
DaveC426913 said:
No it won't. And we'll try very hard not to lose sleep after your devastating coup de grace. :tongue2:

Well put. I assume that Epicurus3 is simply baiting us at some level.

Hey, there are people who don't think man has been to the moon either. No accounting for some folks' beliefs. No need to waste our time here with them either.
 
  • #73
Epicurus3 said:
If you understand superposition - as Bell did - then his Theroem follows from that with a bit of Sudoku level statistics.

The correlation between the states of two entangled particles is random for both particles as is clear from the wave equation. (not that they have secret states that only 'appear' when observed)

Bell realized that this statement (or similar, - better worded than mine) would not make a career for him, so he devised his joke intelligence test after a couple of beers and managed to make a career out of it. One of the guys in the same bar gave him the idea.

I have seen Bell interviewed - it was clear he had nothing more to contribute to physics than his little superposition side bar - and actually had a thin understanding of physics generally.

<flames on>
So let's hear no more about the subtley of Bell's Theorem PLEASE. Try AOP and pattern programming if you like convolution. Bell's is NOTHING MUCH, and follows from superposition directly.
<flames off>

Ok, your next post will have to contain something more of substance than a statement that Bell was a kind of Irish idiot, or I consider that you are just trolling.

You're warned.
 
  • #74
RandallB said:
This makes absolutely no sense at all, and does little to explain your position.

Other than having an irrational bias against the Irish exactly what is your scientific position.
1) Local Realism is correct – and the Bell proofs against Local Realism are simplistic and flawed.

'Local Realism is correct' - It would be impossible to tie a variable to ONE PARTICLE, because by superposition they could either one decohere - so a variable attached to either particle cannot work unless one can 'contact' the other. Such a local variable would tie a particle to one state - which is not the case. Its obvious from superpostition. So what are you trying to say? I believe that you cannot actually understand what a local variable means.(it means no contact between the two).

They used to think something like this, and this is also what you are still thinking, I assume from lack of clarity:

- 'one particle is actually always X state and the other always Y state, but this is hidden, until observed, so there may be a hidden variable in each particle which says what the particle ACTUALLY IS'. So, yes, yes, these particles have real spins and other states - bla bla bla - which is what they used to think in the 30s, and we know better now - particles do not have ANY state, or ALL states at once - they are in superposition, wave packet, coherent. When observed they pick a state at random. They have NOT got a state that is hidden. - what else can I say to you??

To me its obvious and clear - what's your problem? Stuck in sentences you cannot understand because you believe its so subtle. Sorry, its not. At least Bell's insn't. Come on - let's have some truth here. MAybe you are a physics Historian?

Bell's simply follows on from superposition and what's worse about it, it has so many experimental problems that it almost imposssible to handle in practice (see Thomas's endless whingings about the set up)

Bell's is a red herring and simply derived from his knowledge of superposition. It kind of proves superposition, but is experimentally very hard to do conclusively. A waste of peoples energy.
 
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  • #75
vanesch said:
We've been through this already. But in as much that this is a very plausible picture when it is *the same photon* that went through the first polarizer (and hence "got its principle axis turned into the polarizer direction" by interaction with that polarizer) it is not a surprise that when it arrives at the second polarizer, we find a relationship as given by Malus' law which depends on the difference of the axes of the first polarizer (now integrated into the photon itself after interaction) and the second polarizer (next interaction with the modified photon), I don't see how this can be an evident picture for two separate photons - even though they might start out with the same "principle axis" in the source. In what way will the twisting of the photon axis of the first photon by the first polarizer twist and turn the photon axis of the second one which is far away, so that it gets aligned with the orientation of the first polarizer, before it meets its own (second) polarizer ?

We're assuming that the optical disturbance (associated with each pair of detection attributes) between the two polarizers is one and the same thing during any given coincidence interval (although it is assumed to be varying randomly from interval to interval). That is, in a simple optical Bell setup, whatever is incident on polarizer A during a given coincidence interval is identical to what is incident on polarizer B during that same interval. The polariscopic (not necessarily classical per se -- ie. we could be accumulating single photon detections) analog of this assumption is that the optical disturbance transmitted by the first polarizer is identical to the optical disturbance incident on the second, or analyzing, polarizer.

This is the assumption that underlies the qm projection along the transmission axis associated with a detection.

In the polariscopic setup, the intensity of the transmission from the analyzing polarizer is a cos2 function of the difference between the transmission axes of the first and second polarizers. The amplitude of the resultant wave is altered by the rotation of either the second polarizer or the first polarizer.

This is what happens in Bell tests also. For essentially the same reason.
_____________________

Whether the above is acceptable or not, I've thought some more about your comments on the definition of quantum entanglement and I see your point that the only unambiguous meaning of the term quantum entanglement is its formal expression(s) within quantum theory.
 
  • #76
ThomasT said:
This is what happens in Bell tests also. For essentially the same reason.
NO
I suspect you still do not have a proper grasp of what Bell is asking an experiment to show and a LR (Local Realistic) description to explain.

Bell tests measure and summarize results from individual photons – NOT intensirtly levels from large groups of Photons (polariscopic method).
Sure – IF when B is measured at 22.5° while A is at 45° (If both were at 45° it gives 0% intensity correlation) you still saw 100% of the Photons – BUT at 15% of the proper energy of each photons energy level then yes your “Intensity” comparison would work. But the well known facts are quantum photon energy level don’t change intensity! You observe 15% of the photons not 15% of their each photons energy.

With A at 45° a random distribution of 50% V (0°) and 50% H (90°) could correctly predict 50% at H or V for the B test. A random distribution can also correctly predict 100% at 135° and 0% at 45°. BUT at 22.5° a polariscopic style random distribution can only predict 25% NOT 15°.

That is ON the Bell inequity line not a violation of it as is required to match the QM results.
 
  • #77
ThomasT said:
We're assuming that the optical disturbance (associated with each pair of detection attributes) between the two polarizers is one and the same thing during any given coincidence interval (although it is assumed to be varying randomly from interval to interval). That is, in a simple optical Bell setup, whatever is incident on polarizer A during a given coincidence interval is identical to what is incident on polarizer B during that same interval. The polariscopic (not necessarily classical per se -- ie. we could be accumulating single photon detections) analog of this assumption is that the optical disturbance transmitted by the first polarizer is identical to the optical disturbance incident on the second, or analyzing, polarizer.

Well, that's not going to work, as we discussed already. Imagine the polarizers at 90 degrees with respect to one another. As the incident disturbances are random, some will have "45 degrees", no ? Well, a 45 degree disturbance (which will be part of the set of random disturbances sent out) should have 50% probability (intensity whatever) to get through each polarizer. So in half of the cases where we saw something "left", we should also see something "right" for these distrubances, right ?
Well, that doesn't follow the cos^2 law, which tells us that there can NEVER be any click on the left when there is one on the right and vice versa. So even if there's only a small amount of "45 degree disturbances" in the lot that's randomly sent out to the left and the right, and half of them make "common clicks", this would violate the cos^2 law which says that never there can be any common clicks.

Again, the reason why we have a cos^2 law in the *successive* polarizers, is that the disturbance AFTER the first one has been aligned with the orientation of the first polarizer (and has "forgotten" its original incident orientation).
 
  • #78
vanesch said:
Well, that's not going to work, as we discussed already. Imagine the polarizers at 90 degrees with respect to one another. As the incident disturbances are random, some will have "45 degrees", no ? Well, a 45 degree disturbance (which will be part of the set of random disturbances sent out) should have 50% probability (intensity whatever) to get through each polarizer. So in half of the cases where we saw something "left", we should also see something "right" for these distrubances, right ?
I don't think this is the correct way to analyze the situation. Nothing can be said about the orientations of the incident disturbances or for that matter about anything that's qualitatively going on in individual trials independent of instrumental behavior. We only know if a detection is registered or not during a certain interval.

So, let's say that A detects first during some coincidence interval. The projected amplitude wrt the disturbance incident on B is assumed to be the same as A's which means that the probability of detection at B with polarizers aligned is 1, and with polarizers perpendicular to each other it's 0.

As the polarizer at B is rotated away from alignment with the polarizer at A, the amplitude of the transmitted component of the wave incident on the polarizer at B will vary as the cosine of the angular difference. The probability of detection at B is the intensity of the wave transmitted by the polarizer at B, which is the amplitude squared, which is cos2
Theta, which is Malus' Law -- which is the probability of coincidental detection.

Of course this probability has no physical meaning wrt any given individual trial, or coincidence interval or single pair of detection attributes. It's a statement regarding the expected frequency of coincidental detection given a large number of trials.

So, I don't think that your argument above renders the analogy invalid. Something else might, but not that.

vanesch said:
Again, the reason why we have a cos^2 law in the *successive* polarizers, is that the disturbance AFTER the first one has been aligned with the orientation of the first polarizer (and has "forgotten" its original incident orientation).
In either case (Bell test or polariscope), extending between the two polarizers is a disturbance or disturbances with common properties.

In the polariscopic setup, the first polarizer is the analog of polarizer A above (that is, it's the analog of whichever polarizer is associated with the first detection during some coincidence interval in a Bell test). Think of the disturbance between A and B in a Bell test as being transmitted by the polarizer that registers the initial detection in a given coincidence interval and incident on the other polarizer.
 
  • #79
ThomasT said:
I don't think this is the correct way to analyze the situation.
……
….. As the polarizer at B is rotated away from alignment with the polarizer at A, the amplitude of the transmitted component of the wave incident on the polarizer at B will vary as the cosine of the angular difference.
No once again – it is your way of analyzing the situation here that eliminates it from having any meaningful relation to BELL or “Entangelement”.

The only way you can use “amplitude” to allow your polariscopic to work is to select a detection coincidence interval at A that measures 1000 photons, and compare that to the same detection coincidence interval at B and use the B photon count as a measure of amplitude. Sure that will work – But is not an explanation entanglement and not at all what Bell is talking about as Bell requires quantum level (individual photon) comparisions. Until you understand that you will not be on the same page as vanesch or anyone else. Bell requires a Local Realistic description of the individual photon behaviors Not the apparent amplitude changes of groups of large numbers of photons.

The polariscopic Classical description can only be though of as a Classical Non-Local because it does not bring with it enough photon level detail to call it “Local”. And without that level of detail it can hardly be looked to for any help with the formalization of QM entanglement.
Remember Bell only addresses the viability of Local descriptions.
 
  • #80
RandallB said:
NO
I suspect you still do not have a proper grasp of what Bell is asking an experiment to show and a LR (Local Realistic) description to explain.
That's a definite possibility.

RandallB said:
Bell tests measure and summarize results from individual photons – NOT intensirtly levels from large groups of Photons (polariscopic method).
Intensity in Bell tests is the rate of coincidental detection per unit of time. The probabilities of coincidental detection refer to large numbers of individual trials. As the unit of time (and the number of trials) increases, the experimental results more closely approximate the predicted values.

The probability of individual detection at A or B for any and all individual trials remains 1/2.

RandallB said:
But the well known facts are quantum photon energy level don’t change intensity! You observe 15% of the photons not 15% of their each photons energy.
Yes, that's a hypothesis that seems to be supported by certain experimental results. I've got to think about this some more.

Without going into why, I'm confused again. Thanks for your replies, and vanesch and DrChinese and others.
 
  • #81
RandallB said:
No once again – it is your way of analyzing the situation here that eliminates it from having any meaningful relation to BELL or “Entangelement”.
I'm just talking about the similarities between the experimental setups.

RandallB said:
The only way you can use “amplitude” to allow your polariscopic to work is to select a detection coincidence interval at A that measures 1000 photons, and compare that to the same detection coincidence interval at B and use the B photon count as a measure of amplitude. Sure that will work ...
As I mentioned, the polariscopic setup can be one that counts individual photon detections.

RandallB said:
... But is not an explanation entanglement ...
I agree. Even if the analogy is valid, it still doesn't explain what entanglement is. But, if it is valid, then it's a stepping stone to understanding what it probably isn't and what it might be.

RandallB said:
... and not at all what Bell is talking about as Bell requires quantum level (individual photon) comparisions.
I don't know what you mean by this.

RandallB said:
Until you understand that you will not be on the same page as vanesch or anyone else.
Are you talking about the pairing process?

I've got to run -- will get to the rest of this later. Thanks.
 
  • #82
ThomasT said:
So, let's say that A detects first during some coincidence interval. The projected amplitude wrt the disturbance incident on B is assumed to be the same as A's which means that the probability of detection at B with polarizers aligned is 1, and with polarizers perpendicular to each other it's 0.

Why should the projected amplitude wrt to the disturbance incident on B be the same as A's ?
They should be the same, all right. But they should not be the same as the orientation of the polarizer, no ?

You do agree that the orientation of the light *changes* when it gets through a polarizer, right ? That it takes on the orientation of the polarizer ? Or not ? But how could the light at B *change* in accordance with the orientation of the polarizer at A ?

Because, consider the following setup:
one beam, 3 polarizers. The first one at 90 degrees, the second one at 45 degrees, the third one at 0 degrees. Does any light get through this setup or not ?
 
  • #83
ThomasT said:
As I mentioned, the polariscopic setup can be one that counts individual photon detections.
But you disrearded that when you earler said:

"As the polarizer at B is rotated away from alignment with the polarizer at A, the amplitude of the transmitted component of the wave incident on the polarizer at B will vary as the cosine of the angular difference."

Which would require that at some angle that individual photon deterction would require it deleiver only 15% of the energy in that photon.

Nothing Quantum or in Bell can accept that plan.
 
  • #84
peter0302 said:
Hehe, I can't really tell what side you're taking. :) But my answer is collapse can't be a physical process; it's just that quantum statistics don't conform to the laws of macro-statistics.

Wouldn't you rather throw out classical statistics than throw out relativity? :)

hmmm, maybe neither...
...as photons are bosons, can we use Bose-Einstein statistics?
Can we consider an entangled pair of photons as a Bose-Einstein condensate?
If we 'push it too hard' it collapses, but will 'reform' if the conditions are amenable.
 
  • #85
'kay...
...that went down well.

What about this...

...I get two 'bosons' into a 'space' for which they are too big to 'fit' they conform to Bose-Einstein statistics.

...if I make two 'bosons' from something that was in a 'space' in which they are too big to 'fit' (presuming that two photons created as an entanged pair from a single
photon take more 'space' than the original) what happens?

Assuming, presumably, that 'all things are still equal' (i.e. there is conservation of energy/momentum) does this seem correctly time-symmetric?

If so, could we appeal to Noether's theorem? (I'm not quite sure of the logic!...might be the wrong way round... does a conservation law imply a symmetry?)

I'm not fully familiar with the fermionic version of entanglement so this might seem stupid as a result of the mechanics of that.

Is this crackpottery? Probably...
 
  • #86
moving_on said:
'kay...
...that went down well.

What about this...

...I get two 'bosons' into a 'space' for which they are too big to 'fit' they conform to Bose-Einstein statistics.

...if I make two 'bosons' from something that was in a 'space' in which they are too big to 'fit' (presuming that two photons created as an entanged pair from a single
photon take more 'space' than the original) what happens?

Assuming, presumably, that 'all things are still equal' (i.e. there is conservation of energy/momentum) does this seem correctly time-symmetric?

If so, could we appeal to Noether's theorem? (I'm not quite sure of the logic!...might be the wrong way round... does a conservation law imply a symmetry?)

I'm not fully familiar with the fermionic version of entanglement so this might seem stupid as a result of the mechanics of that.

Is this crackpottery? Probably...

Photon not needing space because fermion already in hole wait for him.
 
  • #87
DrChinese said:
I haven't seen a paper which answers this particular question, maybe someone else has... (I have scanned the preprint archive but to no avail so far).

Most Bell tests use polarizing beam splitters (PBS) to check photons at Alice and Bob. Typical are 2 detectors at Alice and 2 at Bob. Results of all 4 are correlated and analyzed. You would normally say the entanglement ends once we know which way the photon goes through the beam splitter.

What if we takes the 2 beams at Alice and merge them back very precisely together again? I.e. such that it is no longer possible to tell which path the photon took through the PBS. I would expect that the resultant reconstructed beam (Alice) is still entangled with Bob. If you tested Alice and Bob at this point, I would expect us to see the perfect correlations and the Bell inequality violations per usual. Is this correct?

So when does the entanglement actually end? If what I am saying is right, the PBS is not actually capable of ending the entanglement itself. Instead, it is the detection of the photon - and what we know about it at that point - which ends the entanglement. I believe this is fully consistent with the QM prediction.



I have come to this thread rather late so I may have missed references to recent literature, if so I apologize for the duplication. There is a short review article Sudden Death of Entanglement by Ting Yu and J. H. Eberly in the 30 January 2009 issue of Science . Note: the article starts on page 598. The table of contents is in error --it says page 602. There are several other relevant articles in previous issues of Science :

J. H. Eberly and Ting Yu, The End of an Entanglement , Science . 27 April 2007, page 555.

M. P. Almeida, et. al. Environment-Induced Sudden Death of Entanglement , same issue of Science , page 579.

The longevity of entanglement is an active area of research because of its importance to the success of quantum computation and communication. One of the initial problems was to settle on a suitable measure of entanglement. It seems that the current consensus is to use a concept called concurrence { W. K. Wootters, Phys. Rev. Lett. 80 , 2245 (1998), see also Phys. Rev. Lett. 78 , 5022 (1997) and arXiv:quant-ph/9709029v2 13 Sep 1997 }

There is a long review article: Measures and dynamics of entangled states in Phys. Repts. 415 , 207 (2005), also arXiv:quant-ph/0505162v1 22 May 2005

Quantum entanglement for composite systems is often defined in terms of the density matix. See Appendix B of David J. Tannor, Introduction to Quantum Mechanics, A Time Dependent Perspective for a lucid synopsis.

I hope that you will find these references useful.
 
  • #88
AEM said:
J. H. Eberly and Ting Yu, The End of an Entanglement , Science . 27 April 2007, page 555.

http://www.sciencemag.org/cgi/content/abstract/316/5824/579
from the above quote:
"Using an all-optical experimental setup, we showed that, even when the environment-induced decay of each system is asymptotic, quantum entanglement may suddenly disappear. This sudden death constitutes yet another distinct and counterintuitive trait of entanglement."

The disappearance of entanglement appears to be unpredictable but sudden, ummmm..
 
  • #89
p764rds said:
http://www.sciencemag.org/cgi/content/abstract/316/5824/579
from the above quote:
"Using an all-optical experimental setup, we showed that, even when the environment-induced decay of each system is asymptotic, quantum entanglement may suddenly disappear. This sudden death constitutes yet another distinct and counterintuitive trait of entanglement."

The disappearance of entanglement appears to be unpredictable but sudden, ummmm..

Please, what's your point? Could you elaborate a little more?
 
  • #90
AEM said:
Please, what's your point? Could you elaborate a little more?

The results of exactly how entanglement ends - i.e. the exact ending time is unpredictable but it ends suddenly - is interesting.

sudden ending:
I would expect the end of entanglement to be instantaneous because otherwise there would be strange states of half entangled, quarter entangled until completely disentangled. There is nothing I know in QM that indicates this could happen. i.e. particles are entangled or not - no half way or phase lags. So it means, probably, that there is no 'inertia', so to speak, in the entangled states ending and occurring. It points to the hypothesis that the process of entanglement would probably be instantaneous too, but that was not tested. Would anyone expect different though?

unpredictable ending time:
This is a problem for quantum computing since it would limit the attainable clock speeds for computing. I would like to know if the entanglement end time has any variables associated with it and why is it unpredictable? It could be as simple as the probabilty of state observables between two particles (observer and observed) is the cause. So that when they approach each other there is no clear trigger point - only combined probabilities to condsider.
Could we also expect the same is true for entanglement starting? Is that also unpredictable but sudden?

Thats my take on it, and its only surmising possiblities from the paper quoted above.
 
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  • #91
As AEM said, the brazilian group of quantum optics at UFRJ has worked on this subject a lot.

Best regards

DaTario
 
  • #92
DaTario said:
As AEM said, the brazilian group of quantum optics at UFRJ has worked on this subject a lot.

Best regards

DaTario

I would expect that the disappearance of entanglement is sudden (instant) but the exact moment unpredictable. If it had been anything else I would be very confused.

Is it a big deal? I mean is it crucial in some way?
 

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