Photon entanglement: why three angles?

[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:
$\frac{1}{2\pi} \int d\alpha (P(a|\alpha) P(b|\alpha) +\bar{P}(a|\alpha) \bar{P}(b|\alpha)<br /> -P(a|\alpha) \bar{P}(b|\alpha) - \bar{P}(a|\alpha) P(b|\alpha))$

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.