Entanglement and teleportation

[DrChinese]

In any case, I don't think the fact that the cos^2 theta
formula works in the standard two-detector optical EPR/Bell
setup, and the fact that it's a 200 year old optics formula is
just a coincidence. (Remember all that stuff about
an "underlying unity to physics" above? :) )

There are definitely TWO ways to look at that statement. Some of the vocal local realists argue that the cos^2 law isn't correct! They do that so the Bell Inequality can be respected; and then explain that experimental loopholes account for the difference between observation and their theory.

Clearly, classical results sometimes match QM and sometimes don't; and when they don't, you really must side with the predictions of QM. Even Einstein saw that this was a steamroller he had to ride, and the best he could muster was that QM was incomplete.

 

[vanesch]

I said the following:

And if you accept THAT, you appreciate the EPR "riddle", and you do not explain it with the old cos^2 theta law, because that SAME cos^2 theta law would also give us SIMULTANEOUS HITS in A+ and A-, which we don't have. The EPR problem is only valid in the case where you do not have simultaneous
YES/NO answers, of course, otherwise you have, apart from a +z and a -z answer, also a (+z AND -z) answer, which changes Bell's ansatz.

and I would like to illustrate WHERE it changes Bell's ansatz.

Consider again 3 directions, a, b and c, for Alice and Bob.

Alice has an A+ and an A- detector, and Bob has a B+ and a B- detector.
Usually people talk only about the A+ hit or the "no-A+ hit" (where it is understood that the no-A+ hit is an A- hit).

We then take as hidden variable a bit for each a, b and c:

If we have a+ this means that Alice will have A+ and bob will have no B+ in the a direction, if we have a b+ that means that Alice will have an A+ and bob will have no B+ in the b direction, and ...

So we can have: a(+/-) b(+/-) c(+/-) as hidden state. But that description already includes the anti-correlation: if A+ triggers, then A- does NOT trigger, and if A- triggers, then A+ does not trigger. When A+ and A- do not trigger, that is then assumed to be due to the finite quantum efficiencies of the detector, which lead to the "fair sampling hypothesis".

But if we accept the possibility that A+ AND A- trigger together, then each direction has, besides the + and - possibility, a THIRD possibility namely X: double trigger. So from here on, we have 27 different possible states. This changes completely the "probability bookkeeping" and Bell's inequalities are bound to change. The local realist cloud even introduces a fourth possibility: A+ and A- do not trigger, and this is not due to some inefficiency, with symbol 0.

So we have a(+/-/X/0), b(+/-/X/0), c(+/-/X/0) which gives us 64 possibilities.
You can then easily show that Bell's inequalities are different and that experiments don't violate them.

The blow to this view is that whenever you make up a detector law as a function of intensity which allows you to consider the 0 case, you also have to consider the X case. The X case is never observed, so there are reasons to think that the 0 case doesn't exist either, especially because QED tells us so, and that you do get out the right results (including the observed number of 0 cases) when applying the quantum efficiency under the fair sampling hypothesis.

cheers,
Patrick.

 

[Sherlock]

That's sort of funny, you know. Application of classical optics' formula cos^2\theta is incompatible with hidden variables but consistent with experiment.

The cos^2 theta formula isn't incompatible with hidden
variables.

For the context of individual results you can write,

P = cos^2 |a - lambda|,

where P is the probability of detection, a is the
polarizer setting and lambda is the variable
angle of emission polarization.

This doesn't conflict with qm. If you knew
the value of lambda, or had any info about
how it was varying (other than just that
it's varying randomly), then you could more
accurately predict individual results (by
individual results I mean the data streams
at one end or the other).

How do we know that there *is* a hidden
variable operating in the individual measurement
context? Because, if you keep the polarizer
setting constant the data stream varies
randomly.

Now, this hidden variable doesn't just
stop existing because we decide to
combine the individual data streams wrt
joint polarizer settings.

However, the *variability* of lambda
isn't a factor wrt determining coincidental
detection.

a b and c are the hypothetical settings you could have IF local hidden variables existed. This is what Bell's Theorem is all about. The difference between any two is a theta. If there WERE a hidden variable function independent of the observations (called lambda collectively), then the third (unobserved) setting existed independently BY DEFINITION and has a non-negative probability.

Bell has nothing to do with explaining coincidences, timing intervals, etc. This is always a red herring with Bell. ALL theories predict coincidences, and most "contender" theories yield predictions quite close to Malus' Law anyway. The fact that there is perfect correlation at a particular theta is NOT evidence of non-local effects and never was. The fact that detections are triggered a certain way is likewise meaningless. It is the idea that Malus' Law leads to negative probabilities for certain cases is what Bell is about and that is where his selection of those cases and his inequality comes in.

Suppose we set polarizers at a=0 and b=67.5 degrees. For the a+b+ and a-b- cases, we call that correlation. The question is, was there a determinate value IF we could have measured at c=45 degrees? Because IF there was such a determinate value, THEN a+b+c- and a-b-c+ cases should have a non-negative likelihood (>=0). Instead, Malus' Law yields a prediction of about -10%. Therefore our assumption of the hypothetical c is wrong if Malus' Law (cos^2) is right.

Bell demonstrated that using the variability of lambda
to augment the qm formulation for coincidental
detection gives a result that is incompatible
with qm predictions for all values of theta
except 0, 45 and 90 degrees.

Now, there's at least two ways to interpret Bell's
analysis. Either (1) lambda suddenly stops existing when we
decide to combine individual results, or (2) the variability
of lambda isn't relevant wrt joint detection.

I think the latter makes more sense, and in fact
it's part of the basis for the qm account which
assumes that photons emitted by the same atom
are entangled in polarization via the emission
process. This is why you have an entangled
quantum state prior to detection. So, all you
need to know to accurately predict the
*coincidental* detection curve is the angular
difference between the polarizer settings. And,
as in all such situations where you're analyzing,
in effect, the same light with crossed linear
polarizers the cos^2 theta formula holds.

 

[vanesch]

The cos^2 theta formula isn't incompatible with hidden
variables.

For the context of individual results you can write,

P = cos^2 |a - lambda|,

where P is the probability of detection, a is the
polarizer setting and lambda is the variable
angle of emission polarization.

Ok, that's the probability for the A+ detector to trigger. And what is the probability for the A- detector to trigger, then ? P = sin^2 |a - lambda| I'd say...

cheers,
Patrick.

EDIT:

I played around a bit with this, and in fact, it is not so easy to arrive at a CORRELATION function which is cos^2(a-b). Indeed, let's take your probability which is p(a+) = cos^2(lambda-a).
Assuming independent probabilities, we have then that the correlation, which is given by p(a+) p(b+) = cos^2(lambda-a) sin^2(lambda-b) for an individual event. (the b+ on the other side is the b- on "this" side)

Now, by the rotation symmetry of the problem, lambda has to be uniformly distributed between 0 and 2 Pi, so we have to weight this p(a+) p(b+) with this uniform distribution in lambda:

P(a+)P(b-) = 1/ (2 Pi) Integral (lambda=0 -> 2 Pi) cos^2(lambda-a) sin^2(lambda-b) d lambda.

If you do that, you find:

1/8 (2 - Cos(2 (a-b)) ) = 1/8 (3-2 Cos^2[a-b])

And NOT 1/2 sin^2(a-b) !

I checked this with a small Monte Carlo simulation in Mathematica and this comes out the same. Ok, in the MC I compared a+ with b+ (not with b-), and then the result is 1/8 (2+cos(2(a-b)))

So this specific model doesn't give us the correct, measured correlations...

cheers,
Patrick.

I attach the small Mathematica notebook with calculation...

 

[DrChinese]

The cos^2 theta formula isn't incompatible with hidden
variables.

For the context of individual results you can write,

P = cos^2 |a - lambda|,

where P is the probability of detection, a is the
polarizer setting and lambda is the variable
angle of emission polarization.

This doesn't conflict with qm. If you knew
the value of lambda, or had any info about
how it was varying (other than just that
it's varying randomly), then you could more
accurately predict individual results (by
individual results I mean the data streams
at one end or the other).

How do we know that there *is* a hidden
variable operating in the individual measurement
context? Because, if you keep the polarizer
setting constant the data stream varies
randomly.

Now, this hidden variable doesn't just
stop existing because we decide to
combine the individual data streams wrt
joint polarizer settings.

Now, there's at least two ways to interpret Bell's
analysis. Either (1) lambda suddenly stops existing when we
decide to combine individual results, or (2) the variability
of lambda isn't relevant wrt joint detection.

I think the latter makes more sense, and in fact
it's part of the basis for the qm account which
assumes that photons emitted by the same atom
are entangled in polarization via the emission
process. This is why you have an entangled
quantum state prior to detection. So, all you
need to know to accurately predict the
*coincidental* detection curve is the angular
difference between the polarizer settings. And,
as in all such situations where you're analyzing,
in effect, the same light with crossed linear
polarizers the cos^2 theta formula holds.

Or Lambda=LHV does not exist, a possibility you consistently pass over. It is a simple matter to show that with a table of 8 permutations on A/B/C, there are no values that can be inserted that add to 100% without having negative values at certain angle settings.

A=___ (try 0 degrees)
B=___ (try 67.5 degrees)
C=___ (try 45 degrees)

Hypothetical hidden variable function: __________ (should be cos^2 or at least close)

1. A+ B+ C+: ___ %
2. A+ B+ C-: ___ %
3. A+ B- C+: ___ %
4. A+ B- C-: ___ %
5. A- B+ C+: ___ %
6. A- B+ C-: ___ %
7. A- B- C+: ___ %
8. A- B- C-: ___ %

It is the existence of C that relates to the hidden variable function. What you describe is just fine as long as we are talking about A and B only. (Well, there are still some problems but there is wiggle room for those determined to keep the hidden variables.) But with C added, everything falls apart as you can see.

You can talk all day long about joint probabilities and lambda, but that continues to ignore the fact that you cannot make the above table work out. If you are testing something else, you are ignoring Bell. After you account for the above table, then your explanation might make sense. Meanwhile, the Copenhagen Interpretation (and MWI) accounts for the facts that LHV cannot.

 

[kleinwolf]

I would like to point out, in a previous round against Vanesh about EPR and many worlds, the following point (1) :

Usual "orthodox Copenhagen QM" contains

1) a local hidden variable that corresponds to the specification of the PRECISE endstate when the latter is degenerate. The "standard" Copenhagen QM is a special configuration of the endstate that corresponds to it's maximum.

However, there is more :

2) a NON-LOCAL hidden variable.

Let see the latter : a non-local measurement is obtained by the operator : \sigma_z\otimes\(\sigma_z\cdot\vec{n}_b)...hence Both side are measured, and there is no 1 operator on the other (non disturbing operator).

Let consider \theta_b=0

Hence : both directions of measurement are the same. The clearly the only 2 possible endstates are :

|+-> or |-+>, with p(+-)=|<+-|\Psi>|^2=\frac{1}{2}=p(-+)

This sounds very like more than intuitive and easy to understand.

However, one can see the things in an other way, by looking that :

M=\sigma_z\otimes\sigma_z=\left(\begin{array}{cccc} 1 &&&\\&-1&&\\&&1&\\&&&-1\end{array}\right)

Hence, then eigenvalues of M are 1,-1 and are both degenerate. 1 corresponds to |A=B> and -1 to |A<>B> (same or different results in A and B).

Here again, the eigenSPACE can be parametrized :

|same&gt;=\left(\begin{array}{c}\cos(\chi)\\0\\\sin(\chi)\\0\end{array}\right)
|different&gt;=\left(\begin{array}{c}0\\cos(\delta)\\0\\\sin(\delta)\end{array}\right)
|\Psi&gt;=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\1\\-1\\0\end{array}\right)

So that : p(different)=|&lt;different|\Psi&gt;|^2=\frac{1}{2}\cos(\delta)^2

p(same)=|&lt;same|\Psi&gt;|^2=\frac{1}{2}\sin(\chi)^2

Where \chi,\delta are GLOBAL HIDDEN VARIABLES...

So that in fact 2p(same)=1 at MAX...what is the interpretation of this, if there is no mistake of course...??

 

[vanesch]

So that in fact 2p(same)=1 at MAX...what is the interpretation of this, if there is no mistake of course...??

To me the interpretation is that your chi and delta are just variables that parametrize the eigenspaces of the operator sigma_z x sigma_z.

However, I don't understand your calculation. When you write out sigma-z x sigma-z, I presume in the basis (++, -+,+-,--), then I'd arrive at a diagonal matrix which is (1,-1,-1,1)... You seem to have taken the DIRECT SUM, no ?


cheers,
Patrick.

 

[kleinwolf]

Yes, you're entirely right...my mistake is unforgivable, since this will change all the afterwards calculation and interpretation of \delta.

Then the result is p(same)=0\quad p(diff)=\frac{1}{2}(1-\sin(2\chi))

However, you admit there are 2 visions of computing the probabilities with your correct M :

locally : p(+-)=p(-+)=1/2

globally, the endstate |->_g=(0,cos(a),sin(a),0), gives the prob :

p(+-)=cos(a)^2, p(-+)=sin(a)^2...hence on average or special values of a, the same as locally...but a infinite of possibilities more are allowed.

Can this be measured on the statistical results in an experiement, and how to find how to change the value of a experimentally ??

 

[vanesch]

Yes, you're entirely right...my mistake is unforgivable, since this will change all the afterwards calculation and interpretation of \delta.

Then the result is p(same)=0\quad p(diff)=\frac{1}{2}(1-\sin(2\chi))

Well, don't you find this funny that the sum of the probabilities for the two possible outcomes don't add up to 1 ? You could think that for each event, you have two possible results: they are the same, or they are different. And if you add up their probabilities, you don't find 1.
It's like: throw up a coin: 25% chance you have head, 30% you have tail :-)

cheers,
Patrick.

 

[kleinwolf]

It's just because we don't understand QM. But QM is omnipotent for everyone, just put : \chi=-\frac{\pi}{4}\Rightarrow p(diff)=1

In the other calculation, the sum add up to 1 in every case...

So what does it mean that the prob of the possible outcomes don't add up to 1 in everycase for the other calculation ?

Just because the correlation, even if measured along the same directions, of the singlet state, is not always perfect, remind : there is a non-local part and a local one...here it's just the non-local one.

Best regards.

 

[vanesch]

It's just because we don't understand QM.

I'd rather say: because the way you want do modify QM doesn't work :-)

But QM is omnipotent for everyone, just put : \chi=-\frac{\pi}{4}\Rightarrow p(diff)=1

Yeah, that's the projection as is proposed in standard QM :-) So then it works...

But you claim that one should have a kind of "equal distribution" or so of outcomes (which clearly is NOT standard QM). And then you get silly results such as that the sum of the probabilities of all possibilities is not equal to 1.

In the other calculation, the sum add up to 1 in every case...

So what does it mean that the prob of the possible outcomes don't add up to 1 in everycase for the other calculation ?

It means that you have been cheating :-) You have in fact used normal quantum mechanics, except for the fact that you have been rotating the |-+> and the |+-> vectors in the "different" eigenspace. When you then calculate the total length (squared) of the original vector, projected on each of those and add it together, you obtain of course the correct QM prediction. Indeed, total length is invariant under a rotation of the basis (in the "different" eigenspace). But that's not what you were proposing in the first place. What you proposed was that the probability of having the "different" result should be the projection on ONE SINGLE arbitrary direction in the "different" eigenspace, not the sum of all the possibilities (which corresponds to finding the total length of the projection, as prescribed by standard QM). And then you're back to your first formula, where the sum of probabilities of all the possible outcomes is not equal to 1. THAT was the technique you used for the EPR stuff. You didn't sum over the different projections (because then you'd have found the same predictions as standard QM: you'd just have been rotating the basis vectors in the eigenspace to calculate the total projection length, something you are of course allowed to do).

cheers,
Patrick.

 

[kleinwolf]

Yes, basically I wrote you, it's just completely normal Copenhagen QM, there is nothing new in what I said...just trying to be more precise.

Anyway, for myself already gave the answer...but this, like always, is only my opinion...you have yours of course, but why say yours is the right one ??

Let's take the definition of the correlation : the following calculation is really old and well-known...but this maybe explains a bit more...I learn like you.

C(A,B)=&lt;AB&gt;-&lt;A&gt;&lt;B&gt;

then we have in fact a correlation operator given by the superposition of non-local and local opertators :

M_{non-local}=\sigma_z\otimes\sigma_z
M_{local A}=\sigma_z\otimes\mathbb{I}
M_{local B}=\mathbb{I}\otimes\sigma_z

So that the correlation operator is :

C=M_{non-local}-M_{local_A}|\Psi\rangle\langle\Psi|M_{local_B}

So that the correlation operator depends on the state we measure, hence this operator is non-linear.

We have also the correspondance : M_{non-local}=M_{local_A}M_{local_B}

Now the fact is that the eigenstate of \mathhbb{I} are degenerate. So if we look at the spectral decomposition of the identity operator, then we are lead to a more general equivalence that can be solved by doing some operations on the parametrization of the eigenstates describing the eigenspace, in other words : we should not only work with orthogonal bases. If we look nearer, then :

Let 2 eigenstates of 1 be :
|\phi\rangle=\left(\begin{array}{c}\cos(\phi)\\\sin(\phi)\end{array}\right)
|\chi\rangle=\left(\begin{array}{c}\cos(\chi)\\\sin(\chi)\end{array}\right)

Hence, this allows for non-orthogonal bases of R^2, the generalized spectral decomposition is :

\mathbb{I}_{decomp}=|\phi\rangle\langle\phi|+|\chi\rangle\langle\chi|=\left(\begin{array}{cc}\cos(\phi)^2+\cos(\chi)^2&amp;\cos(\phi)\sin(\phi)+\cos(\chi)\sin(\chi)\\\cos(\phi)\sin(\phi)+\cos(\chi)\sin(\chi)&amp;\sin(\phi)^2+\sin(\chi)^2\end{array}\right)

Hence we have the relationships :

\mathbb{I}=\langle\mathbb{I}_{decomp}\rangle_{\phi,\chi}

and the other precise decomposition that specifies the parameters :

\exists\phi_0,\chi_0|\mathbb{I}_{decomp}(\phi=\phi_0,\chi=\chi_0)=\mathbb{I}

Now of course we can compute the complete correlation operator, that will give you expressions up to the 4th power in the cos and sin of the parameters...

What I basically want to know is if you consider this exchange about science as a game and you want to win...I feel a kind of something unhealthy in the air...because I don't really see what the game or the competition is...and you ?

Best regards.

 

[Sherlock]

Ok, that's the probability for the A+ detector to trigger. And what is the probability for the A- detector to trigger, then ? P = sin^2 |a - lambda| I'd say...

cheers,
Patrick.

EDIT:

I played around a bit with this, and in fact, it is not so easy to arrive at a CORRELATION function which is cos^2(a-b). Indeed, let's take your probability which is p(a+) = cos^2(lambda-a).
Assuming independent probabilities, we have then that the correlation, which is given by p(a+) p(b+) = cos^2(lambda-a) sin^2(lambda-b) for an individual event. (the b+ on the other side is the b- on "this" side)

Now, by the rotation symmetry of the problem, lambda has to be uniformly distributed between 0 and 2 Pi, so we have to weight this p(a+) p(b+) with this uniform distribution in lambda:

P(a+)P(b-) = 1/ (2 Pi) Integral (lambda=0 -> 2 Pi) cos^2(lambda-a) sin^2(lambda-b) d lambda.

If you do that, you find:

1/8 (2 - Cos(2 (a-b)) ) = 1/8 (3-2 Cos^2[a-b])

And NOT 1/2 sin^2(a-b) !

I checked this with a small Monte Carlo simulation in Mathematica and this comes out the same. Ok, in the MC I compared a+ with b+ (not with b-), and then the result is 1/8 (2+cos(2(a-b)))

So this specific model doesn't give us the correct, measured correlations...

cheers,
Patrick.

I attach the small Mathematica notebook with calculation...

I'm not sure what you're saying above.

The statement in question was that the cos^2 theta formula
is incompatible with hidden variables. It isn't. If you consider just
the individual results, then you can write the probability of
detection as P = cos^2 |a - lambda|, where a is the setting
of an individual polarizer and lambda is the emission polarization.
But of course you don't know the value of lambda ... ever. So,
the actual probability of individual detection is simply .5.
The reason it's .5 is because, presumably, lambda is varying
randomly -- and the data streams indicate that for any set
of n emissions you'll get, in the ideal, .5n detections.
Of course the actual number is modified enormously due
to efficiency considerations.

It can be shown, empirically, that there is a local hidden variable
determining individual detections.

However, while this local hidden variable exists, it does
not determine joint detection.

The combined context, when you're considering detection
at both ends during a given interval, is different in that, while
the local hidden variable is still determining individual results,
its *variability* isn't a factor in determining correlations.
The only thing about the emitted light that matters wrt
joint detection is that during any given detection interval
the light incident on the polarizers is the same at both
ends -- that any two opposite moving photons emitted
by the same atom are polarized identically. This is what
the *entanglement* is based on. If it wasn't assumed
that the two polarizers are analyzing light with the same
physical properties, then what would be the basis for the
projection along the detection axis?

So, during any given coincidence interval, the separated
polarizers are, in effect, analyzing the *same* light.
Hence, the applicability of the cos^2 theta formula in
the joint context involving crossed linear polarizers.

 

[Sherlock]

Or Lambda=LHV does not exist, a possibility you consistently pass over. It is a simple matter to show that with a table of 8 permutations on A/B/C, there are no values that can be inserted that add to 100% without having negative values at certain angle settings.

A=___ (try 0 degrees)
B=___ (try 67.5 degrees)
C=___ (try 45 degrees)

Hypothetical hidden variable function: __________ (should be cos^2 or at least close)

1. A+ B+ C+: ___ %
2. A+ B+ C-: ___ %
3. A+ B- C+: ___ %
4. A+ B- C-: ___ %
5. A- B+ C+: ___ %
6. A- B+ C-: ___ %
7. A- B- C+: ___ %
8. A- B- C-: ___ %

It is the existence of C that relates to the hidden variable function. What you describe is just fine as long as we are talking about A and B only. (Well, there are still some problems but there is wiggle room for those determined to keep the hidden variables.) But with C added, everything falls apart as you can see.

You can talk all day long about joint probabilities and lambda, but that continues to ignore the fact that you cannot make the above table work out. If you are testing something else, you are ignoring Bell. After you account for the above table, then your explanation might make sense. Meanwhile, the Copenhagen Interpretation (and MWI) accounts for the facts that LHV cannot.

I've considered the idea that the lhv
doesn't exist and rejected it.

There's a difference between the lhv not existing
and the lhv being irrelevant in a certain context.
I agree with you that the lhv is not determining
joint detection -- but, that doesn't mean that
it doesn't *exist*.

The above table is irrelevant to the
argument of whether or not the lhv *exists*.

The individual data streams are *direct*
evidence of the existence of the lhv.

 

[Sherlock]

There are definitely TWO ways to look at that statement. Some of the vocal local realists argue that the cos^2 law isn't correct!

Then I think they're wrong about that.

They do that so the Bell Inequality can be respected; and then explain that experimental loopholes account for the difference between observation and their theory.

From what I know of the experiments, they're ok. However,
I think that Bell's analysis and the physical meaning of experimental
violations of the inequality are being misinterpreted.

Clearly, classical results sometimes match QM and sometimes don't; and when they don't, you really must side with the predictions of QM. Even Einstein saw that this was a steamroller he had to ride, and the best he could muster was that QM was incomplete.

I *am* siding with the predictions of qm. Where have I
said otherwise? But it's certainly not a complete description
of the physical reality. It's not designed to be.

 

[vanesch]

I'm not sure what you're saying above.

The statement in question was that the cos^2 theta formula
is incompatible with hidden variables. It isn't. If you consider just
the individual results, then you can write the probability of
detection as P = cos^2 |a - lambda|, where a is the setting
of an individual polarizer and lambda is the emission polarization.
But of course you don't know the value of lambda ... ever.

Well, yes, that's exactly what I did. But apparently now you assume EQUAL polarizations (lambda) at both sides, and not OPPOSITE polarizations. So be it.

For a GIVEN lambda (unknown, I agree), you say that, if we put up a polarizer at Alice in direction a, it has a probability equal to cos^2(a-lambda) of clicking (assuming it "100% efficient" ; we'll come to that later). This means then also, I would think, that we have a probability cos^2(a-lambda) of clicking at Bob's place if he also puts his polarizer in direction a, right ?
And if Bob puts his polarizer in direction b, I assume that his probability of clicking for the same lambda is cos^2(b-lambda), right ?
Of course, specifying individual probabilities doesn't give us the joint distribution, except if you say that these probabilities are independent. But normally, what happens at Bob is independent of what happens at Alice, once lambda is given. So the JOINT PROBABILITY that the detector (in direction a) at Alice clicks, and that within the same time interval, the detector (in direction b at Bob) clicks is then given by

P(a,b,lambda) = cos^2(a-lambda) cos^2(b-lambda)

If that is not the case, then give me your joint probability for a given lambda.

Now, you say that we don't know lambda (which is the random polarization direction of the light sent out to both detectors in any event).
But we know that the distribution, whatever it is, must be rotation-invariant if we consider many events. Indeed, this is the only way to have, on one side, a probability equal to 1/2 averaged over the entire sample for ALL values of a. This means that the DISTRIBUTION of the different lambda values must be uniform over the 0 - 2Pi interval. Otherwise, we'd have on average MORE clicks in one direction than in another (on one single side).

Now, if we know that lambda, over different trials, is distributed uniformly, then we can calculate, over this population, what will be the average correlation P(a,b):
It is simply given by:

P(a,b) = 1/2Pi integral(lambda=0 -> 2 Pi) of P(a,b,lambda) d lambda

And if you do that, well, then you find:

P(a,b) = 1/8 (2 + cos(2 (a-b)) )

What changes now when the detectors are "inefficient" ? Well, this changes normally only the probability of clicking: instead of your "cos^2(a - lambda), we have a scale factor: epsilon cos^2(a - lambda). BTW, that is exactly what you get out of the semi-classical approach for the photo-electric effect.

The only thing it changes for P(a,b) is that you multiply with epsilon^2 (assuming detectors of identical quality on both sides).

The combined context, when you're considering detection
at both ends during a given interval, is different in that, while
the local hidden variable is still determining individual results,
its *variability* isn't a factor in determining correlations.
The only thing about the emitted light that matters wrt
joint detection is that during any given detection interval
the light incident on the polarizers is the same at both
ends -- that any two opposite moving photons emitted
by the same atom are polarized identically. This is what
the *entanglement* is based on. If it wasn't assumed
that the two polarizers are analyzing light with the same
physical properties, then what would be the basis for the
projection along the detection axis?

Well, that's exactly what I do. You give me the individual probabilities for a given value of the hidden variable, and from that, and a symmetry argument, I calculate the joint probability over the entire population of the hidden variable.

And comes out... something that is different than what QM predicts ! Ok, the "right" form is there, namely cos^2, but the "modulation depth" is much lower: you do not reach high enough correlations, but more importantly, you do not reach LOW ENOUGH correlations for certain angles either.
You can intuitively see that too.

If the polarizers at Bob and Alice are perpendicular, then according to the photon picture, you will have perfect ANTICORRELATION. Namely whenever the photon gets through Bob's, it is blocked for sure at Alice's and vice versa. Now, with classical light that cannot happen, because certain light pulses will get in under 45 degrees. That means that there is a reduced, but finite probability at Alice's of clicking, and also at Bob's, so the correlation will not be 0. Nevertheless, experimentally, it is 0, and according to the photon picture, it should also be 0.

cheers,
Patrick.

 

[kleinwolf]

Gosh...I should never have started with non-linear operators...anybody knows about the eigenvalues, or something like that...

For I get that the eigenvalues of the non-linear operator of the local part of the correlation of a bipartite system given by : M_{local}=(\sigma_z\otimes\mathbb{I})|\Psi&gt;&lt;\Psi|(\mathbb{I}\otimes\sigma_z)

are negative definite, continuous apparently, and need to solve a 4th order equation system of the style :

b^4+(2c-1)b^3-2c^2b^2-2c^2b+c^4=0
a^2=c^2-b^2
d^2=-w
a^2+b^2+c^2+d^2=1

where the eigenstate is given by : |\Psi&gt;=(a,b,c,d)

In fact the notation M_{local} is abusive, since the wavefunction is included, so that's why the average correlation is still non-local (I privately exchanged messages with RandallB before about why I came up with a non-perfect correlation for the singlet-state, giving CHSH=2.47 at maximum (hence non-local w.resp. to Bell's Ansatz, but still neare to experimental results 2.25)...

It's clear that if the total correlation operator is taken (with the local and the non-local parts), then, the eigenvalues should be in [-1;1]...but maybe the spectrum is a special one, with an infinite distribution, I cannot make any bet about that...What is your opinion ?

 

[DrChinese]

I've considered the idea that the lhv
doesn't exist and rejected it.

There's a difference between the lhv not existing
and the lhv being irrelevant in a certain context.
I agree with you that the lhv is not determining
joint detection -- but, that doesn't mean that
it doesn't *exist*.

The above table is irrelevant to the
argument of whether or not the lhv *exists*.

The individual data streams are *direct*
evidence of the existence of the lhv.

That isn't so... it is just a question of seeing what you want to see.

Suppose the random value is inserted when the observation is made - supplied by some randomizer which we cannot access and never can. That is a reasonable explanation and completely consistent with the facts. You see the randomness as evidence that there is more to know. Well, maybe that is so. But... that random value DID NOT exist prior to the observation, as Bell clearly tells us.

Now suppose the "random" value is not random at all - it is completely determined by some complex stochastic process having to do with the state of the entire universe. Because it is non-local, the same information is available to the entangled photons. That could be a reasonable explanation and consistent with the facts. Bell would still apply! And thus we know that the observations were still fundamental to the process, and there is no locally hidden variable that explains the observed results.

The observer settings are fundamental to the results; and there are no local hidden variables that completely determine the outcome independent of the observer settings. As Bell states, the purpose of LHV theories is to restore locality and causality to the description, and this cannot be done.

 

[vanesch]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true. But as many people stated that, I believed it for a while myself. However, as I tried to show, if you assume the following:

1) Intensity of (classical) radiation when incident radiation has polarization direction lambda and the polarizer has direction a is given by the original intensity multiplied by cos^2(lambda-a).

2) Identical incident radiation (during one "pulse") at Alice's and Bob's, with identical (or opposite, pick your choice) polarization directions

3) photodetection probability (clicking probability) proportional to intensity, and, for a given intensity, statistically independent of any other photodetection somewhere else.

If we assume that the source sends out pulse trains, each with an (unknown) value of polarization, and equal intensity, then:

The probability at Alice (polarizer at angle a) of clicking is:
P(a,lambda) = eps cos^2(a - lambda)

The probability at Bob (polarizer at angle b) of clicking is:
P(b,lambda) = eps cos^2(b - lambda)

and the probability of clicking together, these probabilities being considered independent, is given by:

P(a,b,lambda) = eps^2 cos^2(a-lambda) cos^2(b-lambda)

From a symmetry argument, one can deduce that lambda must be drawn from a uniform distribution between 0 and 2 Pi, and so, the observed overall probabilities of clicking on N trials are:

P(a) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon \cos^2(a - \lambda) d \lambda = 1/2

Same for P(b)

and for P(a,b):
P(a,b) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon^2 \cos^2(a - \lambda)\cos^2(b - \lambda) d\lambda

and this leads to:
P(a,b) = \frac{\epsilon^2 }{8} (2 + \cos 2 (a - b))

Of course we don't know the absolute number of trials (the true value of epsilon) if we don't consider the photon model, but we can do away with that by calculating:

\frac{P(a,b)}{P(a) P(b)} = \frac{1}{2} (2 + \cos 2(a-b))

Note that this is NOT equal to the quantum prediction, especially for the fact that the above correlation function doesn't go down to 0, when the two polarizers are perpendicular.

cheers,
Patrick.

 

[vanesch]

My previous message may give the impression that one cannot obtain a cos^2(a-b) curve with a hidden variable model. This is not true, and I just made one. It goes as follows:
Imagine that each pair of light pulses that is sent out has two hidden variables: one is lambda, the polarization direction (which will be uniformly distributed in 0-2Pi) and the other one is a one-bit random variable: if the Alice pulse receives the 1 bit, then the Bob pulse receives the 0 bit, and vice versa. This random variable is distributed 50/50, and we call it the tau variable.

Now, a polarizer could work in the following way, upon reception of a light pulse with hidden variables lambda and tau:
If tau = 1, then the intensity that gets through the polarizer under angle a equals the incoming intensity if |a - lambda|< delta (a small angle) and is blocked completely if not.
However, if tau = 0, then the intensity that gets through the polarizer is equal to the incoming intensity times cos^2(a - lambda).

Next, the probability of a detector click is proportional to the incoming intensity.

Applying this model yields a correlation about proportional to cos^2(a-b). Mind you, I say: proportional !

Indeed, for the individual polarizers, we get essentially 1/4 of the total number of trials (half of the time we get a bit 1, so then the probability of letting any intensity through is very rare, because lambda needs to be close to a, and the other half of the time, we get the cos^2 curve, which gives us 1/2 on average).
However, for the coincidence, in order for both to click, one of both will have a bit 1. So we KNOW that we are in one of the rare cases when lambda is close, or to a, or to b. In that case, the OTHER polarizer receives the 0 bit, and hence the probability for the OTHER one of clicking is given by the cos^2 rule. Only, we suppressed seriously the entire population and a very small fraction of the trials do give rise to a correlation. But if we add in arbitrary "efficiency" coefficients, we can say that we have a cos^2 relation.

This ad hoc model suffers of course from a lot of difficulties and is made up for the purpose. First of all, this is not classical optics either. The bit left or right mechanism is totally taken out of thin air. Next, although this model can explain certain aspects of the cos^2 curve, it would fail miserably on energy balances: we wouldn't have conservation of total radiant energy when taking the flux that gets through a polarizer, and that gets to the perpendicular polarizer.
But it is a technique to show that a curve, proportional to cos^2 can eventually be constructed.

cheers,
Patrick.

 

[cartuz]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true. But as many people stated that, I believed it for a while myself. However, as I tried to show, if you assume the following:

1) Intensity of (classical) radiation when incident radiation has polarization direction lambda and the polarizer has direction a is given by the original intensity multiplied by cos^2(lambda-a).

2) Identical incident radiation (during one "pulse") at Alice's and Bob's, with identical (or opposite, pick your choice) polarization directions

3) photodetection probability (clicking probability) proportional to intensity, and, for a given intensity, statistically independent of any other photodetection somewhere else.

If we assume that the source sends out pulse trains, each with an (unknown) value of polarization, and equal intensity, then:

The probability at Alice (polarizer at angle a) of clicking is:
P(a,lambda) = eps cos^2(a - lambda)

The probability at Bob (polarizer at angle b) of clicking is:
P(b,lambda) = eps cos^2(b - lambda)

and the probability of clicking together, these probabilities being considered independent, is given by:

P(a,b,lambda) = eps^2 cos^2(a-lambda) cos^2(b-lambda)

From a symmetry argument, one can deduce that lambda must be drawn from a uniform distribution between 0 and 2 Pi, and so, the observed overall probabilities of clicking on N trials are:

P(a) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon \cos^2(a - \lambda) d \lambda = 1/2

Same for P(b)

and for P(a,b):
P(a,b) = \frac{1}{2 \pi} \int_{\lambda = 0}^{2 \pi} \epsilon^2 \cos^2(a - \lambda)\cos^2(b - \lambda) d\lambda

and this leads to:
P(a,b) = \frac{\epsilon^2 }{8} (2 + \cos 2 (a - b))

Of course we don't know the absolute number of trials (the true value of epsilon) if we don't consider the photon model, but we can do away with that by calculating:

\frac{P(a,b)}{P(a) P(b)} = \frac{1}{2} (2 + \cos 2(a-b))

Note that this is NOT equal to the quantum prediction, especially for the fact that the above correlation function doesn't go down to 0, when the two polarizers are perpendicular.

cheers,
Patrick.

Excuse me for it is hard to read a long English texts for me. Because it is ordinary I read texts with mathematics only.
We can clear see that Bell's Inequalities it is possible to complete by curved and stationary geometry, which play role the non-local hidden variables. Why the geometry must be non-curved? Curved geometry is more suitable because it is the general case.
In that case Bell's Inequalities will be violet!
It is clear from
Correlation factor M of random variables \lambda ^{i} are projections
onto directions A^{\nu } and B^{n} defined by polarizers (all these
vectors being unit) is

\left| M_{AB}\right| =\left| \left\langle AB\right\rangle \right| =\left|<br /> \langle \lambda ^{i}A^{k}g_{ik}\lambda ^{m}B^{n}g_{mn}\rangle \right|

The deferential geometry gives

\cos \Phi =\frac{g_{ik}\lambda ^{i}A^{k}}{\sqrt{\lambda ^{i}\lambda _{i}}\sqrt{A^{k}A_{k}}},

\cos (\Phi +\theta )=\frac{g_{mn}\lambda ^{m}B^{n}}{\sqrt{\lambda<br /> ^{m}\lambda _{m}}\sqrt{B^{n}B_{n}}}.

Here i,k,m,n possesses 0,1,2,3; \theta is angle between polarizers, then
\left| M_{AB}\right| =\left| \frac{1}{2\pi }\int_{0}^{2\pi }\rho (\Phi<br /> )\cos \Phi \cos \left( \Phi +\theta \right) d\Phi \right|

This the case when entanglement explained be by stationary gravitational fields (curved geometry) is perspective, I hope.

 

[DrChinese]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true. But as many people stated that, I believed it for a while myself. However, as I tried to show, if you assume the following:

Patrick,

I don't mean to pick apart words. But there are two ways to interpret the situation you describe. You are of course correct that the application of Malus' Law in the "classical" manner you describe yields a different prediction for joint detection.

But that is not the only way to apply Malus' Law to this case. Since it describes the results only when initial polarizeration is KNOWN, you should wait to apply it until an observation is performed on one or the other of the entangled photons...similar to how it is done in classical application (in which the first polarizer tells us the initial polarization). Then the joint results will match QM exactly.

In other words, if you apply Malus' Law ASSUMING hidden variables, you get a different result (for joint detection) than if you apply it using (what I think of as) a "traditional" application which does not specifically assume HV. Presumably, Malus never thought of hidden variables one way or the other.

So to summarize: if you push a "classical" application of Malus' Law as you describe, you immediately run into problems because the results disagree with experiment (as you pointed out). If you push the hidden variable version and apply as I described, you immediately run into problems with Bell's Theorem. So you know that there is something wrong with the "classical" or "traditional" views (which are just labels as I have used them here) either way you choose to look at it.

 

[vanesch]

But that is not the only way to apply Malus' Law to this case. Since it describes the results only when initial polarizeration is KNOWN, you should wait to apply it until an observation is performed on one or the other of the entangled photons...similar to how it is done in classical application (in which the first polarizer tells us the initial polarization). Then the joint results will match QM exactly.

Because people told this already a few times, intuitively I thought that that was very acceptable, and that's why for a long time I thought that "classical optics" predicted the same correlations as quantum theory.
When you apply it as you state, when you say: ah, one detector clicked, so the light MUST BE in the same polarization direction as that polarizer, YOU ARE IN FACT APPLYING QUANTUM THEORY ! You projected and normalized the ENTIRE state on the direction of the first polarizer, as is typically done in QM (in the Copenhagen view, let's be clear).
But that is NOT what is done in classical optics. In classical optics a polarizer ONLY SELECTS THE COMPONENT of the light in the direction of the polarizer FOR THE LIGHT AT THAT POLARIZER. When you send light polarized under 45 degrees with the polarizer, only HALF of the intensity gets through ; and if you have two "identical copies of light" (as entanglement is seen in classical optics), it is not because on one side, you selected only half of the intensity (because your polarizer made an angle of 45 degrees) that suddenly the light will jump over 45 degrees on the other side to match the detection. That is a typical pure quantum phenomenon.

And that's why I wanted to point out, that, in the case Alice and Bob put their polarizers at 90 degrees, in the classical picture, THERE IS STILL LIGHT COMING THROUGH at both sides: namely all that light that is not exactly polarized at 0 or at 90 degrees. If light is incident under 45 degrees, at both sides, they get half the intensity, so there is a real chance of having coincident clicks.

In other words, if you apply Malus' Law ASSUMING hidden variables, you get a different result (for joint detection) than if you apply it using (what I think of as) a "traditional" application which does not specifically assume HV. Presumably, Malus never thought of hidden variables one way or the other.

I was not really using "hidden variables" ; the "hidden variable" was the random polarization of classical light. We know that the source sent out light pulses which have identical polarization for bob and alice, wavetrain per wavetrain. But this common polarization can fluctuate randomly (as the phase of light can fluctuate randomly outside of the coherence time). There is nothing surprising about that in classical optics.
Then I calculated the light intensity that got through each polarizer individually, using Malus' law, and then I applied a probability law for each photodetector, that gives us a probability of clicking per unit of time which is proportional to the incident intensity (after the polarizer).
I would think that that is exactly what one is supposed to do in classical optics, no ? I didn't "push" anything.

So to summarize: if you push a "classical" application of Malus' Law as you describe, you immediately run into problems because the results disagree with experiment (as you pointed out). If you push the hidden variable version and apply as I described, you immediately run into problems with Bell's Theorem. So you know that there is something wrong with the "classical" or "traditional" views (which are just labels as I have used them here) either way you choose to look at it.

Yes, of course. But it just appeared to me that what I had been taking for granted because so many people said it, namely that in PURELY CLASSICAL OPTICS, you get out the same correlations as in quantum theory, is absolutely not true ! And the most striking aspect is again the pure ANTI correlation when Alice and bob have perpendicular polarizers, which is impossible to obtain in classical optics. (but for which you can build an ad hoc hidden variable model, as I did - without physical plausibility).

cheers,
Patrick.

 

[kleinwolf]

Excuse me for it is hard to read a long English texts for me. Because it is ordinary I read texts with mathematics only.
We can clear see that Bell's Inequalities it is possible to complete by curved and stationary geometry, which play role the non-local hidden variables. Why the geometry must be non-curved? Curved geometry is more suitable because it is the general case.
In that case Bell's Inequalities will be violet!
It is clear from
Correlation factor M of random variables \lambda ^{i} are projections
onto directions A^{\nu } and B^{n} defined by polarizers (all these
vectors being unit) is

\left| M_{AB}\right| =\left| \left\langle AB\right\rangle \right| =\left|<br /> \langle \lambda ^{i}A^{k}g_{ik}\lambda ^{m}B^{n}g_{mn}\rangle \right|

The deferential geometry gives

\cos \Phi =\frac{g_{ik}\lambda ^{i}A^{k}}{\sqrt{\lambda ^{i}\lambda _{i}}\sqrt{A^{k}A_{k}}},

\cos (\Phi +\theta )=\frac{g_{mn}\lambda ^{m}B^{n}}{\sqrt{\lambda<br /> ^{m}\lambda _{m}}\sqrt{B^{n}B_{n}}}.

Here i,k,m,n possesses 0,1,2,3; \theta is angle between polarizers, then
\left| M_{AB}\right| =\left| \frac{1}{2\pi }\int_{0}^{2\pi }\rho (\Phi<br /> )\cos \Phi \cos \left( \Phi +\theta \right) d\Phi \right|

This the case when entanglement explained be by stationary gravitational fields (curved geometry) is perspective, I hope.

Well, in fact in this case, then Bell's idea is not really reached because you have something like :

M_{AB}=\frac{1}{2\pi}\int_0^{2\pi}\cos(\theta_A-\phi)\cos(\theta_B-\phi)\rho(\phi)d\phi
=\int_0^{2\pi}A(\theta_A,\phi)B(\theta_B,\phi)\rho(\phi)d\phi

But your functions A and B ARE NOT THE RESULTS OF MEASUREMENT (because the results can be only 1 and -1)...That's why Bell could deduce a difference between hidden variable and QM...Remind that the results of measurement of the quantum-mechanical operator (\sigma\cdot n_A)\otimes(\sigma\cdot n_B) are +1 or -1 (the eigenvalues)...but nothing is allowed inbetween.

What you could say is that in your case :A(\theta_A,\phi)=\int a(\theta_A,\phi,\lambda)u(\lamda)d\lambda

where a(\theta_A,\phi,\lambda)=+1,-1

So that then you can apply Bell's theorem on a,b...so that your proposition doesn't violate the inequqality (CHSH for example).

 

[kleinwolf]

Excuse me for it is hard to read a long English texts for me. Because it is ordinary I read texts with mathematics only.
We can clear see that Bell's Inequalities it is possible to complete by curved and stationary geometry, which play role the non-local hidden variables. Why the geometry must be non-curved? Curved geometry is more suitable because it is the general case.
In that case Bell's Inequalities will be violet!
It is clear from
Correlation factor M of random variables \lambda ^{i} are projections
onto directions A^{\nu } and B^{n} defined by polarizers (all these
vectors being unit) is

\left| M_{AB}\right| =\left| \left\langle AB\right\rangle \right| =\left|<br /> \langle \lambda ^{i}A^{k}g_{ik}\lambda ^{m}B^{n}g_{mn}\rangle \right|

The deferential geometry gives

\cos \Phi =\frac{g_{ik}\lambda ^{i}A^{k}}{\sqrt{\lambda ^{i}\lambda _{i}}\sqrt{A^{k}A_{k}}},

\cos (\Phi +\theta )=\frac{g_{mn}\lambda ^{m}B^{n}}{\sqrt{\lambda<br /> ^{m}\lambda _{m}}\sqrt{B^{n}B_{n}}}.

Here i,k,m,n possesses 0,1,2,3; \theta is angle between polarizers, then
\left| M_{AB}\right| =\left| \frac{1}{2\pi }\int_{0}^{2\pi }\rho (\Phi<br /> )\cos \Phi \cos \left( \Phi +\theta \right) d\Phi \right|

This the case when entanglement explained be by stationary gravitational fields (curved geometry) is perspective, I hope.

Well, in fact in this case, then Bell's idea is not really reached because you have something like :

M_{AB}=\frac{1}{2\pi}\int_0^{2\pi}\cos(\theta_A-\phi)\cos(\theta_B-\phi)\rho(\phi)d\phi
=\int_0^{2\pi}A(\theta_A,\phi)B(\theta_B,\phi)\rho(\phi)d\phi

But your functions A and B are not the results of measurement (because the results can be only 1 and -1)...That's why Bell could deduce a difference between hidden variable and QM...Remind that the results of measurement of the quantum-mechanical operator (\sigma\cdot n_A)\otimes(\sigma\cdot n_B) are +1 or -1 (the eigenvalues)...but nothing is allowed inbetween.

What you could say is that in your case :A(\theta_A,\phi)=\int a(\theta_A,\phi,\lambda)u(\lambda)d\lambda

where a(\theta_A,\phi,\lambda)=+1,-1

So that then you can apply Bell's theorem on a,b...so that it seems your proposition doesn't violate the inequality (CHSH for example).

The point is that if the correlation is expressed as hidden variables only :

C(A,B)=\int_a^b a(\vec{n}_A,\vec{\lambda})b(\vec{n}_A,\vec{\lambda})\rho(\vec{\lambda})d\lambda_1...d\lambda_n

Then Bell's inequality is respected..(where a,b are constant (e.g. infinity or 0->2pi)...

however, if the correlation contains a visible variable :

C(A,B,\mu)=\int_{f(\mu)}^{g(\mu)}A(n_A,\lambda,\mu)B(n_B,\lambda,\mu)\rho(\lambda,\mu)d\lambda

Then you can violate the inequality, in the sense you choose \mu so that you cannot factorize the results like in Bell's theorem.

 

[cartuz]

Well, in fact in this case, then Bell's idea is not really reached because you have something like :

M_{AB}=\frac{1}{2\pi}\int_0^{2\pi}\cos(\theta_A-\phi)\cos(\theta_B-\phi)\rho(\phi)d\phi
=\int_0^{2\pi}A(\theta_A,\phi)B(\theta_B,\phi)\rho(\phi)d\phi

But your functions A and B are not the results of measurement (because the results can be only 1 and -1)...That's why Bell could deduce a difference between hidden variable and QM...Remind that the results of measurement of the quantum-mechanical operator (\sigma\cdot n_A)\otimes(\sigma\cdot n_B) are +1 or -1 (the eigenvalues)...but nothing is allowed inbetween.

What you could say is that in your case :A(\theta_A,\phi)=\int a(\theta_A,\phi,\lambda)u(\lambda)d\lambda

where a(\theta_A,\phi,\lambda)=+1,-1

So that then you can apply Bell's theorem on a,b...so that it seems your proposition doesn't violate the inequality (CHSH for example).

The point is that if the correlation is expressed as hidden variables only :

C(A,B)=\int_a^b a(\vec{n}_A,\vec{\lambda})b(\vec{n}_A,\vec{\lambda})\rho(\vec{\lambda})d\lambda_1...d\lambda_n

Then Bell's inequality is respected..(where a,b are constant (e.g. infinity or 0->2pi)...

however, if the correlation contains a visible variable :

C(A,B,\mu)=\int_{f(\mu)}^{g(\mu)}A(n_A,\lambda,\mu)B(n_B,\lambda,\mu)\rho(\lambda,\mu)d\lambda

Then you can violate the inequality, in the sense you choose \mu so that you cannot factorize the results like in Bell's theorem.

Thank you for reply and excuse me for delay.Are this mean that I
must to add here \mu which have the sense of distribution of metric g_{ik}?
Yes, I understand that metric allow correlated different quantatives in the space.

 

[Sherlock]

The statement in question was that the cos^2 theta formula
is incompatible with hidden variables. It isn't. If you consider just
the individual results, then you can write the probability of
detection as P = cos^2 |a - lambda|, where a is the setting
of an individual polarizer and lambda is the emission polarization.

Well, yes, that's exactly what I did. But apparently now you
assume EQUAL polarizations (lambda) at both sides, and not
OPPOSITE polarizations. So be it.

For a GIVEN lambda (unknown, I agree), you say that, if we put up a polarizer at Alice in direction a, it has a probability equal to cos^2(a-lambda) of clicking (assuming it "100% efficient" ; we'll come to that later). This means then also, I would think, that we have a probability cos^2(a-lambda) of clicking at Bob's place if he also puts his polarizer in direction a, right ?
And if Bob puts his polarizer in direction b, I assume that his probability of clicking for the same lambda is cos^2(b-lambda), right ?
Of course, specifying individual probabilities doesn't give us the joint distribution, except if you say that these probabilities are independent. But normally, what happens at Bob is independent of what happens at Alice, once lambda is given. So the JOINT PROBABILITY that the detector (in direction a) at Alice clicks, and that within the same time interval, the detector (in direction b at Bob) clicks is then given by

P(a,b,lambda) = cos^2(a-lambda) cos^2(b-lambda)

If that is not the case, then give me your joint probability for a given lambda.

First, the statement I was replying to was that local hidden
variables don't *exist*. I disagreed. They do exist, and
it can be demonstrated by looking at *individual* results.
Lambda is what's being analyzed in the individual context.

Regarding the *joint* probability, it doesn't depend
on a given lambda. That is, lambda isn't what's being
analyzed. What's being analyzed in the combined
context is a global constant. Lambda isn't the
global *constant*, so a description of joint
probability (such as what you evaluated) based on
lambda might give the correct functional form, but a
reduced spread (as you pointed out).

Ok, so, in the joint context, the polarizers, taken
together, are analyzing the degree to which photon_1
and photon_2 of any given pair are polarized identically.

So, if A records a photon detection, then the
probability of detection at B with the polarizers
aligned is 1 -- and the probability of detection
at B varies as cos^2 |a - b|. Isn't this
just standard quantum optics?

But where is the identical polarization produced?
As far as we *know* it *could* be produced at
the polarizers or detectors. But there's
an explanation that fits the observations,
and doesn't require the existence of undetectable
superluminal 'influences'.

Does quantum theory have an answer for the
question of where the identical polarization
is produced?

As far as I can tell, you could
interpret the projection as being due to local
or nonlocal transitions. That is, you *can*
assume that the identical polarization of
photon_1 and photon_2 is produced via the
emission process, and this assumption isn't
contradicted by the data. It also isn't, imo,
contradicted by Bell's analysis which deals
with the variable lambda. It also fits the
original idea of entanglement being due
to past interaction or common source.

Since interpretations of Bell's analysis
seem to be the only thing that superluminality
(wrt to associated experiments) has going
for it, it doesn't seem to me to be the
most reasonable option.

Regarding your statement that, using a wave
picture for light, a detector sitting behind
perpendicularly crossed linear polarizers has a >0
probability of registering a detection:
I don't think that's so. Anyway, I'm not
sure what that has to do with what we're
talking about -- which is, I thought, the
nature of entanglement in general, and in Bell/EPR
experiments in particular.

 

[vanesch]

Regarding the *joint* probability, it doesn't depend
on a given lambda. That is, lambda isn't what's being
analyzed. What's being analyzed in the combined
context is a global constant. Lambda isn't the
global *constant*, so a description of joint
probability (such as what you evaluated) based on
lambda might give the correct functional form, but a
reduced spread (as you pointed out).

But it isn't doing that ! It is doing that only in my second,
ad hoc, model.

Ok, so, in the joint context, the polarizers, taken
together, are analyzing the degree to which photon_1
and photon_2 of any given pair are polarized identically.

So, if A records a photon detection, then the
probability of detection at B with the polarizers
aligned is 1 -- and the probability of detection
at B varies as cos^2 |a - b|. Isn't this
just standard quantum optics?

No, not at all. That's standard optics IF IT IS THE SAME BEAM:

beam in ==> (pol 1) =(beam1)==> (pol2) ==> beam 2.

If pol1 and pol2 are aligned, of course beam 1 is reduced in intensity from beam in, but beam 2 is not reduced anymore, it is equal to beam1. And if pol1 and pol2 are perpendicular, beam2 equals 0. BUT THAT IS BECAUSE POL2 WORKS ON THE BEAM THAT GOT THROUGH POL1, which selected the component of beamin in the pol1 direction. If you then select AGAIN that component, of course everything gets through. It is as if you have a vector in the xy plane, and you project on the x-axis, and then you project the projection AGAIN on the x-axis. You work the second time with the projection.

But if you have:
beam2 <== (pol2) == (beamin) ====(pol1)==> beam1

then beam1 is simply the component of beamin in the pol1 direction, and beam2 is simply the component of beamin in the pol2 direction, and in classical optics these projections are done on the original beam. In the xy analogy, if pol1 and pol2 are resp. on the x and the y axis, beam1 gives you the x-component, and beam2 gives you the y component of the arbitrary vector coming from beamin. It is the special quantum context which makes that the measurement of the component x at pol1 makes that suddenly the beam at pol2 is projected out together in that x-direction (to take up the Copenhagen picture). But that's not what is done in classical optics.

In fact, in classical optics, there is no difference between this setup, and simply an incoming beam that is split on a (non-polarizing) beamsplitter, the transmitted beam arriving at pol1 and the reflected beam arriving at pol2.

That is, you *can*
assume that the identical polarization of
photon_1 and photon_2 is produced via the
emission process, and this assumption isn't
contradicted by the data. It also isn't, imo,
contradicted by Bell's analysis which deals
with the variable lambda. It also fits the
original idea of entanglement being due
to past interaction or common source.

Well, of course I assume that the two beams have identical polarization (lambda)!

Let us assume, for fun, that the source is polarized: it is sending out light always under 45 degrees, at both sides. Now assume that pol1 is set parallel to the x-axis, and pol2 is set parallel to the y axis. What do you think are the intensities at both sides in classical optics ?

Regarding your statement that, using a wave
picture for light, a detector sitting behind
perpendicularly crossed linear polarizers has a >0
probability of registering a detection:
I don't think that's so.

Eh, a detector behind CROSSED linear polarizers doesn't see any light, of course ! I never said that. I said that a detector behind ONE polarizer sees some light, and if ANOTHER BEAM (with identical polarization) falls upon ANOTHER POLARIZER (which is crossed to the first), it will ALSO SEE SOME LIGHT (in classical optics).

Anyway, I'm not
sure what that has to do with what we're
talking about -- which is, I thought, the
nature of entanglement in general, and in Bell/EPR
experiments in particular.

?

cheers,
Patrick.

 

[Sherlock]

But it isn't doing that ! It is doing
that only in my second, ad hoc, model.

I'm not sure what you're referring to here.

 

[Sherlock]

I would like to point out, again, that when people (sherlock) say that classical optics gives us the probability of joint detection equal to cos^2(a-b), then this is NOT true.

Actually, I wasn't arguing that -- but
rather just offering a perspective on why the
qm formulation works. A view of some possible
physical reasons for the correlations in optical Bell/EPR
experiments. These setups *do* have some
important features in common with the classical
setups that first produced the cos^2 theta formula.

The thing about photon detections is that there's
no way to tell whether all or some portion of the
light incident on a polarizer has been transmitted
by the polarizer when a detection occurs.

I suspect that you and I might have somewhat different
conceptions of what the word "photon" might refer to,
aside from it's existence as a theoretical entity and
a recorded detection.

Regarding anticoincidence experiments using
beamsplitters -- it's the same problem. There's
no way to tell if the light incident on a beamsplitter
and subsequently producing a photon detection at
one detector or the other (but never both)
was unevenly split or traveled only one path or
the other.

Do the uncertainty relations associated with these
types of experiments forbid ever knowing the
answers to these questions (given the current
fundamental quantum of action)?