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Jacobian in path integral equal to one?

  1. Nov 18, 2012 #1

    [tex]\int d\phi e^{iS[\phi]}=\int d\phi' J e^{iS'[\phi']} [/tex]

    where J is the Jacobian. If the transformation of variables to phi' is a symmetry of the action [i.e., S'=S], then this becomes:

    [tex]\int d\phi e^{iS[\phi]}=\int d\phi' J e^{iS[\phi']} [/tex]

    But doesn't this imply that the Jacobian has to equal one?

    But surely that doesn't have to be true in general? If the action has a symmetry, and you perform the change of coordinates corresponding to the symmetry transformation, then does the Jacobian of that transformation have to equal one?
    Last edited: Nov 18, 2012
  2. jcsd
  3. Nov 18, 2012 #2
    I think you miss that, in general, [itex]\mathrm{d} \phi \neq \mathrm{d} \phi'[/itex].

    For a simple example, consider [itex]S=\frac{1}{2}\phi^2[/itex]. Obviously, this is invariant under the inversion [itex]\phi \rightarrow \phi'=-\phi[/itex], for which [itex]\mathrm{d}\phi'=-\mathrm{d}\phi[/itex] and [itex] J=\frac{\partial \phi'}{\partial \phi}=-1[/itex].
  4. Nov 18, 2012 #3
    The Jacobian should take care of differences in measure, so what happens to

    ∫ e-x^2 dx from -∞ to ∞

    under y=-x is:

    ∫ e-y^2 dx/dy dy from ∞ to -∞

    which equals ∫ e-y^2 (-1) dy from ∞ to -∞

    so it's the change of the order of the limits in the integration that allows the Jacobian to not have to be equal to 1.

    Maybe a stronger statement is true: If the limits in an integration are unchanged by a transformation, then the Jacobian must equal one?
  5. Nov 19, 2012 #4
    I guess that is the more common way of putting it, yeah.

    I don't have an explicit example, but couldn't we have a local transformation [itex]y=f(x)[/itex] that doesn't change the endpoints (i.e. limits) and [itex]J=\frac{\partial f}{\partial x} \not\equiv 1[/itex]?

    I think your statement should hold for linear functions [itex]f(x)[/itex] though.
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