Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Jacobian in path integral equal to one?

  1. Nov 18, 2012 #1
    Consider:

    [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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Jacobian in path integral equal to one?
  1. Path integrals (Replies: 5)

  2. The Path Integral (Replies: 0)

  3. Path integral (Replies: 2)

  4. The Path Integral (Replies: 15)

Loading...