In particle physics, we say: a particle has +1 helicity (right-handed) if its momentum and spin are parallel, or it has -1 helicity (left-handed) if its momentum and spin are antiparellel. Now, if we suppose, the particle is a photon, then the helicity states correspondence to either left or right circular polarization. My question is, which helicity state corresponds to right circular polarization and which corresponds to left circular polarization?
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$\begingroup$ WP & also. $\endgroup$– Cosmas ZachosCommented Apr 10 at 14:49
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$\begingroup$ WP. Indeed, the optics convention and the helicity convention are opposite... $\endgroup$– Cosmas ZachosCommented Apr 10 at 22:36
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For historical reasons, the optics convention and the helicity convention are opposite. The historical reason is that we didn't originally know that these two things were related, and so the initial labels were completely arbitrary, and happen to not coincide.
In the particle convention, we define the angular momentum using the right-hand rule. For instance the Earth, which rotates to the east, has an angular momentum vector pointing out of its north pole. We say that a "right-handed" photon, or better a "positive helicity" photon, is one whose spin $\vec\sigma$ and momentum $\vec p$ have pseudoscalar product $\vec\sigma \cdot \vec p > 0$.
But in optics, we imagine the circularly-polarized light approaching us, and ask whether the electric field vector at a point in space goes clockwise or counterclockwise from the receiver's perspective. Counter-clockwise, where the motion at the top of the circle is to the left, is "left circularly polarized."
So, take your right hand and make a loose fist. A "positive helicity" particle is one whose momentum goes in the direction of the thumb. Now point that thumb at your face, and observe that the fingers go to your left.
Note that this connection was experimentally confirmed in my favorite classic physics paper: Richard Beth, Mechanical detection and measurement of the angular momentum of Light, Physical Review 50 115 (1936). I've mentioned or briefly summarized this paper many times on Physics Stack Exchange; one example.
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$\begingroup$ So tl;dr, + helicity has $\vec E$ rotating counter clockwise with light coming out of the clock? idk if that is clearer. I did find another paper (researchgate.net/publication/… ), but don't know if it achieves "most favored paper" status. $\endgroup$– JEBCommented Apr 10 at 14:58
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1$\begingroup$ If you hate this inconsistency, you should find an astronomer, get two or three beers in them, and then have them explain the historical process by which we decided a star with "magnitude negative one" is brighter than a star with "magnitude zero." $\endgroup$– rob ♦Commented Apr 10 at 14:59
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$\begingroup$ @JEB Your ResearchGate link is broken. The Beth paper uses circularly polarized light to drive macroscopic mechanical motion in a torsion pendulum. It's a masterpiece. $\endgroup$– rob ♦Commented Apr 10 at 15:00
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$\begingroup$ OK, I truncated the link and it seems to work. Magnitude should have positive values visible to the (average) naked eye, and negative be "optics required". Zero? 50% of ppl can see it. But it's too late for that. At least its not measured in dB...and what wrong with memorizing the fifth root of 100? $\endgroup$– JEBCommented Apr 10 at 15:05
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1$\begingroup$ I mean, the stellar magnitude system makes sense for the visible stars, if you imagine lying on your back and saying "that really bright star is first-rate! That slightly less-bright star is second-rate by comparison. Hey, we could list them. I can distinguish about six brightness levels." It's not a problem until you try to include objects that are even brighter than your first-rate stars. In that particular case, the history is complicated because the stellar magnitude system was used by astronomers who didn't yet know about the number zero. $\endgroup$– rob ♦Commented Apr 10 at 15:18