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Total Angular Momentum of Circularly Polarised Light

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nla7
#1
Feb11-13, 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 :)
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mfb
#2
Feb11-13, 12:33 PM
Mentor
P: 11,890
Unlike electrons, photons have spin 1, so m_s = +-1.
Meir Achuz
#3
Feb11-13, 03:41 PM
Sci Advisor
HW Helper
PF Gold
P: 2,013
The m=+1 or -1 is for L_z of the orbital angular momentum of the electron.

nla7
#4
Feb11-13, 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 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
#5
Feb12-13, 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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