Probabilities of $ \Omega^- $ baryon decay

Question

The $ \Omega^- $-decay can occur in a few different ways. According to this document (page 3, $\Omega^−$ DECAY MODES), the three most probable decays are \begin{align} \Omega^- &\to \Lambda^0 + K^- &(67.8\%);\\ \Omega^- &\to \Xi^0 + \pi^- &(23.6\%);\\ \Omega^- &\to \Xi^- + \pi^0 &(8.6\%). \end{align}

For these decays, I've made the feynman-diagrams below (are they correct?). The three decays all use the $W^-$-boson and thus the weak interaction.

Now my question is: if the three different decays use the same interaction (weak), why is there a difference in the probabilities for each decay. Has it something to do with the masses of the end products?

Masses:

• $\Omega^-\ (sss) = 1672 \,\text{MeV}/c^2$;
• $\Lambda^0\ (uds) = 1116 \,\text{MeV}/c^2$;
• $K^-\ (\bar{u}s) = 494 \,\text{MeV}/c^2$;
• $\Xi^0\ (ssu) = 1314 \,\text{MeV}/c^2$;
• $\Xi^-\ (ssd) = 1322 \,\text{MeV}/c^2$;
• $\pi^-\ (\bar{u}d) = 140 \,\text{MeV}/c^2$ and
• $\pi^0\ (u\bar{u}/d\bar{d}) = 135 \,\text{MeV}/c^2$.

Thanks in advance!

Answer 1

if the three different decays use the same interaction (weak), why is there a difference in the probabilities for each decay.

This statement seems to make the following assumption: "any two decays involving the same force-carrying particle must have the same rate."

There's a sense in which this assumption is not 100% wrong -- we would generically expect the overall timescale of the decays to be similar. And, this is true -- the branching ratios are different by order 1 factors, not orders of magnitude.

But this assumption is certainly false in general if you want more than an order of magnitude estimate, for at least two reasons.

• The coupling between two different particles and the gauge field can be different, leading to different interaction strengths.
• The decay rate depends available phase space for the decay products, which depends on the masses.

I haven't done the calculation of the amplitudes for the specific processes in your question. But the point is that you need to do the calculation, it is not enough to say "there is a $W$ boson so I can compare the rates directly."