Good morning everyone ! I've been reading discussions on PF for a long time, but here I'm stuck on a little problem that really annoys me and I couldn't find answer anywhere, so I guess it was time to register. :>(adsbygoogle = window.adsbygoogle || []).push({});

I've been focusing on quantum electrodynamics for a couple of weeks now as part of my graduate studies (I'm usually more into fluid mechanics, but a little quantum mechanics is nice as well) and I decided to begin with Landau-Lifgarbagez Vol. 4 (Second Edition). I got to finish the first half of the book, but I just need a little explanation on something located at the very beginning (not so important for my overall comprehension, but it's always good to be able to justify everything you read...). :p

In §6 page 18, the authors calculate the number of states avalaible to the photon for a given value of its total angular momentum j = l + s, l and s being respectively its orbital and spin quantum numbers. For a spin-1 particule, there are 3 states, given by

l = j

l = j ± 1

However, and that's where my question lies, it is said that when j = 0, only one state subsists, with l = 1. My question is : Why is l = 0 not an acceptable value ? I understand that l = -1 is not acceptable, but if you relate to the ordinary azimuthal quantum number, 0 is perfectly cool and associated to s orbitals. It's quite a big deal, because it implies later that when you take out the degree of freedom related to the transversality condition, no state remains when j = 0, i.e. j = 0 is not an acceptable value for the photon. For exemple, that means that no 1-photon atomic transition can occur between to states with Δj = 0. Is it because there is no frame where the photon is at rest, meaning that it would always possess at least one orbital momentum quantum ?

I hope my question is sufficiently clear, I'm not really used to formulate "hard" problems in English...

Thank you in advance :)

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# Landau-Lifschitz : Photon helicity states

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