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

1. Aug 8, 2006

### BuckeyePhysicist

How to prove
Code (Text):
$P=(-1)^L$,
$C=(-1)^{L+S}$
for
Code (Text):
$q \bar q$
?

And $P=?$, $C=?$ for tetraquarks
Code (Text):

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

?

2. Aug 10, 2006

### marlon

marlon

3. Aug 10, 2006

### Meir Achuz

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.

4. Aug 10, 2006

### 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.

5. Aug 10, 2006

### 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.

6. Aug 10, 2006

### 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} ?

7. Aug 16, 2006

### 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.