# I Cabibbo angle for coupling of quark to antiquark?

1. Nov 26, 2016

### Kara386

I'm not clear on whether the coupling constant for strange to antidown is modified in the same way as for strange to down, i.e. multiplied by $\sin(\theta_C)$. And are the coupling constants of interactions mediated by Z also dependent on the Cabibbo angle?

And what happens if you have say a charm and down quark interact via weak to produce an up quark and antistrange quark? The coupling constant is modified at both vertices, multiplied by $\sin(\theta_C)$ in both cases. Decay width is proportional to the coupling constant squared, so does that mean it would then depend on $sin^4(\theta_C)$?

I've read a textbook (Griffiths Intro to Elementary Particle Physics) which made me ask these questions rather than helping to answer them!

Last edited: Nov 26, 2016
2. Nov 26, 2016

### Staff: Mentor

Where do you expect a strange/down or strange/antidown coupling? Where do you see couplings where changing a particle to an antiparticle would work? Which interactions happen with the Z, and where do you expect the Cabibbo angle to be relevant?

In general, couplings don't care about time ordering - particles and antiparticles work the same way.
That doesn't work.
But if you fix the quark content, the answer is yes.

3. Nov 26, 2016

### Kara386

I've resolved the Z issue. I was drawing a Feynman diagram for a process I thought could only proceed via Z but actually using a gluon instead worked fine.
In terms of the strange antidown coupling I think I meant up antidown coupling in the decay of a $D^+$ meson to $K^+ + \pi^+ \pi^+$. So on the diagram I've drawn a line with charm on the left and strange on the right. There's a W boson coming off that line and producing an up and antidown. I know that for an up, down and W boson vertex the Cabibbo angle dependence is $\sin^2(\theta_C)$ and I wondered if it's the same for an up and antidown vertex. And whether up, antidown, W is even an allowed vertex.

On a list of allowed W decay modes, I have $W \rightarrow s \bar{u}$ so I thought $W \rightarrow u \bar{s}$ would be allowed. As to fixing the quark content, how do I do that?

4. Nov 26, 2016

### Staff: Mentor

That violates charge conservation, and K+ has an anti-strange, not a strange quark. I guess this should be a K-.
It is not, up and down are in the same generation.
Does it conserve charge?
W+ or W-?
By writing down the correct quarks involved everywhere.

5. Nov 27, 2016

### arivero

GIM effect,does it? I never look at it carefully, but the cham was predicted on grounds of this kinf of cabibbo malabar games.

6. Nov 27, 2016

### Kara386

Yes, apologies, typo.
Now that's a silly mistake on my part.
Not actually specified. But the $s \bar{u}$ would be a $W^-$ to conserve charge, so $u \bar{s}$ would be $W^+$. It's an up type and a down type, it conserves charge, and in terms of crossing generations isn't it an allowed diagonal change? I should also mention that I'm imagining both those quarks to be in the final state, so it's the boson decaying to that pair rather than the up interacting with weak to change flavour to an antistrange. Would that work?

7. Nov 28, 2016

### Staff: Mentor

Right.
In hadron decays, the W is virtual - it is not an actual W decay. But apart from that: right.

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