Is entanglement still considered spooky ?

[lugita15]

EdiT: One question I still have no answer to: is the relationship between the 1 and 0 for a single detector and a non-entangle photon also sinusoidal?

Yes, if a photon that is polarized in some direction, for example 0°, then the probability that it will go through a polarization detector oriented at an angle θ is a sinusoidal function of θ

 

[salvestrom]

Yes, if a photon that is polarized in some direction, for example 0°, then the probability that it will go through a polarization detector oriented at an angle θ is a sinusoidal function of θ

Firstly, my apologies, I didn't see your post about the alcoholic coin tossers until just before this one.

My primary focus has been on whether this actually proves non-locality. Your reply here indicates that in a purely local experiment at 60° the results are 75% 0's and 25% 1's - accepting that you get all 1's at 0° and all 0's at 90°. Therefore the local experiment doesn't even accord to the proposed locality, which was based on a linear increase in deviation.

This would seem to be the best starting point to explain anything. Can we interpret the above numbers as a 75% deviation from the results at 0°?

 

[zonde]

Quantum mechanic's most unusual features are analogous to long perported magical abilities. I'm not saying they are, just that the next time you want to know what I mean by something: ask me and leave the derisery assumptions beside the keyboard.

What I meant was that although the outcomes can only be 1 of 2 possibilities, each potential outcome is weighted depending on the angle. At 45° each is no more likely than the other. Between 0 and 45 the outcome favours toward 1 and above 45 to 90 the outcome favours toward 0. Non-linearly.

Ok, so what do you mean with outcome and what do you mean with angle? There are two analyzers, two angles, two outcomes and then there are relative angle between angles of analyzers and "relative outcome" i.e. two outcomes either match (1/1 and 0/0) or they don't (1/0 and 0/1).

EdiT: One question I still have no answer to: is the relationship between the 1 and 0 for a single detector and a non-entangle photon also sinusoidal?

For linearly polarized photons it is as lugita says. You can take a look at Malus law.

But relationship between the 1 and 0 for a single detector and a single photon from source that produces entangled photon pairs (we just ignore the other one) is not sinusoidal but linear i.e. (for idealized setup) there is no change in outcome whatever rotation of analyzer.

 

[lugita15]

Firstly, my apologies, I didn't see your post about the alcoholic coin tossers until just before this one.

My primary focus has been on whether this actually proves non-locality. Your reply here indicates that in a purely local experiment at 60° the results are 75% 0's and 25% 1's - accepting that you get all 1's at 0° and all 0's at 90°. Therefore the local experiment doesn't even accord to the proposed locality, which was based on a linear increase in deviation.

This would seem to be the best starting point to explain anything. Can we interpret the above numbers as a 75% deviation from the results at 0°?

Here are the facts you need to understand:

1. If you have an unpolarized photon, and you put it through a detector, it will have a 50-50 chance of going through, regardless of the angle it's oriented at.
2. A local realist would say that the photon doesn't just randomly go through or not go through the detector oriented at an angle θ; he would say that each unpolarized photon has its own function P(θ) which is guiding it's behavior: it goes through if P(θ)=1 and it doesn't go through it P(θ)=0.
3. Unfortunately, for any given unpolarized photon we can only find out one value of P(θ), because after we send it through a detector and it successfully goes through, it will now be polarized in the direction of the detector and it will "forget" the function P(θ).
4. If you have a pair of entangled photons and you put one of them through a detector, it will have a 50-50 chance of going through, regardless of the angle it's oriented at, just like an unpolarized photon.
5. Just as above, the local realist would say that the photon is acting according to some function P(θ) which tells it what to do.
6. If you have a pair of entangled photons and you put both of them through detectors that are turned to the same angle, then they will either both go through or both not go through.
7. Since the local realist does not believe that the two photons can coordinate their behavior by communicating instantaneously, he concludes the reason they're doing the same thing at the same angle is that they're both using the same function P(θ).
8. He is in a better position than he was before, because now he can find out the values of the function P(θ) at two different angles, by putting one photon through one angle and the other photon through a different angle.
9. If the entangled photons are put through detectors 30° apart, they have 25% chance of not matching.
10. The local realist concludes that for any angle θ, the probability that P(θ±30°)≠P(θ) is 25%, or to put it another way the probability that P(θ±30°)=P(θ) is 75%.
11. So 75% of the time, P(-30)=P(0), and 75% of the time P(0)=P(30), so there's no way that P(-30)≠P(30) 75% of the time.
12. Yet when the entangled photons are put through detector 60°, they have a 75% chance of not matching, so the local realist is very confused.

What step do you not agree with?

 

[salvestrom]

The outcome is not 50/50 regardless of angle. During calibration it is firmly established that at 0° all outcomes are 1 and at 90° all outcomes are 0. At 45° all outcomes are 50/50. At 60° the outcome is 75% 0's and 25% 1's.

In a single detector, non-entangled photon experiment the results are sinusoidal. At 60° the deviation from the 0° result is 75%. This is the result of a purely localised experiment.

Introducing Bob and entangled photons allows us to do something special. We can now know the outcome of two settings for Alice at the same time. Bob can be set to show us what Alice would show at any given angle, such as -30° while Alice can be set to show another set of results at 30°. This is a 60° split and will show a 75% deviation, exactly as you get in a localisied experiment. There is nothing non-local implied about this relationship. Alice is effectively showing a deviation from her own potential results, if we could actually record both angles at once purely using her detector.

The only spookyness is in the fact both detectors return pricesly the same results at the same angular setting which could potentially be explained at the point of entanglement.

My argument is solely that non-locality cannot be in inferred because of the 75% deviation over 60°, because that deviation occurs in a purely localisied version of the experiment as well.

Edit: at this point it's safe to say the step I don't agree with is the first one. It is contrary to the facts established during the calibration.

 

[lugita15]

The outcome is not 50/50 regardless of angle. During calibration it is firmly established that at 0° all outcomes are 1 and at 90° all outcomes are 0. At 45° all outcomes are 50/50. At 60° the outcome is 75% 0's and 25% 1's.

For a photon that is polarized in a particular direction, and you send it through a detector oriented at an angle θ, then the probability it will go through the detector is a sinusoidal function of θ. But for an unpolarized photon, the probability that it goes through a detector is always 50-50 regardless of the angle. And the same is true for entangled photons: if you put one of the entangled photons through a detector, no matter what the angle it will have a 50-50 chance of going through. So even though each of the two photons has a 50-50 chance of going through or not, when you look at their results together you find that they are guaranteed to do the same thing, either both going through or both not going through, if they are detected at the same angle.

In a single detector, non-entangled photon experiment the results are sinusoidal.

Again, you're talking about polarized photons, which are irrelevant for this discussion.

At 60° the deviation from the 0° result is 75%. This is the result of a purely localised experiment.

I'm not sure what you mean by the "deviation from the 0° result". Regardless of whether you're dealing with an ordinary unpolarized photon or an entangled photon, the probability that it goes through the detector is 50% regardless of whether the detector is oriented at 0°, at 60°, or any other angle.

Introducing Bob and entangled photons allows us to do something special. We can now know the outcome of two settings for Alice at the same time. Bob can be set to show us what Alice would show at any given angle, such as -30° while Alice can be set to show another set of results at 30°.

This is all true.

This is a 60° split and will show a 75% deviation, exactly as you get in a localisied experiment. There is nothing non-local implied about this relationship.

No, you're talking about polarized photons. The mystery is not just that there is a 75% probability of a mismatch when the detectors are set 60° apart. The mystery is the contradiction between the following three facts:
1. The probability that P(0) is different from P(30) is 25%
2. The probability that P(-30) is different from P(0) is 25%
3. The probability that P(-30) is different from P(30) is 75%

(Of course, you can switch -30,0,30 to 0,30,60 if you want, it doesn't make a difference.)

Alice is effectively showing a deviation from her own potential results, if we could actually record both angles at once purely using her detector.

That's true.

The only spookyness is in the fact both detectors return pricesly the same results at the same angular setting which could potentially be explained at the point of entanglement.

Well, it could have potentially been explained in that way, but I showed the problems with such an explanation.

My argument is solely that non-locality cannot be in inferred because of the 75% deviation over 60°, because that deviation occurs in a purely localisied version of the experiment as well.

But the kind of experiment with a polarized photon you're discussing doesn't have anything to do with the phenomenon of entanglement, where any single photon will have a 50-50 chance but when you compare the results of the two entangled photons you get a sinusoidal effect.

 

[salvestrom]

Thanks for pursuing this. I'm going to have to investigate more clearly the detector and what it's doing, it seems. Having said that, it still seems clear, at present, that the sinusoidal shows that the quantity of 1's and 0's at any given angle is not always 50/50. I mean quite literally that the outcome is weighted toward one or the other as you turn the detector. But, as I say, I will take a good long look at the setup description again.

Aside from that, your one query was what I meant by "deviation from the 0° result". The 0° result during calibration was entirely 1's. At 60° it's 75% 0's. That's entirely all I'm referring to. The change in the balance of 1 and 0. While mulling over the whole thing, I came to prefer the term deviation between the balance of 1's and 0's rather than error, because the latter implied one of the detectors was somehow 'wrong', as opposed to 'different'.

No photon left behind.

 

[salvestrom]

It's clear now where the confusion has arisen: I was carrying over the calibration description using polarised light and treating the photons in the experiment as polarised. Well, they are, but we don't know in what way til they hit something. I was also treating it as if every single photon had the same polarisation as the previous one. I realize now that during the experiment there are no angles for which all results are 1 or all results are 0.

Although I wish to think on this some more in general, pursuing any arguements of locality or non-locality at this point would require a largely unanswerable philosophical discussion about QM and the real world, I'd say we're done. Thank you again for the help. Zonde, too. /hug

-
Virtual gnomes: they're in your black hole, radiating your energy!

 

[lugita15]

Well, they are, but we don't know in what way til they hit something.

That can't be right for the following reason. The probability that a polarized photon goes through a detector at an angle θ is a sinusoidal function of θ. The probability that an unpolarized photon (or an entangled photon) goes through a detector is 50-50 independent of θ. So even without knowing in advance what angle a photon may be polarized in, we can determine experimentally whether a photon is polarized or not.

Although I wish to think on this some more in general, pursuing any arguements of locality or non-locality at this point would require a largely unanswerable philosophical discussion about QM and the real world, I'd say we're done. Thank you again for the help. Zonde, too. /hug/

I think you'd find it useful to do more research on Bell's theorem. That's what reveals the heart of the nature of quantum entanglement. For a good popular explanation of all this, you can read Quantum Reality by Nick Herbert, or for a more recent book Amir D'Aczel's Entanglement. If you can't get your hands on these, just searching this forum for threads about Bell's theorem would be an excellent first step.

 

[salvestrom]

That can't be right for the following reason. The probability that a polarized photon goes through a detector at an angle θ is a sinusoidal function of θ. The probability that an unpolarized photon (or an entangled photon) goes through a detector is 50-50 independent of θ. So even without knowing in advance what angle a photon may be polarized in, we can determine experimentally whether a photon is polarized or not

I made the assumption because I'm under the impression that unpolarised light is due to large numbers of photons with differing angles of oscillation. I've not found anything that explains how a single photon can be unpolarised. Surely every photon has an angle of oscillation? Or am I missing something (again), i.e. that a photons polarisation isn't the same as its angle of oscillation? Help me, Lugita15, you're my only hope!

 

[lugita15]

I made the assumption because I'm under the impression that unpolarised light is due to large numbers of photons with differing angles of oscillation. I've not found anything that explains how a single photon can be unpolarised. Surely every photon has an angle of oscillation? Or am I missing something (again), i.e. that a photons polarisation isn't the same as its angle of oscillation? Help me, Lugita15, you're my only hope!

An unpolarized photon is just a photon which is in a superposition of polarization states. I assume you know how wave functions work: particles usually have a wave function which is a superposition of a bunch of different states, which means that they have probability amplitudes of being measured in anyone of them. In the case of entangled photons, we have a wave function for the whole two-photon system, and this wave function is in a superposition of polarization states.

 

[zonde]

Although I wish to think on this some more in general, pursuing any arguements of locality or non-locality at this point would require a largely unanswerable philosophical discussion about QM and the real world, I'd say we're done. Thank you again for the help. Zonde, too.

Actually there is one argument defending local realism that is not philosophical but rather very practical. Argument is about why that theoretical model I gave does not describe real experiments. It says that fair sampling assumption in photon experiments does not hold.

 

[lugita15]

Actually there is one argument defending local realism that is not philosophical but rather very practical. Argument is about why that theoretical model I gave does not describe real experiments. It says that fair sampling assumption in photon experiments does not hold.

But hasn't the fair sampling loophole been closed by this experiment? I know we haven't succeeded in doing a single Bell test that closes all loopholes, but hasn't each experimental loophole been closed by at least one experiment or the other (other than superdeterminism, which can never be closed by experiment)?

 

[zonde]

An unpolarized photon is just a photon which is in a superposition of polarization states. I assume you know how wave functions work: particles usually have a wave function which is a superposition of a bunch of different states, which means that they have probability amplitudes of being measured in anyone of them. In the case of entangled photons, we have a wave function for the whole two-photon system, and this wave function is in a superposition of polarization states.

Have to say you are assuming a lot.
Do you know what is "superposition of states" and what is "probability amplitude"?

 

[lugita15]

Have to say you are assuming a lot.
Do you know what is "superposition of states" and what is "probability amplitude"?

Yes, I do; I've been studying QM and QFT for years. I assume you're asking this because strictly speaking my comment to salvestrom wasn't technically accurate; an unpolarized photon is in a mixed state, not a pure state. I was just simplifying things a bit to key into the important point that a local measurement of an entangled photon always gives a 50-50 result, while the sinusoidal graph shown in Nick Hebert's page comes from comparing the results of both photons. Even if what I said is not technically true of ordinary unpolarized photons, it is true of entangled photons, but explaining that would have muddied the discussion.

 

[zonde]

But hasn't the fair sampling loophole been closed by this experiment? I know we haven't succeeded in doing a single Bell test that closes all loopholes, but hasn't each experimental loophole been closed by at least one experiment or the other (other than superdeterminism, which can never be closed by experiment)?

Oh, but I said that it does not hold in photon experiments.
Or do you want to argue that we can apply one to one results of ion experiment to photon experiment?

Besides there are nearly 100% efficient photon detectors. So it should be easy to close fair sampling loophole in photon experiment. This would avoid dubious comparison between photon and ion experiments.

 

[zonde]

Yes, I do; I've been studying QM and QFT for years. I assume you're asking this because strictly speaking my comment to salvestrom wasn't technically accurate; an unpolarized photon is in a mixed state, not a pure state.

No, I was asking this because there is no consensus about what is "superposition of states" and what is "probability amplitude".

 

[lugita15]

Oh, but I said that it does not hold in photon experiments.
Or do you want to argue that we can apply one to one results of ion experiment to photon experiment?

Besides there are nearly 100% efficient photon detectors. So it should be easy to close fair sampling loophole in photon experiment. This would avoid dubious comparison between photon and ion experiments.

Does the kind of experiment performed make a difference? Isn't the important thing to demonstrate that the local realist cannot hide behind the fair sampling loophole?

 

[salvestrom]

No, I was asking this because there is no consensus about what is "superposition of states" and what is "probability amplitude".

All states at once and the odds?

That's generally what I take them to mean.

I've been poking around at Bell's theorum all day, He may have enjoyed it. I actually have more questions, but it's late. Play nice.

 

[zonde]

Does the kind of experiment performed make a difference?

Well, if this ion experiment could stand on it's own then of course there is not much difference. But this is not the case. Possibility that Bell inequalities are violated because of measurement cross-talk is not tested in this kind of experiments.

Basically you have to assume that Bell inequality violations appear due to the same (unknown) physical mechanism in ion experiments and photon experiments only then it means something. Obviously it is much more preferable to avoid such assumptions.

Isn't the important thing to demonstrate that the local realist cannot hide behind the fair sampling loophole?

Absolutely not. Important thing is to demonstrate that non-local mysticist cannot hide behind the fair sampling assumption.


If you will continue this discussion I have to apologize in advance as I won't be responding for couple of weeks. But I can pick it up later.