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A Invariant combination of SU(3) states

  1. Feb 7, 2017 #1
    Hi everyone,
    this is something i know because i saw it many times, but i have never fully understand it. Suppose i have a quark field (singlet under SU(2) let's say) ##q## and i would like to build an invariant term to write in the Lagrangian. The obvious choice is to write a mass-term ##\bar{q}q##. This is both Lorentz invariant and SU(3) invariant, since

    $$q'=Uq \\ \bar{q}'=\bar{q}U^\dagger$$

    This is okay, but sometimes i encounter other terms, that involve 3 quarks and a lepton. I heuristically know that to make a color singlet you need a anti-symmetric combination of 3 quarks or 3 anti-quarks. So we can write a term like:

    $$\epsilon_{\alpha\beta\gamma}\bar{u}^c_\alpha \, u_\beta \, \bar{d}^c_\gamma \, e$$
    where e is the electron field and c means that it is the charge conjugated field ##u^c=C(\bar{u})^T##. Why this term is gauge invariant under SU(3)? Why do we have to combine them like that to obtain an invariant, with the anti-symmetic tensor?

    Also there is something else that i think it is related to this, but with SU(2). If i have an object L that is a doublet, i know that the invariant is something like ##L^\dagger L##. However, in the Standard Model for example, we have the Higgs field ##H## but we also need the field ##\tilde{H}_i=\epsilon_{ij}H^*_j## that is a field with opposite hypercharge. So for example, we can build the yukawa term

    $$\bar{q} \, \tilde{H} \, u_R$$

    and the first part is basically from an SU(2) point of view ##\bar{q}_i H^*_j \epsilon_{ij}## that looks exactly like the counterpart of the SU(3) singlet above.

    So i guess, my question is: why can we build these kind of invariant terms in SU(N), basically taking the complex conjugated of a field and then contracting them with the anti-symmetric tensor?

    If you also have some references that explain in detail why we can build such kind of terms in the Lagrangian it could be very helpful.

    Thanks in advance for the help!
     
  2. jcsd
  3. Feb 7, 2017 #2

    Orodruin

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    In general, the product of the SU(3) irreps under which the quarks transform can be decomposed into irreps. When you take the product of three fundamental SU(3) representations, the resulting representation contains a singlet representation which is the completely anti-symmetric product (there is only one way of taking the completely anti-symmetric product and so that representation is one-dimensional and must be the singlet representation).
     
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