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

Invariant measures

  1. Jun 24, 2008 #1
    I am reading through sidney colemans lectures on QFT and I am stuck on what seem to be a silly question: He talks about the fact that the measure used in a calculation should be invariant in order to prove unitarity and later on that operators transform properly. He uses the example of rotational invariance.

    [tex]U^{-1}(R)\int d^{3}k|k><k| U(R)[/tex]
    [tex]\int d^{3}k |R^{-1}k><R^{-1}k|[/tex]

    Change of variable: Rk' = k
    Now this is the part I dont get (I must be confused)
    [tex]d^{3}k' = d^{3}k[/tex]
    It seems to me like there should be a factor of R or something. However, the strange thing is that R is a matrix (isn't it?) so I dont really get it. Can someone explain what is going on here?
  2. jcsd
  3. Jun 24, 2008 #2


    User Avatar
    Science Advisor

    When in doubt, use brute force. Grind it out. That is, here, compute the appropriate Jacobean, and thus prove the equality. (Done in countless texts.)
    Reilly Atkinson
  4. Jun 24, 2008 #3


    User Avatar
    Science Advisor

    In general, when you perform a change of variable in an integral, you
    must include the Jacobian determinant of the transformation. In this case
    the Jacobian is [itex]| \partial k'_i/\partial k_j |[/itex].
    (Consult Wiki for more detail.)

    In the current case, [itex]d^3k[/itex] is an infinitesimal volume element
    in 3-momentum space, and are preserved by rotation transformations,
    so the Jacobian turns out to be 1. But that's not necessarily so in more
    general transformations.
  5. Jun 25, 2008 #4
    From the point of view of group theory, what he used is the so-called rearrange lemma. It's most easily to be understood in finite dimensional.
    For example, you have a cyclic group [tex] G = \{e,a,a^2\}[/tex] with [tex]a^3 = e[/tex]. Let [tex]g\in G[/tex], for example, say [tex] g = a[/tex], then [tex]g\{e,a,a^2\} = \{a,a^2,e\}[/tex], meaning, [tex]g[/tex] operates on the all group element would reproduce all group elements. Hence, if we consider a summation over the group elements, say [tex]\sum_{g\in G}f(g'g)[/tex] where [tex]f[/tex] is a function of the group element. By the rearrangement lemma, we may safely rewrite the summation as [tex] \sum_g f(g) [/tex]

    In the continuous group case, the rearrangement lemma is somewhat more involved. Since we have infinitely many ways to parametrize a Lie group, so we have to be more careful. We define that, a parametrisation [tex] g(\xi) [/tex], together with a weight function [tex] \rho_g(\xi) [/tex] such that the following equation holds
    [tex] \int dg f(g) = \int dg f(s^{-1}g) [/tex]
    where [tex]s\in G[/tex] and [tex] dg = \rho_g(\xi)d\xi [/tex] is called to provide an invariant measure. And one can prove that the weight function for SO(3) group is just 1. (It happens to be the Jacobian factor).
  6. Jun 25, 2008 #5
    Is this the Haar measure?
  7. Jun 25, 2008 #6
    looks like it, if you nomalize it to 1 when you integrate 1 over the whole manifold (group).
  8. Jun 25, 2008 #7
    Thanks everyone for the help. I was forgetting about the fact that the determinant of the rotation matrix was 1... oops. Thanks for the further insight ismaili that is very interesting and I will have to look further into your comment.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Invariant measures
  1. Gauge invariance (Replies: 4)

  2. Lorentz invariance (Replies: 3)

  3. Rotational invariance (Replies: 4)

  4. Translation invariant (Replies: 5)