Why is entanglement necessary for understanding quantum mechanics?

[Jabbu]

http://people.umass.edu/blaylock/personal/WringingJohnBell-Clark-17Feb10.ppt

Look at page 8.

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.

 

[stevendaryl]

Let us remove most of the details and focus on the absolutely basic stuff and a strongly simplified scenario. We have a light source emitting two photons and the polarizations have to be orthogonal in the end.

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.

 

[Nugatory]

But it is constant. I believe that's how the setup is calibrated. You rotate the polarizer until the detector starts reading all zeros or all ones, where zeros mean the polarizer is at 90 degrees relative to photons polarization, and all ones means they are perfectly aligned and theta = 0, so there you set the zero point, deliberately, not arbitrarily.

No, that is not how it works. The source does not have any preferred direction, so no matter how you rotate the source or one of the directors you will get a random stream of zeroes and ones at that detector as half the photons pass and half don't (which is, BTW, consistent with Malus's law for photons of random orientation reaching the polarizer).

If photon polarization was random, then when we change polarizer angle from vertical to horizontal, why does it matter? Photons with random polarization is unpolarized light, they have 50% chance to go through regardless of the polarizer absolute rotation.

Changing the angle of the polarizer does not matter, because you get a 50% pass rate no matter what angle you choose. However, the angle between the two polarizers does matter, and indeed that's what the whole thing is about. If the two polarizers are oriented at 90 degrees relative to one another, then every time we get a 1 at the left-hand polarizer we'll get a 1 at the right-hand polarizer; if they are oriented at zero degrees relative to each other, then every time we get a 1 at the left-hand polarizer we get a 0 at the right-hand polarizer; but if we look at the results from only one of them, we'll see a random stream of ones and zeroes.

For any in-between angle: If the A detector reads 1, the B detector will give a 0 or 1 result with the probability predicted by Malus's law if $\omega_B$ is at right angles to $\theta_A$; and if the A detector reads 0 the B detector will give a 0 or 1 result with the probability predicted by Malus's law if $\omega_B$ is equal to $\theta_A$.

 

[Cthugha]

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.

Oh, both are possible, depending on whether you have a type I or type II spdc crystal.

My main point was just to remove a bit of the discussion on the physical details of entanglement and steer Jabbu towards the point of the realism assumption. The fact that it makes a real measurable difference whether you assume that particles are emitted with well defined properties right from the start or whether you assume that these properties become well defined only during the first measurement made.

 

[stevendaryl]

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, it does not say that. On the previous page, it says:

What if we measured polarizations at arbitrary angles \theta_1, \theta_2?

\theta_1 and \theta_2 are the orientations of the filters. That's clearly what's being shown in the figure on page 8.

Read through to page 13. They clearly say that the QM prediction is different from what would be expected classically.

 

[Jabbu]

That point was already covered. The relative angle between the polarizers and the relative angle of polarization between the two photons matter. This relative angle is well defined. The single photon polarizations are not.

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?

Let us remove most of the details and focus on the absolutely basic stuff and a strongly simplified scenario. We have a light source emitting two photons and the polarizations have to be orthogonal in the end. Now we set the polarizers to some fixed settings (say 0°/90°) and compare two basic scenarios:

1) The source emits two photons of well defined polarization. Sometimes 0°/90°. Sometimes 45°/135°. Sometimes 157°/247°. You can get the expected correlation by applying Malus' law independently to each photon and polarizer setting.

2) The source emits two photons, but the polarization is undefined. If the first measurement at a polarizer results in transmission, the polarizer forces the photon to acquire a polarization matching that of the polarizer setting and the other photon instantaneously "jumps" to the orthogonal polarization.

Leaving all details on real physics aside, do you agree that these two scenarios will give different coincidence count rates?

The simplest case is to take a single setup where for example theta_A = -30 and theta_B = +30 degrees. QM predicts 25% correlation (75% discordance), and Malus law is supposed to give different result, right? So what result Malus law predicts and how do you arrive at it?

 

[Jabbu]

I think you're getting two different experiments confused. In the EPR experiment, both Alice and Bob detect 50% of the photons, regardless of \theta_A or \theta_B. What's of interest are the correlations. If \theta_A = \theta_B, then there is 100% correlation. if \theta_B = \theta_A + 90^o, then there is 100% anti-correlation.

If photons always had 50% chance it would always yield the same correlation. Correlation here is a simple relation between two probabilities. If both probabilities are constant the correlation stays constant.

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

These two sequences represent two probabilities: 100% and 0%, how do you calculate correlation between them?

 

[stevendaryl]

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?

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)

 

[stevendaryl]

If photons always had 50% chance it would always yield the same correlation. Correlation here is a simple relation between two probabilities.

No, what I've been talking about is the probability that Alice and Bob get the same result (either 0 or 1). For the runs that you gave:

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

That number would be 0. The quantum prediction is cos^2(\theta_A - \theta_B)

 

[Jabbu]

No, it does not say that. On the previous page, it says:

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.

 

[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?