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

  1. Aug 8, 2006 #1
    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. jcsd
  3. Aug 10, 2006 #2
    Err, a bit more info please...

    marlon
     
  4. Aug 10, 2006 #3

    Meir Achuz

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    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.
     
  5. Aug 10, 2006 #4

    Meir Achuz

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    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.
     
  6. Aug 10, 2006 #5

    Meir Achuz

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    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.
     
  7. Aug 10, 2006 #6
    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} ?
     
  8. Aug 16, 2006 #7

    Meir Achuz

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    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.
     
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