Can grandpa understand the Bell's Theorem?

[DrChinese]

...It is why I prefer to use the interpretation of EPR as it was stated in the original paper (1935) according to which correlated particles after separation:

“… are no longer interact, no real change can take place in the second system in consequence of anything that may be done to the first system… thus it is possible to assign two different wave functions… to the same reality …”

And that was a wrong statement from EPR, as we now know (but was NOT clear then). For particles in a state of superposition, there is only one wave function. Please recall that the whole point was that the results must be predetermined (FAPP) if Alice and Bob get the same answers to the same questions every time (with "certainty") without otherwise communicating.

 

[harrylin]

Please recall that the whole point was that the results must be predetermined (FAPP) if Alice and Bob get the same answers to the same questions every time (with "certainty") without otherwise communicating.

Isn't "predetermined" a special, limiting case of "determinism"? As we know from Bell's "Bertlmann's socks", perfect certainty is not claimed and determinism is not the essential point. The essential point is locality, as EPR also stated.

 

[miosim]

SpectraCat,


I took some time to better undersatnd the following post.

…

I would like to verify that I understand you correctly prior to continue. I think that now I am on the same page with you regarding therminology of CORRELATION and ANTI-CORRELATION that applies to the SAME or OPPOSITE (90 deg shift) polarisation. Also, the “cos^2” isn’t a correlation (as I called it before) but the “joint detection probability” that allows to calculate the “correlation coefficient.”

…you CAN get the cos^2 correlation for gedanken #2, but ONLY for one choice of detection basis, namely the original polarization basis for the photons. In other words, if you keep the A detector set at 0º relative to the original polarization basis (the |H> and |V> polarization defining the entangled state as it was emitted from the source), then you will get cos^2 as you rotate the B detector angle from 0º to 90º.

Do you mean that in this case we simply abserve the Malus’ law?

…However, if you start with A and B both at the same random angle (say 79º) with respect to the original polarization basis, and then rotate B from 79º to 169º then you will not get a simple cos^2 relation anymore.

Is it beause in the gedanken experiment #2 the photons are unentangled (polarisation corelated but fully independent)?

 

[SpectraCat]

SpectraCat,


I took some time to better undersatnd the following post. I would like to verify that I understand you correctly prior to continue. I think that now I am on the same page with you regarding therminology of CORRELATION and ANTI-CORRELATION that applies to the SAME or OPPOSITE (90 deg shift) polarisation. Also, the “cos^2” isn’t a correlation (as I called it before) but the “joint detection probability” that allows to calculate the “correlation coefficient.”

Yes.

…you CAN get the cos^2 correlation for gedanken #2, but ONLY for one choice of detection basis, namely the original polarization basis for the photons. In other words, if you keep the A detector set at 0º relative to the original polarization basis (the |H> and |V> polarization defining the entangled state as it was emitted from the source), then you will get cos^2 as you rotate the B detector angle from 0º to 90º.

Do you mean that in this case we simply abserve the Malus’ law?

Yes, for measurements in the polarization basis in which the non-entangled pair was initially generated, and for that basis ONLY, you can get the Malus' Law relationship. On the other hand, for entangled photons, you see the Malus' Law relationship in whatever measurement basis you choose. That was the experimental observation of Aspect which led to the Bell's inequality violation, which has subsequently been re-tested and improved upon by other groups.

…However, if you start with A and B both at the same random angle (say 79º) with respect to the original polarization basis, and then rotate B from 79º to 169º then you will not get a simple cos^2 relation anymore.

Is it beause in the gedanken experiment #2 the photons are unentangled (polarisation corelated but fully independent)?

Yes, precisely.

 

[miosim]

And that was a wrong statement from EPR, as we now know (but was NOT clear then). For particles in a state of superposition, there is only one wave function. Please recall that the whole point was that the results must be predetermined (FAPP) if Alice and Bob get the same answers to the same questions every time (with "certainty") without otherwise communicating.

http://plato.stanford.edu/entries/qt-epr/#1.3 (Einstein's versions of the EPR argument):

“… The central point of EPR was to argue that in interpreting the quantum state functions the real states of spatially separate objects are independent of each other …”

 

[miosim]

SpectraCat,

According to you, in the gedanken experiment #2 we will not get a simple cos^2 correlation coefficient because their photons are non-entangled. According to my prediction the gedanken experiment #2 does produce the cos^2 correlation coefficient, because in my interpretation this result is nothing to do with entanglement (influence over distance). This is why this experiment questions a role (and probably the existence) of the influence over distance that is the most important conclusion of the Bell theorem. This what I mean by saying that the gedanken experiment #2 can falsify the Bell's theorem.

 

[SpectraCat]

SpectraCat,

According to you, in the gedanken experiment #2 we will not get a simple cos^2 correlation coefficient because their photons are non-entangled. According to my prediction the gedanken experiment #2 does produce the cos^2 correlation coefficient, because in my interpretation this result is nothing to do with entanglement (influence over distance). This is why this experiment questions a role (and probably the existence) of the influence over distance that is the most important conclusion of the Bell theorem. This what I mean by saying that the gedanken experiment #2 can falsify the Bell's theorem.

Well, with all due respect, I can only say that is because you do not understand the underlying mathematics. There is simply no way that the observed cos^2 dependence can be reproduced (for arbitrary choice of measurement basis), by photons that are not entangled. Your prediction above is not even consistent with classical physics, let alone quantum mechanics. It would be fairly straight forward to set up an experiment with polarized light beams to demonstrate that .. I don't even think you need coincidence counting. Just have two independent sources with polarizers .. one polarizer is set to 0º and the other is set to 90º, and have them synchronized to flip their orientations at random time intervals. Then set up the A & B measurement stations with independent polarizers and simple intensity detectors (much cheaper than the single photon counters used for experiments with entangled photons).

Such a setup is basically identical to your gedanken #2, because you claim that the photons are independent, so being able to pair specific photons by coincidence counting won't make any difference. Comparison of the angle-dependent intensities at detectors A and B will reproduce the local realistic predictions for unentangled photons. You certainly will not see a cos^2 relationship that depends only on the relative angle between A and B. If you think differently, please explain why in the context of this experiment, which is much easier to understand and analyze.

 

[miosim]

Well, with all due respect, I can only say that is because you do not understand the underlying mathematics. There is simply no way that the observed cos^2 dependence can be reproduced (for arbitrary choice of measurement basis), by photons that are not entangled.

It is true that I do not understand the underlying mathematics of the cos^2 correlation coefficient predicted by the QM.
However I treat QM as an empirical theory that predicts the outcome of the measurements; everything else is interpretation. In this particular examples, the claim that cos^2 correlation coefficient is caused by entanglement is just an interpretation of this result.
Can you rule out that it is possible to produce the same prediction (cos^2 correlation coefficient) using the QM model of the two independent wave functions?

Your prediction above is not even consistent with classical physics, let alone quantum mechanics.

I don’t expect that my prediction is consistent with classical physics, because, as I understand, Gedanken experiment #2 is within QM domain (regardless that I am not in an agrement with the traditional views on the wave function collapse).

It would be fairly straight forward to set up an experiment with polarized light beams... I don't even think you need coincidence counting. Just have two independent sources with polarizers ... one polarizer is set to 0º and the other is set to 90º, and have them synchronized to flip their orientations at random time intervals…

Agree.

… Then set up the A & B measurement stations with independent polarizers and simple intensity detectors (much cheaper than the single photon counters used for experiments with entangled photons). Such a setup is basically identical to your gedanken #2, because you claim that the photons are independent, so being able to pair specific photons by coincidence counting won't make any difference. Comparison of the angle-dependent intensities at detectors A and B will reproduce the local realistic predictions for unentangled photons. You certainly will not see a cos^2 relationship that depends only on the relative angle between A and B. If you think differently, please explain why in the context of this experiment, which is much easier to understand and analyze.

In general I agree with you. I do expect that the simple intensity detectors may be enough. However, because I don’t understand the underlying mathematics, I can’t be confident that the intensity measurement is equivalent to a correlation coefficient. It is why in my gedanken experiment #2 I try to limit the difference between Aspect’s and my experiments to the entanglement only.

 

[miosim]

P.S.
Actually in the setup you described, if the source of photon beam is polarized, the intensity at each detectors will follow the cos^2 of the angle of the corresponding detectors, independently of the detector angle on other side. But this wouldn't prove anything.

 

[DrChinese]

http://plato.stanford.edu/entries/qt-epr/#1.3 (Einstein's versions of the EPR argument):

“… The central point of EPR was to argue that in interpreting the quantum state functions the real states of spatially separate objects are independent of each other …”

Yes, that was their incorrect argument.

 

[DrChinese]

SpectraCat,

According to you, in the gedanken experiment #2 we will not get a simple cos^2 correlation coefficient because their photons are non-entangled. According to my prediction the gedanken experiment #2 does produce the cos^2 correlation coefficient, because in my interpretation this result is nothing to do with entanglement (influence over distance). This is why this experiment questions a role (and probably the existence) of the influence over distance that is the most important conclusion of the Bell theorem. This what I mean by saying that the gedanken experiment #2 can falsify the Bell's theorem.

Basically, you are making an incorrect prediction for an experiment that has already been performed and yielded different results. Also, yours is NOT the classical prediction. It is just something you made up out of thin air with not the slightest justification or mathematical derivation. You may as well claim that your prediction is .758401 and state that performing the experiment will falsify Bell. Which it wouldn't. It would "only" falsify QM.

 

[miosim]

http://plato.stanford.edu/entries/qt-epr/#1.3 (Einstein's versions of the EPR argument):
“… The central point of EPR was to argue that in interpreting the quantum state functions the real states of spatially separate objects are independent of each other …”

Yes, that was their incorrect argument.

Can you collaborate about disprove of EPR argument?

Basically, you are making an incorrect prediction for an experiment that has already been performed and yielded different results…

Can you please describe this experiment?

… Also, yours is NOT the classical prediction.
What does classical prediction mean? …

It is just something you made up out of thin air with not the slightest justification or mathematical derivation.

My prediction is based on the same mathematical derivation as prediction for Aspect's experiment, but on the different interpretation of events; in my interpretation the cos^2 correlation coefficient is nothing to do with non-local interactions.

Can you demonstrate that this mathematical derivation is based on non-local interactions?

 

[DrChinese]

Can you collaborate about disprove of EPR argument?Can you please describe this experiment?

This has already been done many times in this circular thread. After 400 posts, if you don't know that Bell's Theorem is a disproof of EPR, I really don't know what to tell you. As to the experiment, I have told you previously that I don't have a reference and don't know any reason there would be one. Maybe you will be the first to write up this null result.

 

[JDoolin]

\Psi=(|H_A>\otimes|V_B> + |V_A>\otimes|H_B>)

Is there anywhere online where I could get an introductiton to that notation? (or if not online, a good introductory text?) I had one professor that called it "bra" "ket" notation, and usually <A|, the "bra" represented an operator, and |B> represented a "ket" which was some numerical value or vector or matrix.

But I can't find anything under "bra ket notation," so I guess that terminology isn't in common usage.

Anyway, when I last posted to this thread, I was wondering whether somehow the |H_A&gt; , |H_B&gt; notation referred somehow to matrices, for instance

|H_A>=\begin{pmatrix}
cos(\theta_A )\\
sin(\theta_A)
\end{pmatrix}?

But with the multiplication |H_A&gt; |H_B&gt; of course, matrices can't be multiplied that way, but you introduced a new operator here: \otimes which might resolve it?

 

[SpectraCat]

Is there anywhere online where I could get an introductiton to that notation? (or if not online, a good introductory text?) I had one professor that called it "bra" "ket" notation, and usually <A|, the "bra" represented an operator, and |B> represented a "ket" which was some numerical value or vector or matrix.

But I can't find anything under "bra ket notation," so I guess that terminology isn't in common usage.

Anyway, when I last posted to this thread, I was wondering whether somehow the |H_A&gt; , |H_B&gt; notation referred somehow to matrices, for instance

|H_A>=\begin{pmatrix}
cos(\theta_A )\\
sin(\theta_A)
\end{pmatrix}?

But with the multiplication |H_A&gt; |H_B&gt; of course, matrices can't be multiplied that way, but you introduced a new operator here: \otimes which might resolve it?

It is called Dirac notation, or bra-ket notation. The wikipedia page is a decent place to start for an intro. The \otimes is just used to indicate that the state is composed of two "kets" (i.e. vectors) from different vector spaces. For example, if you treat a molecule in the Born-Oppenheimer approximation, you could write it's total wavefunction as a composition of the electronic and vibrational wavefunctions (there are other contributions which might also be important), which are solved independently.

It does not make sense to write the product of two kets from the same vector space (i.e. |m>|n> is non-sensical), so the \otimes is useful to make explicit that the two states come from different vector spaces. Note that the notation for the dot product of a bra and a ket vector (<m|n>) does NOT combine vectors from the same space. The space of bra vectors is "dual" to the space of "ket" vectors, but is not identically the same. For example, you might write the position space representation of some ket |n> as the wavefunction \psi_n(x) ... in that case, the position space representation of the bra <n| that is dual to |n> would be the complex conjugate of the same wavefunction ... i.e. \psi_n^*(x).

Hope this helps.