Find projection-valued measure associated with parity operator

Question

Let's define parity operator as follows: $$\pi:L^2(\mathbb{R})\to L^2(\mathbb{R})$$ $$\psi(x)\mapsto \psi(-x)$$ It's easy to show that $\pi$ is a self-adjoint operator and its spectrum is just $\sigma(\pi)=\{-1,+1\}$. According to spectral theorem there is a unique projection-valued measure $P_{\pi}$ such that: $$\pi=\int_{\mathbb{R}}\lambda \,dP_{\pi}(\lambda)$$ How do I find explicitly $P_{\pi}$?

Answer 1

Observe for any $\psi\in L^2(\mathbb{R})$ we see that \begin{align} \psi(x) = \frac{\psi(x)+\psi(-x)}{2}+ \frac{\psi(x)-\psi(-x)}{2}=: \psi_\text{even}(x)+\psi_\text{odd}(x) \end{align} then \begin{align} \pi(\psi)(x) = \psi_\text{even}(-x)+\psi_\text{odd}(-x)=\psi_\text{even}(x)-\psi_\text{odd}(x). \end{align} In short, we have that \begin{align} \pi = P_\text{even}-P_\text{odd} \end{align} where $P_\text{even}\psi = \psi_\text{even}$ and $P_\text{odd}\psi=\psi_\text{odd}$.

Edit: Note that \begin{align} P_\pi(\lambda) = \delta(\lambda-1)P_\text{even}+\delta(\lambda+1)P_\text{odd} \end{align} i.e. $P_\pi:\mathcal{B}_\mathbb{R}\rightarrow \mathcal{L}(L^2(\mathbb{R}),L^2(\mathbb{R}))$. Hence \begin{align} \int_{\mathbb{R}}\lambda\cdot dP_\pi(\lambda) = P_\text{even}-P_\text{odd}. \end{align}