Why is entanglement necessary for understanding quantum mechanics?

[stevendaryl]

But when light polarization is constant those angles are still relative to light polarization even if you change them randomly and the result is the same, the meaning stays the same.

It doesn't make any sense to assume that the light polarization is constant. That's because no matter what angle is chosen for \theta_1, only 50% of the photons are transmitted, and 50% are blocked. If the polarization were constant, then there would be some angle that would produce 100% transmission.

 

[Jabbu]

Those are not realistic runs for EPR. As I said, in the photon version of EPR, both Alice and Bob get 50% transmission rates. So a more likely run would look like:


For \theta_A - \theta_B = 0:

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1

(They agree 100% of the time)

For \theta_A - \theta_B = 45^o:

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0

(They agree 50% of the time)

So that's where we disagree, very subtle difference. How do you calculate correlation? Sequence length is 20, there is 10 matching pairs of ones and 10 matching pairs of zeros, how do you get 100% out of that?

 

[stevendaryl]

So that's where we disagree, very subtle difference. How do you calculate correlation? Sequence length is 20, there is 10 matching pairs of ones and 10 matching pairs of zeros, how do you get 100% out of that?

That's 50%. That's what I said. Once again:

Case 1: \theta_A - \theta_B = 0^o

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1

That's 100% agreement.

Case 2: \theta_A - \theta_B = 45^o

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0

That's 50% agreement.

 

[DrChinese]

I think I may have gotten a detail wrong in the two-photon EPR case. Or maybe you did. What I thought was that the two photons had the SAME polarizations. rather than polarizations that differ by 90 degrees.

As Cthugha says, it is dependent on the type.

Type I: The entangled pairs are of the same polarization. It takes 2 crystals oriented perpendicular to get the desired effect.

Type II: The entangled pairs are of the orthogonal (opposite) polarization. Only 1 crystal is needed for this.

(I get this backwards all the time.)

 

[Jabbu]

Now you're getting to what's strange about EPR. It seems as if Bob's result depends on Alice's filter setting.

If there is no any information about photons in that equation then the equation doesn't know whether photons are entangled or not, would it not make the same prediction?

 

[DrChinese]

They say probability is cos^2(theta_2 - theta_1), where theta 1 and 2 are light polarization and polarizer angle. That's what I've been saying, it's Malus's law.

No no no! It is true that the formula looks the same, but that is somewhat superficial. Please note that all formulas based on cos^2() are NOT the same as Malus. The area of a square drawn on the adjacent side of a right triangle is proportional to the product of the hypotenuse^2 times cos^2(theta). But we don't say that is an application of Malus, do we?

Malus is applied for light of KNOWN polarization going through a polarizer. Entangled photons are NOT such an application. Please note that 50% of entangled photons will go through ANY polarizer. That is certainly a different prediction than Malus!

 

[DrChinese]

If there is no any information about photons in that equation then the equation doesn't know whether photons are entangled or not, would it not make the same prediction?

Entangled photon pairs yield different results for match percentages than photons that are not entangled. So the answer is NO.

 

[DrChinese]

How do you calculate correlation?

Correlation ranges from -1 to 1. Match percentage (which is the easiest and clearest thing to discuss) ranges from 0 to 1.

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0

The above has 0 correlation, and 50% match ratio.

 

[Jabbu]

That's 50%. That's what I said. Once again:

Case 1: \theta_A - \theta_B = 0^o

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1

That's 100% agreement.

Case 2: \theta_A - \theta_B = 45^o

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0

That's 50% agreement.

cos^2(0) = 100% correlation, and cos^2(45) = 50% correlation is QM theoretical prediction. Correlation is calculated differently from experimental data, in terms of matching and mismatching pairs. I think DrChinese was using something like that few pages ago.

 

[DrChinese]

I was talking about only one polarizer. Detector readings are recorded separately for each polarizer and streams of data, when theta = 0 and correlation = 100%, look like this:

A: 1 1 1 1 1 1 1 1 1 1...
B: 1 1 1 1 1 1 1 1 1 1...

Can angle between photons A polarization and polarizer A be anything else but zero degrees?

Since you will never get that pattern from entangled photons, it is a meaningless question. In reality, entangled photons give you a 50% rate through polarizer A for your example. According to you, it should be 100% since cos^2(0)=100%. At least one of us is wrong.

 

[DrChinese]

cos^2(0) = 100% correlation, and cos^2(45) = 50% correlation is QM theoretical prediction.

Change that to the following and you will be a bit more precise:

cos^2(0) = 100% match rate, and cos^2(45) = 50% match rate is QM theoretical prediction.

 

[stevendaryl]

If there is no any information about photons in that equation then the equation doesn't know whether photons are entangled or not, would it not make the same prediction?

The way that quantum mechanics works, in the case of entangled photons: Fix \theta_A, Alice's filter angle, and \theta_B, Bob's filter angle.

1. The probability that Alice's photon has polarization \theta_A is 1/2.
2. The probability that Bob's photon has polarization \theta_B given that Alice's photon has polarization \theta_A is cos^2(\theta_A - \theta_B)
3. So the probability that both are true is \frac{1}{2} cos^2(\theta_A - \theta_B)

With randomly polarized unentangled photons:

1. The probability that Alice's photon has polarization \theta_A is 1/2.
2. The probability that Bob's photon has polarization \theta_B given that Alice's photon has polarization \theta_A is still 1/2.
3. So the probability that both are true is 1/4.

With photons that are polarized at a polarization half-way between Alice's angle and Bob's angle, you have:

1. The probability that Alice's photon has polarization \theta_A is cos^2(\frac{1}{2} (\theta_A - \theta_B)).
2. The probability that Bob's photon has polarization \theta_B given that Alice's photon has polarization \theta_A is cos^2(\frac{1}{2} (\theta_A - \theta_B)).
3. So the probability that both are true is cos^4(\frac{1}{2} (\theta_A - \theta_B)).

 

[Jabbu]

Malus is applied for light of KNOWN polarization going through a polarizer. Entangled photons are NOT such an application. Please note that 50% of entangled photons will go through ANY polarizer. That is certainly a different prediction than Malus!

Malus can also be applied to random polarization, in which case it's always 50% chance.

No no no! It is true that the formula looks the same, but that is somewhat superficial. Please note that all formulas based on cos^2() are NOT the same as Malus. The area of a square drawn on the adjacent side of a right triangle is proportional to the product of the hypotenuse^2 times cos^2(theta). But we don't say that is an application of Malus, do we?

I know what you're talking about, the conflict is elsewhere. Basically I think when theta_A = 0 and theta_B = 0, the data will look precisely like this:

A: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
B: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

... and you think it will look something like this:

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1

Right?

 

[DrChinese]

I know what you're talking about, the conflict is elsewhere. Basically I think when theta_A = 0 and theta_B = 0, the data will look precisely like this:

A: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
B: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

... and you think it will look something like this:

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1

Right?

Assuming you mean that theta_A is the polarizer angle setting for the A entangled photon stream, yes.

 

[DrChinese]

Malus can also be applied to random polarization, in which case it's always 50% chance.

If you integrate across 360 degrees, you can use the cos^2()/Malus term and get 50% as the answer. And Malus *can* be somewhat similarly used to get the right answer for the entangled photon match percentage IF you assume a non-local model.

But that is not really how the cos^2() formula is arrived at for QM's prediction. The more correct process (which uses textbook QM) is per the reference I provided.

We all understand that you can get from Rome to France a variety of ways. But we don't reference every route as following Caesar (who traveled that way too). As important as Malus is, the applications we are discussing are quite different. You are doing yourself a disservice to think in those terms.

 

[Jabbu]

Since you will never get that pattern from entangled photons, it is a meaningless question. In reality, entangled photons give you a 50% rate through polarizer A for your example. According to you, it should be 100% since cos^2(0)=100%. At least one of us is wrong.

Ok. So if entangled photons always have 50% chance, what chance is for not-entangled photons? Do both entangled and not-entangled photons have random polarization, but entangled photons polarization is the same for the two photons in each pair?

 

[Nugatory]

Ok. So if entangled photons always have 50% chance, what chance is for not-entangled photons? Do both entangled and not-entangled photons have random polarization, but entangled photons polarization is the same for the two photons in each pair?

If two photons are not entangled, then the results of a measurement of one photon has no bearing whatsoever on the results of a measurement of the other one - they are two independent and unrelated probability distributions.

In this situation, not only do we have a 50% probability than any given photon will pass a polarizer set to a randomly selected angle, but also (and this is the difference between two entangled photons and two non-entangled photons) no matter what the difference between the detector angles, there will be zero correlation between the result of the two measurements. 0-0, 0-1, 1-0, and 1-1 will occur with equal probability.

 

[Jabbu]

Change that to the following and you will be a bit more precise:

cos^2(0) = 100% match rate, and cos^2(45) = 50% match rate is QM theoretical prediction.

Please confirm if I got this right now.

1.)
A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1

Example setup:
theta_A = 0, theta_B = 0
theta_A = +30, theta_B = +30
theta_A = +90, theta_B = +90

Sequence length is 20, there is 10 matching zeros and 10 matching ones
Correlation = (match_0 + match_1) / sequence length = 20/20 = 100% ?


2.)
A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0

Example setup:
theta_A = 0, theta_B = +45
theta_A = -22.5, theta_B = +22.5
theta_A = +45, theta_B = +90

Sequence length is 20, there is 5 matching zeros and 5 matching ones
Correlation = (match_0 + match_1) / sequence length = 10/20 = 50% ?


3.)
A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
B: 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

Example setup:
theta_A = 0, theta_B = +90
theta_A = -45, theta_B = +45
theta_A = -90, theta_B = 0

Sequence length is 20, there is 0 matching zeros and 0 matching ones
Correlation = (match_0 + match_1) / sequence length = 0/20 = 0% ?

 

[DrChinese]

Please confirm if I got this right now.

For entangled photon pairs: Looks great!

Entangled photon pairs exhibit some very puzzling properties that highlight some of the strange elements of QM. Tremendous work has been done in the past few decades to highlight this.

a. Using a single Type I PDC crystal, it is possible to create pairs of photons that are entangled but are NOT polarization entangled. They come out as HH every time.

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0

Example setup:
theta_A = +45, theta_B = +45
Match % = 50%

b. However: at angles other than 45 degrees, there is correlation. The general formula for the match % is:

(cos^2(theta_A) * cos^2(theta_B)) + (sin^2(theta_A) * sin^2(theta_B))

c. It is also possible to create pairs of photons that are entangled but are NOT polarization entangled, and that are in a statistical mixture of HH or VV.

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 0 0 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0

Example setup:
theta_A = theta_B = any angle
Match % = 50%
There is no correlation in the outcomes.

The reason I mention a. and b. is that it is otherwise hard to find pairs of photons emitted at the same time, and these are good examples of how that is done. It is easy to see with these examples HOW the polarization entanglement is created and the differences between polarization entangled pairs and pairs that are not. If you have any questions about how these setups occur, I or one of the other may be able to answer it.

 

[Jabbu]

The reason I mention a. and b. is that it is otherwise hard to find pairs of photons emitted at the same time, and these are good examples of how that is done. It is easy to see with these examples HOW the polarization entanglement is created and the differences between polarization entangled pairs and pairs that are not. If you have any questions about how these setups occur, I or one of the other may be able to answer it.

Thank you. But since I only just realized what the mystery is supposed to be I'd like to stick with the setup we were talking about before a little bit more. The strangest thing I see is not correlation when there are matching pairs, but correlation of completely unmatched pairs in the 3rd case scenario:

theta_A = 0, theta_B = +90
A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
B: 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

This sequence that is said to have 0% correlation is not even random, it's perfectly symmetric. Not only the pairs are correlated between the two sequences, albeit reversely, but each sequence is perfectly correlated within itself. It's one thing when something has 50% chance, but if that chance is precisely split into odd and even position in a sequence that's not probability any more, it's inevitability and determinism. How can this be accounted for?

So when we start with those two sequences and Alice brings her polarizer to 90 degrees, not only her sequence will change for some reason, but also Bob's sequence will change for no reason?

I see there is no reason for Bob's sequence to change when Alice rotates her polarizer, but isn't it just as mysterious why would Alice's sequence vary relative her own polarizer given photons polarization is random? Are there experiments with only a single entangled photon and a single polarizer? Are those two different mysteries or are they ultimately supposed to have the same explanation?

 

[Cthugha]

theta_A = 0, theta_B = +90
A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
B: 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

This sequence that is said to have 0% correlation is not even random, it's perfectly symmetric.

Let us get the terminology clear: The sequence shows 0% matches, not 0% correlation. If both sequences were statistically independent and random you would get a higher percentage of matches. That would be the uncorrelated case. As the number of matches is smaller here, the sequences are anticorrelated and have a negative correlation coefficient.

 

[DrChinese]

Thank you. But since I only just realized what the mystery is supposed to be I'd like to stick with the setup we were talking about before a little bit more.

... but isn't it just as mysterious why would Alice's sequence vary relative her own polarizer given photons polarization is random?

OK, fine.

Alice's photon is randomly polarized, true enough. The difficult part is to accept that a) it is not predetermined and it is not well-defined; and b) it maintains a relationship to Bob, which is in a similar state, even though Alice and Bob are no longer in the vicinity of each other.

This is the mystery of entanglement: how does this occur?

 

[Jabbu]

Let us get the terminology clear: The sequence shows 0% matches, not 0% correlation. If both sequences were statistically independent and random you would get a higher percentage of matches. That would be the uncorrelated case. As the number of matches is smaller here, the sequences are anticorrelated and have a negative correlation coefficient.

Rather than correlation between the two sequences I'm now paying attention to regularity of each sequence in itself.

A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

This sequence is supposed to come out if A and B polarizers are orthogonally aligned. Why in the world would supposedly random polarized photons arrange themselves in such perfect order?

Now, let's go and smash Bob's polarizer with a hammer. What sequence does Alice get now and what is it relative to when there is no other polarizer?

 

[DrChinese]

Rather than correlation between the two sequences I'm now paying attention to regularity of each sequence in itself.

A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

This sequence is supposed to come out if A and B polarizers are orthogonally aligned. Why in the world would supposedly random polarized photons arrange themselves in such perfect order?

Nothing like this happens with polarization entangled photon pairs. For entangled photons, each sequence is always completely random when viewed by itself.

I didn't look closely enough at your post #118, example 3 above to notice the sequence you presented as being a pattern.

 

[DrChinese]

Just as a reminder: Nothing Alice chooses to do will show up in any manner for Bob on his own. Otherwise you could send an FTL signal. If only...

 

[Jabbu]

Nothing like this happens with polarization entangled photon pairs. For entangled photons, each sequence is always completely random when viewed by itself.

I didn't look closely enough at your post #118, example 3 above to notice the sequence you presented as being a pattern.

A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
B: 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

They are perfectly symmetric patterns because otherwise correlation is not zero, and for orthogonally aligned polarizers it is supposed to be zero. The smallest change in that sequence pair will lead straight to 10% correlation. How could you possibly arrange the two sequences in a random way and still get zero correlation?

 

[Jabbu]

Just as a reminder: Nothing Alice chooses to do will show up in any manner for Bob on his own. Otherwise you could send an FTL signal. If only...

I'm not sure what do you mean by "on his own". If Alice changes her polarizer angle, and Bob's polarizer angle stays as before, does Bob's sequence change or not?

 

[DrChinese]

I'm not sure what do you mean by "on his own". If Alice changes her polarizer angle, and Bob's polarizer angle stays as before, does Bob's sequence change or not?

Bob's sequence always looks random. If Alice changes her setting, it is possible that Bob's outcomes will in fact be different. However, it may merely change from one random sequence to another.

So no one can actually answer this question definitively.

 

[Jabbu]

Bob's sequence always looks random. If Alice changes her setting, it is possible that Bob's outcomes will in fact be different. However, it may merely change from one random sequence to another.

So no one can actually answer this question definitively.

The sequence for orthogonally aligned polarizers has zero correlation and is thus supposed to look like this:

A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
B: 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

Right? That's very unique and recognizable pattern, so if Alice's polarizer angle vary we could tell with certainty whether Bob's sequences is changing along or not.

 

[DrChinese]

The sequence for orthogonally aligned polarizers has zero correlation and is thus supposed to look like this:

A: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
B: 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

Right?

Sorry, not sure where you got that idea. More like this for theta=90 degrees and entangled pairs:

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 1 1 0 1 0 0 0 0 0 1 1 0 1 0 1 1 1 0 1 0

Entangled photons will always yield a random pattern. The pattern from entangled photons only emerges when the results from both streams are compared.

 

[Jabbu]

Sorry, not sure where you got that idea. True, there is no correlation (as you say). More like this for theta=90 degrees and entangled pairs:

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 1 1 0 1 0 0 0 0 0 1 1 0 1 0 1 1 1 0 1 0

Entangled photons will always yield a random pattern. The pattern from entangled photons only emerges when the results from both streams are compared.

I see now, somehow it wasn't obvious to me before. Is photon energy anyhow related to photons interaction with polarizers? Equally polarized red beam of light looses the same intensity as blue beam of light when passed through the same polarizer? Do photons loose or maybe gain energy after going through a polarizer?

 

[PeterDonis]

More like this for theta=90 degrees and entangled pairs:

A: 0 0 1 0 1 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1
B: 1 1 0 1 0 0 0 0 0 1 1 0 1 0 1 1 1 0 1 0

This doesn't look to me like no correlation; it looks to me like perfect anti-correlation. ISTM that no correlation would look something like this:

A: 0 0 1 0 1 0 1 1 1 0 1 1 0 0 0 0 1 1 0 1
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0

I.e., on average, 50 percent matches, 50 percent non-matches (and each individual sequence also with a 50-50 chance, on average, of 0 or 1).

 

[DrChinese]

This doesn't look to me like no correlation; it looks to me like perfect anti-correlation. ISTM that no correlation would look something like this:

A: 0 0 1 0 1 0 1 1 1 0 1 1 0 0 0 0 1 1 0 1
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0

I.e., on average, 50 percent matches, 50 percent non-matches (and each individual sequence also with a 50-50 chance, on average, of 0 or 1).

Peter, in going back and forth on this, I misspoke myself on that point. Thanks for catching it, and I have edited to correct.

 

[DrChinese]

I see now, somehow it wasn't obvious to me before. Is photon energy anyhow related to photons interaction with polarizers? Equally polarized red beam of light looses the same intensity as blue beam of light when passed through the same polarizer? Do photons loose or maybe gain energy after going through a polarizer?

Not all polarizers work for all wavelengths of light. Generally, energy is not relevant in any way and there is no theoretical basis for any energy transfer that I am aware of.

 

[Jabbu]

This doesn't look to me like no correlation; it looks to me like perfect anti-correlation. ISTM that no correlation would look something like this:

A: 0 0 1 0 1 0 1 1 1 0 1 1 0 0 0 0 1 1 0 1
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0

I.e., on average, 50 percent matches, 50 percent non-matches (and each individual sequence also with a 50-50 chance, on average, of 0 or 1).

That's different formula then.

theta_A = 0, theta_B = +90
A: 1 0 1 0 1 0 1 0 1 0 1 1 0 0 0 0 1 1 0 1
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0

sequence length = 20
matching pairs (00 or 11) = 10
mismatching pairs (01 or 10) = 10
Correlation = (match - mismatch) / sequence length = 0/20 = 0% ?

 

[PeterDonis]

Correlation = (match - mismatch) / sequence length = 0/20 = 0% ?

Yes, that's basically how a correlation coefficient is computed. Note that if there were all matches, the correlation would be 1, and if there were all mismatches, the correlation would be -1.

 

[Jabbu]

Yes, that's basically how a correlation coefficient is computed. Note that if there were all matches, the correlation would be 1, and if there were all mismatches, the correlation would be -1.

Ok, all together it should now look like this:


theta_A = 0, theta_B = 0
A: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0

sequence length = 20
matching pairs (00 or 11) = 20
mismatching pairs (01 or 10) = 0
Correlation = (match - mismatch) / sequence length = 20/20 = 100%


theta_A = 0, theta_B = 45
A: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0
B: 1 0 0 0 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 1

sequence length = 20
matching pairs (00 or 11) = 15
mismatching pairs (01 or 10) = 5
Correlation = (match - mismatch) / sequence length = 10/20 = 50%


theta_A = 0, theta_B = 90
A: 1 0 1 0 1 0 1 0 1 0 1 1 0 0 0 0 1 1 0 1
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0

sequence length = 20
matching pairs (00 or 11) = 10
mismatching pairs (01 or 10) = 10
Correlation = (match - mismatch) / sequence length = 0/20 = 0%

 

[Jabbu]

Not all polarizers work for all wavelengths of light.

Would it be possible then emitted photons vary in energy so that 50% probability is not only a function of photon polarization axis, but also its wavelength?

 

[Nugatory]

Would it be possible then emitted photons vary in energy so that 50% probability is not only a function of photon polarization axis, but also its wavelength?

Not according to any experiment that has ever been done, including the classical ones that led to Malus's law.

 

[Jabbu]

Not according to any experiment that has ever been done, including the classical ones that led to Malus's law.

Do you say that because it's known that all photons have the same wavelength, or is there some other reason?

 

[Nugatory]

Do you say that because it's known that all photons have the same wavelength, or is there some other reason?

It is not true that "all photons have the same wavelength" (or even that they have a physically meaningful wavelength, although that digression will take us far off-topic).

Wavelength $\lambda$ and frequency $\nu$ of a wave traveling at speed $c$ are related by $c=\lambda\nu$, so to the extent that polarizers work the same way for a range of frequencies, they must also work the same way for a range of wavelengths - and that's what we observe. Because the energy of a photon is related to the frequency by $E=h\nu$, and the polarizers work the same way for a range of frequencies, they must also work the same way for the corresponding range of energies.

Also some of the Bell-type experiments that measure the polarization of polarization-entangled photons use the calcium atomic cascade (google for "calcium cascade photon") to generate their photons, and this process generates photons at specific frequencies. Thus, even if there were some dependency on frequency and wavelength, it wouldn't affect these experiments.

 

[Jabbu]

It is not true that "all photons have the same wavelength" (or even that they have a physically meaningful wavelength, although that digression will take us far off-topic).

Wavelength $\lambda$ and frequency $\nu$ of a wave traveling at speed $c$ are related by $c=\lambda\nu$, so to the extent that polarizers work the same way for a range of frequencies, they must also work the same way for a range of wavelengths - and that's what we observe. Because the energy of a photon is related to the frequency by $E=h\nu$, and the polarizers work the same way for a range of frequencies, they must also work the same way for the corresponding range of energies.

Suppose photon energy is proposed to be a "hidden variable". How do you show there can not exist a function involving this unknown which would duplicate experimental results?

Also some of the Bell-type experiments that measure the polarization of polarization-entangled photons use the calcium atomic cascade (google for "calcium cascade photon") to generate their photons, and this process generates photons at specific frequencies. Thus, even if there were some dependency on frequency and wavelength, it wouldn't affect these experiments.

theta_A = 0, theta_B = 0
A: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0

Ok. So finally now I'm staring at the naked mystery face to face. Entangled photons polarization is random between pairs, but same for each pair. Their overall chance to pass through any polarizator angle remains 50% as it should. However, unexpectedly, both photon twins have the same chance at different locations as if it was one event with a single probability and not two separate events each with its own independent probability. Would this be a good description of the essence of the mystery, or is there even more to it?

 

[Nugatory]

Suppose photon energy is proposed to be a "hidden variable". How do you show there can not exist a function involving this unknown which would duplicate experimental results?

There's Bell's theorem: no local hidden variable theory can match the quantum mechanical prediction in all cases. Then there are the experiments of Alain Aspect and many others, showing that the quantum mechanical predictions are correct in these cases. Put them together, and we know that there is no possible local hidden variable theory which matches experiment.

Ok. So finally now I'm staring at the naked mystery face to face. Entangled photons polarization is random between pairs, but same for each pair. Their overall chance to pass through any polarizator angle remains 50% as it should. However, unexpectedly, both photon twins have the same chance at different locations as if it was one event with a single probability and not two separate events each with its own independent probability. Would this be a good description of the essence of the mystery, or is there even more to it?

That's a pretty good summary.

 

[Jabbu]

There's Bell's theorem: no local hidden variable theory can match the quantum mechanical prediction in all cases.

Original Bell's inequality... This simple form does have the virtue of being quite intuitive. It is easily seen to be equivalent to the following elementary result from probability theory. Consider three (highly correlated, and possibly biased) coin-flips X, Y, and Z, with the property that:

X and Y give the same outcome (both heads or both tails) 99% of the time
Y and Z also give the same outcome 99% of the time,

then X and Z must also yield the same outcome at least 98% of the time. The number of mismatches between X and Y (1/100) plus the number of mismatches between Y and Z (1/100) are together the maximum possible number of mismatches between X and Z (a simple Boole–Fréchet inequality).
http://en.wikipedia.org/wiki/Bell's_theorem


This looks like a nice simple explanation, I just don't get it. Why are they even talking about some third coin, wouldn't two photons correspond to only two coins?

 

[bhobba]

That's a pretty good summary.

Yes it is.

But I really am scratching my head about all the angst this has engendered.

Its simply the weirdness of the underlying vector space structure of QM which is the essence of entanglement.

Such is totally inexplicable classically so its hardly surprising its not explainable by classical analogues like Bertlmanns socks.

Bell cottoned onto it, and his paper is clear about it.

Thanks
Bill

 

[Jabbu]

Such is totally inexplicable classically so its hardly surprising its not explainable by classical analogues like Bertlmanns socks.

I'm still trying to put it into some sensible terms before I even begin thinking about what actual explanation there is for this sorcery.

A: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0...
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0...

For example, this is like every time two of us flip a coin and it's always a match. It's already impossible. What is there to think? It's like trying to divide by zero, it does not compute. The premise must be wrong, surely?

 

[StevieTNZ]

May I recommend the recently published book by Nicolas Gisin - http://www.springer.com/physics/quantum+physics/book/978-3-319-05472-8

 

[Nugatory]

I'm still trying to put it into some sensible terms before I even begin thinking about what actual explanation there is for this sorcery.

A: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0...
B: 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 1 1 1 0...

For example, this is like every time two of us flip a coin and it's always a match. It's already impossible. What is there to think? It's like trying to divide by zero, it does not compute. The premise must be wrong, surely?

It might be worth going back to the beginning of this thread, reading through it again... It started with johan0001 asking why there was anything strange going on at all, how measuring the polarization of two entangled photons and finding a correlation is any different than putting two gloves in two boxes then finding that the presence of a left-handed glove in one box is correlated with finding a right-handed glove in the other.

 

[bhobba]

I'm still trying to put it into some sensible terms before I even begin thinking about what actual explanation there is for this sorcery.

Mate - that's why we have Bells Theorem

Note that keyword THEOREM. It shows that classical analogues are not possible where classical means naive reality.

Why? In math I gave up aeons ago looking for the why behind theorems. What's the why behind the fundamental theorem of algebra? What's the why behind a Weiner process being continuous and non differentiable everywhere - even such a function exists is downright weird - little alone some physical process is modeled by such? Blowed if I know - the reason is the math.

Thanks
Bill

 

[Nugatory]

This looks like a nice simple explanation, I just don't get it. Why are they even talking about some third coin, wouldn't two photons correspond to only two coins?

We have two photons, entangled so that if one of them passes a filter at a given angle, the other one definitely will not pass a filter at that angle.

The three coins correspond to the possible results of measuring the polarization on any of three possible angles, A, B, and C. For example, heads/heads/tails corresponds to "the left hand-photon would pass a filter at angle A while the right-hand photon would not; the left hand-photon would pass a filter at angle B while the right-hand photon would not; the left hand-photon would not pass a filter at angle C while the right-hand photon would".