Xi decay quarks conserved but flavour not conserved

[Howlin]

Hi

I have been looking into particle physics and i came across this question but i can not find the answer,

the quetsion is as follows:

the xi particle decays into the following
\Xi -> \wedge + \Pi^{-}
dds -> sud + \overline{u}d
the quark numbers are conserved but the flavours are not conserved.

I cannot understand why the flavours are not conserved. Can anyone help me?

 

[mfb]

the flavours are not conserved.

Where does that come from?
Looks conserved to me.

The long lifetime (~10^-10s) is suspicious.

 

[Howlin]

It says on it why isn't it conserved and is it decayed by strong or weak forces.
I would say weak because flavours are not conserved but by looking at it, it seems to be conserved

 

[mfb]

Well, it is easy to draw a diagram for the strong interaction.

The long lifetime indicates that those diagrams are problematic in terms of parity, or cancel each other partially, or something else.

Edit @Bill_K's post: Oh, that explains everything. And $\Sigma^-$ with dds is too light to decay into $\Lambda + \pi$

 

[Bill_K]

The Ξ^- (aka cascade particle) is not dds, it's dss. The decay is weak because it does not conserve strangeness.