Why do massless particles have only two degrees of freedom?

[Lapidus]

Why do massless particles have only two degrees of freedom, i.e. the two helicity states.

Why do massless spin-j particles do not have 2j+1 degrees of freedom like the massive particles?

thanks

 

[Bill_K]

Truth is, massless particles have one spin state, not two. A massless neutrino, for example, would have only one spin state. Two states arise only if you include the action of the parity operator.

The full answer involves some fairly deep group theory, so forgive me, I have to lie a little bit here. :-) The spin operator L acts on the subspace orthogonal to the momentum vector. For a massive particle this space is a normal 3-d Euclidean space and the spin states form a representation of the three-dimensional rotation group SO(3). You know the rest: L(L+1) and all that.

For massless particles the Lorentz contraction compresses this 3-space into a 2-d plane, and consequently the spin states are representations of the two-dimensional rotation group SO(2). This group is Abelian, and consequently every representation is one-dimensional, characterized by a single eigenvalue M and a single state exp(iMφ).

 

[tom.stoer]

For gauge bosons the argument is clear already in classical electrodynamics, and it remains valid after quantization:

The photon field A^m has four components m=0,1,2,3; but A⁰ is not a dynamical degree of freedom (d.o.f.) as there is no time derivative acting on A⁰, which means there is no canonical conjugate momentum. Instead A⁰ acts as a Lagrange multiplier generating a constraint. Now one fixes the time gauge A⁰=0 and keeps the corresponding field equation which is the Gauss law constraint. The Gauss law generates time-independent gauge transformations which means there is still a residual gauge symmetry to be fixed - related to a second unphysical d.o.f. (the residual gauge symmetry allows for time independent gauge transformations leaving the equation A⁰=0 invariant and therefore respecting the time gauge). Implementing the Gauss law eliminates a second unphysical d.o.f. which results in 4-1-1 = 2.

The same reasoning works in QCD as well, but due to the non-abelian gauge symmetry the math is rather involved. The benefit is that the result is suitable for low-energy physics, gauge symmetry is explicitly fixed i.e. no approx. will ever break it - and it comes w/o any ghost fields i.e. is formulated entirely in physical d.o.f.

 

[Lapidus]

Thank you both!

 

[dextercioby]

For massless particles the Lorentz contraction compresses this 3-space into a 2-d plane, and consequently the spin states are representations of the two-dimensional rotation group SO(2). This group is Abelian, and consequently every representation is one-dimensional, characterized by a single eigenvalue M and a single state exp(iMφ).

Good answer. It helps to also post references to known books on this matter. Of course, the first volume of Weinberg's QFT text (1995) and then, of course, Fonda and Ghirardi's 1970 text and some chapters on the Poincare group in Cornwell's group theory text (2nd volume) or Barut and Raczka's 1977 masterpiece.

 

[tom.stoer]

If I remember correctly Ryder does explain the SO(2) game for spin (helicity) of massless particles as well.

 

[dextercioby]

Upon reading Ryder's argument, I'm somewhat disappointed with his writing. The helicity is not a number, but an operator whose spectrum we seek to find out.

 

[Lapidus]

A photon has two helicity states and two transverse polarizations, both correspond to the two the two degrees of freedom, right? How and why is the helicity connected to the possible polarizations? If we had also longitudinal polarization, how would the helicity look like?

thanks

 

[tom.stoer]

The longitudinal polarization of photons is only due to the unphysical gauge degree of freedom which is eliminated using the construction mentioned above. For other massless particles w/o gauge symmetry this question is not applicable.

 

[Lapidus]

I just read that the circular polarization states are the eigenstates of the helicity.

So you are saying, we can only rewrite the two transverse polarization states into two circular polarization states. If there where a third, a longitudinal polarization state, rewriting these now three states into three circular polarizations states is not possible.

 

[tom.stoer]

Yers, this is what I am saying (the longitudinal state exists in the vector potential, but has to be eliminated as it is unphysical = not observed in nature)