Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Decomposition of SU(3) and particles

  1. Dec 20, 2011 #1

    Roumpedakis

    User Avatar
    Gold Member

    As we know the algebra of SU(3) consist of two Cartan generators and 6 raising and lowering operators. We define the eigenstates of the Cartan operators as u,d,s, correspoding to the three lightest quarks.
    Now when we study the [itex] 3\otimes 3[/itex] tensor product we can show that the Hilbert space of these states decompose as
    [itex]3\otimes 3 = 8\oplus 1[/itex]
    Which means that we can devide this Hilbert into two invariant subspaces.
    And similar for baryons
    [itex]3\otimes 3\otimes 3=10\oplus 8 \oplus 8\oplus 1[/itex]
    My question is following. We can categorize the mesons in two multiplets of 8 particles and 1 singlet, and the baryons in one multiplet of 10 particles, one of 8 particles and one singlet. So is not obvious to me which is the correspondence between these groups of particles and the above Hilbert spaces. If the decomposition of [itex] 3\otimes 3[/itex] has one 8-plet why we have two mesons diagrams and for baryons why we have only one 8-plet of particles?
     
  2. jcsd
  3. Dec 20, 2011 #2

    samalkhaiat

    User Avatar
    Science Advisor

    Look at the [itex]J^{P}[/itex] quantum numbers. Mesons have [itex]0^{-}[/itex] and [itex]1^{-}[/itex] octets; the vector meson octet ([itex]1^{-}[/itex]) carries a single space-time index; transforms as a Lotentz vector hence the name “vector meson”. In terms the [itex]SU_{f}(3)[/itex] quark fields, the vector mesons have the following combinations
    [tex]\{8\}_{\mu}^{i} \sim \bar{q}(x) \lambda^{i}\gamma_{\mu} q(x),[/tex]
    [tex]\{1\}_{\mu} \sim \bar{q}(x)\gamma_{\mu}q(x),[/tex]

    compared with pseudoscalar mesons ([itex]0^{-}[/itex]) singlet and octet;
    [tex]\{1\} \sim \bar{q}(x)q(x) = \bar{u}u + \bar{d}d + \bar{s}s,[/tex]
    [tex]\{8\}^{rs} \sim \bar{q}(x)R_{(a)}( \lambda ) q(x), \ \ r,s = 1,2,3[/tex]
    where
    [tex]
    R_{(a)}(\lambda) = \lambda^{3}, \lambda^{8}, (\lambda^{i} \pm \lambda^{i+1}); i = 1,4,6
    [/tex]

    Sam
     
    Last edited: Dec 20, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Decomposition of SU(3) and particles
  1. SU(3) decomposition (Replies: 9)

  2. QCD and SU(3) (Replies: 3)

  3. Roots of SU(3). (Replies: 2)

Loading...