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Continuous symmetries (Srednicki)

  1. Sep 14, 2010 #1
    Hi,

    In ch22, Srednicki considers the path integral [tex] Z(J)=\int D\phi \exp{i[S+\int d^4y J_a\phi_a]} [/tex]

    He says the value of Z(J) is unchanged if we make the change of var [tex] \phi_a(x)\rightarrow\phi_a(x)+\delta\phi_a(x)[/tex], with [tex] \phi_a(x)[/tex] an arbitrary infinitesimal shift that leaves the mesure [tex] D\phi[/tex] invariant by assumption.

    Why does this guarantee that Z(J) is unchanged, i.e [tex] \delta Z(J)=0[/tex]. I would have presumed that only changes that are symmetries of the Lagrangian, would leave the action and therefore Z(J) invariant, not just arbitrary changes. Does is have something to do with our assumption that the measure is invariant?

    Thanks
     
  2. jcsd
  3. Sep 14, 2010 #2
    I think you can always change the variables of integration. Then the invariance of the measure does the job. Much like you prove that
    [tex]\int_{-\infty}^{\infty}\exp(-(x-a)^2)dx = \int_{-\infty}^{\infty}\exp(-x^2)dx[/tex]
     
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