# What's the chance that each photon of an entagled photon pair passes through two polarizers?

This is a pretty basic question I think. But it's quite hard to find actual experimental results on the web (or maybe I don't know the right keywords).

I'm new to quantum mechanics and want to understand its effects by seeing the results of experiments.

Assume we have an entangled photon source that emits photons in two directions so their polarization plane is the same*. Each of them pass through a polarizer. And the polarizers have an angle $\alpha$ between their filter planes.

I calculated the chance in classical way. Due to the Malus' law the chance that the photon passes the polarizer is $\mathrm{cos}^2 x$, where $x$ is the angle between the polarizer filter's and the photon's polarization plane.

So the chance that both photons pass through their respective polarizers is: $(\mathrm{cos}^2 x)(\mathrm{cos}^2 (x+\alpha))$. $x$ is a random angle. So to get the general chance I must integrate over $x$ between $0$ and $2\pi$, and take the average by dividing by $2\pi$. I got:

$$\frac{\mathrm{cos}(2\alpha)+2}{8}$$

So this would mean 3/8 chance of pass when the two polarizers have the same and angle have 1/8 chance when they are perpendicular to each other.

Interestingly I also calculated what's the chance that photons fail both filters, so doing the previous computation on $(1-\mathrm{cos}^2 x)(1-\mathrm{cos}^2 (x+\alpha))$, I got the very same formula (I guess not without a reason).

Since these kinds of experiments are performed using coincidence counters and polarizing prisms (so we can count the cases when both of them pass or fail). My prediction would suggest a chance of coincidence that varies between 1/4 and 3/4 with the polarization angle.

That was the classical approach.

Now my questions:

• What probabilities does quantum mechanics predict? Is there 100% match when the polarizers are aligned?
• What did we actually observed when this measurement was performed experimentally?

* For simplicity. I guess in real life the planes would probably be perpendicular. But we can bias the polarizer's angle accordingly.

• A correction: x is the angle between the direction transmitted by the polarization filter and the photon polarization direction (assuming that the photon is linearly polarized. Commented Dec 7, 2014 at 21:59

I understand that initially the photons have both the same polarization, but no fixed polarization relative to the filters. If so, as the relative polarization is random, there should be full symmetry between passing the filters and being stopped by them.

1. For the polarization singlet, (whose state is $$(|x>|x> + |y>|y>)/\sqrt{2}$$ ) and if the two polarizers are aligned, either both pass, either both fail.

2. What was performed experimentally, was the test of a Bell-type inequality named in short CHSH inequality (after its discoverers, Clauser, Horne, Shimony, and Holt). The experiment is done s.t. the each polarizer transmits in another direction, e.g. polarizer for one photon transmits if the photon is x polarized, and the polarizer for the other photon transmits if that photon is polarized along x'.

You have to know that in the photon singlet the two photons have always the same polarizations, but which is the direction of this polarization, it's undefined.

It's a long story to develop the CHSH equality, you can read in Wikipedia the proof of Bell's inequality , and then about the CHSH inequality.

These inequalities say, in short, that for the polarization singlet, one cannot assume that the responses of the photons at the polarizers (passing or not passing) are not predetermined. In other words, we cannot assume that there exists a distribution of properties of the photons, such that the responses of the photons are given according to this distribution.

If such a distribution would exist, a certain expression calculated in base of the results would be smaller than 2 , this is what the CHSH inequality says. But, from the experimental results we obtain that the expression can be bigger than 2.

I hope that I helped.