# Excited states and selection rules

1. Feb 18, 2013

### Gregg

If you have a meson in the states

$^3S_1$ and $^1S_0$ this means that $J^P = 1^+$ and $0^+$ doesn't it?

But if you have excited states

$^1P_1$ this is $J^P=1^-$ but isn't $^3P_1$ supposed to be $J^P = 1^-$? Does this matter?

$^3P_0$, $^3P_1$ and $^3P_2$ for the triplet with $|L-S| \le J \le L+S$ what are the values $J=0,1,2$ referring to physically. $L$ is the orbital angular moment of the two quarks, and S is their spin, so for them to couple in a way that gives $J=2$ what does that mean?

Last edited: Feb 18, 2013