## W Decays and Color

*January 2, 2010 at 2:28 pm* *
4 comments *

Tommaso Dorigo posted three physics questions on his blog. They’re rather easy and I hope any particle physics student could answer them correctly. His third question touches upon a favorite bit of phenomenology, so let me expand upon it a bit here.

The W boson decays to a pair of fermions nearly all of the time. I will not worry about radiative corrections – i.e., the “extra” photons and gluons that may be emitted in the process W→f+fbar (where “fbar” means an anti-fermion). As Tommaso points out, the weak interactions are universal – the probability of the W to decay into one particular f+fbar pair is the same as the probability for any other f+fbar pair. More precisely, the coupling constant is the same – the phase space will be smaller for heavier fermions than for lighter fermions, which reduces the likelihood that heavier fermions will materialize in a W decay. I will neglect these mass effects here.

So predicting a branching ratio such as BR(W→e+ν) amounts to counting all of the possible f+fbar pairs that a W boson can decay to. The branching ratio is then just one over that total number of possible final states.

How many such states are there, in the standard model – i.e., in the real world as we know it? For leptonic final states, we have (e,ν_{e}), (μ,ν_{μ}) and (τ,ν_{τ}) – this is quite clear. (Forgive me for not putting bars where they belong – it is hard to do it with this editor and it does not matter for the present discussion.) The quark final states require a little more care. Clearly, the top quark (mass = 172 GeV) is too heavy, but the other five quarks are not. You might think you have these six states: (u,d), (c,s), (u,s), (c,d), (u,b) and (c,b), based on electric charge. But the last four of these six states are not weak doublets. More to the point, the CKM matrix, which allows quarks from different weak doublets to couple to the W boson, is nearly diagonal, meaning that the (u,b) and (c,b) final states make a very small, even negligible contribution. Furthermore, the 2×2 sub-matrix which governs the (u,d) and (c,s) couplings is nearly unitary, so whatever part of (u,d) is reduced is picked up by (u,s), so to speak. In the end, because of this important and unexplained feature of the standard model, it is fine to just take the naive set (u,d) and (c,s) and ignore the mixing of weak doublets allowed by the CKM matrix.

If you’re quick and not careful, you’ll conclude that the W can decay only to the five states (e,ν_{e}), (μ,ν_{μ}), (τ,ν_{τ}), (u,d) and (c,s), and you would predict that BR(W→e+ν) = 1/5 = 0.2. This prediction is wrong, as measurements give BR(W→e+ν) = (10.75±0.13)% (see the Particle Data Group web page).

**Color** is the key to the calculation. Remember that quarks come in three colors (the conserved charge of the strong interaction), so when we consider W→u+dbar, there are three distinct channels, corresponding to u(red)+dbar(anti-red), u(blue)+dbar(anti-blue) and u(green)+dbar(anti-green). Notice that the W boson is a color singlet, so if we choose the color of the u-quark, then the color of the d-anti–quark is determined.

Revising our calculation, we have three leptonic states plus __six__ quark states, so the naive prediction is BR(W→e+ν) = 1/9 = 11%, which is quite good indeed. The agreement with the experimental value is clear proof that there are three colors of quarks, and that the W couples to all fermion doublets with equal strength, modulo the factors incorporated in the CKM matrix. I find this a really very nice piece of physics.

This kind of simple phenomenological calculation is at the heart of basic experimental particle physics. It is nice to cast it as an exercise for the student, but in truth we do this kind of work whenever a new particle is observed. For example, a crude measurement of BR(W→e+ν) told us in the 1980s that the top quark mass must be at least M_{W}, else a smaller BR would have been observed. (What is that number, by the way? Take a look again at Tommao’s post.) It came as a bit of a surprise that M_{t}≈172 GeV, which of course is much too heavy to allow W→t+b, which is part of the reason why single-top production is so interesting.

In the 1990s, a parallel line of reasoning led to the conclusion that there are only three species of light neutrinos, through measurement of Z→ν+νbar. This is one of the most important and most beautiful of the results from LEP 1.

In the spirit of Tommaso’s post, let me pose a question for the reader. Suppose there were a hidden lepton charge, similar to color, so that there were __two__ kinds of electrons, muons and taus. What would be the prediction for BR(W→e+ν_{e}), and to what degree is this excluded by the measured value?

An additional question: what can we conclude about W decays to exotic particles this way? I stated that W bosons decay only decay to fermion pairs. Why not to boson pairs??

Entry filed under: Particle Physics.

1.A quantum diaries survivor | January 2, 2010 at 4:09 pmCiao Michael,

great post, and I like the slant you gave to the information on the partial width pointing at colour. You should write more often!

Cheers,

T.

2.Michael Schmitt | January 2, 2010 at 4:31 pmHi Tommaso, thanks a lot. I have resolved to keep blogging this year, and the encouragement of people like you helps me to keep this resolution. There certainly will be a lot to blog about in 2010!

ciao

Michael

3.carlbrannen | January 2, 2010 at 7:10 pmDoubling up the leptons would give you 9+3 = 12 ways the reaction could go. Each lepton would then end up with 2/12 = 1/6 or 16.67%. This is (16.67 -10.75) /.13 = 45 which is (forgive me if I get this wrong) 45 sigma excluded. I’ll leave part b for the next guy.

The fact that the W goes into leptons equally is part of the reason I looked around for another way of parameterizing the MNS matrix. In density matrix formalism, you use pure density matrices to represent particles. Then if you have three orthogonal states (say electron, muon, tau), their pure density matrices will sum to the unit matrix. And that all sort of implies that you should assume that there is a natural choice for their relative phases. The one I chose is one where the sums of the entries of a row or column of the MNS matrix are all 1.

Another way of saying this is that stuff like magnitudes of amplitudes, and the Jarlskog invariants are related to Berry-Pancharatnam phase, they do not depend on choice of phase of the state vectors. So they are more neatly described in pure density matrix form, as these matrices also eliminate arbitrary phases.

For example, letting e, u, t be the charged leptons, and 1, 2, 3 be the mass eigenstates for the neutrinos, the product of pure density matrices:

|e)(e| |1)(1| |e)(e|

is equal to a real multiple of |e)(e|, and that real multiple is the square of the MNS matrix’s U_e1 entry, which is measurable.

Jarlskog invariants can be written similarlly:

|e)(e| |1)(1| |u)(u| |2)(2| |e)(e|

is a complex multiple of |e)(e|, and that multiple is a Jarlskog invariant. The imaginary part of that complex number is supposed to give you J_cp, though I haven’t quite convinced myself that this works yet. (Got to write the computer program and play with the numbers to really understand it.)

4.Michael Schmitt | January 3, 2010 at 1:01 pmHi Carl,

thanks a lot for your comment. It escaped me temporarily that the MNS matrix would also play a role. I have seen your discussion of the CKM matrix on your web site but I am not expert enough to offer any intelligent comments. 😦

In the past I have used the W width as a way to constrain the CKM matrix (specifically Vcs). Of course that exercise ignored any mixing in the lepton sector. I wonder how it would turn out if we applied measurements of the W width to constrain / check the lepton sector. Probably the constraint would be rather weak compared to other constraints, unless one expects something odd for the taus. (Sorry for the pun.)

regards,

Michael