# How can I construct linear polarization states of a photon from circular polarization states?

I'm a second-year math student and one of the courses is physics. We just started quantum mechanics, and the lecturer briefly explained the Bra-ket formalism and gave us students the following assignment:

Photon is a quantum state of electromagnetic field and can have various polarizations in the xy plane, which is perpendicular to the photon propagation direction z. If we focus exclusively on the polarization, then there are only two quantum states: the right-handed polarization corresponds to state vector $$|r \rangle$$ in the Hilbert space of polarizations, and the left-handed polarization corresponds to state vector $$|l \rangle$$, these states being perpendicular to each other. A linear polarization corresponds to a superposition of these two states with equal weights. Construct two mutually perpendicular states corresponding to linear polarizations along x and y.

My first guess is $$$$| x \rangle = (| r \rangle + | l \rangle)/\sqrt{2},$$$$ $$$$| y \rangle = (| r \rangle - | l \rangle)/\sqrt{2},$$$$ as these states are mutually perpendicular and include $$|r \rangle$$ and $$|l \rangle$$ with equal weights as required in the formulation of the problem. I included the factor $$1/\sqrt{2}$$ to ensure that $$\langle r|r\rangle = 1$$ and $$\langle l|l\rangle = 1$$.

However, as far as I remember the material of the previous year on classical electrodynamics, a circular polarization is a superposition of two mutually perpendicular linear polarizations with one of them being $$\pi/2$$ behind in the phase. This suggests the following answer for the above problem: $$$$| x \rangle = (| r \rangle + i | l \rangle)/\sqrt{2},$$$$ $$$$| y \rangle = (| r \rangle - i | l \rangle)/\sqrt{2},$$$$

What is the correct answer for the above problem? Both (1)-(2) and (3)-(4), or only (3)-(4)? Or something different?

If right and left are given by $$|r\rangle = \frac{1}{\sqrt{2}} (|x\rangle + i |y\rangle )$$ $$|l\rangle = \frac{1}{\sqrt{2}} (|x\rangle - i |y\rangle )$$
$$|x\rangle = \frac{1}{\sqrt{2}} (|r\rangle + |l\rangle )$$ and subtracting them will give you: $$|y\rangle = \frac{1}{\sqrt{2}i} (|r\rangle - |l\rangle )$$ But the $$i$$ in the denominator just adds a global phase, so you get exactly (1)-(2)