# How does parity opeartor act on the integral measurel term $dp$?

Suppose we have a function $$f(x)$$ and its fourier transformation $$f(x)=\int_{-\infty}^{\infty}dpf(p)e^{-ipx}.$$

And let's define $$\hat{P}$$ as space inversion opeartor that maps $$x \rightarrow-x$$ and also $$p \rightarrow -p$$. Meaning $$\hat{P}f(x)=f(-x)$$

Now when I act with this operator to $$f(x)$$

$$\hat{P}f(x)=\hat{P}\int_{-\infty}^{\infty}dpf(p)e^{-ipx}=\int_{-\infty}^{\infty}dpf(-p)e^{-ipx}$$

and by taking $$p\rightarrow -p$$

$$\int_{\infty}^{-\infty}-dpf(p)e^{ipx}=\int_{-\infty}^{\infty}dpf(p)e^{ipx}=f(-x)$$

To get the result I want I ignored how parity transforms the integral. How should the integral boundaries and $$dp$$ term change under parity operation? From an ad-hoc point of view, I can find a reason for both cases.

1. Either $$dp$$ maps into $$-dp$$ but also the integral boundaries map into each other so the parity transformation leaves the integral invariant
2. Or since we are looking at infinitesimal portions $$dp$$ the directions do not matter.

So what would happen if I had non-symmetric boundaries.

$$\hat{P}\int_{-L}^{L/2}dxf(x)$$

You need to be a bit careful here. Let me go through this step-by-step.

$${\hat P}$$ is an operator that acts on states in the Hilbert space. In particular, when working in the $$x$$ basis, it acts on wave-functions. It does not act on every object that has $$x$$ dependence. For instance, if $$f(x)$$ is a function [often referred to as a $$c$$-number] and $$\Psi(x)$$ is a wave-function, then we have $${\hat P} [ f(x) \Psi(x)] = f(x) {\hat P} [ \Psi(x)] = f(x) \Psi(-x).$$ Here, we used that the fact that $${\hat P}$$ is a linear operator, so that $${\hat P} [ a \Psi] = a {\hat P} \Psi$$ for all $$c$$-numbers $$a$$.

Now, let's turn to your example. You want to determine the action $${\hat P}\,\Psi(x) = {\hat P} \int_{\mathbb R} dp e^{- i p x} \Psi(p)$$ For linear operators $${\hat P} (\Psi_1+\Psi_2) = {\hat P} \Psi_1 + {\hat P} \Psi_2$$ so we can move $${\hat P}$$ past the integral sign (which is just glorified summation), $${\hat P}\,\Psi(x) =\int_{\mathbb R} dp {\hat P} [ e^{- i p x} \Psi(p)]$$ Next $$e^{-ipx}$$ is just a $$c$$-number so we can move it past $${\hat P}$$ as well \begin{align} {\hat P}\,\Psi(x) &= \int_{\mathbb R} dp e^{- i p x} {\hat P} [ \Psi(p)]\\ &= \int_{\mathbb R} dp e^{- i p x} \Psi(-p) \end{align} We can now change integration variable $$p \to -p$$ in the usual way and we find \begin{align} {\hat P}\,\Psi(x) &= \int_{\mathbb R} dp e^{ i p x} \Psi(p) \\ &= \int_{\mathbb R} dp e^{ - i p (- x) } \Psi(p) \\ &= \Psi(-x) \end{align}