Momentum operator

Remarks

For $\psi$ in the Schwartz space $\mathcal{S}(\R)$, $\widehat{\psi'}(k) = ik\hat\psi(k)$, so $P\psi = -i\hbar\, \psi'$. More generally $D(P)$ is the Sobolev space $H^1(\R)$ and $P\psi = -i\hbar\, \psi'$ with the weak derivative. Since $\mathcal{F}$ is unitary, $P = \mathcal{F}^{-1} (\hbar \hat X) \mathcal{F}$, where $\hat X$ is the position operator in the variable $k$. So $P$ is self-adjoint, an observable, with spectrum $\R$ and no eigenvalues. In $\R^n$, $P_j = -i\hbar\, \partial/\partial x_j$ is defined in the same way with $k_j$.

Example: $\psi(x) = \pi^{-1/4} e^{-x^2/2}$ is a Schwartz function, and $P\psi = -i\hbar \psi' = i\hbar x\psi$. So $\ip{\psi}{P\psi} = i\hbar \int x \pi^{-1/2} e^{-x^2}\, dx = 0$.

Non-example: the indicator $\psi = \mathbf{1}_{[0, 1]}$ is in $L^2(\R)$, with $\hat\psi(k) = (1 - e^{-ik})/(ik\sqrt{2\pi})$ for $k \ne 0$. Then $k^2 \abs{\hat\psi(k)}^2 = \abs{1 - e^{-ik}}^2/(2\pi) = (1 - \cos k)/\pi$, which is not integrable over $\R$, so $\psi \notin D(P)$.