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Currents / Multiplets

  1. Jan 12, 2006 #1
    Dear Gents,

    I have one question:

    Can I say that if I I have some currents, charges form these currents (integrals over time components) and these charges form a close algebra of some group SU (N), then I say that the particles sitting in these currents (all possible incoming/outgoing particles) form a multiplet of some representation of this group SU (N).

    If I am correct based on what can I state that?

    If not correct what do I say wrong?

    Thanks a lot
  2. jcsd
  3. Jan 17, 2006 #2


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    Science Advisor

    Yes, you are right. However, I would not use the phrase "particles sitting in currents".
    In QFT, particles are represented by vectors in the Fock space of states (remember the one-particle state [itex] |p>={a}^\dagger (p)|0> [/itex]), while currents (like the fields) are operators.
    The reason why you are right is trivial, because we know this fact even before deriving the form the currents.
    As you might know, we derive the currents from a Lagrangian. We costruct this Lagrangian to be, for example, SU(n)-invariant. And to do this, the fields must belong to a certain SU(n) multiplets.
    For example, to construct SU(2) invariant Lagrangian, we use fields belong to;
    1) the fundamental representation (the {2}=SU(2)-doublet), for example; N=(p,n) or q=(u,d).
    2) the adjoint representation (the {3}=SU(2)-triplet), examples; the pions or vector bosons.
    So the total SU(2)-invariant Lagrangian will be;
    L({2}) + L({3}) + L(int.of {2} & {3}),
    and when you derive the total current from this Lagrangian, you will find the {2} and the {3} "sitting" in it.


    Last edited: Jan 17, 2006
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