Photon entanglement: why three angles?

[stevendaryl]

From the Wikipedia article I quoted it looks like locality in EPR experiments is defined exactly by the independence of the two data streams, which translates into prediction they should be completely random (50-50%) regardless of any absolute or relative polarizers rotation.

If that's the impression you got from the Wikipedia article, then it needs to be rewritten, because that's absolutely not true.

If random polarization is uniform then integrated average of cos^2 over 360° is 1/2, that's the same 50-50% Wikipedia is talking about. If the ratio is not 50-50% for any arbitrary (a-b) angle combination, then the theory is not local, or the experiment is not rotationally invariant.

That's not correct. Here's a local realistic model: You generate a pair of photons that are polarized at angle $\alpha$, where $\alpha$ is chosen randomly. Then, the probability of passing through a filter is $cos^2(\alpha - \theta)$ where $\theta$ is the orientation of the filter. Then the correlation $E(a,b)$ will be given by:

$E(a,b) = \frac{1}{2\pi}\int d\alpha (cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b) - cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b))$

The positive terms, $cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b)$, give the probability of both filters having the same result--either they both pass, or they both are blocked. The negative terms, $cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b))$ give the probability that the two filters get different results--one passes and the other is blocked.

You can go through it yourself, if you know trigonometry. The answer is:

$E(a,b) = \frac{1}{2} cos(2(a-b))$

which is definitely not zero, except in the case where $a-b = \frac{\pi}{4}$

 

[johana]

So in the classical case, one doesn't have to get that $P_{AB}(+ and +) = P_{AB}(- and -) = P_{AB}(+ and -) = P_{AB}(- and +) = 0.25$. In the case you calculated the results are correlated, because both photons were prepared in the same state.

Yes, but it depends on rotational invariance which is prerequisite for local theory prediction of 50-50%. So if rotational invariance is assumed (theory), or guaranteed (experiment), then cos^2(θ) is uniformly integrated over 360° which averages out to 50-50%, regardless of any absolute or relative polarizers rotation, which then combined with the other side splits up into 25% for each of 4 possible combinations.

 

[atyy]

Yes, but it depends on rotational invariance which is prerequisite for local theory prediction of 50-50%. So if rotational invariance is assumed (theory), or guaranteed (experiment), then cos^2(θ) is uniformly integrated over 360° which averages out to 50-50%, regardless of any absolute or relative polarizers rotation, which then combined with the other side splits up into 25% for each of 4 possible combinations.

The integration needs some work. So let's first try a simpler scenario to get the 50-50. Let's say both polarizers are oriented at 0°. For half the pairs, both photons are polarized at 0°; the other half of the pairs have both photons polarized at 90°. What are the probabilities P(++), P(--), P(+-), P(-+)?

 

[atyy]

Yes, but it depends on rotational invariance which is prerequisite for local theory prediction of 50-50%. So if rotational invariance is assumed (theory), or guaranteed (experiment), then cos^2(θ) is uniformly integrated over 360° which averages out to 50-50%, regardless of any absolute or relative polarizers rotation, which then combined with the other side splits up into 25% for each of 4 possible combinations.

If you want to try the integration, under the assumption that all polarizations are equally likely, but that both photons always have the same polarization, and that both polarizers are oriented at the same constant angle, then to get P(++) you should integrate cos²(θ)cos²(θ) = cos⁴(θ) over 360°, where θ is the angle between a photon and a polarizer. Similarly to get P(--) you should integrate sin⁴(θ) over 360°. I'm not sure I did this right, but I get the following.

P(+) = 1/2
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(cos(x))^2+dx+from+x+=+0+to+2*pi

P(-) = 1/2
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(sin(x))^2+dx+from+x+=+0+to+2*pi

P(++) = 3/8
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(sin(x))^4+dx+from+x+=+0+to+2*pi

P(--) = 3/8
http://www.wolframalpha.com/input/?i=integrate+(1/(2*pi))+*+(sin(x))^4+dx+from+x+=+0+to+2*pi

P(+-) = P(-+) = 1/8
http://www.wolframalpha.com/input/?...*+(cos(x))^2+(sin(x))^2+dx+from+x+=+0+to+2*pi

So P(++) + P(--) - P(+-) - P(-+) = 6/8-2/8 = 1/2, which is another example of classical correlations built in at the source producing correlated outcomes.

 

[Avodyne]

A function which outputs an average, such as Malus' law, is not deterministic but probabilistic.

Étienne-Louis Malus died in 1812. He never heard of quantum mechanics, and certainly thought of his law as a deterministic formula for the intensity of EM radiation passing through a polarizer.

It is true that, in quantum electrodynamics, Malus' Law becomes probabilistic. According to quantum electrodynamics, this probability has a purely quantum mechanical origin, and does not arise "due to microscopic differences between the polarizers and the point of impact relative to the polarizer molecular structure".

That's the conventional QM story. In hidden-variable theory, some underlying hidden variables are supposed to determine what happens to each and every photon, in a fully deterministic way. But no such theory can reproduce the predictions of QM.

 

[johana]

That's not correct. Here's a local realistic model: You generate a pair of photons that are polarized at angle $\alpha$, where $\alpha$ is chosen randomly. Then, the probability of passing through a filter is...

http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties

A beam of unpolarized light can be thought of as containing a uniform mixture of linear polarizations at all possible angles. Since the average value of $cos^2 \theta$ is 1/2, the transmission coefficient becomes $\frac {I}{I_0} = \frac {1}{2}$.

$\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2$

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

 

[atyy]

http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties$\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2$

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

Sure, that's just P(+) and P(-). To get P(++), you have to multiply the probability the probability to get + on each side for each angle and integrate over all angles, ie. integrate cos⁴(x) in the case that both polarizers are set to the same angle. In fact stevendaryl gave the general answer in the same post you quoted:

That's not correct. Here's a local realistic model: You generate a pair of photons that are polarized at angle $\alpha$, where $\alpha$ is chosen randomly. Then, the probability of passing through a filter is $cos^2(\alpha - \theta)$ where $\theta$ is the orientation of the filter. Then the correlation $E(a,b)$ will be given by:

$E(a,b) = \frac{1}{2\pi}\int d\alpha (cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b) - cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b))$

The positive terms, $cos^2(\alpha - a) cos^2(\alpha - b) + sin^2(\alpha - a) sin^2(\alpha - b)$, give the probability of both filters having the same result--either they both pass, or they both are blocked. The negative terms, $cos^2(\alpha - a) sin^2(\alpha - b) - sin^2(\alpha - a) cos^2(\alpha - b))$ give the probability that the two filters get different results--one passes and the other is blocked.

You can go through it yourself, if you know trigonometry. The answer is:

$E(a,b) = \frac{1}{2} cos(2(a-b))$

which is definitely not zero, except in the case where $a-b = \frac{\pi}{4}$

.

 

[stevendaryl]

http://en.wikipedia.org/wiki/Polarizer#Malus.27_law_and_other_properties


$\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2$

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

And what do you think that number is showing? That's the probability of any single filter passing a photon. It doesn't tell you anything about the correlation between two different filters.

To compute the correlation of two filters, one oriented at angle $a$, and one oriented at angle $b$, you have to consider the following four numbers:

1. $P(a|\alpha) = cos^2(a-\alpha)$ the probability that a photon with polarization $\alpha$ passes through a filter at angle $a$.
2. $P(b|\alpha) = cos^2(b-\alpha)$ the probability that a photon with polarization $\alpha$ passes through a filter at angle $b$.
3. $\bar{P}(a|\alpha) = sin^2(a-\alpha)$ the probability that a photon with polarization $\alpha$ does not pass through a filter at angle $a$.
4. $\bar{P}(b|\alpha) = sin^2(b-\alpha)$ the probability that a photon with polarization $\alpha$ does not pass through a filter at angle $b$.

Then the correlation $E(a,b)$ is given by:
$\frac{1}{2\pi} \int d\alpha (P(a|\alpha) P(b|\alpha) +\bar{P}(a|\alpha) \bar{P}(b|\alpha) -P(a|\alpha) \bar{P}(b|\alpha) - \bar{P}(a|\alpha) P(b|\alpha))$

That number is $E(a,b) = \frac{1}{2}cos(2(a-b))$

 

[Nugatory]

Johana, I said this in you other thread, but repeating it here:

Here is a question for you (and it is not a rhetorical question):

Have you read and understood the EPR paper and Bell's paper? If you haven't read them, you're wasting your time and ours. If you have read them, and there are parts of the arguments that you don't follow, ask and we can have a more focused and productive discussion.

 

[DrChinese]

$\int_0^{2\pi} \frac{cos^2(x)}{2\pi} dx = 1/2$

Wolfram: integrate 1/(2pi) * cos^2(x) dx, x = 0 to 2pi

As stevendaryl and Nugatory and billschnieder have been saying: the reason things are going in circles is because the compass has been lost. What relevance is the above?

We all are familiar with the math of Bell, entangled photons, etc. There are a lot of very similar looking formulae, and the key is to keep things labeled and moving in a direction.

The issue in this thread is that it takes 3 angles, a/b/c, to get the Bell outcome. There are a variety of different candidate local realistic theories that can be tested against this backdrop, and then shown not to match the predictions of QM. As we have said repeatedly, the approach you are taking gives a prediction that is substantially at odds with QM (and experiment). No surprise there, that's Bell. The part none of us follow is: do you see why? Because it doesn't matter if you present a formula and integrate it if you don't know where you are going.

Fact 1: all entangled photon pairs will yield 100% correlated (or anti-correlated depending on type) results when measured at the same angle.

Fact 2: entangled photon pairs act and are best described as single systems of 2 particles, not 2 systems of 1 particle. QM and experiment match.

Fact 3: all local realistic theories are predicated on the idea that entangled photons are fully independent and separable entities, and there is no ongoing physical connection. Bell says no such local realistic theory can yield predictions consistent with QM.


Do you understand these 3 things? If you do not, please let us know which you don't.

 

[Nugatory]

Closed - this discussion is no longer adding any value.