Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In non-abelian gauge theories self interaction of gauge fields is permitted, allowing coupling such as $WWZ$ (i.e. $Z$-boson decaying to $W^+W^-$) or ggg (i.e. gluon splitting into two new gluons).

The aforementioned particles are spin-1 particles, thus I'd assume that the above processes are prohibited from spin considerations. Consider for example a spin +1 gluon splitting in two new gluons, their total spin can either add to +2 or 0.

I am well aware that spin itself is not conserved in these interactions but only the total angular momentum $J=S+L$, thus I'd assume that the two gluons have a nonzero relative angular momentum $L$.

My question is whether that is the correct explanation for the phenomenon. My QM skills are a little bit rusty, so any help is appreciated!

share|cite|improve this question

$Z \rightarrow WW$ only works if the $Z$ is 'virtual', i.e. appears as an intermediate particle in the Feynman diagram and only for (propagator) masses $ \gtrsim 2 \cdot M_W$ (i.e. very 'off-shell').

Be careful not to confuse the following:

  • spin as in property of a particle. $W$, $Z$, $\gamma$ and the gluon are spin 1 particles
  • spin as a state (component with respect to some axis). $W$ and $Z$ bosons can have a spin state -1,0 and 1. Photons and gluons can only have a spin 0 state if they are 'off-shell'.

So from conservation of angular momentum point of view it is possible to have a virtual $Z$ with spin state zero couple to a pair of W bosons (having spin state -1 and +1). The existence of this coupling has been verified e.g. at LEP (see e.g. this example).

Triple gluon couplings (at least one gluon being off-shell which thus can have the spin state 0) are important in jet fragmentation and the calculation of the evolution of parton distribution functions.

share|cite|improve this answer

The massive verse massless cases are different.

Massive vector bosons are a bit more 'honest' in their representation of the Lorentz group in that they have all 3 DOF implied by the $j=1$ representation. That is, they have $2j+1 = 2(1)+1=3$ states with $j = -1,0,+1$ angular momentum. So now you can see that you can have 2 massive vector bosons one with $j = +1$ , the other with $j = -1$ that goes to a $j= 0$ state (with other possibilities as well).

Massless vector bosons only have the two $j = \pm 1$ DOF, the spin 0 component can be gauged away, and even if you keep it around, it is just a gauge redundancy in our description anyway. In this case your problem seems a bit more real. However, the scattering amplitude vanishes for 2 gluons to 1 gluon. You can see this by momentum conservation. Go to a frame where the gluons have equal and opposite momenta, so that the final state has zero momentum, which implies the final state gluon has zero momentum, which is a contradiction for massless particle. That is, there is a 3 point vertex in the Lagrangian, and so there is an off-shell amplitude, but once you put it on shell, the amplitude vanishes.

share|cite|improve this answer
I am only 90% sure about this, so please feel free to downvote. But PLEASE let me know where I am wrong with a comment. – DJBunk Feb 14 '13 at 15:18

If you are worried about angular momentum in your interaction, then you should look at the Lorentz structure of your vertices. For the triple couplings you are talking about, the Lorentz part of the couplings looks like:


Here $k_i$ are the momenta of the interacting particles and the space-time indices $\mu_i$ are going to be convoluted with polarization vectors (or with some propagators).

So even naively one can see that there is an interplay between polarization ($S$) and momentum ($L$) of the "participants". From here I'd conclude that your explanation is, indeed, correct. But...

There is a statement called "The Landau-Yang Theorem". Which says that two massless vector particles -- cannot be in a state with $J = 1$. So you actually cannot have a vector boson decaying into two massless vector bosons. But...

That works only if you hav an on-shell decaying particle. If it is off-shell (aforementioned convolution with a propagator) -- then you still can have that process.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.