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Gauge transformation of Yang-Mills field strength

  1. May 8, 2013 #1
    Hi. I'm reading about non-abelian theories and have thus far an understanding that a gauge invariant Lagrangian is something to strive for. I previously thought that the Yang-Mills gauge boson free field term ##-1/4 F^2 ## was gauge invariant, but now after realizing that the field strength transform homogeneously under gauge transformation U(x) s.t

    $$ F \to UFU^\dagger$$

    it seems like this term also transform like

    $$F^2 \to UF^2U^\dagger.$$

    Am I right there? If so isn't this a problem for the physics? Why, why not?
     
  2. jcsd
  3. May 8, 2013 #2
    The actual lagrangian term is the trace of ##F^2##:

    ##{\rm tr} (F^{\mu \nu} F_{\mu \nu})##

    This is gauge invariant. Sometimes this gets written as

    ##F^{a \mu \nu} F^a_{\mu \nu}##

    where ##F^a_{\mu \nu}## is defined by ##F_{\mu \nu} \equiv F^a_{\mu \nu} T^a##. Here ##a## is a gauge group adjoint representation index, the ##T^a## are the generator matrices of the adjoint representation and ##F^a_{\mu \nu}## is a set of numbers (not matrices). This is just decomposing the field strength matrix as a linear combination of the generators.

    The second form of the Lagrangian is equivalent because ##{\rm tr} (T^a T^b) = \delta^{a b}##. So if you plug the decomposition of ##F^{\mu \nu}## into the first form of the Lagrangian, you get the second form.
     
    Last edited: May 8, 2013
  4. May 8, 2013 #3
    Ah, that's true. So the trace is actually essential. I've heard that it was nothing but compact notation :)
     
  5. May 9, 2013 #4
    It should most probably be seen like this after gauge transformation
    [itex] UFU^\dagger UFU^\dagger= UF^2U^\dagger [/itex]
     
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