Possible decay states strong interaction, parity conservation

[binbagsss]

The question is for which of the $1P$ meson states - $1^{1}P_{1}, 1^{3}P_{0},1^{3}P_{1}, 1^{3}P_{2}$ $D_{s}$ states decaying to a $1S$ state is the decay: $D_{s}**^{+} -> D_{s}^{+}\pi^{0}$ possible?

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 !

 

[mfb]

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

You can look that up (and one can argue that "spin zero" is part of the definition of those particles).

Parity is multiplicative - you multiply all parity values to get the total quantum number.

So a particle has an intrinsic parity. Is this a particle or a quark?

Both descriptions work, but as you deal with hadrons here it is easier to consider those.

 

[binbagsss]

You can look that up (and one can argue that "spin zero" is part of the definition of those particles).

As in from a table of mesons?
And there's no other way of knowing?

 

[binbagsss]

Okay, I've had a look a the next part of the question which is to do the same (find the possible states) for the decay $D_{s}^{+}** -> D_{s}*^{+} \pi ^{0}$.

And I follow the solution , if it is the case that the $D_{s}*^{+}$ has $s=1$ , (it doesn't actually state this but I'm pretty sure it's being used).

Is this the case in general, the unexcited version of the particle has $s=0$ and the excited $s=1$?

Thanks .

 

[mfb]

As in from a table of mesons?

Right.

And there's no other way of knowing?

You have to find out what the symbol "$\pi^0$" means - you either know it or you have to look it up.

What is the s=0, s=1 now?

 

[binbagsss]

What is the s=0, s=1 now?

Sorry? I'm unsure what you are asking?

 

[mfb]

Where does it come from and what does it mean?

 

[binbagsss]

Where does it come from and what does it mean?

Sorry in post 4 $s$ stands for spin.

 

[mfb]

Sometimes spin 1 states are called excited states, sometimes they have their own particle name. That is not completely consistent.