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Charges and lattices

  1. May 8, 2010 #1
    Hi...I do not understand why the charges of a QFT belong to the root lattice of the gauge group...can somebody explain it please??
  2. jcsd
  3. May 9, 2010 #2


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    I am not sure what you mean.

    The charges are the q.m. generators of the gauge group. Look at a simple example, the n-dim. harmonic oscillator with


    [tex][T^a, T^b] = if^{abc}T^c[/tex]

    If you define the charges (in an SU(N) gauge theory you can derive them from the Noether theorem instead of introducing them by hand)

    [tex]Q^a = a^\dagger_i (T^a)_{ik} a_k[/tex]

    you can check easily that they generate the same algebra.

    [tex][Q^a, Q^b] = if^{abc}Q^c[/tex]

    Therefore you can classify all states according to the su(N) algebra and all equations that are valid on the algebraic level carry over to the q.m. states.

    Quantum gauge theory gives you additional equations, e.g. the requirements that in an unbroken SU(N) gauge theory all physical states are SU(N) singulets, but these constraints are due to additional dynamical considerations and cannot be derived frome purely algebraic reasoning.
  4. May 9, 2010 #3
    Hi...thanks for your answer. Well, i have read several times that charges in a gauge theory lie in the root system of the gauge group. For a quick check of the above statement, you can google> "charge wikipedia" and you will find such a claim. Of course, I'm referring to the wiki just because it is easy to check, but this statement is done in several places elsewhere.
  5. May 9, 2010 #4


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    Do you know what a root system of a Lie algebra is?
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