Why is entanglement necessary for understanding quantum mechanics?

[Jabbu]

Another indication that there is something weird going on, experimentally, is just to pick a fixed angle, $A$ for both Alice's and Bob's filter settings. What you will find is that

1. 50% of the time, both photons will pass through their respective filters.
2. 50% of the time, neither photon will pass through.

According to Malus it's 45 degrees relative angle that gives 50% chance: cos^2(45) = 50%, so if both photons have the same 45 degrees polarization relative to their polarizer, or at least same on average, then overall they will both have the same 50% chance to pass through. Where is this different than what actually happens?

What never happens, if Alice and Bob have their filters at the same setting, is that it passes through one filter but not the other.

I think those experiments measure large numbers of both matching and mismatching pairs, and that it is only after some average is taken over many measurements that we can see some kind of overall correlation or discordance.

This is only consistent with Malus' law if you assume that 50% of the time, the photons are polarized in the direction of Alice's filter setting. But Alice can change her setting in-flight. So how could the photons already be polarized in the direction that Alice will choose?

Can you give more specific description of what are you talking about it the last two sentences?

 

[Nugatory]

but if you say that, you also have to say that photon B's polarization is determined in part by the angle at which you choose to measure photon A's polarization

I just don't see how the bold part follows and what is the reasoning behind it.

When two photons are entangled in this way, their polarizations will always be perpendicular to one another. Suppose that while the two photons are in flight, I choose a particular orientation $\Theta$ for the left-hand polarizer, and the left-hand photon passes through. No matter what orientation I choose for the right-hand polarizer, the right-hand photon will pass or not according to Malus's law for a photon polarized at right angles to $\Theta$.

That's both the quantum mechanical prediction and the experimental result... But if you look at the way that the right-hand photon obeys Malus's law and conclude that the right-hand photon is in fact polarized at right-angles to $\Theta$ you've just concluded the part in bold.

 

[stevendaryl]

According to Malus it's 45 degrees relative angle that gives 50% chance: cos^2(45) = 50%, so if both photons have the same 45 degrees polarization relative to their polarizer, or at least same on average, then overall they will both have the same 50% chance to pass through. Where is this different than what actually happens?

If you created two photons polarized at angle 0 degrees, and Bob and Alice both had their filters set at 45 degrees, then:

1. 25% of the time, Alice and Bob would both see photons pass through their filters.
2. 25% of the time, Alice would see a photon pass through, but not Bob.
3. 25% of the time, Bob would see a photon pass through, but not Alice.
4. 25% of the time, neither would see a photon pass through.

But in actual experiments, possibilities 2 and 3 never happen.

 

[Jabbu]

When two photons are entangled in this way, their polarizations will always be perpendicular to one another. Suppose that while the two photons are in flight, I choose a particular orientation $\Theta$ for the left-hand polarizer, and the left-hand photon passes through. No matter what orientation I choose for the right-hand polarizer, the right-hand photon will pass or not according to Malus's law for a photon polarized at right angles to $\Theta$.

That's both the quantum mechanical prediction and the experimental result... But if you look at the way that the right-hand photon obeys Malus's law and conclude that the right-hand photon is in fact polarized at right-angles to $\Theta$ you've just concluded the part in bold.

I don't see what difference does it make whether polarizer angle is set yesterday or just before some photon is supposed to arrive. Is probability for the photon to pass through not always equal to cos^2(theta), where theta is relative angle between photon polarization and polarizer rotation angle that happens to be in the moment in time when, and where, they interact?

 

[stevendaryl]

If you created two photons polarized at angle 0 degrees, and Bob and Alice both had their filters set at 45 degrees, then:

1. 25% of the time, Alice and Bob would both see photons pass through their filters.
2. 25% of the time, Alice would see a photon pass through, but not Bob.
3. 25% of the time, Bob would see a photon pass through, but not Alice.
4. 25% of the time, neither would see a photon pass through.

But in actual experiments, possibilities 2 and 3 never happen.

I should say: in actual experiments involving entangled pairs of photons, possibilities 2 and 3 never happen. If you create unentangled photons, then it works the same way as classically.

 

[stevendaryl]

I don't see what difference does it make whether polarizer angle is set yesterday or just before some photon is supposed to arrive. Is probability for the photon to pass through not always equal to cos^2(theta), where theta is relative angle between photon polarization and polarizer rotation angle that happens to be in the moment in time when, and where, they interact?

Yes, there are two photons, one interacting with Alice's filter, and the other interacting with Bob's filter. The important angle is the angle between Alice's and Bob's filters at the time the two photons interact with the filters. So $\theta$ is not known at the beginning of the experiment (if Alice and Bob change their filters in-flight).

 

[DrChinese]

Aren't photons emitted with some specific polarization, both same or opposite?

Not, not if they are polarization entangled. Entangled photons are in a superposition of states.

As mentioned by stevendaryl and Nugatory, if you measure Alice and Bob at the same angle setting - regardless of what it is - they will match (or mismatch depending on the type of entanglement as Type I or Type II) 100% of the time. Photons that are not entangled will not display this characteristic.

Have you read the EPR and Bell arguments? If you haven't seen those, I would strongly recommend that you do.

 

[DrChinese]

I don't see what difference does it make whether polarizer angle is set yesterday or just before some photon is supposed to arrive. Is probability for the photon to pass through not always equal to cos^2(theta), where theta is relative angle between photon polarization and polarizer rotation angle that happens to be in the moment in time when, and where, they interact?

There are 2 basic ideas usually advanced to provide a classical explanation to entangled particles.

1. The photons are polarized as per usual, but the angle is unknown. As explained by stevendaryl and Nugatory, this argument fails immediately because you do not get the "perfect correlations" which are common to entangled particles (when each is measured at the same angle). Instead of 100% mismatches, you would get more like 50%.

2. The photons are polarized via so-called "hidden variables" that give the answer to the polarization "question" at all angles. This was the EPR hypothesis (1935). It seems reasonable. For example, it might yield + at 0 degrees, + at 120 degrees, and - at 240 degrees. And so on for any angle. On the average, it would reproduce Malus.

But Bell discovered a fatal flaw in this scheme in 1964. Read about Bell's Theorem to see why.

 

[Jabbu]

If you created two photons polarized at angle 0 degrees, and Bob and Alice both had their filters set at 45 degrees, then:

1. 25% of the time, Alice and Bob would both see photons pass through their filters.
2. 25% of the time, Alice would see a photon pass through, but not Bob.
3. 25% of the time, Bob would see a photon pass through, but not Alice.
4. 25% of the time, neither would see a photon pass through.

I should say: in actual experiments involving entangled pairs of photons, possibilities 2 and 3 never happen. If you create unentangled photons, then it works the same way as classically.

Are you saying there will be 100% matching pairs with 45 degrees? Don't both QM and Malus's law say it's cos^2(45) = 50% correlation?

 

[Jabbu]

Yes, there are two photons, one interacting with Alice's filter, and the other interacting with Bob's filter. The important angle is the angle between Alice's and Bob's filters at the time the two photons interact with the filters. So $\theta$ is not known at the beginning of the experiment (if Alice and Bob change their filters in-flight).

The experiments I was reading about didn't need to randomly shuffle angles and they still confirmed QM prediction. I don't see what's the point of involving even more randomness. The only question I'm asking is how QM prediction differs from Malus's law prediction, especially since both equations seem to be the same.

 

[Jabbu]

Not, not if they are polarization entangled. Entangled photons are in a superposition of states.

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?

Have you read the EPR and Bell arguments? If you haven't seen those, I would strongly recommend that you do.

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

 

[Nugatory]

I don't see what difference does it make whether polarizer angle is set yesterday or just before some photon is supposed to arrive. Is probability for the photon to pass through not always equal to cos^2(theta), where theta is relative angle between photon polarization and polarizer rotation angle that happens to be in the moment in time when, and where, they interact?

Yes, but what is the photon polarization? Somehow, it always to turns out to be exactly ninety degrees off of the angle that the other polarizer, the one we used to measure the other photon.

Let's assume that the photons were created with the left-hand photon polarized at zero degrees and the right-hand photon polarized at ninety degrees, and then while the photons are in flight we set the left-hand polarizer to 45 degrees. Fifty percent of the time the left-hand photon will clear its by polarizer (Malus's law).

Now, what is the probability that the right-hand photon will clear a polarizer set at 135 degrees? 90 degrees? 45 degrees? The observed results and the quantum mechanical prediction are 100%, 50%, and 0% for those three values - and that's what you get by applying Malus's law to a photon polarized at 135 degrees, not the 90 degrees that we assumed.

If we had set the left-hand polarizer to 30 degrees, then 75% of the time the left-hand photon would have cleared the polarizer (Malus's Law again) and in those cases the right-hand photon would obey Malus's law for a 120-degree photon.

It's as if measuring the left-hand photon at a given angle changes the polarization of its right-hand partner to that angle plus ninety degrees. That's the unique feature of entanglement.

 

[Nugatory]

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 an example of theta equal to 60. Other examples would be A=0 and B=60, A=11 and B=71, A=44 and B=-16... All that matters is that there is sixty degrees between them.

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.

 

[stevendaryl]

Are you saying there will be 100% matching pairs with 45 degrees? Don't both QM and Malus's law say it's cos^2(45) = 50% correlation?

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

 

[stevendaryl]

The experiments I was reading about didn't need to randomly shuffle angles and they still confirmed QM prediction. I don't see what's the point of involving even more randomness. The only question I'm asking is how QM prediction differs from Malus's law prediction, especially since both equations seem to be the same.

No, they are not the same. If you have light that is polarized at 0 degrees, and you have two filters, Alice's and Bob's, that are oriented at 45 degrees, then 50% of the light will pass through Alice's filter, and 50% will pass through Bob's. That's true both classically (Malus' law) and quantum mechanically. But now lower the intensity so low that you see individual photons. You will find that randomly, half the photons reaching Alice's filter pass through, and half the photons reaching Bob's filter pass through. But there is no correlation between the two. Sometimes both Bob and Alice will see a photon pass. Sometimes neither. Sometimes one and not the other. That's the prediction for unentangled photons.

But now, if the photons are entangled (that is, Alice's photon and Bob's photons are both produced in an atomic decay so that they are correlated), then you see Alice's results correlated with Bob's results. If they both have their filters at the same setting, then they will get the same results: either both will see the photon pass, or neither will see it pass. This is not a consequence of Malus' law. Malus' law doesn't say anything about the correlation between Alice's and Bob's result. (Malus' law is about light intensity, not about numbers of individual photons).

 

[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".