Landau-Lifschitz : Photon helicity states

[Takeshin]

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. :>

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 possesses 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 :)

 

[DrDu]

The general rule for the addition of two angular momenta $j_1$ and $j_2$ is $|j_1-j_2|\le J \le j_1+j_2$. This is called "triangle" relation as it holds for the vectorial sum of two sides of a triangle. It is valid for any pairs of vectors with given length, not specifically related to QM.

 

[andrien]

Is it because there is no frame where the photon is at rest

you got that right.There is no rest frame for photon.A photon has only two polarization states if there will be a rest frame for photon then photon can have a longitudinal polarization which will violate the condition that photon is massless.gauge invariance prevents photon from having a mass.

 

[Takeshin]

The general rule for the addition of two angular momenta $j_1$ and $j_2$ is $|j_1-j_2|\le J \le j_1+j_2$. This is called "triangle" relation as it holds for the vectorial sum of two sides of a triangle. It is valid for any pairs of vectors with given length, not specifically related to QM.

That I know, it's also closely related to the calculation rules of the fine and hyperfine states, but I don't see the point here, nothing would (in appearance I guess) prevent me from stating l = 0 and s = 0, I would still have j = 0...

you got that right.There is no rest frame for photon.A photon has only two polarization states if there will be a rest frame for photon then photon can have a longitudinal polarization which will violate the condition that photon is massless.gauge invariance prevents photon from having a mass.

I can imagine (even if it seems hazardous to me because it's not what spin really is) a photon "spinning" while moving along its wavevector and thus never having a zero angular momentum because it's never at rest, but then, according to the calculation rules given by Laudau-Lifschitz, j = 1 would give me two states, not three, because l = j - 1 would be equal to zero ? Or is only the (l;s) = (0;0) state forbidden ?

 

[DrDu]

That I know, it's also closely related to the calculation rules of the fine and hyperfine states, but I don't see the point here, nothing would (in appearance I guess) prevent me from stating l = 0 and s = 0, I would still have j = 0...

Certainly, but you don't have s=0 but s=1.

 

[Takeshin]

Well, I found my mistake, and you are right. Very dumb mistake, by the way. I was adding the magnetic quantum number $m_s$ posing $- s \leq m_s \leq s$ and adding l and $m_s$. I did again the calculation with the right tools and without being completely dumb, and found Landau-Lifschitz result.

My deepest thanks for your answers, I'm now in peace with physics. :rofl:

 

[Takeshin]

While I'm at it, I'd have another question (I promise it's the last one).

I know about R. Beth experiment in 1936 which confirmed the photon spin value of 1, although technical details are far too difficult for my understanding, but is there another experiment which confirmed this value while staying in a "typical" quantum optics experimental set-up ? I highly doubt this is the only experimental verification that has been done, but I was unable to find anything relevant with "pure" quantum optics, only experimental set-ups involving liquid crystals and lots of things I'm really not familiar with. :uhh: