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**Solution**

So the strong interaction conserves parity. Parity of meson is given by ## (-1)^{l+1} ##, for the ##1s## states, ##l=0## and so ##p=1##.

The solution than states as both the decay products have zero spin the total angular momentum of the decaying particle must be equal to the orbitial angular momentum of ##D_{s}^{+}\pi^{0}## system.

**So I agree with this last comment, but I have no idea why both decay products have ##0## spin????**

I know that a meson is a quark and its antiquark, and so the spin adds to either ##1## or ##0##. But how do we know which it is?

The solution then uses parity of the ##D_{s}^{+}\pi^{0}## system is ##P = (−1)^{l} × −1 × −1 ## *2.

**Im confused where this comes from -**so I know that the parity of a meson in it's lowest states - i.e- ##l=0## is ##-1##. And I know that for a system of particles ##P=(-1)^{l}##, but, I've never seen how to consider the parity of two particles decayed. Is this how you 'add' the parities ?

And so how exactly should you think of the system of decay products. So a particle has an intrinsic parity. Is this a particle or a quark? I.e is ##P=(-1)^{l}## coming from thinking of a system of quarks or a system of the two particles ##D_{s}^{+}\pi^{0}##?

(From which the solution follows from the fact that we require ##P=1##,

**again I'm okay with this once I understand expression *2**.)

**Thanks, your help is really appreciated !!**