Redefinitions of Lagrangians using EOM I am trying to understand an statement of this paper. In section 2 this Lagrangian is introduced
$$\mathcal{L}_4=-|D_{\mu}\phi|^2-\lambda_{\phi}|\phi|^4-\frac{1}{4g^2}F_{\mu\nu}^2\qquad{}\mathcal{L}_6=\frac{1}{\Lambda^2}[c_r\mathcal{O}_r+c_6\mathcal{O}_6+c_{FF}\mathcal{O}_{FF}],$$
where $\Lambda$ is an energy scale suppresing the dimension 6 operators and
$$\mathcal{O}_r=|\phi|^2|D_{\mu}\phi|^2\qquad{}\mathcal{O}_6=|\phi|^6\qquad{}\mathcal{O}_{FF}=|\phi|^2F_{\mu\nu}F^{\mu\nu}.$$
On the second paragraph of section two it i said

Integrating by parts and uing the EOM, we can elimiate $\mathcal{O}_r$ in favor of $\mathcal{O}'_r=(\phi{}D_{\mu}\phi^*)^2+h.c$

Now, I would like someone made this computation explicit. It puzzles me what role the equation of motion of $\phi$ play in all this (since it must be a mess involving the gauge field). Also, is it legitimate to use equations of motion to make redefinitions in Lagrangians?
 A: This is a particular example of a general theorem in effective field theory: if you have an operator that is proportional to the lowest order equations of motion, you can push that operator to higher order in perturbation theory by a field redefinition. This is especially useful if you are working to a fixed order in perturbation theory, in which case you might be able to eliminate that operator from your calculation completely (by pushing it to a higher order in perturbation theory than you are considering). In this example, after integration by parts you find a dimension six operator proportional to the lowest order equations of motion. The general theorem tells you that you can do a field redefinition to replace the dimension six operator with operators of higher dimension.
Let's start by re-writing your operator by integrating by parts
\begin{equation}
\mathcal{O}_r = |\phi|^2 |D \phi|^2 =- \frac{1}{4} (\phi D_\mu \phi^*)^2 - \frac{1}{4} |\phi|^2 \phi (D_\mu D^\mu \phi^*) + h.c.
\end{equation}
Where equality here really means up to total derivatives. Now the lowest order equation of motion is $D_\mu D^\mu \phi = 0$. We should not use this equation of motion directly in the lagrangian. However, we can take advantage of this fact to do a field redefinition to push the operator proportional to $D_\mu D^\mu \phi$ to higher order, let's see how that works.
The only terms in the Lagrangian we need to keep are the kinetic term and the operator proportional to the lowest order equations of motion (the technical name for this is a `redundant operator'). 
\begin{equation}
\mathcal{L} \supset \phi (D_\mu D^\mu \phi^*) - \left( \frac{1}{4\Lambda^2} |\phi|^2 \phi (D_\mu D^\mu \phi^*) + h.c.\right)
\end{equation}
Now the field redefinition we need to do is
\begin{equation}
\phi \rightarrow \phi + \frac{1}{4\Lambda^2} |\phi|^2 \phi
\end{equation}
(and similarly for $\phi^*$). You can check that this field redefinition pushes the redundant operator to higher order. Even though the redundant operator looks like it is a dimension 6 operator, this field redefinition pushes it to higher order. We can always remove redundant operators in this way, for a proof of the theorem in general see for example the review of EFT by Cliff Burgess, hep-th/0701053, especially around equation 21.
