Why massless particle can only have two helicity states?

liucl78
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why massless particle, such as photon, can only have two helicity states?
Photon's helicity is 1,-1. Helicity zero is forbidden. why?
 
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Because it's massless...
A massive spin1 particle introductively for QFT, can be studied in Proca's Lagrangian.

The symmetry reason is because of the possible unitary transformations you can have on your fields. These transformations (which will lead you in a specific gauge), such as the Lorentz gauge, will drop you initial degrees of freedom... In the photon's case, as massless, you have the abilitiy to do twice this trick, dropping the degrees of freedom 4 (A_{\mu} is a real 4 component field) to the physical degrees of freedom which is 2...
In other words you are killing components which can be "related" to the other components via a transformation, meaning they are not physical...
 
Apart from the QFT approach, the reason can go even further ahead. For example,for a massless particle you cannot really define 3 polarization vectors perpendicular to the momentum vector.
In the case of massive particle, in order to construct the polarization vectors, you can always find the 2 vectors with only spatial components which are perpendicular to the momentum vector... However for a massless particle, you cannot build a 3rd vector perpendicular to momentum ... it will be either longitudial or it will be a linear combination of the previous two...
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

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