Why is entanglement necessary for understanding quantum mechanics?

[stevendaryl]

No, they are not the same.

Okay, the same factor, cos^2(\theta) is used in both the Malus formula and the quantum mechanical formula. However, in the case of Malus' formula, \theta is measured relative to the polarization of the light at the source. In the case of QM, \theta is the relative angle between the detectors. Those are very different things. The only way to try to reconcile them is the "collapse" interpretation, whereby when the photon passes through Alice's filter, the other photon suddenly changes so that it is polarized in the same direction as Alice's filter. So that's a nonlocal effect.

 

[Jabbu]

It occurs to me that you may be being confused by the way that the symbol $\theta$ is being used into two different ways. There are four angles in the experiment: The angle we set the left-hand polarizer to ($\theta_L$), the angle we set the right-hand polarizer to ($\theta_R$), the polarization angle of the left-hand photon ($\omega_L$), and the polarization angle of the right-hand photon ($\omega_L$).

The thetas that appear in Malus's law are the angles $\theta_R-\omega_R$ for the right-hand photon interacting with the right-hand polarizer and $\theta_L-\omega_L$ for the left-hand photon interacting with the left-hand polarizer.

The theta that appears in the QM calculation of the correlation probabilities is $\theta_L-\theta_R$, the difference in the detector angles.

That's what I was thinking. If QM theta is relative angle between L and R polarizers then Malus's law theta has to be QM theta/2, because it applies to each photon separately, so QM theta = 60 is Malus's law theta L = -30, and theta R = +30.

 

[DrChinese]

When we measure correlation with angle theta of 60 degrees, does it mean:

1. polarizer A = +60, polarizer B = +60
2. polarizer A = -60, polarizer B = +60
3. polarizer A = -30, polarizer B = +30

Or something else?

3. is a theta of 60 degrees. First is 0, second is 120. As it happens, theta of 60 and 120 are the same as far as cos^2 and sin^2 go.

 

[DrChinese]

Yes, I'm trying to clarify things that don't make sense to me.

The EPR argument is in favor of hidden variables, what they call a more "complete" specification of the system. Bell says that in a theory with additional parameters (i.e. a more complete specification of the system, or hidden variables), that there must be an appropriate relationship between all angle settings. The cos^2(theta) relationship is obviously not linear, so you cannot really have a consistent relationship between hypothetical values across the 360 degree spectrum.

You can't really see that with a single photon, but it is just as true: even a polarized photon does not possesses pre-determined values for other polarization angles. But it becomes very clear when you have a pair of entangled photons. (If you are familiar with Bell, of course. )

 

[Jabbu]

If both filters are at the same angle, then there is perfect, 100% correlation--either both filters pass the photons, or both do not.

That's theta = 0 then, so Malus's law is cos^2(0) = 100% as well. I just don't understand where QM comes in play, is there anything non-local about Malus's law itself?

 

[stevendaryl]

That's theta = 0 then, so Malus's law is cos^2(0) = 100% as well. I just don't understand where QM comes in play, is there anything non-local about Malus's law itself?

No, you're confusing different angles. Nugatory explained the difference:
(1) There's the orientation of Alice's filter, \theta_A
(2) There's the orientation of Bob's filter, \theta_B
(3) There's the hypothetical polarization of Alice's photon, \omega_A
(4) There's the hypothetical polarization of Bob's photon, \omega_B

Malus' law says that the intensity of light passing through Alice's filter is
cos^2(\theta_A - \omega_A)
The intensity of light passing through Bob's filter is
cos^2(\theta_B - \omega_B)

QM says that the probability of Alice and Bob getting the same result (either both photon's pass, or both are blocked) for entangled photon pairs is
cos^2(\theta_A - \theta_B)

Even though both formulas involve cos^2, they are not at all the same. They aren't about the same angles. The only way to interpret the QM result using Malus is if you assume that:

If Alice's filter passes her photon, then immediately afterwards, \omega_B = \theta_A.

If Alice's filter blocks her photon, then immediately afterwards, \omega_B = 90^o - \theta_A
​

 

[Jabbu]

No, you're confusing different angles. Nugatory explained the difference:
(1) There's the orientation of Alice's filter, \theta_A
(2) There's the orientation of Bob's filter, \theta_B
(3) There's the hypothetical polarization of Alice's photon, \omega_A
(4) There's the hypothetical polarization of Bob's photon, \omega_B

I think we were talking about the same thing just before, and still are, but why do you say "hypothetical" polarization? When theta = 60, didn't we establish that specifically mean \theta_A - \omega_A = -30 and \theta_B - \omega_B = +30?

Malus' law says that the intensity of light passing through Alice's filter is
cos^2(\theta_A - \omega_A)
The intensity of light passing through Bob's filter is
cos^2(\theta_B - \omega_B)

Exactly.

QM says that the probability of Alice and Bob getting the same result (either both photon's pass, or both are blocked) for entangled photon pairs is
cos^2(\theta_A - \theta_B)

Yes. Although, isn't it more commonly written as sin^2(\theta_A - \theta_B)?

Even though both formulas involve cos^2, they are not at all the same. They aren't about the same angles.

Sure, they are not the same. QM equation calculates correlation/discordance straight away, and with Malus's law we only have two probabilities for two separate polarizers. We can't compare them directly, shouldn't we first calculate "correlation" for those two probabilities?

 

[DrChinese]

QM equation calculates correlation/discordance straight away, and with Malus's law we only have two probabilities for two separate polarizers. We can't compare them directly, shouldn't we first calculate "correlation" for those two probabilities?

Sure, try it. Assume the (unknown) polarization is 30 degrees for photon A and 30+90=120 degrees for photon B (or use any angle if you don't like 30 degrees).

Now assume Alice sets her polarizer at 75 degrees and Bob sets his at 75+90=165 degrees (or replace 75 degrees with anything you like as long as it is not 30). Now apply your formula -which you think is the same as cos^2(theta) but is not. The QM answer is 100% match and yours will be something different.

The true answer for your version is separately calculated for each side, with 2 outcomes matching (++ or --). For Alice matching Bob: (cos^2(75-30) x cos^(165-120)) + (sin^2(75-30) x sin^(165-120)) = .25 + .25 = 50%.

So Malus doesn't really apply as a local realistic (classical) explanation, does it? Since it gives a WRONG prediction, as you have been told repeatedly. You should step back and look at your premises a little closer. Entangled photons do not have a classical polarization until collapse occurs. And I am not sure they have a classical polarization even then, although this has not been proven rigorously.

 

[stevendaryl]

I think we were talking about the same thing just before, and still are,

No, I don't think so. If we have the case where \omega_A = \omega_B = 0,
\theta_A = \theta_B = 45^o, then Malus' formula does not predict that Alice and Bob will get the same results. It doesn't predict 100% correlation betwen Alice and Bob. It only predicts that both Alice and Bob will get 50% of the intensity to pass through their filters.

but why do you say "hypothetical" polarization?

Because in the EPR experiment, you don't know the photon polarizations when the photons are produced. You can hypothesize that the photons have an unknown polarization, but that's just a hypothesis.

When theta = 60, didn't we establish that specifically mean \theta_A - \omega_A = -30 and \theta_B - \omega_B = +30?

No. All that we can measure in the EPR experiment is \theta_A and \theta_B. You can't measure the hypothetical values \omega_A and \omega_B.

Yes. Although, isn't it more commonly written as sin^2(\theta_A - \theta_B)?

There are two different EPR experiments: one uses photons, and the other uses electron-positron pairs. For the photon case, the QM prediction is that the probability that Alice and Bob get the same result is cos^2(\theta_A - \theta_B). For the electron-positron case, the probability that they get the same result is sin^2(\dfrac{\theta_A - \theta_B}{2})

Sure, they are not the same. QM equation calculates correlation/discordance straight away, and with Malus's law we only have two probabilities for two separate polarizers. We can't compare them directly, shouldn't we first calculate "correlation" for those two probabilities?

Yes, and that's what Bell did, to show that the QM case cannot be explained using Malus' law (or any other local law).

 

[Jabbu]

Sure, try it. Assume the (unknown) polarization is 30 degrees for photon A and 30+90=120 degrees for photon B (or use any angle if you don't like 30 degrees).

Now assume Alice sets her polarizer at 75 degrees and Bob sets his at 75+90=165 degrees (or replace 75 degrees with anything you like as long as it is not 30). Now apply your formula -which you think is the same as cos^2(theta) but is not. The QM answer is 100% match and yours will be something different.

The true answer for your version is separately calculated for each side, with 2 outcomes matching (++ or --). For Alice matching Bob: (cos^2(75-30) x cos^(165-120)) + (sin^2(75-30) x sin^(165-120)) = .25 + .25 = 50%.

Can't we simply take just one angle and compare? That looks like the equation I was asking you about before. What formula is that?

 

[Jabbu]

Because in the EPR experiment, you don't know the photon polarizations when the photons are produced. You can hypothesize that the photons have an unknown polarization, but that's just a hypothesis.

No. All that we can measure in the EPR experiment is \theta_A and \theta_B. You can't measure the hypothetical values \omega_A and \omega_B.

I'm not saying it's measured, but a part of the set up, that's what (\theta_A - \theta_B) implies. When we have theta = 60 in QM equation, can it translate for Malus's equation into anything other than: \theta_A - \omega_A = -30, \theta_B - \omega_B = +30? What, for example?

 

[DrChinese]

Can't we simply take just one angle and compare? That looks like the equation I was asking you about before. What formula is that?

No you need to work it through for 2 particles, separable. Because classically, entanglement does not even exist!

 

[Jabbu]

No you need to work it through for 2 particles, separable. Because classically, entanglement does not even exist!

Yes, of course. I should have said I was talking about "master" theta angle, relative angle between polarizers, the one used in QM equation: master \theta = (\theta_A - \theta_B). That one angle involves both photons and both polarizers.

 

[stevendaryl]

I'm not saying it's measured, but a part of the set up, that's what (\theta_A - \theta_B) implies. When we have theta = 60 in QM equation, can it translate for Malus's equation into anything other than: \theta_A - \omega_A = -30, \theta_B - \omega_B = +30? What, for example?

You're saying that the polarization of the photon is always half-way between \theta_A and \theta_B? How could that happen? The photon doesn't know what settings Alice and Bob are going to use.

 

[stevendaryl]

Yes, of course. I should have said I was talking about "master" theta angle, relative angle between polarizers, the one used in QM equation: master \theta = (\theta_A - \theta_B). That one angle involves both photons and both polarizers.

That expression doesn't mention the photon polarization at all. That's why I said that the QM result is not the same as the Malus' formula, even though they both mention cos^2

 

[Nugatory]

I'm not saying it's measured, but a part of the set up, that's what (\theta_A - \theta_B) implies. When we have theta = 60 in QM equation, can it translate for Malus's equation into anything other than: \theta_A - \omega_A = -30, \theta_B - \omega_B = +30? What, for example?

One of many examples would be $\theta_A=10$, $\theta_B=70$, $\omega_A=-35$, $\omega_B=55$.

 

[DrChinese]

Yes, of course. I should have said I was talking about "master" theta angle, relative angle between polarizers, the one used in QM equation: master \theta = (\theta_A - \theta_B). That one angle involves both photons and both polarizers.

Note that is the QM predictions for an entangled pair of photons, and lacks a term for the photons' separate polarizations (since there isn't such).

QM also makes a prediction for a pair of normal UNentangled photons which is different and DOES consider individual photon polarizations. Further, this equation can be enhanced to consider an average of a bunch of random initial polarizations. Even in this case, the formula will NOT match the entangled case.

In each scenario, experiment closely matches the relevant QM prediction. So clearly, classical models have serious constraints since there are no ongoing connections between entangled particles as there is in QM (which models the pair as being a single system, not 2 separate ones).

 

[Jabbu]

You're saying that the polarization of the photon is always half-way between \theta_A and \theta_B? How could that happen? The photon doesn't know what settings Alice and Bob are going to use.

Because \theta_A and \theta_B are relative angles themselves, they are angles between the light's initial polarization and the axis of the polarizer. If they weren't relative to photon polarization photons wouldn't even be a part of the equation.

When we say we emitted a single photon through a single polarizer and \theta = 30 degrees, it means polarizer absolute angle can be anything, as long as photon is either 30 degrees clockwise or 30 degrees anticlockwise polarized relative to the polarizer absolute axis. Right?

 

[stevendaryl]

Because \theta_A and \theta_B are relative angles themselves, they are angles between the light's initial polarization and the axis of the polarizer.

No, \theta_A has nothing to do with any "initial polarization". The filter can turn about an axis. There is one point on the axis that is labeled 0, and the angle is measured relative to that. Of course, to be meaningful, \theta_B and \theta_A have to be measured relative to the same zero point. But the zero point has nothing to do with the polarization of the light.

As I said earlier, you can change \theta_A and \theta_B in the middle of the experiment, and it makes no difference (in the entangled case). All that matters is \theta_A - \theta_B at the time the photons pass through Alice's and Bob's filter.

If they weren't relative to photon polarization photons wouldn't even be a part of the equation.

That would be true if Malus' law worked for individual photons, but it doesn't. That's what I've been telling you---Malus' law does not work. The QM prediction (for entangled photons) only involves \theta_A - \theta_B. It does NOT involve any polarization angle for the photons. (Because there is no such angle.)

 

[stevendaryl]

When we say we emitted a single photon through a single polarizer and \theta = 30 degrees, it means polarizer absolute angle can be anything, as long as photon is either 30 degrees clockwise or 30 degrees anticlockwise polarized relative to the polarizer absolute axis. Right?

In an EPR experiment, we don't produce polarized photons. There is some atomic reaction (radioactive decay) that produces two photons. One photon goes through (or is blocked by) Alice's filter, and the other goes through (or is blocked by) Bob's filter. Alice's and Bob's filter angles are measured relative to an arbitrary zero point, not relative to any initial photon polarization. When a particle decays, there is no associated "polarization angle".

 

[Nugatory]

When we say we emitted a single photon through a single polarizer and \theta = 30 degrees, it means polarizer absolute angle can be anything, as long as photon is either 30 degrees clockwise or 30 degrees anticlockwise polarized relative to the polarizer absolute axis. Right?

Yes, in the sense that we can choose to call any angle we wish "zero" and (for example) $\theta_A=10$, $\theta_B=70$, $\omega_A=-35$, $\omega_B=55$ is the same situation with a different zero point $\theta_A=20$, $\theta_B=80$, $\omega_A=-25$, $\omega_B=65$.

But you're still missing the point. We set one polarizer to angle $\theta_A$ and second polarizer to angle $\theta_B$. There are no values $\omega_A$ and $\omega_B$ such that for all $\theta_A$ and $\theta_B$:
1) Malus's law for $\theta_A-\omega_A$ matches the experimental results at polarizer A AND
2) Malus's law for $\theta_B-\omega_B$ matches the experimental results at polarizer B AND
3) $\omega_A$ and $\omega_B$ are at right angles to one another.

That's how we know that the photons do not have definite polarizations $\omega_A$ and $\omega_B$ when they're created.

 

[Jabbu]

No, \theta_A has nothing to do with any "initial polarization". The filter can turn about an axis. There is one point on the axis that is labeled 0, and the angle is measured relative to that. Of course, to be meaningful, \theta_B and \theta_A have to be measured relative to the same zero point. But the zero point has nothing to do with the polarization of the light.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/polcross.html
http://en.wikipedia.org/wiki/Malus's_law#Malus.27_law_and_other_properties
http://en.wikipedia.org/wiki/Photon_polarization

Theta is always defined as the angle between the light's initial polarization and the axis of the polarizer, otherwise it would be arbitrary. Where did you ever see theta refers to some absolute polarizer angle unrelated to light polarization?

As I said earlier, you can change \theta_A and \theta_B in the middle of the experiment, and it makes no difference (in the entangled case). All that matters is \theta_A - \theta_B at the time the photons pass through Alice's and Bob's filter.

Of course you can change \theta_A and \theta_B whenever you want, just keep photons polarization constant and rotate the polarizers. Yes, all that matters is \theta_A - \theta_B at the time the photons pass through Alice's and Bob's filter. But does it mean the result is predetermined exclusively by the polarizers and photons don't even matter at all?

That would be true if Malus' law worked for individual photons, but it doesn't. That's what I've been telling you---Malus' law does not work. The QM prediction (for entangled photons) only involves \theta_A - \theta_B. It does NOT involve any polarization angle for the photons. (Because there is no such angle.)

Why do you think Malus law would work for a light beam with 10 billion photons but not for a single photon? If each one of 10 billion vertically polarized photons have 50% chance to pass through polarizer rotated at 45 degrees relative angle, then a single photon has to have that same chance as well. I see you're saying \theta_A and \theta_B are absolute polarizer angles, but I don't understand why would you think something like that.

 

[stevendaryl]

Theta is always defined as the angle between the light's initial polarization and the axis of the polarizer

That is not correct. Malus' law does not apply in EPR experiments. In EPR experiments, there IS no "initial polarization"

 

[stevendaryl]

Why do you think Malus law would work for a light beam with 10 billion photons but not for a single photon?

You're comparing two different experiments:

(1) You start with light having a known polarization. In that case, the photons are NOT entangled. In that case, Malus' law works.

(2) You start with a pair of entangled photons. In that case, Malus' law does NOT work.

Both cases are described by QM, but only the first case is described by Malus' law. The two theories make the same prediction in the first case, but not the second case.

Once again, Malus' law does not describe entangled photons. The case where you are talking about, with a light beam consisting of 10 billion photons, does not involve entangled photons.

 

[DrChinese]

If each one of 10 billion vertically polarized photons have 50% chance to pass through polarizer rotated at 45 degrees relative angle, then a single photon has to have that same chance as well. I see you're saying \theta_A and \theta_B are absolute polarizer angles, but I don't understand why would you think something like that.

No one is disputing that for polarized photons, and that is entirely the point. But entangled photons lack a specific polarization until they are measured. As a result, their statistics are different. Because you keep mentioning Malus, you are confusing things. Malus applies to the example above just fine.

a. Take a stream of photons polarized at 0 degrees (call that vertical), and send the stream through a polarizer at 45 degrees. Malus (applied to classical light) and QM make the same predictions: cos^2(45)= .5.

b. Take 2 vertical photons and run them through polarizers at 45 degrees, and the chance they will match is: .25 (++ case) + .25 (-- case) = .5. Malus (applied to classical light) and QM make the same predictions again.

c. Take 2 Type I polarization entangled photons and run them through polarizers at 45 degrees, and the chance they will match is: 0 (++ case) + 0 (-- case) = 0. Note that by coincidence (although it is related), the result is per the formula 1-cos^2(theta) where theta is 0 degrees. Yes, it seems to be the same formula as per a. but we arrive at that point by a completely different logic set. And you won't get the correct answer using a classical model a la Malus. You cannot just say "I will use the QM formula on my classical model" - that is called hand-waving.

So this is the point we are making, the statistics are different for entangled photons than what you are modeling. Work it out and you will conclude the same.

 

[atyy]

You're comparing two different experiments:

(1) You start with light having a known polarization. In that case, the photons are NOT entangled. In that case, Malus' law works.

(2) You start with a pair of entangled photons. In that case, Malus' law does NOT work.

Both cases are described by QM, but only the first case is described by Malus' law. The two theories make the same prediction in the first case, but not the second case.

Once again, Malus' law does not describe entangled photons. The case where you are talking about, with a light beam consisting of 10 billion photons, does not involve entangled photons.

For the entangled photons, couldn't I take the reduced density matrix for one, so that its polarization would be a local observable, and effectively equivalent to a statistical mixture of photons with definite polarization, to which I can then apply Malus's Law?

 

[stevendaryl]

For the entangled photons, couldn't I take the reduced density matrix for one, so that its polarization would be a local observable, and effectively equivalent to a statistical mixture of photons with definite polarization, to which I can then apply Malus's Law?

Yes, that's equivalent (as far as the answer you get) to the "collapse" interpretation, where after a measurement, the photons are in a definite polarization state. The point is that you can't get the right answer from Malus' law alone. The thread started with someone asking what's weird about QM, when the predictions for EPR are the same as the classical predictions (Malus' law).

 

[DrChinese]

For the entangled photons, couldn't I take the reduced density matrix for one, so that its polarization would be a local observable, and effectively equivalent to a statistical mixture of photons with definite polarization, to which I can then apply Malus's Law?

Not unless you first run it through a polarizer to reduce it. Otherwise, you just get what appears to be a statistical mixture of H and V photons.

 

[atyy]

Not unless you first run it through a polarizer to reduce it. Otherwise, you just get what appears to be a statistical mixture of H and V photons.

The idea is if I have a statistical mixture only of H photons, then I can treat the mixture with Malus's Law. Similarly, if I have a mixture only of V photons, I can treat the mixture with Malus's Law. Since I have effectively a statistical mixture of H and V photons, the result should be a statistical mixture of Malus's Law applied to H photons and Malus's Law applied to V photons.

 

[atyy]

Yes, that's equivalent (as far as the answer you get) to the "collapse" interpretation, where after a measurement, the photons are in a definite polarization state. The point is that you can't get the right answer from Malus' law alone. The thread started with someone asking what's weird about QM, when the predictions for EPR are the same as the classical predictions (Malus' law).

Yes, I agree. I just thought maybe it's confusing to Jabbu to have been told both that Malus's Law does and does not apply to single photons in the quantum case (I think Nugatory said that in post #22, and he was probably thinking of the reduced density matrix). Anyway, I thought maybe this idea that one can effectively apply Malus's Law to single photons even in the entangled quantum case, can help make the entangled quantum case closer to classical case, so that one can see how they really diverge.

 

[stevendaryl]

Yes, I agree. I just thought maybe it's confusing to Jabbu to have been told both that Malus's Law does and does not apply to single photons in the quantum case (I think Nugatory said that in post #22, and he was probably thinking of the reduced density matrix). Anyway, I thought maybe this idea that one can effectively apply Malus's Law to single photons even in the entangled quantum case, can help make the entangled quantum case closer to classical case, so that one can see how they really diverge.

Yes. I guess there are two different facts about quantum polarization experiments: What happens at individual filters can be explained by Malus' law. But correlations between two different filters is an additional piece of information that is not explained by Malus' law (unless you add the "collapse" interpretation).

 

[Jabbu]

In an EPR experiment, we don't produce polarized photons. There is some atomic reaction (radioactive decay) that produces two photons. One photon goes through (or is blocked by) Alice's filter, and the other goes through (or is blocked by) Bob's filter. Alice's and Bob's filter angles are measured relative to an arbitrary zero point, not relative to any initial photon polarization. When a particle decays, there is no associated "polarization angle".

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.

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.

 

[stevendaryl]

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.

That is not what is done in EPR-type experiments. Once again, in the EPR experiment, you don't start with polarized light. You start with some particle decaying.

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.

The point is that in the case of an entangled pair of photons, the polarization seems random, in the sense that Alice's filter passes 50% of the photons that reach it, and Bob's filter also passes 50% of its photons. But in another sense it is not random: If Alice and Bob have the same filter settings, and a photon passes Alice's filter, it will, with 100% probability, also pass Bob's filter. If a photon is blocked by Alice's filter, it will, with 100% probability, also be blocked by Bob's filter.

 

[Jabbu]

That is not correct. Malus' law does not apply in EPR experiments. In EPR experiments, there IS no "initial polarization"

Absolute theta is arbitrary by itself no matter what experiment. It would be like entering 38 every time into the equation and still get different results. Do you have any reference that defines theta as absolute polarizer angle and unrelated to light polarization?

 

[stevendaryl]

Absolute theta is arbitrary by itself no matter what experiment. It would be like entering 38 every time into the equation and still get different results.

Alice's angle \theta_A is arbitrary, and Bob's angle \theta_B is arbitrary, but the difference \theta_A - \theta_B is the relevant quantity that enters in the QM prediction:

Probability that Alice's result is the same as Bob's result = cos^2(\theta_A - \theta_B)

Do you have any reference that defines theta as absolute polarizer angle and unrelated to light polarization?

As I said, in EPR, you don't start with polarized light. The formula can't possibly be related to light polarization, because the light isn't polarized.

 

[Cthugha]

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.

In that case your belief is wrong. The quantity which is entangled is necessarily ill defined. polarization entangled photons are umpolarized on their own. Momentum entangled photons have a broad distribution of individual momenta. Energy-entangled photons are spectrally broad.

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.

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.

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?

 

[stevendaryl]

Do you have any reference that defines theta as absolute polarizer angle and unrelated to light polarization?

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

Look at page 8.

 

[Jabbu]

Alice's angle \theta_A is arbitrary, and Bob's angle \theta_B is arbitrary, but the difference \theta_A - \theta_B is the relevant quantity that enters in the QM prediction:

Probability that Alice's result is the same as Bob's result = cos^2(\theta_A - \theta_B)

So if the equation carries no information about photons, does it not mean the result is predetermined by polarizers alone?

As I said, in EPR, you don't start with polarized light. The formula can't possibly be related to light polarization, because the light isn't polarized.

From all the papers I've seen I don't remember anyone suggested anything like that.

When detectors A and B read:

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

relative angle between photons A polarization and polarizer A is theta_A = 0?
relative angle between photons B polarization and polarizer B is theta_B = 90?
Correlation = ?


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

relative angle between photon A polarization and polarizer A is theta_A = 45?
relative angle between photon B polarization and polarizer B is theta_B = ?
Correlation = ?

 

[stevendaryl]

So if the equation carries no information about photons, does it not mean the result is predetermined by polarizers alone?

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

From all the papers I've seen I don't remember anyone suggested anything like that.

Did you ever read anything about EPR that suggested that the twin photons are polarized?

 

[stevendaryl]

When detectors A and B read:

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

relative angle between photons A polarization and polarizer A is theta_A = 0?
relative angle between photons B polarization and polarizer B is theta_B = 90?
Correlation = ?

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.

 

[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.