How to prove P=(-1)^L, c=(-1)^{L+S} for q \bar q

[BuckeyePhysicist]

How to prove

$P=(-1)^L$, 
$C=(-1)^{L+S}$

for

$q \bar q$

?

And $P=?$, $C=?$ for tetraquarks

$[(q_1 q_2)_{S_1}(\bar q_1 \bar q_2)_{S_2}$

?

 

[marlon]

Err, a bit more info please...

marlon

 

[Meir Achuz]

How to prove

$P=(-1)^L$, 
$C=(-1)^{L+S}$

for

$q \bar q$

?
?

P=(-1)^L for two particles. The relative parity of a fermion and its antiparticle is negative. So P=(-1)^(L+1) for q-qbar.
e.g., the spin zero pion has negative parity.

 

[Meir Achuz]

C=(-1)^(L+S) is correct for spin 1/2 fermion-antifermion.
e.g., the pi0 has C=+1.
To get the C eigenvalue, you have to reverse the spatial positions (-1)^L,
and the spin positions, which gives (-1)^(S+1) in adding 1/2+1/2=S.
There is another factor of (-1) when two fermiions are exchanged.

 

[Meir Achuz]

For tetraquarks, P=(-1)^(L1+L2+L3), because there are three internal orbital angular momenta. C depends on how you couple the quarks and antiquarks. Of course, C only applies to a neutral particle.
C is simpler when you first couple q to qbar. Then C for each pair is as above.

 

[BuckeyePhysicist]

What if I couple q_1 q_2 first with total angular momentum quantum number: J_1, and couple \bar q_1 and \bar q_2 together with J_2. And at last couple the diquark with the anti-diquark[or di-antiquark] with orbital momentum L. What I get for C for this tetraquark?

Is it (-1)^2 * (-1)^{J_1+1}* (-1)^{J_2+1} * (-1)^{L} ?

 

[Meir Achuz]

Coupling q-qbar first is simpler and more reasonable (since the q-qbar attraction is greater).
Your formula for C is wrong.
For your coupling C=(-1)^L, and you need the internal J,L,S to be the same for each diquark to have an eigenstate of C.