Using Euler-Lagrange equations show that the following EoM can be written as $D_\mu D^\mu\phi+m^2\phi + \lambda\left(\phi^\ast\phi\right)\phi=0$

Question

This post is a follow-up to this previous question.

Consider the following Lagrangian:
$$\mathcal{L}=\frac14\left(\partial_\mu A_\nu-\partial_\nu A_\mu\right)\left(\partial^\mu A^\nu-\partial^\nu A^\mu\right)+\partial_\mu\phi^\ast\partial^\mu\phi$$
$$+ieA^\mu\left(\phi\partial_\mu\phi^\ast – \phi^\ast\partial_\mu\phi\right)
+e^2A_\mu A^\mu\phi^\ast\phi-m^2\phi^\ast\phi-\frac12\lambda\left(\phi^\ast\phi\right)^2\tag{1}$$
The equation of motion obtained from the Lagrangian, $(1)$, for the scalar field $\phi$ is
$$\frac{\partial \mathcal{L}}{\partial \phi^\ast}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi^\ast}=-ieA^\mu\partial_\mu\phi+e^2A_\mu A^\mu\phi-m^2\phi-\lambda\left(\phi^\ast\phi\right)\phi$$
$$-\partial_\mu\left(\partial^\mu\phi+ieA^\mu\phi\right)=0\tag{2}$$
which can be written compactly as
$$D_\mu D^\mu\phi+m^2\phi + \lambda\left(\phi^\ast\phi\right)\phi=0\tag{3}$$

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I don't understand how in the last line of the quote above the author managed to write the equation of motion in this compact form, $(3)$, in particular, it is the $D_\mu D^\mu\phi$ term I cannot seem to prove, where $D_\mu\phi\equiv \partial_\mu\phi + ieA_\mu\phi$.

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So as an attempt to arrive at eqn. $(3)$ I start by finding the Euler Lagrange equations,

$$\frac{\partial \mathcal{L}}{\partial \phi^\ast}=-ieA^\mu\partial_\mu\phi+e^2A_\mu A^\mu\phi-m^2\phi-\lambda\left(\phi^\ast\phi\right)\phi$$

$$\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi^\ast}=\partial_\mu\left(\partial^\mu\phi+ieA^\mu\phi\right)=\partial_\mu\partial^\mu\phi+ie\partial_\mu \left(A^\mu\phi\right)$$
$$=\partial_\mu\partial^\mu\phi+ie\left(\partial_\mu A^\mu\right)\phi+ie A^\mu\partial_\mu\phi$$
where I have used the product rule in the last line.

then
$$\frac{\partial \mathcal{L}}{\partial \phi^\ast}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi^\ast}=-ieA^\mu\partial_\mu\phi+e^2A_\mu A^\mu\phi-m^2\phi-\lambda\left(\phi^\ast\phi\right)\phi$$
$$-\partial_\mu\partial^\mu\phi-ie\partial_\mu A^\mu\phi-ie A^\mu\partial_\mu\phi=0$$
$$=-\left(\partial_\mu\partial^\mu\phi+ie\left(\partial_\mu A^\mu\right)\phi+2ieA^\mu\partial_\mu\phi-e^2A_\mu A^\mu\phi\right)$$
$$-m^2\phi-\lambda\left(\phi^\ast\phi\right)\phi=0$$

$$=-\left(\partial_\mu\partial^\mu\phi+\color{red}{ie\left(\partial_\mu A^\mu\right)\phi}+\color{red}{2ieA_\mu\partial^\mu\phi}-\color{red}{e^2A_\mu A^\mu\phi}\right)$$
$$-m^2\phi-\lambda\left(\phi^\ast\phi\right)\phi=0\tag{4}$$

In the last equality I lowered and raised the indices of the third term in parentheses so that $A^\mu\partial_\mu\phi\to A_\mu\partial^\mu\phi$.

So consistency demands that the terms in the parentheses of $(4)$ are equal to $D_\mu D^\mu\phi$ which I will write as $D_\mu\phi D^\mu\phi$.

But,
$$D_\mu\phi D^\mu\phi=\left(\partial_\mu\phi+ieA_\mu\phi\right)\left(\partial^\mu\phi+ieA^\mu\phi\right)$$
$$=\partial_\mu\partial^\mu\phi+ie\partial_\mu\phi A^\mu\phi+ieA_\mu\phi\partial^\mu\phi-e^2A_\mu A^\mu\phi^2\tag{5}$$

and in the above equation, $(5)$, the last three terms are not the same as the terms in red of equation $(4)$ as there is a $\partial_\mu A^\mu$, an extra factor of $2$ and a missing factor of $\phi$, respectively, in the three red terms.

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My attempt at trying to prove $(2)=(3)$ has failed. Can anyone please show that $(2)$ can be written as $(3)$ or provide hints/tips on where I have gone wrong with my attempt?

Answer 1

You wrote $D_\mu \phi D^\mu \phi$ instead of $D_\mu D^\mu \phi$.

You also put $\phi$ in the definition of $D_\mu$ but it is something that acts on $\phi$. $D_\mu$ does not have $\phi$ in it's expression yet.

\begin{eqnarray*} D_\mu D^\mu \phi &=& D_\mu (\partial^\mu \phi + ie A^\mu \phi)\\ &=& (\partial_\mu + ieA_\mu) (\partial^\mu \phi + ie A^\mu \phi)\\ &=& \partial_\mu (\partial^\mu \phi + ie A^\mu \phi)\\ &+& ieA_\mu (\partial^\mu \phi + ie A^\mu \phi)\\ &=& \partial_\mu \partial^\mu \phi + \partial_\mu (ie A^\mu \phi)\\ &+& ieA_\mu \partial^\mu \phi - e^2 A_\mu A^\mu \phi\\ &=& \partial_\mu \partial^\mu \phi + ie \partial_\mu A^\mu \phi + ieA^\mu \partial_\mu \phi\\ &+& ieA_\mu \partial^\mu \phi - e^2 A_\mu A^\mu \phi)\\ &=& \partial_\mu \partial^\mu \phi + ie\partial_\mu A^\mu \phi + 2 ieA_\mu \partial^\mu \phi - e^2 A_\mu A^\mu \phi \end{eqnarray*}

Used parentheses when $\partial_\mu$ was acting on something when it was acting on a product of two things that were spacetime dependent. Otherwise just left the parentheses out and it acted on only the thing immediately after it.