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I'm studying the Lagrangian $$ \mathcal{L} = \frac{1}{2}\partial_\mu\phi \partial^\mu\phi+\lambda\phi\partial_\mu\phi\partial^\mu\phi~=~\frac{1}{2}(1+2\lambda \phi)\partial_\mu\phi \partial^\mu\phi.\tag{1} $$ We can take the field redefinition to make it looks like free field theory: $$ \frac{1}{2}(\partial\phi')^2 = \frac{1}{2}(\partial\phi)^2+\lambda\phi(\partial\phi)^2\tag{2} $$ and we can solve and obtain that $$ \phi' = \frac{1}{3\lambda}(1+2\lambda\phi)^{3/2}.\tag{3} $$ My question is since this redefinition is non-linear, how can we see $$ \int\mathcal{D}\phi\rightarrow\int\mathcal{D}\phi'\tag{4} $$ produce the same theory? I heard about this would give an additional Jacobi factor, which vanishes in dimensional regularization because it’s scaleless, but I don't know how to make sense of this. Also, how can we tell if a field could be redefined?

Qmechanic
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IGY
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    Page 11 of these notes https://courses.edx.org/c4x/MITx/8.EFTx/asset/notes_EFT.pdf discusses the "Representation Independence Theorem" that is probably what you are looking for, with some details. I don't understand it well enough to write more, I'll try to think about it though – QCD_IS_GOOD Feb 09 '23 at 01:41
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    How did you manage to solve that equation? In any case, the allowed field redefinitions (i.e. those for which the physics remains unchanged) are of the form $\phi \to f(\phi)$ where $f(\phi) = \phi + {\cal O}(\phi^2)$ near $\phi = 0$. Your field redefinition does not satisfy this. – Prahar Feb 09 '23 at 02:26
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    @Prahar Thanks for the comment! I got $\partial_\mu\phi'=\partial_\mu\phi\sqrt{1+2\lambda\phi}$, this could be obtained by adding $\partial_\mu$ on both sides of that redefinition. – IGY Feb 09 '23 at 02:42

1 Answers1

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  1. Yes, OP's field redefinition (3) can be made perturbative by a simple shift: $$ \phi^{\prime}~=~\frac{1}{3\lambda}(1+2\lambda\phi)^{3/2}-\frac{1}{3\lambda}~=~\phi+{\cal O}(\phi^2).\tag{3'}$$

  2. Moreover, in dimensional regularization the Jacobian is 1 due to the presence of $\delta^d(0)$, cf. e.g. my Phys.SE answer here.

Qmechanic
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