Possible decay states strong interaction, parity conservation

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  • #1
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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 !!
 

Answers and Replies

  • #2
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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.
 
  • #3
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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 theres no other way of knowing?
 
  • #4
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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 doesnt 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 .
 
  • #5
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As in from a table of mesons?
Right.
And theres 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?
 
  • #6
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What is the s=0, s=1 now?
Sorry? I'm unsure what you are asking?
 
  • #7
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Where does it come from and what does it mean?
 
  • #8
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Where does it come from and what does it mean?
Sorry in post 4 ##s## stands for spin.
 
  • #9
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Sometimes spin 1 states are called excited states, sometimes they have their own particle name. That is not completely consistent.
 

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