# Linear polarization and phase

In this discussion of polarization, it says

For linearly polarized light, the $x$- and $y$-components are exactly in phase and therefore rise and fall together

Can it therefore be assumed, then, that if light is linearly polarized, it is in phase? More specifically, if two photons, separate from each other, are linearly polarized, are they in phase with each other?

I apologize if this exhibits a misunderstanding of the topic; I'm rather unfamiliar with optics.

• It's not that light is in phase. When talking about quantities being in phase, it (in your question's case) means the phase difference between the 2 quantities: the $x$ and $y$ components of $\textbf{E}$. What I think you're probably trying to ask is if light has the same polarization throughout the beam, which is true, which in the quantum picture is because every photon has the same polarization state vector. – Avantgarde Jul 23 '17 at 3:28
• It took me a while, when I was learning, to recognize just how widespread the concept of 'phase' (and thus of being in or out of phase) is in physics. It shows up in endless applications, and this is just one of several contexts where it can be applied in more than one way in a single description. It is useful to pay attention to any hints in the text about what things are being compared (in this case the electric and magnetic fields rather than different parts of a light front or different fronts). – dmckee --- ex-moderator kitten Jul 23 '17 at 18:10

Linearly polarized light is linearly polarized because the x and y components of the electric field are in phase with each other. All this means is that the the x and y components both have the same factor of $\cos(kz-\omega t+\phi)$. For light that is circularly polarized, they have different $\phi$s that differ by $90^\circ$.