Is action at a distance possible as envisaged by the EPR Paradox.

[DrChinese]

...
Malus Law applies in this situation. We can denote the individual detection rates as,

P(A) = cos²(|a - L|) and
P(B) = cos²(|b - Ll|)

where a and b are polarizer settings and L is the polarization angle of the optical disturbances incident on a and b.

Since the average angular difference between the polarizer setting and L is 45^o, then the expected normalized individual detection rates are,

P(A) = .5 and
P(B) = .5

which agrees with QM prediction and experiment.

For the joint detection situation Malus Law also applies since we have crossed polarizers analyzing identically polarized optical emissions.

The relevant independent variable is the angular difference of the polarizer settings, |a-b|, which can be expressed as (||a-L| - |b-L||).

So, we can denoted the joint detection rate as

P(A,B) = cos²(||a-L| - |b-L||)

which agrees with QM prediction and experiment.

Sorry, this most definitely does NOT agree with experiment and the math is wrong. In fact the coincidence rate varies between .25 and .75 per your example. Experiment has it varying between 0 and 1.

See where your P(A)=.5 and p(B)=.5? That is correct. But it does not lead to your result. For example, where A=0 and B=0, you will NOT get correlation of 100% UNLESS you have entangled photons. Unentangled photons would have the same math as you describe but yield Product statistics. They are NOT the same. Yet they should be, according to you.

This is a frequent mistake that folks make in attempting to come up with local hidden variable models. You must model BOTH of these following cases successfully:

a) Polarization entangled photon pairs yield perfect correlations and entangled state statistics (the cos^2(theta) rule);

b) Polarization UNentangled photon pairs - coming from the same PDC crystal as in a) - yield Product State statistics (.25 + .5(cos^2(theta)). With these, the polarization is known, which is why they are not entangled.

Your example models b) and not a). If I need to, I will be glad to derive this for you in a later post. Or you can do it, but you must put in the correct values for the expansion.

 

[ThomasT]

Sorry, this most definitely does NOT agree with experiment and the math is wrong. In fact the coincidence rate varies between .25 and .75 per your example.

I think you've got it backwards.

As the angular difference, |a - b| = ||a - L| - |b - L||, goes from 0^o to 90^o, the cos² of that angular difference goes from 1 to 0.

Experiment has it varying between 0 and 1.

The experiments I've seen have it varying between 0 and .5, and it never quite reaches those limits. It can't go to 1 if the individual detection rates never exceed .5 .

The correct expression for an idealized setup, normalized joint detection rate is, then,

.5 (cos²(|a-b|) ,

which would be modified depending on whatever inefficiencies are involved in the actual production of the data.

By the way, the normalization is wrt the detection rate with no polarizers present.

See where your P(A)=.5 and p(B)=.5? That is correct. But it does not lead to your result.

It can't. Isn't that what we've been discussing?

For example, where A=0 and B=0, you will NOT get correlation of 100% UNLESS you have entangled photons. Unentangled photons would have the same math as you describe but yield Product statistics. They are NOT the same. Yet they should be, according to you.

If you mean A and B to refer to detection rates, then the normalized individual rates with polarizers in place at both ends is always, in the ideal, A = .5 and B = .5 . If not, then I don't know what you're saying.

a) Polarization entangled photon pairs yield perfect correlations and entangled state statistics (the cos^2(theta) rule);

b) Polarization UNentangled photon pairs - coming from the same PDC crystal as in a) - yield Product State statistics (.25 + .5(cos^2(theta)). With these, the polarization is known, which is why they are not entangled.

Your example models b) and not a).

It accounts for a) in the assumptions underlying the application of Malus Law and the model(s) of the production of the entanglement via the emission process. Keep in mind that the perfect correlation/anticorrelation corresponding to angular differences of 0^o and 90^o have always been explainable assuming a local common cause for the entanglement.

 

[RUTA]

AFAICS, RoS is a non-issue for this case.

Correct, RoS doesn't bear on nonseparability. It does, of course, bear on the nonlocality issue, as I pointed out in post #140.

There is no question of causality here, only comparison of measurement. There is no logical inconsistency in a frame where B was measured before A or vice-versa, because the results are perfectly correlated.

QM predicts the correlations between space-like separated experimental outcomes that violate Bell inequalities, that's not the issue. The question QM doesn't answer is, "What is the nature of reality such that correlations between space-like separated experimental outcomes violate Bell inequalities?" For some people, the answer is FTL communication between the measurement events. For those people, RoS becomes an issue.

IOW, if you measure B first, you determine the value of A, if you measure A first, you determine the value of B. This holds in all frames, for all observers, so IMO there is no problem.

The problem with "if you measure B first, you determine the value of A" when A and B are space-like separated is that, per RoS, there is no absolute temporal ordering of A and B. So, what does "if you measure B first" really mean?

Furthermore, when observers in different frames compare answers (as is required for a Bell test), they may disagree on the ordering of events, but they will always agree that there is a Bell violation for the results.

Correct.

 

[RUTA]

To reiterate, from a QM point of view, there are two measurements performed on the members of an entangled pair. Since the results are always perfectly correlated, it fundamentally does not matter which measurement comes first, at least for the purposes of Bell tests. All observers in all frames agree on the results of the measurements, once they have communicated them by normal sub-lightspeed channels for comparison.

The strange nature of the correlated outcomes is not illucidated by the measurements for which there is 100% correlation alone. One can imagine there is a "fact of the matter" for the outcomes corresponding to those measurement settings (like settings). But, if you assume there is a "fact of the matter" for the outcomes corresponding to like settings (in order to account for the 100% correlation), then you don't obtain the correct (QM predicted) correlation rate for outcomes in unlike settings. Thus, some people invoke FTL communication of settings between measurements so the particles can "give the right outcomes" no matter how they're measured.

 

[DrChinese]

I think you've got it backwards.

As the angular difference, |a - b| = ||a - L| - |b - L||, goes from 0^o to 90^o, the cos² of that angular difference goes from 1 to 0.

The experiments I've seen have it varying between 0 and .5, and it never quite reaches those limits. It can't go to 1 if the individual detection rates never exceed .5 .

The correct expression for an idealized setup, normalized joint detection rate is, then,

.5 (cos²(|a-b|) ,

which would be modified depending on whatever inefficiencies are involved in the actual production of the data.

By the way, the normalization is wrt the detection rate with no polarizers present.

It can't. Isn't that what we've been discussing?

If you mean A and B to refer to detection rates, then the normalized individual rates with polarizers in place at both ends is always, in the ideal, A = .5 and B = .5 . If not, then I don't know what you're saying.

No. It models a). Keep in mind that the perfect correlation/anticorrelation corresponding to angular differences of 0^o and 90^o have always been explainable assuming a local common cause for the entanglement.

OK, first let's agree that there are 2 types of PDC. We can discuss either one, but I personally think it is easier to discuss Type I. With Type I, both photons have identical polarization. With Type II, the photons have anti-symmetric/orthogonal/crossed/opposite polarization.

It makes no difference to the result whether we are going from 0 to 1 or 1 to 0, when moving from 0 to 90 degrees apart. So I will use the Type I example.

Second: the joint detection rate is NOT cos^2(theta) in your example. It is actually .25+.5(cos^2(theta)) which is obviously different. This is the Product State statistics.

Remember: we are talking about your example. IF you take a pair of PDC photons at any KNOWN angle (the hypothesized L in your example) and actually check their polarizations at any angle settings A and B, THEN you will get .25+.5(cos^2(A-B)). That is an experimental fact.

What may be confusing is that PDC photon pairs can come out either polarization entangled or not. Either way, they have identical polarization! If the polarization is known, it is not entangled. You hypothesize that there is a definite polarization L associated with the photons. If there were such, then the expected statistics are PRODUCT state.

If, on the other hand, you specify that there is no definite polarization L, then you would have entanglement. But of course, then you would be accepting that no LHV can reproduce the predictions of QM - exactly what you wish to avoid. Of course, everyone else (save a diehard few) already accepts this.

 

[DrChinese]

Keep in mind that the perfect correlation/anticorrelation corresponding to angular differences of 0^o and 90^o have always been explainable assuming a local common cause for the entanglement.

Yes and No. YES: there have been models that could explain some situations such as these special cases. But NO: your model is NOT one of those. Those models are different. In effect, they postulate that there are a large (and perhaps infinite) number of hidden variables associated with the range of polarization settings. For simplicity, imagine that there is: HV(1), HV(2),... HV(359, HV(360).

Of course, this model "fixes" the problem with your model. But at a cost. Because now you just fell prey to Bell and the problem of being able to provide a realistic resultset for the 3 angles I specified (0/120/240 a la Mermin). You cannot do it - but please feel free to try!

 

[RUTA]

Ok, let's use a two photon setup where polarization entanglement is produced when two photons are emitted in opposite directions by the same atom. Due to conservation of angular momentum, they're polarized identically.

For the purpose of discussion we can assume an ideal setup (perfect efficiency, all loopholes closed).

The hidden variable is the polarization angle, and it's the same for each member of any pair of entangled photons (though it varies randomly from pair to pair).

Malus Law applies in this situation. We can denote the individual detection rates as,

P(A) = cos²(|a - L|) and
P(B) = cos²(|b - Ll|)

where a and b are polarizer settings and L is the polarization angle of the optical disturbances incident on a and b.

Since the average angular difference between the polarizer setting and L is 45^o, then the expected normalized individual detection rates are,

P(A) = .5 and
P(B) = .5

which agrees with QM prediction and experiment.

For the joint detection situation Malus Law also applies since we have crossed polarizers analyzing identically polarized optical emissions.

The relevant independent variable is the angular difference of the polarizer settings, |a-b|, which can be expressed as (||a-L| - |b-L||).

So, we can denoted the joint detection rate as

P(A,B) = cos²(||a-L| - |b-L||)

which agrees with QM prediction and experiment.

Even though this expression for the joint expectation, per se, doesn't explicate locality (as, say, Bell's ansatz purports to but actually doesn't), it is nonetheless a local hidden variable account of entanglement insofar as it (1) incorporates the hidden variable, and (2) the assumptions underlying it are in accord with the principle of local causality.

I don't see how you obtained your P(A,B) other than by fiat.

Suppose you say the photons pass a polarizer (vertically polarized wrt the setting, typically denoted V) if their polarization L is within 45 deg of the setting. Between 45 deg and 90 deg the photons are blocked (horizontally polarized wrt the setting, typically denoted H). This is a reasonable assumption and leads to an overall 50% rate for V at each detector. Now, what is the probability of a VV outcome for settings A and B? The answer is 0.5 - |A - B|/pi, which you can obtain by simply drawing the 45-deg cones about settings A and B and looking at their overlap. For a detailed explanation, see equation 19, p 907, of section VIII. Local Realistic Hidden Variable Theory, in "Entangled photons, nonlocality, and Bell inequalities in the undergraduate laboratory," Dietrich Dehlinger and M.W. Mitchell, Am. J. Phys. 70 (9), Sep 2002.

 

[ThomasT]

OK, first let's agree that there are 2 types of PDC. We can discuss either one, but I personally think it is easier to discuss Type I. With Type I, both photons have identical polarization. With Type II, the photons have anti-symmetric/orthogonal/crossed/opposite polarization.

It makes no difference to the result whether we are going from 0 to 1 or 1 to 0, when moving from 0 to 90 degrees apart. So I will use the Type I example.

Second: the joint detection rate is NOT cos^2(theta) in your example. It is actually .25+.5(cos^2(theta)) which is obviously different. This is the Product State statistics.

Remember: we are talking about your example. IF you take a pair of PDC photons at any KNOWN angle (the hypothesized L in your example) and actually check their polarizations at any angle settings A and B, THEN you will get .25+.5(cos^2(A-B)). That is an experimental fact.

What may be confusing is that PDC photon pairs can come out either polarization entangled or not. Either way, they have identical polarization! If the polarization is known, it is not entangled. You hypothesize that there is a definite polarization L associated with the photons. If there were such, then the expected statistics are PRODUCT state.

If, on the other hand, you specify that there is no definite polarization L, then you would have entanglement. But of course, then you would be accepting that no LHV can reproduce the predictions of QM - exactly what you wish to avoid. Of course, everyone else (save a diehard few) already accepts this.

If you reread the post where I introduced this you'll see that I wasn't talking about SPDC photons.

The counter-propagating photons emitted by the same atom in my example are always entangled in polarization due to conservation of angular momentum. This entanglement means that members of an entangled pair are polarized identically. However, the value of L, the polarization angle of any given pair, is varying randomly.

 

[ThomasT]

I don't see how you obtained your P(A,B) other than by fiat.

Not by fiat. Malus Law is applied and the angular difference is simply rendered in terms of the hidden variable.

 

[RUTA]

Not by fiat. Malus Law is applied and the angular difference is simply rendered in terms of the hidden variable.

Why did you apply Malus Law?

 

[DrChinese]

If you reread the post where I introduced this you'll see that I wasn't talking about SPDC photons.

The counter-propagating photons emitted by the same atom in my example are always entangled in polarization due to conservation of angular momentum. This entanglement means that members of an entangled pair are polarized identically. However, the value of L, the polarization angle of any given pair, is varying randomly.

Sorry, you are making an important mistake here. Yes, it is true that the photons you describe from the atom are entangled. However, the model you describe is NOT the same. Instead, it matches the PDC polarization unentangled situation I described above. You cannot say your model works if you apply it to the wrong situation. There is a GIANT different in Entangled State stats and Product State stats. Your example - where there is a definite polarization L - only matches the Product State situation. This is a very important distinction and you need to understand this. It is probably the reason you have had trouble seeing some of the arguments we have provided in the past.

 

[ThomasT]

Why did you apply Malus Law?

Wrt each trial, two identically polarized optical disturbances are being analyzed by crossed linear polarizers.

Why is Malus Law applied in the QM treatment?

 

[DrChinese]

Not by fiat. Malus Law is applied and the angular difference is simply rendered in terms of the hidden variable.

As RUTA is also trying to tell you: Application of Malus as you are trying will NOT yield Entangled State stats. Please note that it is true that Malus is a cos^2 function, and so are the Entangled State statistics. But how you apply these are different for different experimental situations.

a) Application of Malus to (A-L) and (B-L) does NOT lead to Malus for (A-B) as you imagine. It leads to different stats, as I have already told you.

b) The reason you apply Malus to entangled pairs is because of the superposition of states: HH> + VV>. When you then apply rotation to the superposition, rotating by some A or B, you end up with an expression that simplifies to cos^2(A-B).

The Dehlinger reference derives this.

 

[ThomasT]

Suppose you say the photons pass a polarizer (vertically polarized wrt the setting, typically denoted V) if their polarization L is within 45 deg of the setting. Between 45 deg and 90 deg the photons are blocked (horizontally polarized wrt the setting, typically denoted H). This is a reasonable assumption and leads to an overall 50% rate for V at each detector. Now, what is the probability of a VV outcome for settings A and B? The answer is 0.5 - |A - B|/pi, which you can obtain by simply drawing the 45-deg cones about settings A and B and looking at their overlap. For a detailed explanation, see equation 19, p 907, of section VIII. Local Realistic Hidden Variable Theory, in "Entangled photons, nonlocality, and Bell inequalities in the undergraduate laboratory," Dietrich Dehlinger and M.W. Mitchell, Am. J. Phys. 70 (9), Sep 2002.

All we have are averages. Photon counts per run are accounted for. There are no probabilities for the results of individual trials.

 

[ThomasT]

Yes and No. YES: there have been models that could explain some situations such as these special cases. But NO: your model is NOT one of those. Those models are different. In effect, they postulate that there are a large (and perhaps infinite) number of hidden variables associated with the range of polarization settings. For simplicity, imagine that there is: HV(1), HV(2),... HV(359, HV(360).

Of course, this model "fixes" the problem with your model. But at a cost. Because now you just fell prey to Bell and the problem of being able to provide a realistic resultset for the 3 angles I specified (0/120/240 a la Mermin). You cannot do it - but please feel free to try!

The perfect correlation/anticorrelation corresponding to angular differences of 0^o and 90^o are accounted for by the identical polarizations of the members of each entangled pair.

 

[DrChinese]

Wrt each trial, two identically polarized optical disturbances are being analyzed by crossed linear polarizers.

Why is Malus Law applied in the QM treatment?

See the Dehlinger paper. Although it really doesn't matter for your particular model, because yours gives completely different results than experiment even for the perfect correlations cases.

On the other hand, Dehlinger describes a DIFFERENT LHV than yours - as an example - and shows how it falls apart. But his works for the perfect correlation cases, which is more or less what EPR envisioned.

 

[DrChinese]

The perfect correlation/anticorrelation corresponding to angular differences of 0^o and 90^o are accounted for by the identical polarizations of the members of each entangled pair.

No, they are not. Try it and you will see.

L=30 degrees
A=0 degrees
B=90 degrees

cos^2(A-L) * cos^2(B-L) = [Not 0 or 1 as QM predicts]

 

[ThomasT]

As RUTA is also trying to tell you: Application of Malus as you are trying will NOT yield Entangled State stats. Please note that it is true that Malus is a cos^2 function, and so are the Entangled State statistics. But how you apply these are different for different experimental situations.

Malus Law applies because of the physical setup.

Application of Malus to (A-L) and (B-L) does NOT lead to Malus for (A-B) as you imagine.

Malus Law applies in the individual trials because we have an optical disturbance with an unknown polarization L being analyzed by a linear polarizer with a certain setting. The interaction of the incident disturbance with the polarizer yields a resultant disturbance with intensity proportional to cos²|a - L|.

I've already explained the physical reasoning behind the application of Malus Law applies in the joint context.

 

[DrChinese]

Malus Law applies because of the physical setup.

Malus Law applies in the individual trials because we have an optical disturbance with an unknown polarization L being analyzed by a linear polarizer with a certain setting. The interaction of the incident disturbance with the polarizer yields a resultant disturbance with intensity proportional to cos²|a - L|.

I've already explained the physical reasoning behind the application of Malus Law applies in the joint context.

You may as well say dogs quack*. Malus does not apply to your setup. Please work through the example I gave above and you will immediately see that the math fails. Sorry, there is nothing gray about your example. If you apply Malus to A-L and to B-L, you don't also get Malus for A-B.

*Although I should add that my dog is so weird, she may as well quack.

 

[RUTA]

All we have are averages. Photon counts per run are accounted for. There are no probabilities for the results of individual trials.

I derived the probability per trial using a particular lhv model. The assumption that such a probability will correspond to the frequencies of outcomes in actual experiments is a particular (and still debated in philosophical circles) interpretation of the meaning of "probability." In physics, we take this for granted and it works, so ... .

DrC has answered your other questions. I suggest you read the AJP paper I referenced. Work through all the calculations to make sure you understand them. If you need help, send me your questions. I have my intro QM students supply all the missing calcs in that paper as an exercise. Caveat: There are some typos in his equations but if you understand what he's doing, you'll catch those easy enough.

 

[Frame Dragger]

ThomasT, what is your fascination with Malus' Law? This is sounding more and more like a pet theory or an article of faith.

 

[DrChinese]

If you need help, send me your questions. I have my intro QM students supply all the missing calcs in that paper as an exercise. Caveat: There are some typos in his equations but if you understand what he's doing, you'll catch those easy enough.

A very kind offer.

I hope I remember this thread the next time ThomasT brings this subject up. Now I understand better where he is coming from.

 

[ThomasT]

No, they are not. Try it and you will see.

L=30 degrees
A=0 degrees
B=90 degrees

cos^2(A-L) * cos^2(B-L) = [Not 0 or 1 as QM predicts]

You're evaluating the expression incorrectly. But, yes, there's still a problem. Maybe |a - b| can't be expressed in terms of the hidden variable or maybe there's some simple fix. In either case, cos²|a - b| does give the correct result for all values of a, b and L.

Since this is an expression of Malus Law, then the question is: does Malus Law actually apply simply because we're analyzing identically polarized optical disturbances with crossed linear polarizers?

 

[DrChinese]

You're evaluating the expression incorrectly.

Ok, say we have:

A=0
B=0
L=45 degrees

p(A, heads) = cos^2(A-L) = .5
p(B, heads) = cos^2(A-L) = .5
p(A and B, heads) = .25

p(A, tails) = cos^2(A-L) = .5
p(B, tails) = cos^2(A-L) = .5
p(A and B, tails) = .25

Per your model, the matches will be .25 + .25 or 50% of the time.

But entangled A and B give both as matches 100% of the time.

 

[ThomasT]

You may as well say dogs quack*. Malus does not apply to your setup.

Actually, it applies whenever you're analyzing polarization.

If you apply Malus to A-L and to B-L, you don't also get Malus for A-B.

The application of Malus Law gives the correct results in the individual as well as joint contexts. Now, can we express the angular difference in terms of the hidden variable? If we can then we have a local hidden variable account. If not then we just have a local account.

 

[DrChinese]

Actually, it applies whenever you're analyzing polarization.

The application of Malus Law gives the correct results in the individual as well as joint contexts. Now, can we express the angular difference in terms of the hidden variable? If we can then we have a local hidden variable account. If not then we just have a local account.

If you are not going to do any work on this issue yourself, you won't see me continuing to try to assist you. Work the math yourself following my example and read the Dehlinger paper too.

 

[RUTA]

Actually, it applies whenever you're analyzing polarization.
The application of Malus Law gives the correct results in the individual as well as joint contexts.

I still don't understand where you're getting this result because you're not providing a state or a physical mechanism. You're simply claiming that Malus Law applies to the analysis of polarization experiments, but that's not true in general. The correlation outcome is state dependent, so the result you're calling Malus Law (.5cos^2(A - B)) is only true for a specific QM situation. You're claiming to obtain this formula without specifying the context, so there are certainly many experiments that would not agree with your prediction. For example, in Dehlinger and Mitchell's experimental set up they find the probability of a VV outcome for settings A and B is:

sin^2(A)*sin^2(B)*cos^2(theta) + cos^2(A)*cos^2(B)*sin^2(theta) + .25*sin(2A)*sin(2B)*sin(2theta)*cos(phi).

They don't get the simple .5cos^2(A - B) because their equipment doesn't produce the state (|HH> + |VV>)/sqrt(2). Instead, their equipment produces the state

cos(theta)*|HH> + exp[i*phi]*sin(theta)*|VV>.

That's why I explained the lhv equation (section VIII of their paper) in such detail. It's not true that Malus Law applies to all polarization experiments. You have to derive what is true and sometimes it is Malus Law, but you can't simply posit Malus Law as providing the correlation rate in all polarization experiments, b/c it's not true in general.

 

[DrChinese]

That's why I explained the lhv equation (section VIII of their paper) in such detail. It's not true that Malus Law applies to all polarization experiments. You have to derive what is true and sometimes it is Malus Law, but you can't simply posit Malus Law as providing the correlation rate in all polarization experiments, b/c it's not true in general.

Yes, in fact I was tripped up sadly on that once. For example, suppose you have 3 polarization entangled photons. They do not follow the cos^2(theta) rule. Tez had to correct me on that one.

 

[ThomasT]

Ok, say we have:

A=0
B=0
L=45 degrees

p(A, heads) = cos^2(A-L) = .5
p(B, heads) = cos^2(A-L) = .5
p(A and B, heads) = .25

p(A, tails) = cos^2(A-L) = .5
p(B, tails) = cos^2(A-L) = .5
p(A and B, tails) = .25

Per your model, the matches will be .25 + .25 or 50% of the time.

No. Wrt the setup I described there are no tails. Wrt my account (I wouldn't call it a model, per se) P(A,B) = cos²(|a - b|).

The problem is in expressing |a - b| in terms of the hidden variable. It might not be possible. But even if not, it's still a local account due to my Malus Law rationalization.

 

[DrChinese]

No. Wrt the setup I described there are no tails. Wrt my account (I wouldn't call it a model, per se) P(A,B) = cos²(|a - b|).

What do you mean there are no tails? All actual experiments give you a 1/0, Y/N, Heads/tails boolean result.

And if you are not putting forth a model, then what are you arguing? The whole point of this discussion is to convince you that you CANNOT construct such a model.

 

[ThomasT]

What do you mean there are no tails? All actual experiments give you a 1/0, Y/N, Heads/tails boolean result.

We're counting photons, detections. Not nondetections.

And if you are not putting forth a model, then what are you arguing? The whole point of this discussion is to convince you that you CANNOT construct such a model.

It's only a model if |a - b| can be expressed in a way that includes L without contradiction. I think there might be a way to do it.

 

[DrChinese]

We're counting photons, detections. Not nondetections.

Most Bell tests no longer use that technique because it is inferior. By using a polarizing beamsplitter (PBS) with separate detectors for both outputs, you get a more definite statement.

Obviously, there is a way to adjust the counts to match your setup. And you still get the wrong answer.

 

[ThomasT]

I still don't understand where you're getting this result because you're not providing a state or a physical mechanism. You're simply claiming that Malus Law applies to the analysis of polarization experiments, but that's not true in general. The correlation outcome is state dependent, so the result you're calling Malus Law (.5cos^2(A - B)) is only true for a specific QM situation. You're claiming to obtain this formula without specifying the context, so there are certainly many experiments that would not agree with your prediction. For example, in Dehlinger and Mitchell's experimental set up they find the probability of a VV outcome for settings A and B is:

sin^2(A)*sin^2(B)*cos^2(theta) + cos^2(A)*cos^2(B)*sin^2(theta) + .25*sin(2A)*sin(2B)*sin(2theta)*cos(phi).

They don't get the simple .5cos^2(A - B) because their equipment doesn't produce the state (|HH> + |VV>)/sqrt(2). Instead, their equipment produces the state

cos(theta)*|HH> + exp[i*phi]*sin(theta)*|VV>.

That's why I explained the lhv equation (section VIII of their paper) in such detail. It's not true that Malus Law applies to all polarization experiments. You have to derive what is true and sometimes it is Malus Law, but you can't simply posit Malus Law as providing the correlation rate in all polarization experiments, b/c it's not true in general.

That's why I used an experimental setup where Malus Law does clearly apply.

 

[RUTA]

That's why I used an experimental setup where Malus Law does clearly apply.

I haven't seen you derive your coincidence rate. If you've done that, please tell me the post number. If not, please derive it now.

 

[Frame Dragger]

Most Bell tests no longer use that technique because it is inferior. By using a polarizing beamsplitter (PBS) with separate detectors for both outputs, you get a more definite statement.

Obviously, there is a way to adjust the counts to match your setup. And you still get the wrong answer.

When is it alright to accuse someone of not knowing what the hell they're talking about?! I know that this is a civilized forum, but come on...

@ThomasT: Show some work, that WORKS, and until then stop taking the painfully long way around to finding that your assumptions are baseless. PLEASE.