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The massless scalar field action on Minkowski background is given by \begin{equation} S[\phi]=\int_{\mathbb{R}^D}d^Dx~\eta^{\mu\nu}\partial_\mu\phi\partial_\nu\phi. \end{equation} This action is often referred as one of the easiest CFT one can have so I would like to show that is it conformally invariant. If we only consider dilatations $x'^\mu\mapsto x^\mu+\epsilon x^\mu$ for example, the scalar field transforms as \begin{equation} \delta\phi(x)=\epsilon x^\mu\partial_\mu\phi(x). \end{equation} I can't manage to show that the action only changes up to a boundary term. How would one show that ?

[EDIT]

I already saw a lot of proofs of this (e.g. here) but they always transform the metric or the partial derivatives. This is equivalent of considering the trivial action of a diffeomoprhism that acts on each object in the integral and therefore trivially leaves it invariant. Am I wrong ?

xpsf
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  • I suggest that you use the relation $\eta_{\hat{\mu}\hat{\nu}} = \Omega^2\eta_{\mu\nu}$ where $\Omega^2$ is a function of the space-time co-ordinates and also use the transformation $\partial_{\hat{\mu}} = \frac{\partial x^{\alpha}}{\partial x^{\hat {\mu}}}\frac{\partial,}{\partial x^{\alpha}},.$ – Physics_Et_Al Dec 16 '21 at 14:05
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    Does this answer your question? Proof that free scalar field is conformally invariant – Connor Behan Dec 16 '21 at 14:07
  • I don't want to transform the metric or the partial derivatives because this would result in a trivial coordinate transformation, under which any integral is invariant. Am I wrong ? – xpsf Dec 16 '21 at 14:08
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    That is not how the scalar field transforms under dilations. Rather, the transformation has the form $\delta \phi(x) = \epsilon ( x^\mu \partial_\mu + \Delta ) \phi(x)$ for a particular choice of $\Delta$. You can work out what this choice is. – Prahar Dec 16 '21 at 14:25
  • Why is that ? Conformal transformations are just a generalization of Poincaré transformations no ? And also performing the expansion of $\phi(x+\epsilon x)$, we find that $\delta\phi=\epsilon x^\mu\partial_\mu\phi$. – xpsf Dec 16 '21 at 15:23
  • @xpsf - conformal transformations are NOT a generalization of Poincar'e transformation. They involve a Weyl transformation in ADDITION to the conformal diffeomorphism you are referring to. There is a reason you've seen a lot of proofs involving metric transformations. It's because that's what a conformal transformation IS! – Prahar Dec 16 '21 at 16:40
  • Ok, I think that the terminology was not clear to me indeed. Thank you for this clarification, I now understand. – xpsf Dec 16 '21 at 18:28

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