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Total Angular Momentum of Circularly Polarised Light 
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#1
Feb1113, 09:53 AM

P: 2

I am considering the occurence 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 :) 


#2
Feb1113, 12:33 PM

Mentor
P: 11,900

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



#3
Feb1113, 03:41 PM

Sci Advisor
HW Helper
PF Gold
P: 2,015

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



#4
Feb1113, 04:03 PM

P: 2

Total Angular Momentum of Circularly Polarised Light
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 spinorbit 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. 


#5
Feb1213, 04:50 AM

P: 1,020

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



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