- #1
geoduck
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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=[itex]\pm 1[/itex], 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., [itex]\alpha |+1\rangle+\beta | -1\rangle[/itex] is an eigenvector of spin along an axis not equal to the z-axis?
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=[itex]\pm 1[/itex], 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., [itex]\alpha |+1\rangle+\beta | -1\rangle[/itex] is an eigenvector of spin along an axis not equal to the z-axis?