I understand how one associates the spin of a quantum particle, e.g. of a photon, with intrinsic angular momentum. And in electromagnetism I have always understood the polarization of an EM wave as the oscillations of the E and M field, not necessarily being aligned with the direction of propagation of the wave.
Questions:
But when one talks about the polarization of a photon in Quantum Mechanics, how does it really differ from its spin?
Are they somehow related and what's the physical idea behind photon polarization in contrast to photon-spin? Feel free to use mathematical reasonings as well if you see it fit!
Answer
The short answer is that the spin states of a photon come in two kinds, based on helicity, how the circular polarization tracks with the direction of the photons momentum. You can think of them as circularly polarized in the sense that we can define the relative relationship between the different polarizations the same way we do for classical electromagnetic waves (even though a single photon is not a classical electromagnetic wave), but we'll use the same math and the same terminology.
So I'll talk about polarization of classical electromagnetic waves just because you've already seen it. Imagine a wave travelling in the $z$ direction with the electric field always pointing in the same direction, say $\pm x$. This is called a linearly polarized wave. Same if the wave traveled in the $z$ direction and the electric field was in the plus or minus y direction. If those two waves were in phase and had the same magnitude, then their superposition would be a wave that oscillates at the same frequency/wavelength as the previous waves, and is still linearly polarized but this time not in the $x$ or $y$ direction but instead in the direction $45$ degrees (halfway) between them. Basically if the electric field always points in plus or minus the same direction, then that's linear polarization, and it could in theory be in any direction by adjusting the relative magnitude of an $x$ polarized one and a $y$ polarized one (that are in phase with each other).
OK, what if they aren't in phase, what if they they are a quarter of a period out of phase, then when the x direction is big the y direction is zero, so it points entirely in the x direction, then later it is entirely in the y direction, and so its direction moves in a circle (if the magnitudes of the out of phase fields in the x and y direction are the same magnitude the head does move in a circle, otherwise the head moves in an ellipse). If instead you put them three quarters of a a period out of phase, they will go in a circle in the opposite direction. The waves where the head of the electric field move in a circle are called circularly polarized waves.
OK, that's it for classical waves. You could discuss how photons make up classical waves, but that's not really what the question is about. The question is about spin for photons. And spin states for the photon come in two kinds, and the names for the positive spin $|+\hbar\rangle$ and the negative spin $|-\hbar\rangle$ are plus $|+\rangle$ and minus $|-\rangle$ and you can treat them just like the circularly polarized states.
Now we're going to steal some math and some terminology. Think of multiplying by $i$ as changing the phase of the wave by a quarter period, then we built up a circular polarization by $X+iY$ and the other circular polarization by $X+iii Y=X-iY$ so given two circular polarization you see that we can add them to get a linearly polarized state $|+\rangle + |-\rangle$ gives one of the linearly polarized states and $-i(|+\rangle - |-\rangle)$ gives a linearly polarized state orthogonal to the other one. We can borrow all the math and terminology from the classical waves, and associate the spin states of the photon with the right and left circularly polarized waves.
We are stealing the math and stealing the terminology, but the fact is that we have two vectors $|+\rangle$ and $|-\rangle$ and they span a (complex) two space of possibilities and the basis $$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$$ would work just as well. We could also use $$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$$ which are two more linearly polarized states. Mathematically the spin states are like the left and right circularly polarized waves, so their sum and difference are like the $x$ and $y$ polarized waves but one of them shifted by a phase, and the $45$ degrees tilted ones really are literal sums and differences of the $x$ and $y$ (in phase) waves.
So $\{ |+\rangle , |-\rangle \}$ is one basis,
$\left\{(|+\rangle + |-\rangle), -i(|+\rangle - |-\rangle) \right\}$ is another basis and
$\left\{((|+\rangle + |-\rangle) - i(|+\rangle - |-\rangle)),((|+\rangle + |-\rangle) +i(|+\rangle - |-\rangle))\right\}$ is a third basis.
Each basis can the property that it is equal parts any one from the other two basis sets. And that's what the key distribution is based on. Just having multiple basis for a two dimensional set of states. All I've done above is write everything in terms of the spin states. Mathematically any basis is fine, and all three of these are equally nice in that within a basis the two are orthogonal to each other, and if you pick one from one basis it has equal sized dot products with either of the ones from the other sets.
Worrying about how these relate to classical waves is a distraction since it is the borrowing of the math and the terminology that is going on.
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