Total Angular Momentum of Circularly Polarised Light

[nla7]

I am considering the occurrence of an incident circularly polarised EM wave on a ground state hydrogen atom. The result is that the final state of the atomic electron transition will be at m= ±1 depending on the orientation of polarisation (LCP or RCP).

I understand that this is due to the conservation of angular momentum. The EM wave carries total angular momentum J of ± hbar. J also equals L+S (orbital angular momentum & spin angular monetum respectively). However the angular momentum of circularly polarised light comes solely from the spin angular momentum, which would imply that L=0 and S= ± hbar.

My problem is tho, is that S=m_s hbar and I understood that m_s could only equal ± 1/2. So how in the instance of my light, can S be an integer value of hbar?

Any assistance would be much appreciated. Thanks :)

 

[mfb]

Unlike electrons, photons have spin 1, so m_s = +-1.

 

[Meir Achuz]

The m=+1 or -1 is for L_z of the orbital angular momentum of the electron.

 

[nla7]

Thank you. That makes sense now :)

I'm thinking that although it is the S component of the photon's J that has a value, because of spin-orbit interaction, it is only J that needs to be conserved and not the individual components S or L. Which is why it can be the electron's L that carries the conservation.

 

[andrien]

one more thing,photons are massless then don't have a zero value for m.