# Homework Help: Kaon decay to a lepton and associated neutrino

1. Apr 19, 2012

### Ai52487963

1. The problem statement, all variables and given/known data

Draw the lowest-order Feynman diagram for the decay: $$K^+ \rightarrow l^+\nu_l$$

2. Relevant equations

Kaon structure is given as $$u\bar{s}$$

3. The attempt at a solution

So I know that you can get a lepton and an associated neutrino from a W decay, where the strange quark would flip over to a charm quark in the kaon. The question doesn't say there's a pion produced, though, only the lepton and neutrino. Is there an actual decay here that vaporizes the kaon completely into a lepton and neutrino without leaving the up quark just sitting around?

I drew a diagram (dunno how to represent in TeX) that shows a USbar going to a UCbar where the Sbar decays into a W+ which decays into a lepton and neutrino. Am I missing something? Does the UCbar go away or is it just ignored?

2. Apr 19, 2012

### Steely Dan

If such a decay exists, then something needs to happen to the quarks to produce virtual particles other than the lepton and neutrino, and those virtual particles need to be used up in the process of making the end state products so that they too don't appear at the end (since they're virtual, after all). So try finding a more complicated internal set of interactions that accomplishes this.

3. Apr 20, 2012

### fzero

The CKM matrix directly couples all up type quarks to all down type quarks (in particular u and s). There's no need to involve a charm quark.

4. Apr 20, 2012

### Ai52487963

I see. I was getting confused with a table I have as a reminder that: when you change particular flavor states (like from C to S or U to D) you emit a W+ and a W- for the reverse operation. I thought in order to go from like a U -> D, D -> C, C -> S you had to emit a W+ each time.

So the CKM matrix takes care of that?

5. Apr 20, 2012

### fzero

By couple I mean there is a W involved. But there are direct couplings $u\leftrightarrow d$, $u\leftrightarrow s$, $u\leftrightarrow b$, as well as the analogous couplings for the other up type quarks.

This is usually explained in terms of the weak eigenstates not being diagonal in the mass eigenstates. Let $u,c,t,d,s,b$ be the mass eigenstates. Then the weak eigenstates are $u,c,t,d',s',b'$. In terms of the CKM matrix $V_{ij}$, we can write

$$d' = V_{ud} d + V_{us} s + V_{ub}b.$$

The couplings to the W bosons are of the schematic form

$$\bar{u} W d' = V_{ud} \bar{u} W d + V_{us}\bar{u} W s + V_{ub}\bar{u} W b.$$

In particular, there will be a coupling between $u$ and $\bar{s}$ that depends on $V_{su}^*$.