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Linear Sigma Model Invariance Under O(N)

  1. Jan 19, 2013 #1
    In addition to my Faddeev-Popov Trick thread, I'm still tying up a few other loose ends before going into Part III of Peskin and Schroeder.

    I was able to show that the other Lagrangians introduced thus far are indeed invariant under the transformations given. But, I am hung up on what I think should probably be the easiest - the linear sigma model from page 349, Chapter 11:

    L[itex]_{LSM}[/itex] = (1/2) ( [itex]\partial_{\mu}[/itex] [itex]\phi^{i}[/itex] )^2 + (1/2)[itex]\mu[/itex]^2 ( [itex]\phi^{i}[/itex] )^2 - ([itex]\lambda/4![/itex]) ( [itex]\phi^{i}[/itex] )^4

    which is invariant under

    [itex]\phi^{i}[/itex] --> R[itex]^{ij}[/itex] [itex]\phi^{j}[/itex],

    or, the Orthogonal Group O(N).

    To show this, I've been using:

    [itex]\phi^{j}[/itex] ^2 --> R[itex]^{ij}[/itex] R[itex]^{ik}[/itex] [itex]\phi^{j}[/itex] [itex]\phi^{k}[/itex]
    = [itex]\delta^{j}_{k}[/itex] [itex]\phi^{j}[/itex] [itex]\phi^{k}[/itex]
    = [itex]\phi^{j}[/itex] ^2

    but, I guess I haven't convinced myself. Seems contrived (with the indices)

    Any help/clarification would be greatly appreciated.
    Last edited: Jan 19, 2013
  2. jcsd
  3. Jan 20, 2013 #2
  4. Jan 20, 2013 #3
    Thanks andrien.

    Looks like that's exactly what I have above in the OP, so I guess you're confirming that's correct.

    Don't know why it still leaves me uneasy. I'll probably work out some explicit examples next, as that usually clears things up.
  5. Jan 20, 2013 #4
    You can see it in a "vectorial way". O(N) are rotations and you know that these kind of transformations leave the value of the square of the vector unchanged. That's the same thing.
  6. Jan 20, 2013 #5
    Thanks Einj. I totally get it in a conceptual way like that.

    It was just the notation with the math. Wasn't quite sure I had it right!
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