# Parity Invariance Complex Scalar Field Lagrangian

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

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 $$\mathcal{S}$$.
• Actually I think that the last line is incorrect since the measure transforms with the jacobian of the transformation. In this case $d^4x \xrightarrow[]{P} +1 d^4x_p$. But the overall meaning should be the same Commented Jan 16, 2020 at 14:31