Do Photons Spin in Our Reference Frame?

[scope]

hi,

i have read that photons must have an angular momentum because they have no rest mass. does anyone there understand why?

 

[Vanadium 50]

Where did you read this? And what exactly did it say?

 

[scope]

sorry, i don't know, for example the two possible values of the spin projection of photon in QFT are h/2pi and -h/2pi, and not 0.

i do wonder about the spin of the photon. in OUR reference frame, do we see the photon spinning or not, if we apply special relativity?
by the way how does spin transform under Lorentz transformations?

I would be grateful for any reply

 

[haael]

sorry, i don't know, for example the two possible values of the spin projection of photon in QFT are h/2pi and -h/2pi, and not 0.

i do wonder about the spin of the photon. in OUR reference frame, do we see the photon spinning or not, if we apply special relativity?

Consider a photon flying in the direction of z-axis. Any vector massive particle can have 3 spin states - however one of them requires the z component of momentum to be 0 (particle in rest). Photon can not be in rest, so only 2 spin states remain.

 

[tom.stoer]

i do wonder about the spin of the photon. in OUR reference frame, do we see the photon spinning or not, if we apply special relativity?

The polarization parallel to the momentum is ruled out because photons are massless. You cannot go to their rest frame.
This can be understood in QED in the following way: QED has a U(1) gauge symmetry which allows you to eliminate one gauge degree of freedom. Let's keep things as simple as possible and set A°=0= use the (time-gauge; now we have eliminated one polarization.

A°=0 is a rather good choice as A° is not a dynamical degree of freedom b/c its conjugate momentum is zero; therefore it acts as a Lagrange multiplier genarating a constraint. This constraint is the Gauss law G which again must be set to zero, i.e. G=0. In addition the Gauss law generates time-independent gauge transformations which respect the A°=0 gauge, that means gauge transgformations where the gauge function is time-independent.

Via this residual gauge symmery or (which is more or less the same) via the solution of the Gauss law constraint another degree of freedom can be eliminated leaving us with to physical polarizations of the photon.

If we would introduce a mass term in the Lagrangian gauge invariance would be broken and the A°=0 gauge would no longer be allowed (so the first elimination does no longer work). But A° is still a Lagrange multiplier, the - now modified - Gauss law G'=0 would still be solved for and the second elimination would go through leaving us with three degrees of freedom.

by the way how does spin transform under Lorentz transformations?

The Lorentz group has a rather complicatedstructure. In principle it is something like SO(3,1), but one must use the complexification / the universal covering group Spin(4) which is (locally) the same as SU(2)*SU(2). Now every SU(2) has integer and half-integer representations leaving us with two series like 0, 1/2, 1, 3/2, ... E.g. the Dirac equation uses 4-spinors (which are bi-spinors) transforming as (1/2, 1/2) using two 2-spinors, one from each series.

For massless particles the situation is slightly different and I think it's etter not to use spin but helicity.