1
$\begingroup$

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 $\langle \psi,f(A)\rangle = \langle \varphi,f(A)\varphi\rangle$.

$\endgroup$
4
  • $\begingroup$ What about $\psi = - \varphi$? $\endgroup$ Mar 29, 2023 at 17:31
  • $\begingroup$ You mean taking $f= -1$? $\endgroup$
    – Idontgetit
    Mar 29, 2023 at 17:31
  • $\begingroup$ No, prior to apply Riesz-Markov theorem try to see what happens when $\psi = - \varphi$ $\endgroup$ Mar 29, 2023 at 17:39
  • $\begingroup$ Perhaps you should reformulate: are these vectors linearly dependent ? $\endgroup$ Mar 29, 2023 at 17:46

1 Answer 1

1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .