Photon Polarization

1. Aug 18, 2011

BWV

Peres, p47 has an thought experiment where:

1 million linearly polarized photons in the x-direction and another million polarized in the y-direction are placed in a box where they travel in the +- z direction with no change in the polarizations. No record is kept of the order in which they are introduced, only the total numbers are recorded. A second box repeats this but with circularly polarized photons - a million clockwise and a million counter-clockwise. You are given one of the boxes at random and test each photon with a perfect detector. what is the probability of making a wrong guess?

The author lists it as approximately $(4\pi * 10^6)^{-1/2}$

My thinking is that testing for linear vs. circular polarization are non-complementary, pure states so if you get the linear box and set up a test for circular polarization, then you will get a random answer - a 50/50 chance of being either clockwise or counterclockwise. If you test for linear polarization on the linear box with a perfect detector you will recover the original polarization.

So I get the probability of making a wrong guess of which box you received is just smidgen under 50% : the prob of getting the right box less the prob of getting the wrong box and getting a random outcome where exactly 1,000,000 photons are detected in one of the polarized states.

Obviously, I am way off in my thinking - where am I going wrong here?

2. Aug 18, 2011

xts

Let's forget about all QM (and assume the photons are not coherent, so they are described by simple statistics of independent particles rather than by Bose statistics) and take it in a purely Bayesian probability approach.

Let's utilise the detector: the one able to pick just one photon at time and measure its polarisation in H/V base. As a result we get a series of 2 million bits.

For a box with circular photons the series should be absolutely random, exhibiting no autocorelations.

For a box with mix of H/V photons, you must get exactly 1,000,000 H photons and 1,000,000 V.

The same measurement on circular box would give an outcome of binomial distribution, which may be approximated by Gaussian. Thus every outcome different than (1000000, 1000000) gives you certainity that it is a circular box. If you get the result (1,000,000, 1,000,000) you claim you got the H/V box and probability of false claim is $$\left({2,000,000\atop 1,000,000}\right)\frac{1}{2^{2,000,000}}\approx {\cal N}(0, \frac{1}{4}2,000,000)= \frac{1}{\sqrt{2\pi}\sqrt{500,000}}=\frac{1}{\sqrt{4\pi\cdot 10^6}}$$

- what Peres claims.

Last edited: Aug 18, 2011
3. Aug 18, 2011

BWV

Thanks,

so it looks like my understanding of the physics was OK, I just interpreted the question as the probability of making an error as to the original polarization of the particles, not the probability of guessing the wrong box.

So after testing the linear box with the circular detector is it correct that you could not then use the linear detector on the box and recover the original polarizations?