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 $f(x)$


and by taking $p\rightarrow -p$


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.



1 Answer 1


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}


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