# Is there a source for this equation for a Bell's Theorem test?

In this, he gives an example of entangled photons spin being measured – one photon sent to a detector at $$0^\circ$$, and the other photon sent to a detector at $$20^\circ$$. He says the odds of both detectors triggering a positive result is $$5.8\%$$, and he gives this as the formula for calculating that:

$$\frac12\sin^2 (20^\circ)$$

I would like to find an external source for this function, but there isn't one provided in that article. I think the example he gives in this article is really brilliant and illustrative and I would like to use a similar example, I just need a reliable source for this piece of it.

• I assume it's this picture that they are describing Aug 14, 2021 at 19:45
• I'd like to point out as a side comment that while Eliezer is a great writer, he's not a physicist and one might want to be a bit wary while reading him if they are not already familiar with the subject. For example, he confuses decoherence with many-worlds, locality with special relativity, etc. in the article you linked (which was a pleasant read but might sow seeds of unnecessary confusion if you are learning the subject for the first time).
– ACat
Aug 14, 2021 at 20:43
• @DvijD.C. oh wow, from a website titled "less wrong" Aug 15, 2021 at 0:03

The idea is that if the photon spin is "$$0^\circ$$ then it has a probability $$\cos^2\theta$$ of being transmitted through a detector oriented at "$$\theta$$". Photon spin is really known as polarization, with "$$0^\circ$$" being labeled as a horizontally polarized photon and $$90^\circ$$ being labeled as a vertically polarized photon. We can write the states as $$|H\rangle$$ and $$|V\rangle$$.
A general state (well, almost the most general but disregarding complex numbers) can be expanded in this basis as $$|\psi\rangle=\cos x|H\rangle+\sin x|V\rangle$$ for some value of $$x$$. The likelihood of it passing through a detector oriented at $$\theta$$ is $$\cos^2(x-\theta)$$. A horizontally polarized photon has $$x=0^\circ$$, so the probability of transmission is $$\cos^2\theta$$ as desired, which goes to zero when the detector is perpendicular to this incident photon.
Now, for the entangled state in question (known as a singlet state), we have a state that is a superposition of two different combinations for the respective polarizations of each photon (I label the photons by $$1$$ and $$2$$): $$|\mathrm{singlet}\rangle=\frac{|H\rangle_1 |V\rangle_2-|V\rangle_1|H\rangle_2}{\sqrt{2}}.$$ When we send the first photon of this joint state through a detector oriented at $$0^\circ$$, only the branch $${|H\rangle_1 |V\rangle_2}$$ of the superposition can make it through. The photons have the same probability of being in this branch as in the other branch, so the probability thus far is $$50\%$$. Next, the second photon needs to make it through a detector (polarizer) oriented at $$\theta$$. The second photon is in state $$|V\rangle$$, which is like $$|\psi\rangle$$ with $$x=90^\circ$$. The probability of success for this second photon is thus $$\cos^2(90^\circ-\theta)=\sin^2\theta$$, which can be proven using trigonometry.
Putting these together, the joint probability is $$P=\frac{1}{2}\times \sin^2\theta$$ as desired.