Little group and photon polarizations

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From what I understand, the little group for a particle moving at the speed of light, has 3 generators. 2 generators generate gauge transformation, and 1 generator rotates the particle about its axis of motion.

I have 3 questions:

1) Do all particles moving at the speed of light (not just photons) have gauge transformations?

2) Since gauge transformations are Lorentz transformations, if someone asks you what a gauge transformation is, can you say it's what the photon looks like in a different Lorentz frame?

3) How exactly does it follow that the photon has only two polarizations from the fact that the only generator in the little group not involved in gauge transformations is rotation about a single axis, instead of 3 possible axis? Call this axis the z-axis. Why can't there be a mz=0 polarization? I would like to argue that mz=0 means the photon is spinning along some other axis, either the x or y axis, and that type of rotation is not part of the little group, hence mz=0 is not allowed . However, since you can have a linear combination of mz=\pm 1, can't you choose your coefficients in your linear combination such that it's spinning about an axis that's not the z-axis, i.e., \alpha |+1\rangle+\beta | -1\rangle is an eigenvector of spin along an axis not equal to the z-axis?
 
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ad 1) No. Consider a massless spin 1/2 particle described by the dirac equation.

ad 3) combinations of eigenstates with m_z =1 and -1 aren't eigenstates of J_x or J_y.
 
The little group of massless particles is ISO(2,R), i.e., the symmetry space of the Euclidean 2-dimensional plane. It is generated by rotations around an arbitrary point and by the translations in arbitrary directions in the plane. That's indeed a three-dimensional Lie group.

Now, as you know from quantum theory, the translations have as irreducible representations only the trivial representation and the one representing momentum, i.e., with continuous spectrum. The latter realization implies that you'd have some continuous spin-like degrees of freedom, something that yet has never been observed. Thus, for massless particles with spin you have to make sure that the translations of the little group are trivially represented. This implies for spin 1 and higher spins that you have a gauge theory. A massless particle with spin s has only two helicity degrees of freedom, \lambda = \pm s. In the case s=1/2 there are thus no redundant degrees of freedom, and thus it's not implying a local gauge group. In all other cases that indeed happens. This is the group-theoretical foundation of the gauge structure of massless spin-1 particles, as we know them in the Standard Model as photons and gluons.

For more details about the group-representation theory of the Poincare group, relevant for relativistic QFT, see my manuscript:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf
 
2) Gauge transformations are not Lorentz transformations; you made wrong implications from (unspecified) wording in a textbook. A subgroup of the little group branded as “gauge transformations” means that actions of its elements on the specified photon state are equivalent to gauge transformations, i.e. that subgroup effectively preserves the state. An analogy: after one sidereal day, Earth’s orientation in the space doesn’t change. We can say that evolution of Earth for a sidereal day is a spatial translation. But it is a thing utterly different from saying such nonsense as “spatial translations are evolutions of Earth for multiples of sidereal day”.

3) Look at http://physics.stackexchange.com/qu...ave-only-two-possible-eigenvalues-of-helicity
 
Incnis Mrsi said:
2) Gauge transformations are not Lorentz transformations;

I think the point which is really of interest here is the fact that already the representation theory of the Lorentz group urges us to introduce gauge degrees of freedom in the representation of massless spin 1 particles.
 
By the way, can anybody recommend a reputable source that considers relationship between Poincaré sphere and photon’s little group? It happened that Ī began to learn it only recently.
 
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