Is a Lagrangian term $L_{kin}=(\partial^{\mu}\phi^{*})(\partial_{\mu}\phi)$ equivalent to $L_{kin}=\phi^{*}\partial^{\mu}\partial_{\mu}\phi$?

Question

In looking at the Lagrangian of a (free for simplicity) complex scalar field $\phi$, we have a kinetic term that goes like:

$$L_{kin}=(\partial^{\mu}\phi^{*})(\partial_{\mu}\phi)$$

Given instead, a kinetic term of the form:

$$L_{kin}=\phi^{*}\partial^{\mu}\partial_{\mu}\phi$$

Is this not equivalent to the first term? Just by simply varying each term with respect to $\phi^{*}$ and it's derivative we arrive at the same equation of motion (up to some arbitary multiplacative factor) namely:

$$\partial^{\mu}\partial_{\mu}\phi$$

I'm sure this can be expressed in terms of formal adjoints more precisely, but if they are equivalent, why do we always write scalar fields in terms of first order lagrangians (boundary conditions maybe)?

Answer 1

Almost, you're off by a minus sign, but otherwise yes. You're supposed to use (higher-dimensional) integration by parts on the action $$ S = - \int_{\Omega} d^4 x\ \big( \partial^\mu \phi^\ast(x) \big) \big( \partial_{\mu} \phi(x) \big) = - \int_{\partial \Omega} d^3 \Sigma\ \phi^\ast(x) n^{\mu} \partial_{\mu} \phi(x) \big) + \int_{\Omega} d^4 x\ \phi^\ast(x) \partial^\mu \partial_{\mu} \phi(x) $$ where $\Omega \to \mathbb{R}^{4}$ and $n^{\mu}$ is the outward unit normal to the boundary $\partial \Omega$ (and $d \Sigma$ is the volume form on the boundary). The standard thing to do is assume that the boundary term vanishes and you find that $\partial^\mu \phi^\ast \partial_\mu \phi$ and $-\phi^\ast\partial^\mu \partial_{\mu} \phi$ can be used interchangebly.