Parity formulae, orbital angular momentum, mesons

1. Mar 14, 2015

binbagsss

So a particle has intrinsic parity $\pm 1$ .
The parity of a system of particles is given by product of intrinsic parities and the result is: $(-1)^l$ (1).

Questions:

1) How does this result follow?
and what exactly is $l$ here? so it's the orbital angular momentum, so say a particle is made up of 3 quarks, then it's described to be in a certain state, $1p$ , $1s$ states etc, so $l$ is the orbital angular momentum of this state?

2) For a meson $p=(-1)^{l+1}$
The reasoning in my book being that the quark and antiquark have opposite intrinsic parities,

So I assume this followss from (1), although I am not seeing how??

But, in the result (1) , this is a general result for any system of particles right? So quarks and antiquarks of the same type could be included in any system yet $(-1)^l$ still holds?

2. Mar 18, 2015

binbagsss

bump.

3. Mar 18, 2015

ChrisVer

$l$ is the angular momentum number. The particles in general will be described by spherical harmonic functions $Y_{lm}(\theta, \phi)$ which have the given parity:
$\hat{P} Y_{lm} (\theta, \phi) \equiv Y_{lm} (\pi-\theta, \phi+ \pi)= (-1)^l Y_{lm} (\theta, \phi)$

Take for example a meson.
The meson will have a quark and antiquark $q, \bar{q}$.
So the intrinsic parity of the one will be +1 and the parity of the other will be -1.
On the other hand, the meson can have some spatial angular momentum number $l$.
So the total parity will be: $(+1) \times (-1) \times (-1)^l = (-1)^{l+1}$

4. Mar 18, 2015

ChrisVer

So I guess you make a wrong assumption before your eq.1.
"The parity of a system of particles is given by product of intrinsic parities "
This is wrong (probably a misconception).

Again take a meson... it is a $q \bar{q}$ state with some angular momentum $l$ due to the quarks. Since the meson has an orbital angular momentum $l$, its angular distributions are described by the spherical harmonics $Y_{lm}$. It's from this relation that you get the eq.1 $(-1)^l$, and not by getting the intrinsic parities.
http://pdg.lbl.gov/2008/reviews/quarkmodrpp.pdf
Sec. 14.2

5. May 2, 2015

binbagsss

Is this always the case with a particle and antiparticle, that they have opposite parities?

6. May 2, 2015

Envelope

For fermions, yes. For bosons, no.