Continuous symmetries (Srednicki)

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SUMMARY

The forum discussion centers on the path integral formulation presented in Chapter 22 of Srednicki's text, specifically regarding the invariance of the functional integral Z(J) under arbitrary infinitesimal shifts of the field variable φ_a(x). The key conclusion is that the invariance of the measure Dφ under these shifts guarantees that Z(J) remains unchanged, leading to the result δZ(J) = 0. This principle parallels the established mathematical proof of the invariance of Gaussian integrals, demonstrating that the measure's invariance is crucial for maintaining the integral's value despite variable transformations.

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LAHLH
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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
 
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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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