I am studying the position operator $(\hat{q},D_{\hat{q}})$ on $L_2(/R)$ with: $$\hat{q}\psi=x\psi$$ $$ D_{\hat{q}}={\psi\in L_2(/R):\hat{q}\psi \in L_2(/R)}$$ As far as I know $(\hat{q},D_{\hat{q}})$ is not bounded but it is densly defined and self-adjoint.
When we try to study the point spectrum of $(\hat{q},D_{\hat{q}})$ we stumble in the equation: $$\hat{q}\psi_\lambda=x\psi_\lambda=\lambda \psi_\lambda$$
Which is true iff $\psi_\lambda=0 \space a.e.$ , hence we can conlude that the point spectrum of $(\hat{q},D_{\hat{q}})$ is empty.
Since $(\hat{q},D_{\hat{q}})$ is self adjoint we also know that the residual spectrum is empty too.
Now, the problem comes when I try to find the continuous spectrum. I have read that $\lambda$ belongs to the continuous spectrum of $(\hat{q},D_{\hat{q}})$ iff there exist a succession $\hat{\psi}_n\in D_{\hat{q}}$ such that: $$\lim\limits_{n\to\infty}||\hat{q}\hat{\psi}_n-\lambda\hat{\psi}_n||=0.$$ I wanted to show that a succession of Gaussian functions of the kind: $$\phi_n=\frac{1}{\sqrt{2\pi}1/n}e^\frac{-(x-\lambda)^2}{2/n^2}$$ respect such condition, but when i try to solve the integral I get stuck. Is there a easier way to find the continuous spectrum ?
According to wikipedia https://en.wikipedia.org/wiki/Position_operator the spectrum should be empty but I have been told that the continuous spectrum is the whole real line could someone clarify this point ?