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I understand the classical model of light polarization in terms of two complex numbers, known as Jones vector.

In the quantum case, for example, consider photons sent through two polarizers, the first one at an angle $\theta$ with the second one, aligned on the $x$-axis (perpendicular to the $y$-axis).

According to The Feynman Lectures on Physics, vol III, § 11-4, the state vector of a photon passing the first polarizer is $$\cos\theta\ |x\rangle+\sin\theta\ |y\rangle,$$ where $|x\rangle,|y\rangle$ are the state vectors of photons polarized along $x,y$-axes.

Similarly, the state vector of a right polarized photon is $$\cos\theta\ |x\rangle+ \exp(i\ \pi/2)\ \sin\theta\ |y\rangle,\theta=\frac{\pi}{4}$$

Why is it so?

The moduli of coordinates are the square roots of the probabilities of transmission, $\cos^2\theta,\ \sin^2\theta$, set by classical electrodynamics.

But how are the coordinate phase differences ($y$ less $x$, $0$ in the first case, $\pi/2$ in the second case) determined?

They are just taken from the classical case: the quantum state vector is identified with the classical Jones vector.

Can it be justified from symmetries in the 2D Euclidean space $(x,y)$, where the electrical field lies?

Is it a theorem in quantum electrodynamics?

Pierre ALBARÈDE
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  • It is not clear what the second expression means. – flippiefanus Apr 10 '23 at 04:08
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 10 '23 at 05:04
  • For starters: a polarizer is one of three things: it is either absorbing part of the energy of the field OR it is changing the total angular momentum of the field OR both. I hope that Feynman's book doesn't actually contain a phrase that says that "a photon" has a wave vector. A multi-photon wave function may have a wave vector in the classical limit, but a single photon has only an energy, a momentum and an angular momentum. – FlatterMann Apr 10 '23 at 07:32
  • The second expression is equivalent to $\frac{1}{\sqrt{2}}(|x\rangle+i\ |y\rangle)$ but shows $\theta$ and the phase $\pi/2$. – Pierre ALBARÈDE Apr 10 '23 at 07:45
  • In short, the problem is that phases are taken from the classical case to the quantum case without any justification, not even an hypothesis. – Pierre ALBARÈDE Apr 10 '23 at 07:49
  • To avoid some confusions, I have replaced "wave vector" in the question by the more common "state vector" (which is even more confusing, but this is another problem). – Pierre ALBARÈDE Apr 10 '23 at 08:20
  • This may useful https://physics.stackexchange.com/questions/305609/specifying-the-state-of-polarization-of-a-photon-a-classical-polarized-beam-of?rq=1 – Pierre ALBARÈDE Apr 10 '23 at 22:30
  • You can show that linear-optical components, such as polarizers and linearly birefringent crystals, transform single-photon states and classical states of light the same way. That's why you get an equivalence with the Jones vector. You shouldn't think of the amplitudes as square roots of the probabilities, it's the other way around. The amplitudes correspond to the transmission coefficients for the classical electric fields. – fulis Apr 21 '23 at 08:40
  • I can´t show this. Can you? The problem is only with the phases. The theory to be used is quantum electrodtnamics, I guess. – Pierre ALBARÈDE May 09 '23 at 10:34

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