D Meson Decay: Neutral D-Meson Branching Ratios

[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.

 

[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.

 

[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

 

[Vanadium 50]

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

 

[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.

 

[Vanadium 50]

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

 

[mfb]

$D^0 \to \gamma \gamma$ 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).

 

[kokolovehuh]

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

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.

 

[Vanadium 50]

I was trying to point you in the right direction - or at least away from the wrong one.

 

[mfb]

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

LHCb was a bit quicker than I expected.

9.1 sigma observation of $D^0 \overline{D^0}$ mixing