Parity formulae, orbital angular momentum, mesons

[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?Thanks in advance.

 

[binbagsss]

bump.

 

[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}

 

[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

 

[binbagsss]

l
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.

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

 

[Envelope]

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

For fermions, yes. For bosons, no.