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Functional determinants and gauge invariance

  1. Apr 10, 2010 #1
    Hi all,

    I've been studying the path-integral quantisation of gauge theories in Zee III.4. My understanding is roughly as follows: that one can think of the differential operator in the quadratic tems in the lagrangian as a linear operator between infinite dimensional spaces (morally equivalent to a quadratic form on the cartesian product of the spaces); that the solution to the free theory looks like the reciprocal square root of the corresponding functional determinant; that the functional determinant will be zero if the linear operator annihilates a non-zero field, and consequently our operator will have no inverse (i.e. no propagator); and that the addition of a field thus annihilated to another effectively constitutes a gauge transformation.

    Given this (possibly erroneous) understanding, I'm struggling to see why the simple non-interacting scalar klein-gordon isn't a gauge theory, as the vanishing of a field under application of the operator [tex](\partial^2+m^2)\phi=0[/tex] is just the equation of motion, and the "gauge transformation" wouldn't connect physically equivalent states, but would be a restatement of the principle of superposition.

    As this seems to be true more or less in general for scalar field theories, I'm thinking that it relates specifically to the fact that Zee is treating a formal matrix of differential operators, and it's actually this level of structure that introduces the non-invertibility. However, Peskin and Schroeder still explicitly identify [tex](const.)(\partial^2+m^2)^{-\frac{1}{2}}[/tex] as the functional determinant, so I'm not sure why the logic doesn't extend to the "1-dimensional" case. If this the reason, is there a similar feature that singles out a gauge symmetry in the scalar case (e.g. conformal invariance of massless phi-fourth theory, or U(1) transformations of complex scalar fields)?

    Thanks in advance.
     
  2. jcsd
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