On a Hilbert space $\cal H$, consider an essentially self-adjoint operator $A\colon Dom(A)\to {\cal H}$, and a vector $\psi\in\bigcap_{n=1}^\infty Dom(A^n)$. Without further assumptions, can we say that the associated spectral measure $\mu_\psi$ is the unique Borel measure satisfying $$\langle \psi,A^n\psi\rangle = \int\lambda^n\mu_{\psi}(d\lambda),$$ for all $n\ge 0$ ?
1 Answer
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The answer is no. There are random variables with all moments finite, whose distributions are not determined by their moments (e.g. the log-normal distribution).
Let $X$ be such a random variable and let $\mu$ be its distribution. Then the operator of multiplication by $x$ on the space $L^{2}(\mathbb{R},\mu)$ has $\psi:= 1$ (the constant function) in the domain of all of its powers, but the spectral measure in this case is $\mu$ and it is not determined by moments.