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What does SU(3) x SU(2) x U(1) means?

  1. Jun 14, 2009 #1
    Hi,

    1: I just want to ask that what does SU(3) x SU(2) x U(1) means?
    and when a lagrangian is invariant under SU(3) x SU(2) x U(1)
    what does that mean?

    Does it mean that lagrangian L is invariant under SU(3) and then SU(2) and
    then U(1)? so if one wants to check SU(3) invariance of L then one operates
    with an element of SU(3) and leave the rest? doesent it operate on electro-weak part of L?


    2: Suppose that I have total standard-model Lagrangian for fermions as:

    L(sm) = L(qcd) + L(Electro-weak)............(left handed)

    where fermions in L(qcd) are triplets and fermions in L(electro weak) are
    doublets SEPARATELY
    which means that the generator of SU(3) = 3x3 matrices and generators of
    SU(2) = 2x2 matrices SEPARATELY.

    How can I arrange these different dimension matrices in the above lagrangian? which would be
    valid for leptons as well as quarks?

    Actually my main confusion is that how can the same lagrangian be valid for
    leptopns in doublets, quakrs in triplets and generators in 3x3 and 2x2 matrices.
    But what about the U(1) matrix?

    If I take only QCD SU(3) lagrangian and write it next to Electro-weak SU(2) x U(1) lagrangian,
    is this OK for SU(3) x SU(2) x U(1) complete lagrangian? there will be a mismatch
    in dimension of matrices though or not?

    At least, I need the dimensions of different terms in a simple QCD -Elec-Weak lagrangian (as above)
    for left handed leptons and then left handed quarks to remove my confusions.

    many thanks.
     
  2. jcsd
  3. Jun 14, 2009 #2

    malawi_glenn

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    There will be no colour index for the leptons...
     
  4. Jun 15, 2009 #3

    nrqed

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    Yes, it means that we can operate with each group separately and the Lagrangian is invariant under each.
    Physicists use a very condensed notation when describing the Standard Model. You can think of all the quarks and leptons (as well as the Higgs) as being in six-dimensional matrices. Take the left-handed electron and neutrino weak doublet, for example. They are color singlets. You can think repeat them three times to form a six-dimensional vector. When acting on them, a SU(3) transformation will simply be an identity matrix since they are color singlets. When an SU(2)_L transformation is applied, it will act the same way on each of the three electron and neutrino doublets, so it will be representend by a block diagonal matrix with the SU(2) repeated three times in block diagonal form.

    Since the SU(3) transformation is trivial, and since the SU(2) transformation is three copies of the same thing, you see that we don't gain much by working in a six-dimensional vector space. This is why people cheat a bit and just write instead a single, two-dimensional, copy of the weak doublet.

    By the way, a U(1) transformation is not represented by a matrix, but by a simple complex phase [itex] e^{i \theta} [/itex]


    Hope this helps.
     
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