1
$\begingroup$

Let $T$ be an unbounded selfadjoint operator and let $P_T$ denote it's spectral measure such that $T= \int_\mathbb{R}\lambda dP_T (\lambda)$. Suppose $\psi$ is an eigenvector of $T$ such that $T\psi=\lambda \psi$ for $\lambda \in \mathbb{R}$. I want to compute the corresponding spectral measure $\mu_\psi(\Omega):=\langle \psi|P_T(\Omega)\psi \rangle$. Obviously it has to be $\mu_\psi=\delta_\lambda$, where $\delta_\lambda$ denotes the Dirac measure centered at $\lambda$, when $||\psi||=1$. But I've so far failed to prove that. How do you do that? I don't want to use any explicit formula for $P_T$, and want to only use the property $T= \int_\mathbb{R}\lambda dP_T (\lambda)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.