# Measuring polarization - problem with understanding

Let's assume that we have 2 polarizing filters. First with vertical (1) orientation and second with horizontal (0). I want to measure probability that photon passes through those 2 filters.

I have: ${|\psi\rangle = cos\theta|1\rangle + sin\theta|0\rangle}$

The probability that the photon passes the first filter is $cos^2\theta$ and after that it's polarized vertically so there is probability = 0 that the photon passes second filter which is polarized horizontally. That's easy and understandable for me.

But the problem is when we add third filter between previous two. This filter will be oriented at 45 degrees. It blocks and passes photon with the same probability 1/2 and it's said that after passing this filter, the new polarization of photon is: ${\frac{1}{\sqrt2}|1\rangle + \frac{1}{\sqrt2}|0\rangle}$. So I have probability 1/2 for passing first filter and probability 1/2 that photon will be polarized horizontally before last filter. Total probability is 1/4.

But how we can say that photon pass through the second filter if we know that after passing the first one it's polarized vertically? For me there is probability 0 to passing second filter if that filter isn't polarized vertically.

But how we can say that photon pass through the second filter if we know that after passing the first one it's polarized vertically? For me there is probability 0 to passing second filter if that filter isn't polarized vertically.

Polarizing filters don't just discard photons, they change the polarization of photons. This is simply true, by experiment.

If a photon just went through a vertical filter, the photon is now vertically polarized regardless of what the photon's polarization was beforehand. The beforehand polarization determines the probability of making it through the filter, but otherwise has no effect on the afterwards polarization. If that goes against your intuition... well, your intuition is wrong and you need to learn to ignore it in this case.

1. Initial setup with vertical photon heading rightward towards vertical then diagonal then horizontal polarizers.

                 |||            ///              ___
photon|V| ---->  |||           ///               ___
|||          ///                ___

2. Photon reaches first polarizer. 100% chance of making it through.

                 |||            ///              ___
--|V|->||           ///               ___
|||          ///                ___

3. Photon passed through.

                 |||            ///              ___
||| --|V|->   ///               ___
|||          ///                ___

4. Photon reaches diagonal polarizer. 50% chance of transmission.

                 |||            ///              ___
|||     --|V|->//               ___
|||          ///                ___

5. Phew, we won the coin flip! Photon made it through, but now it's diagonally polarized.

                 |||            ///              ___
|||           /// --/D/->       ___
|||          ///                ___

6. Photon reaches horizontal beam splitter. 50% chance of transmission.

                 |||            ///              ___
|||           ///         --/D/->__
|||          ///                ___

7. We won the coin flip again! Only had a 25% chance of making it this far. Photon exits rightward, but horizontally polarized.

                 |||            ///              ___
|||           ///               ___  --_H_->
|||          ///                ___


And if you think that's weird, read up on the quantum Zeno effect. As you use more and more polarizers, with finer and finer changes in angle going from vertical to horizontal, you end up with a thing that rotates vertical photons into horizontal photons with negligible loss!

$$\lim_{n \rightarrow \infty} \cos^{2n} \frac{90^\circ}{n} = 100\%$$

In the small-angle limit, the turning effect beats the filtering effect! So maybe calling them "filters" wasn't ideal.

• Isn't it 25% chance that photon passes through 3 polarizers but when isn't polarized before first polarizer? – bawq Jul 17 '16 at 14:48
• @user No, that'd be 12.5%. – Craig Gidney Jul 17 '16 at 16:53
• And last question. Is it always that unpolarized photon passes through any polarizer with probability 50%? – bawq Jul 17 '16 at 17:52
• @user Right. If you don't know the polarization then you expect 50/50 transmission. – Craig Gidney Jul 17 '16 at 18:01

Take a look online at the "Dirac three polarizer experiment" for a more comprehensive answer.The key points are well explained by the following illustrations from "http://alienryderflex.com/polarizer/" keep in mind that these illustrations are based upon an initial horizontal polarizer, not a vertical one as stated above, but the concept is the same either way:

before / after middle filter http://alienryderflex.com/polarizer/13.gif

before / after last filter http://alienryderflex.com/polarizer/14.gif