# Do vectors uniquely determine their spectral measures?

Let $$\mathscr{H}$$ be a Hilbert space, $$\psi \in \mathscr{H}$$ and $$A$$ be a densely-defined self-adjoint operator. From the Borel functional calculus, for each measurable function $$f: \mathbb{R} \to \mathbb{C}$$, there exists a bounded linear operator $$f(A)$$ on $$\mathscr{H}$$. For each bounded continuous function $$f: \mathbb{R}\to \mathbb{C}$$, define: $$\omega(f) = \langle \psi, f(A)\psi\rangle$$ This is a positive linear functional, so by Riesz-Markov Theorem there exists a Borel measure $$\mu_{\psi}$$ on $$\mathbb{R}$$ such that: $$\langle \psi, f(A)\psi \rangle = \int_{\mathbb{R}}f(x)d\mu_{\psi}(x).$$

Question: Suppose $$\mu_{\psi} = \mu_{\varphi}$$. Is it true that $$\psi = \varphi$$? I could not conclude this from $$\langle \psi,f(A)\rangle = \langle \varphi,f(A)\varphi\rangle$$.

• What about $\psi = - \varphi$? Mar 29, 2023 at 17:31
• You mean taking $f= -1$? Mar 29, 2023 at 17:31
• No, prior to apply Riesz-Markov theorem try to see what happens when $\psi = - \varphi$ Mar 29, 2023 at 17:39
• Perhaps you should reformulate: are these vectors linearly dependent ? Mar 29, 2023 at 17:46

The claim is not true even for two-dimensional space with orthonormal basis $$e_1,e_2$$ and the identity operator. Then the measures corresponding to $$e_1$$ and $$e_2$$ are equal, although the vectors are linearly independent. I guess the conclusion may hold for operators with simple spectrum