Parity Invariance Complex Scalar Field Lagrangian

Question

I am trying to prove the parity invariance of some terms in a complex scalar field Lagrangian, for example $m^2 \; \phi^* \phi$ or $\partial_{\mu} \phi \;\partial^{\mu} \phi^*$. So what I want to prove is that

$\delta S=0$

for every term I am considering, where $S$ is the action of the theory. So I start from the basic definition of parity

$x^{\mu}=(x_0, \vec{x})\equiv x$,

$x^{\mu} \xrightarrow[]{P} x_{\mu}=(x_0,-\vec{x})\equiv x_p$,

from this follows that the derivative transforms as

$\partial_{\mu} \xrightarrow[]{P} \partial^{\mu} $.

At this point let's consider the complex scalar field $\phi$, if I am correct we have

$\phi(x) \xrightarrow[]{P} \phi(x_p) $

$\phi^*(x) \xrightarrow[]{P} \phi^*(x_p)$

At this point we just have to consider the action $S$ for the mass term

$S \supset \int d^4x \; m^2 \phi^*(x) \phi(x)$

and see how it changes with parity. My doubt is regarding the measure. I think it transforms under parity like

$d^4x \xrightarrow[]{P} d^4x_p$

therefore the transformation of the action term is

$ \int d^4x \; m^2 \phi^*(x) \phi(x) \xrightarrow[]{P} \int \; d^4x_p m^2 \phi^*(x_p) \phi(x_p)$

therefore since the $x_p$ is a dummy variable we proved the invariance. Am I correct? Did I make any mistake?

P.S. $d^4x \xrightarrow[]{P} dx_0 (-dx_1)(-dx_2)(-dx_3) = d^4x_p$

Answer 1

Seems correct. You can think of the integration measure as integrating from $-\infty$ to $+\infty$ so a transformation from (+) to (-) leaves the integration invariant. As a general rule of thumb, if the Lagrangian density $\mathcal{L}$ is invariant, so is the action $\mathcal{S}$.