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

[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.

 

[DevilsAvocado]

...didn’t we run this debate previously – about the 'detection' loophole...??

Anyhow for myself and any other layman out there, let’s reconcile:

http://en.wikipedia.org/wiki/Malus'_law#Malus.27_law_and_other_properties"
Malus' law, which is named after Etienne-Louis Malus, says that when a perfect polarizer is placed in a polarized beam of light, the intensity, I, of the light that passes through is given by ... (yada, yada, yada) ... In practice, some light is lost in the polarizer and the actual transmission of unpolarized light will be somewhat lower than this, around 38% for Polaroid-type polarizers but considerably higher (>49.9%) for some birefringent prism types.

Who is Etienne-Louis Malus? Well, he’s this guy:

A participant in Napoleon's expedition into Egypt (1798 to 1801)​

Can we start a poll... if this Napoleon-guy is going to win the battle between QM and ... and ... the Waterloophole Theory Fernwirkung (!?WTF!?) ...

 

[Frame Dragger]

...didn’t we run this debate previously – about the 'detection' loophole...??

Anyhow for myself and any other layman out there, let’s reconcile:

Who is Etienne-Louis Malus? Well, he’s this guy:

A participant in Napoleon's expedition into Egypt (1798 to 1801)​

Can we start a poll... if this Napoleon-guy is going to win the battle between QM and ... and ... the Waterloophole Theory Fernwirkung (!?WTF!?) ...

Oooh... for a guy who's said that English is not your first language, you have a keen sense of wielding it for the sake of humour and making a point. I think yours was a fairly... direct way of explaining the "apples and oranges" concept to ThomasT. Malus' Law is certainly useful (sort of)... it just has nothing at all to do with the issue at hand!

 

[DevilsAvocado]

Hehe, of course Etienne-Louis Malus is completely innocent – he’s just a victim to his "apples" being used to prove that "oranges" do not exist, by "someone"...

English is not your first language

Correct, but the "Swedish Chef" has taught me almost everything there is to know!

Thanks!

 

[SpectraCat]

..

Correct, but the "Swedish Chef" has taught me almost everything there is to know!

Bork Bork Bork!

At his finest:

 

[DevilsAvocado]

Bork Bork Bork!

Hahaha! "After that – I’m running away!"


(but I’ll be back to reply the rest ASAP)