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Helicity is different form Spin for massless particle(photon)? 
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#1
May2614, 02:00 AM

P: 3

As we know photon's helicity are [itex]\pm[/itex]1. Helicity is the projection of the spin S onto the direction of momentum, p, which is considered as S_{z}.
What about S_{x} and S_{y}? They are both ZERO? 


#2
May2614, 02:07 AM

P: 3

Is it reasonable to say that massless particles have no spin but just helicity?



#3
May2614, 05:47 AM

Sci Advisor
Thanks
P: 4,160

What we mean by the spin of a particle is the subgroup of the Lorentz group that commutes with its 4momentum (the "little group"). For a particle with mass, go to its rest frame where the 4momentum is P_{μ} = (0,0,0,1) and the spin operators are the rotations in 3space, S_{x}, S_{y} and S_{z}. They form SO(3).
For a massless particle there is no rest frame, so take the 4momentum in the zdirection, k_{μ} = (0,0,1,1), and its spin operators are the three operators that preserve k_{μ}. The first one is a rotation in the (x,y) plane. This is the helicity. It acts on the components of the particle's 4potential as A_{x} ± iA_{y} → ±(A_{x} ± iA_{y}). The other two are null rotations, x → x + εk and y → y + εk. These operations just add a multiple of k to the 4potential. But this is just a gauge transformation. So helicity is the only observable part. 


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