# D-meson decay

1. Nov 4, 2012

### kokolovehuh

Hi all,
I have a question regarding to the branching ratio of neutral D-meson decay.
Give three decays:
1. D --> negative kaon + positive pion
2. D --> positive pion + negative pion
3. D --> positive kaon + negative pion

The one with highest branching ratio is 1. due to its non-cross generation mixing. However, my question is: why isn't 2 more likely since D-meson can decay into two photons and split into two pairs of quarks? It's because Intermediate photons are involved with EM interactions, which dominate over weak interactions (1 and 3 b/c they don't conserve strangeness).

Thanks :\

O.

2. Nov 4, 2012

### Hepth

The first thing to notice is you MUST have a weak transition. So the leading order will be a weak-tree diagram, with NO photons (as that would be next order in QED).

These are the weak transitions right?

c-> s + dbar u
ubar-> ubar

= K- pi+

c->d + dbar u
ubar-> ubar
= pi- pi+

c->d + sbar u
ubar-> ubar
= K+ pi-

c->s + sbar u
ubar -> ubar
= K+ K-

So in this order we have:
Vcs Vud ~ 1 * 1
Vcd Vud ~ (-lambda) * 1
Vcd Vus ~ (-lambda) * (lambda)
Vcs Vus ~ 1 * (lambda)

where lambda ~ 0.2257

I imagine this is the first reason why they're ordered thusly.

3. Nov 4, 2012

### kokolovehuh

Yea, I see how this works like that.
But as I mentioned before, since 'positive pion + negative pion' can be produced through two photons why wouldn't it be produced more often than the ones from weak interaction?

Also a stupid question, when do you take the Vxx to be negative? Vxx = any element in KM matrix

4. Nov 5, 2012

Staff Emeritus
If you want #2 to proceed via two virtual photons, why would you not have an even larger branching fraction into two real photons?

5. Nov 5, 2012

### kokolovehuh

Vanadium 50, it's probably because of some CP conservation; so we have pions as products. This is why the intermediate photons are only virtual and we need quarks to be produced.

6. Nov 5, 2012

Staff Emeritus
No, that's not your problem. A D0 is not in a CP eigenstate.

7. Nov 5, 2012

### Staff: Mentor

$D^0 \to \gamma \gamma$ is not observed yet, with 2.5*10-5 as upper limit on the branching fraction (~1/5 of the WS $D^0 \to K^+ \pi^-$ BF). It needs weak and electromagnetic interaction at the same time. Why do you want photons in the diagram? Gluons would do the same, just with a stronger coupling. It cannot compete with the singly cabibbo suppressed tree diagram, however.

If you look at the -> KK, Kpi, pipi branching fractions (and similar -> K pi pi pi, -> K K K pi and so on), powers of lambda dominate.
At least roughly:
BF $D^0 \to \pi^+ \pi^-$ is 0.14%
BF $D^0 \to K^+ K^-$ is 0.40%

And of course there is mixing... HCP2012 will have some news about that (14. November).

Last edited: Nov 5, 2012
8. Nov 5, 2012

### kokolovehuh

mfb, it does make whole lot more sense now; thanks alot

btw Vanadium 50, (correct me if I'm wrong) you don't need a particle decay to be in CP eigenstate to conserve charge conjugation/parity in any non-weak interactions.

9. Nov 5, 2012