Can grandpa understand the Bell's Theorem?

[DrChinese]

It helps a lot, because now I can propose the experiment that hopefully will directly prove the existence of an “influence over distance” in spirit of Bell’s theorem.

In this experiment we will rotate the polarization of one entangled photon in hope to observe the symmetrical rotation of another entangled photon.

We may start with the experimental setup very similar to Aspect’s, but with the polarized light source of entangled photons. We also will use two polarizers A and B set in parallel with each other and parallel with the light source polarization. However instead of correlation we will simply measure a light intensity on both sides.

Now, let’s place the wave plates (to rotate photon’s polarization) between light source and polarizer A. Because light beam that passes wave plates isn’t parallel any more to the polarizer A the intensity of the beam that passes this polarizer is changed according to Malus’ law.

If the light source is closer to the polarizer A (for the photon a to collapse first) we should observe the intensity of light at another side is changing in sync with side A that will be a direct proof of an “influence over distance”

What is your prediction of this experiment?

The intensity stays the same at all times on both sides at all polarization angles, as I thought I had indicated previously. This is because an entangled stream consists of randomly polarized photons (actually a superposition of H and V so I am being loose in my language) and 50% of those will pass the polarizer at any orientation. Just as would occur with a stream in a mixed state, which by definition is random/unknown. (These states are indistinguishable unless coincidences are determined.)

Same is true when wave plates are added. No change in intensity with an entangled source. When particle A meets a wave plate, it does not change particle B in any way. And vice versa.

 

[miosim]

The intensity stays the same at all times on both sides at all polarization angles, as I thought I had indicated previously. This is because an entangled stream consists of randomly polarized photons...

I agree with you that the intensity stays the same at all times on both sides at all polarization angles if an entangled stream consists of RANDOMLY POLARIZED photons.
However, the key requirement for the proposed in the post #310 experiment is that the light source consists not of RANDOMLY POLARIZED photons, but is a polarized light that consists of POLARIZED ENTANGLED PHOTONS. I probably should emphasize this critical condition. My bad.

Could you please provide your prediction for this experiment taking in account my clarification?

Thank you

 

[DrChinese]

I agree with you that the intensity stays the same at all times on both sides at all polarization angles if an entangled stream consists of RANDOMLY POLARIZED photons.
However, the key requirement for the proposed in the post #310 experiment is that the light source consists not of RANDOMLY POLARIZED photons, but is a polarized light that consists of POLARIZED ENTANGLED PHOTONS. I probably should emphasize this critical condition. My bad.

Could you please provide your prediction for this experiment taking in account my clarification?

Thank you

If they are polarization entangled, then they are not polarized. They are in a superposition of polarization states. (By definition.) They will not be entangled as to polarization if you know their polarization.

They could be polarized but NOT entangled as to polarization, is that what you mean? You get that from a single Type I PDC crystal. If so, the intensity will simply follow Malus. Of course, these will not meet the criteria of EPR (i.e. their polarizations will not lead to perfect correlations).

 

[DrChinese]

...

If the light source is closer to the polarizer A (for the photon a to collapse first) we should observe the intensity of light at another side is changing in sync with side A that will be a direct proof of an “influence over distance”

What is your prediction of this experiment?

Just to be clear: there is no setup in which a change in A leads to a change in intensity at B. This is true regardless of the ordering of arrival at the polarizers.

 

[miosim]

If they are polarization entangled, then they are not polarized. They are in a superposition of polarization states. (By definition.) They will not be entangled as to polarization if you know their polarization

As I understood you correctly, it is theoretically impossible to produce the light source of entangled protons that will behave as polarized light during interaction with polarizer. I would like to better understand this phenomenon. Can you please explain the difference between a photon of polarized light and the entangled photon from the QM wave function point of view?

Thanks

 

[SpectraCat]

As I understood you correctly, it is theoretically impossible to produce the light source of entangled protons that will behave as polarized light during interaction with polarizer. I would like to better understand this phenomenon. Can you please explain the difference between a photon of polarized light and the entangled photon from the QM wave function point of view?

Thanks

If you have a polarized photon it is in a state of definite polarization, i.e. either |H> or |V> ... those are the basis states for polarization. A polarization-entangled photon is in a superposition state, e.g. 0.707(|H> + |V>) or 0.707(|H> -|V>). Thus when a stream of polarization-entangled photons interacts with a polarization sensitive detector, half of the time the |H> state is registered, and the other half of the time the |V> state is registered. That is why DrC told you that you would see 50% intensity in both beams (for ideally efficient instrumentation) in the example you asked about earlier. Rotating the polarization of one beam prior to detection makes no difference. You can think of that as changing the basis to |H'> and |V'>, but now half the photons will be detected in |H'> and the other half in |V'>, so the intensity is still 50%.

 

[DrChinese]

As I understood you correctly, it is theoretically impossible to produce the light source of entangled protons that will behave as polarized light during interaction with polarizer. I would like to better understand this phenomenon. Can you please explain the difference between a photon of polarized light and the entangled photon from the QM wave function point of view?

Thanks

SpectraCat said it nicely above. To that I would add:

The state of an individual photon can be either known (pure, which occurs when it passes a polarizer or is produced coherently such as from a laser) or a superposition (entangled). However, a photon polarized at 45 degrees can rightfully be considered a superposition of H and V as well. It is impossible to experimentally distinguish that from an entangled photon on the H/V basis (without resorting to coincidence counting). Further, you could have a group of photons in unknown but pure states (i.e. a statistical distribution). Those are similarly indistinguishable on the H/V basis. And yet all of these streams have somewhat different properties in other bases.

 

[San K]

SpectraCat said it nicely above. To that I would add:

However, a photon polarized at 45 degrees can rightfully be considered a superposition of H and V as well. It is impossible to experimentally distinguish that from an entangled photon on the H/V basis (without resorting to coincidence counting).

how would you distinguish (between 45 and entangled photon) via co-incidence counting?

the photon polarized at 45 degree would not give the exact opposite spin? but the entangled photon would?

 

[SpectraCat]

how would you distinguish (between 45 and entangled photon) via co-incidence counting?

the photon polarized at 45 degree would not give the exact opposite spin? but the entangled photon would?

Yes .. if you have two entangled photons, and you compare coincidence counts for them against coincidence counts for unentangled photons with 45 degree polarization, you will find that the entangled photons show coincidence counts with a certain degree of correlation (you can make it theoretically perfect if you choose the detector angles appropriately), whereas the unentangled are uncorrelated (i.e. random coincidences). It takes some work, but this is essentially the control experiment that folks like Aspect and Zeilinger run to make sure they have entangled photons for their experiments.

 

[San K]

Yes .. if you have two entangled photons, and you compare coincidence counts for them against coincidence counts for unentangled photons with 45 degree polarization, you will find that the entangled photons show coincidence counts with a certain degree of correlation (you can make it theoretically perfect if you choose the detector angles appropriately), whereas the unentangled are uncorrelated (i.e. random coincidences). It takes some work, but this is essentially the control experiment that folks like Aspect and Zeilinger run to make sure they have entangled photons for their experiments.

thanks spectracat...i am reading up on coincidence counter on Wikipedia...

spectraCat wrote "coincidence counts with a certain degree of correlation" ...correlated via timing or quantum state (i.e. spin?) or both?

 

[miosim]

I would like to try another Gedanken Experiment.
Say we have the classical Aspect experiment with both polarizers set in parallel (is it antiparallel?) to achieve the 100% correlation.
Now let’s insert one wave plate at any side between photon source and the polarizer to rotate photon’s polarization on this side by 90 degree. As I understand, according to Bell paradigm the result of the Aspect’s experiment shouldn’t change because both entangled photons exibit the same antiparallel polarization. However according to EPR paradigm the polarization between both entangled photons will be shifted by 90 degree and the Aspect experiment should yield zero correlation.

Does it make sense?

 

[DrChinese]

I would like to try another Gedanken Experiment.
Say we have the classical Aspect experiment with both polarizers set in parallel (is it antiparallel?) to achieve the 100% correlation.
Now let’s insert one wave plate at any side between photon source and the polarizer to rotate photon’s polarization on this side by 90 degree. As I understand, according to Bell paradigm the result of the Aspect’s experiment shouldn’t change because both entangled photons exibit the same antiparallel polarization. However according to EPR paradigm the polarization between both entangled photons will be shifted by 90 degree and the Aspect experiment should yield zero correlation.

Does it make sense?

Yes, it makes sense. But you label the outcomes incorrectly. Inserting a 90 degree wave plate on Alice's side makes the correlations go from 100% to 0%. This is the QM prediction.This does not collapse the wave function, it just rotates the polarization on that side ONLY. Doing something to one does not cause the same action to occur on the other. However: learning something about one causes us to learn something about the other within the limits of the HUP.

 

[miosim]

...Inserting a 90 degree wave plate on Alice's side makes the correlations go from 100% to 0%. This is the QM prediction. This does not collapse the wave function, it just rotates the polarization on that side ONLY...

But what happens when the wave function collapses? Where is the “action over the distance” that according to QM and Bell paradigm must produce antiparallel polarization and 100% correlation?

It seems to me that this Gedanken Experiment refutes QM and Bell paradigms and embraces EPR argument.

 

[DrChinese]

But what happens when the wave function collapses? Where is the “action over the distance” that according to QM and Bell paradigm must produce antiparallel polarization and 100% correlation?

They are still correlated after the wave plate, because the wave function has not collapsed. The collapse occurs when we finally learn something definite about one or the other. Actually the collapse occurs as the number of outcomes is restricted, ultimately down to one in the typical case.

The action at a distance, if there is such, is that what is measured at point A restricts the outcomes at point B to those that are compatible.

 

[miosim]

They are still correlated after the wave plate, because the wave function has not collapsed. The collapse occurs when we finally learn something definite about one or the other. Actually the collapse occurs as the number of outcomes is restricted, ultimately down to one in the typical case.

The action at a distance, if there is such, is that what is measured at point A restricts the outcomes at point B to those that are compatible.

So, if the correlated photons are remained entangled after the wave plate and the outcome at point A restricts the outcomes at point B -- that means that both photons will exhibit the anntiparalel polarization. This means that QM and Bell’s paradigm should predict 100% correlation while EPR paradigm predicts 0% correlation.

This is the very simple Gedanken Experiment and I don't expect an ambiguous prediction and explanation.

 

[DrChinese]

So, if the correlated photons are remained entangled after the wave plate and the outcome at point A restricts the outcomes at point B -- that means that both photons will exhibit the anntiparalel polarization. This means that QM and Bell’s paradigm should predict 100% correlation while EPR paradigm predicts 0% correlation.

And I keep telling you that entangled photons go through wave plates and rotate accordingly. If two were parallel to begin with and then one is rotated 90 degrees, they will be anti-parallel. This is in accordance with QM and, as far as I know, most realistic predictions as well.

You might want to learn the rules regarding manipulation of optics BEFORE making further assessments. If you would care to discuss your reasoning, I would be happy to point out where you are going wrong.

 

[miosim]

And I keep telling you that entangled photons go through wave plates and rotate accordingly. If two were parallel to begin with and then one is rotated 90 degrees, they will be anti-parallel. This is in accordance with QM and, as far as I know, most realistic predictions as well.

As I understand you contradict yourself. Just recently you told that photons could be either polarized or entangled, but not both:

"... If they are polarization entangled, then they are not polarized. They are in a superposition of polarization states. (By definition.) They will not be entangled as to polarization if you know their polarization. ..."

But regardless what you said, the QM paradigm is clear that you can’t change one entangled particle without affecting another. This is the main difference between QM and EPR that views correlated photons as fully independent after separation.

In my Gedanken Experiment the entangled photon are independent, based on the result you predicted, and therefore QM and Bell’s theories are in conflict with this prediction.

 

[DrChinese]

Just recently you told that photons could be either polarized or entangled, but not both:

"... If they are polarization entangled, then they are not polarized. They are in a superposition of polarization states. (By definition.) They will not be entangled as to polarization if you know their polarization. ..."

But regardless what you said, the QM paradigm is clear that you can’t change one entangled particle without affecting another. This is the main difference between QM and EPR that views correlated photons as fully independent after separation.

In my Gedanken Experiment the entangled photon are independent, based on the result you predicted, and therefore QM and Bell’s theories are in conflict with this prediction.

Polarization entangled photons do not have a definite polarization because they are in a superposition of polarization states. Nonetheless, a wave plate WILL rotate that superposition. This transformation is NOT communicated to its twin (because the wave state is still a superposition and will remain so until a measurement collapses them). That is traditional QM. EPR does NOT predict that a change to Alice will be communicated to Bob either.

I will remind you that most Bell tests involve a series of filters, wave plates, fiber, mirrors, etc. which have certain optical properties. These are respected by entangled photons as well as those in a pure state. For example, it is not unusual to send entangled photons through coiled fiber. Doesn't change anything.

So I will repeat: there is nothing about QM that leads to the prediction you are making. It would be pretty obvious since you could simply play with the wave plates to locate any unusual or unexpected effects. But there aren't any.

 

[DrChinese]

As I understand you contradict yourself.

I think you misunderstood my comment about entangled photons being parallel or anti-parallel (crossed, orthogonal, perpedicular). Perhaps the following recap will assist:

Type I PDC entangled photons are parallel (0 degrees of theta). They do NOT have a definite polarization because they are in a superposition of states. They yield perfect EPR correlations at identical angle settings. Without rotation, that will be 100% correlated.

Type II PDC entangled photons are crossed (90 degrees of theta). They do NOT have a definite polarization because they are in a superposition of states. They yield perfect EPR correlations at identical angle settings. Without rotation, that will be 0% correlated.

You can rotate either or both with wave plates, and adjust the theta stats accordingly. Use cos^2 rule as before.

 

[harrylin]

I think you misunderstood my comment about entangled photons being parallel or anti-parallel (crossed, orthogonal, perpedicular).

Probably so, as it also made me wonder. You wrote:

"And I keep telling you that entangled photons go through wave plates and rotate accordingly."

That sounded like a "realist" description. If a photon does not yet have a polarisation, what do you imagine to "rotate accordingly"?

You can rotate either or both with wave plates, and adjust the theta stats accordingly. Use cos^2 rule as before.

Assuming that you are right here, this rotation that changes the correlation is an interesting aspect that I had not thought of before. Doesn't that effectively kill the concept of magical action at a distance? For the measured photon should then instantly inform the entangled photon, not of its polarisation state, but of its history - and according to QM a photon does not have memory built in.

 

[DrChinese]

Probably so, as it also made me wonder. You wrote:

"And I keep telling you that entangled photons go through wave plates and rotate accordingly."

That sounded like a "realist" description. If a photon does not yet have a polarisation, what do you imagine to "rotate accordingly"?
Assuming that you are right here, this rotation that changes the correlation is an interesting aspect that I had not thought of before. Doesn't that effectively kill the concept of magical action at a distance? For the measured photon should then instantly inform the entangled photon, not of its polarisation state, but of its history - and according to QM a photon does not have memory built in.

And I keep repeating what QM says and what actually happens: the entangled photon can be bounced off mirrors, go through color filters, be rotated through wave plates, follow optical fiber - all without losing its polarization entanglement and without any way altering its polarization twin.

If the media affects the polarization by rotating it or otherwise altering it, you would need to consider that when predicting the results of correlation measurements. In other words: I can rotate the superposition of H> + V> by 90 degrees and it becomes V> + H>. And I can do that without causing collapse.

I think you and grandpa have a confused idea of what entangled particles do. A general change to one does NOT affect the other by action at a distance (as far as I know anyway). No one here has implied that. A measurement outcome causing wave function collapse on one will cause a suitable wave function collapse on the remainder of the entangled system regardless of spacetime distance. That is as far as it goes. If the outcomes are not restricted as a result of a measurement, there is no collapse. The rule is that you cannot use entangled particles to gain more information about one than the Heisenberg Uncertainty Principle allows. I hope this clarifies things.

 

[miosim]

I think you and grandpa have a confused idea of what entangled particles do. A general change to one does NOT affect the other by action at a distance (as far as I know anyway). No one here has implied that. A measurement outcome causing wave function collapse on one will cause a suitable wave function collapse on the remainder of the entangled system regardless of spacetime distance. That is as far as it goes. If the outcomes are not restricted as a result of a measurement, there is no collapse. The rule is that you cannot use entangled particles to gain more information about one than the Heisenberg Uncertainty Principle allows. I hope this clarifies things.

Actually I get more confused. First, I have no idea what “A general change … ” is, because I am talking about very specific change in one of polarization entangled photons that must affect (according to QM) both photons; otherwise these photons will loose their entanglement.

By saying that "A general change to one does NOT affect the other ..." you seem to contrdict yourself, as folows:

... the entangled photon can be bounced off mirrors, go through color filters, be rotated through wave plates, follow optical fiber - all without losing its polarization entanglement and without any way altering its polarization twin…

…that in reference to my Gedanken Experiment means that correlation in the Aspect experiment shouldn’t change (remains 100%) after introduction of the wave plate while the EPR interpretatation expects that this correlation will drop to 0%.

 

[DrChinese]

Actually I get more confused. First, I have no idea what “A general change … ” is, because I am talking about very specific change in one of polarization entangled photons that must affect (according to QM) both photons; otherwise these photons will loose their entanglement.

By saying that "A general change to one does NOT affect the other ..." you seem to contrdict yourself, as folows:

…that in reference to my Gedanken Experiment means that correlation in the Aspect experiment shouldn’t change (remains 100%) after introduction of the wave plate while the EPR interpretatation expects that this correlation will drop to 0%.

If it makes it more clear, just take out the word "general". You keep saying that QM predicts X when it actually predicts Y.

I have told you that a wave plate rotates the polarization state of any photon, entangled or not. It does not cause collapse of the wave function for entangled photons. It does not change the polarization in any way for other photons. Is there any element of that which is not specific or clear? This is the same predicted result for both QM and classical setups.

 

[DrChinese]

And a note for miosim: an EPR state is one in which the polarization of Alice can be predicted with certainty by reference to Bob. This is the starting point of Aspect style experiments. It is not assumed, it is demonstrated so that things begin where EPR left off - there are elements of reality present.

That in and of itself rules out an entire class of local realistic theories. I.e. those which assume you could only obtain a Product state correlations from a seemingly random distribution of outcomes. In other words, if any of the randomness is introduced at the time of observation, you would have Product State statistics and the EPR state would NOT be achieved.

 

[harrylin]

And I keep repeating what QM says and what actually happens: the entangled photon can be bounced off mirrors, go through color filters, be rotated through wave plates, follow optical fiber - all without losing its polarization entanglement and without any way altering its polarization twin.

If the media affects the polarization by rotating it or otherwise altering it, you would need to consider that when predicting the results of correlation measurements. In other words: I can rotate the superposition of H> + V> by 90 degrees and it becomes V> + H>. And I can do that without causing collapse.

Ok, so when you wrote that "entangled photons go through wave plates and rotate accordingly", you imagine a superposition of H and V as something "real", relative to which a photon (or its future polarisation?) can rotate - thanks for the clarification.

I think you and grandpa have a confused idea of what entangled particles do. A general change to one does NOT affect the other by action at a distance (as far as I know anyway). No one here has implied that.

I told you that I assume that you are right about this. And that we agree on this doesn't mean that we are confused!

A measurement outcome causing wave function collapse on one will cause a suitable wave function collapse on the remainder of the entangled system regardless of spacetime distance. That is as far as it goes. If the outcomes are not restricted as a result of a measurement, there is no collapse. The rule is that you cannot use entangled particles to gain more information about one than the Heisenberg Uncertainty Principle allows. I hope this clarifies things.

A "suitable" wave function collapse implies that the two photons have polarisations that may be not anti-parallel. I'm afraid that you still do not realize the consequence for information... If you believe in action at a distance from the one photon to the other, where do you propose that the information about the correct ("suitable") phase difference exists? Where is the superposition stored?

 

[miosim]

… I have told you that a wave plate rotates the polarization state of any photon, entangled or not. ... Is there any element of that which is not specific or clear?

It is very confusing that a wave plate can rotate a polarization that doesn’t exist! As you stated before for photons …

If they are polarization entangled, then they are not polarized.

 

[DrChinese]

It is very confusing that a wave plate can rotate a polarization that doesn’t exist! As you stated before for photons …

The polarization is a superposition of states. As to the phrase "doesn't exist", this is purely semantics and is not fully representative. All I can really say is that is not in a "definite" polarization in the context of QM. A wave plate will rotate the superposition state, and that is a simple observational fact. The EPR state survives.

 

[miosim]

... A wave plate will rotate the superposition state, and that is a simple observational fact. ...

How in your opinion a rotation of the "…superposition state…" affects observable mutual polarization of the photon’s pair?
I give you a hint: without this “rotation” the photon’s pair would exhibit observable antiparallel polarization.

 

[SpectraCat]

How in your opinion a rotation of the "…superposition state…" affects observable mutual polarization of the photon’s pair?
I give you a hint: without this “rotation” the photon’s pair would exhibit observable antiparallel polarization.

I think part of the problem is that the states under discussion are not really the Bell states, at least not as I understand them. The Bell states are defined as superpositions in the PRODUCT space of the two possible outcomes at detectors A & B:

$$\Psi=(|H_A>\otimes|V_B> + |V_A>\otimes|H_B>)$$

So, if you rotation the polarization of the, say, the A beam by some arbitrary angle, you change the detection basis at detector A to |H'> and |V'>. That means now you are in the superposition:


$$\Psi=(|H'_A>\otimes|V_B> + |V'_A>\otimes|H_B>)$$

So what? You could have started out in that basis if you wanted to ... the vector spaces representing the polarizations of the two photons are independent. In my understanding, this is why non-destructive manipulations of the two different beams going to A and B can be made without destroying the entanglement.

 

[miosim]

... the vector spaces representing the polarizations of the two photons are independent. In my understanding, this is why non-destructive manipulations of the two different beams going to A and B can be made without destroying the entanglement.

So what is the observable mutual polarization of the photon’s pair you expect after one of them passed 90 deg wave plate?

 

[SpectraCat]

So what is the observable mutual polarization of the photon’s pair you expect after one of them passed 90 deg wave plate?

Please define mutual polarization ... I don't know what you mean.

 

[miosim]

Please define mutual polarization ... I don't know what you mean.

I mean the angle between observable photons’ polarization (parallel, antiparallel or anything in between).

I just realized that I made a mistake that causes a major confusion.
While asking about observable polarization of the photon’s pair after one of them passed 90 deg wave plate I thought that antiparallel polarization means 180 degree, while 90 degree represents the middle point between parallel and antiparallel polarization. But I was wrong because 90 degree is difference between parallel and untiparalell polarization so 45 degree is a middle point.

I am sorry for this confusion.

Let me correct my previous question. What happens if one of the entangled photons passed 45 deg wave plate? Would the observable polarization of this photon's pair remain antiparallel or will be shifted by 45 degree?

 

[harrylin]

I [..] You could have started out in that basis if you wanted to ... the vector spaces representing the polarizations of the two photons are independent. In my understanding, this is why non-destructive manipulations of the two different beams going to A and B can be made without destroying the entanglement.

Yes, DrChinese also clarified that. And this raised a question, as we now established that the detected polarization of the entangled photon is irrelevant for the polarization of its twin. Perhaps you can answer it? if one believes that "collapse of the wave function" corresponds to an instantaneous physical signal to the other photon, then where is the relevant information of the wave function stored?

Thanks,
Harald

 

[DrChinese]

I mean the angle between observable photons’ polarization (parallel, antiparallel or anything in between).

I just realized that I made a mistake that causes a major confusion.
While asking about observable polarization of the photon’s pair after one of them passed 90 deg wave plate I thought that antiparallel polarization means 180 degree, while 90 degree represents the middle point between parallel and antiparallel polarization. But I was wrong because 90 degree is difference between parallel and untiparalell polarization so 45 degree is a middle point.

I am sorry for this confusion.

Let me correct my previous question. What happens if one of the entangled photons passed 45 deg wave plate? Would the observable polarization of this photon's pair remain antiparallel or will be shifted by 45 degree?

It will shift it by 45 degrees, corresponding to the wave plate.

 

[DrChinese]

Yes, DrChinese also clarified that. And this raised a question, as we now established that the detected polarization of the entangled photon is irrelevant for the polarization of its twin. Perhaps you can answer it? if one believes that "collapse of the wave function" corresponds to an instantaneous physical signal to the other photon, then where is the relevant information of the wave function stored?

Please, I beg you to be careful with the words you use. They can often get in the way, and we end up "debating" items which are not physics so much as semantics. There IS a correlation between entangled Alice and Bob as to their polarization state, as I have mentioned.

Where is the information stored? It "could" be with the particle. I don't think a firm (meaningful) answer to this question is possible.