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How does U(1) guage symmetry lead to Maxwell's equations?

  1. Jan 13, 2012 #1
    Here are some questions that have been puzzling me about symmetry and charge. Any answers to any of these questions would be very helpful. Thank you.

    What does U(1) gauge symmetry mean? Does anyone have a simple explanation?

    Can Maxwell's equations be derived from the premise of U(1) gauge symmetry? If so, how?

    Do Maxwell's equations imply U(1) symmetry?

    Is there an operator to obtain the electric charge of a particle from its quantum state?
  2. jcsd
  3. Jan 13, 2012 #2
    U(1) gauge symmetry means your physics doesn't change if you add a term of the form exp(i*f(x)) (this is a local gauge symmetry, for a global one f(x) is a constant). If one thinks of the simplest terms one can write down in a Lagrangian which are unchanged by such a transformation (and are Lorentz invariant) you get what turns out to be Maxwell's Equations.

    Also, you could make a trivial operator but charge is always conserved in QM (or rather charge-parity-time reversal is for QFT as well) so it hardly matters.
  4. Jan 16, 2012 #3
    In physical terms it means that the partial differential equations are insensitive to boundary values of the potentials. Since the Maxwell equations are founded on U(1) symmetry they give you gauge freedom for the potentials.

    Equations which comply to SU(2) symmetry for example, as often used in QM, are sensitive to boundary conditions for potentials.
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