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The $\mathrm{U(1)}$ QED case has two physical degrees of freedom, which is easy to understand because the free electromagnetic field must be transverse to the direction of propagation. But what are the physical degrees of freedom the $\mathrm{SU(N)}$ Yang-Mills theory?

I believe that the $\mathrm{SU(N)}$ Y.M. theory has four degrees of freedom, and it is easy to see that by gauge fixing (such as the Lorentz condition or the Coulomb condition) that we can always remove one redundant degree of freedom. But then we still haven't completely fixed the redundant degrees of freedom. Thus my question is:

How many physical degrees and how many are redundant degrees of freedom does the Y.M. theory have?

I would be interested to understand how we can determine this mathematically, and also to understand what the physical intuition behind this is?

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Since I am new here and I cannot comment above I just write my comment as a small answer. I think in your question there is a small mistake. You ask how many degrees of freedom are there in a Yang-Mills theory although you want to ask how many polarizations does the gluon have. First of all, the number of generators of the gauge group will be the number of gauge bosons you will get. $SU(N)$ groups have indeed $N^2-1$ generators, except $U(1)$ that has one. In QED this corresponds to the photon, in $SU(2)$ to three gauge bosons (not exactly the W's and the Z since these are produced by mixing $SU(2)_L$ with $U(1)_y$), and in $SU(3)$ there are 8 gluons as Siva already told you.

Now to your question, how many polarizations do these gauge bosons have. Let us start with a massive vector boson. We can define its helicity states at the rest frame and then of course boost to the frame that the particle moves. Under this boost, the longitudinal polarization has a term $E/m$ (where $E$ the energy and $m$ the mass of the particle) that goes to infinity when the mass of the particle goes to zero. This would cause problems concerning the unitarity of the theory since the contrubutions in the matrix elements would be huge. What one can do is arrange the interactions of the theory such that the contributions from longitudinal polarizations are suppressed by a factor of the order $m/E$. In the case now the vector boson is strictly massless, the longitudinal contributions decouple completely. Therefore, massless vector bosons have two physical states of maximal helicity while massive one have three. Gluons are massless and so are photons, while W's and Z are massive.

I hope I helped. I am pretty sure that you can find more details even in wikipedia if you want.

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  • $\begingroup$ Thanks! This was exactly my question, but I think I phrased it badly. $\endgroup$ – Hunter Mar 11 '14 at 11:20
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$SU(N)$ has $N^2 -1$ gluons, one corresponding to each generator (this is called the adjoint representation). That "quantum number" (which could be called an index for the adjoint rep) could be tensored together with the Lorentz representation for a massless vector particle, which should have 2 polarizations (just like QED). So the theory will have $2(N^2 -1)$ physical degrees of freedom.

I don't understand why you say that the gauge bosons should be massive -- if they were, then the gauge symmetry would be "broken".

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  • $\begingroup$ Yeah, sorry, I edited my message (about the mass of bosons), I think my brain has stopped working. $\endgroup$ – Hunter Mar 11 '14 at 4:12
  • $\begingroup$ I'm not sure I can see why the $\mathrm{SU(N)}$ Y.M. theory has $N^2-1$ degrees of freedom. I'm assuming the $N^2-1$ has to with the fact that $\mathrm{SU(N)}$ has $N^2-1$ generators. I don't understand what this has to do with adjoint representation of the gluons? Also are there $N^2-1$ physical, redundant or total degrees of freedom? $\endgroup$ – Hunter Mar 11 '14 at 4:15
  • $\begingroup$ No wait, I'm really confused now about your comment about the masses of bosons. If they are massless, then they travel with the speed of light and so they must have 2 physical degrees of freedom (like all electromagnetic waves), right? (But this disagrees with your statement about there being $N^2-1$ degrees of freedom.) Thus, if I'm right, my question is only interesting if the symmetry is broken and the bosons have gained mass. $\endgroup$ – Hunter Mar 11 '14 at 4:40
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    $\begingroup$ I've reworded the answer -- is it clear now? It is exactly like you say in the comments. $\endgroup$ – Siva Mar 11 '14 at 4:57
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    $\begingroup$ @Hunter In order for the theory to be invariant under gauge transformations, the gauge field has to transform in the adjoint representation of the gauge group. This representation comes with $N^2-1$ generators. $\endgroup$ – Frederic Brünner Mar 11 '14 at 8:18

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