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Assume $H$ is a Hilbert space and $A=A^*\in L(H)$. Let $x$ be a cyclic vector of $A$. Then there is a unitary operator $U$ from $L^2(\sigma(A), E_x)$ to $H$, where $E_x$ is the spectral measure of $A$. Then

$$U^*AUx(t)=tx(t)$$

for all $x\in L^2(\sigma(A), E_x)$.

Is this unitary operator $U$ unique in some sense i.e. up to some isomorphism?

Here $E_x$ is just a positive and finite Borel measure.

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  • $\begingroup$ Yes, I think it should be unique, cf. for example the following paper: arxiv.org/pdf/2212.13953.pdf (Thm. II.2) $\endgroup$
    – LSK21
    Commented Aug 15, 2023 at 13:44
  • $\begingroup$ Thank you very much. The author seem to make further assumtptions on $U$. I wonder if this holds also with no further assumptions. $\endgroup$ Commented Aug 15, 2023 at 14:26
  • $\begingroup$ You can certainly replace $U$ by $e^{i\theta}U$ for some $\theta\in [0,2\pi)$ to find an infinite family of such unitaries. Note that is a special case of the answer below. $\endgroup$ Commented Aug 15, 2023 at 17:19

2 Answers 2

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If $u:\sigma(A)\to \{z\in\Bbb{C}: |z|=1\}$ is a Borel function and $M_u$ is the operator on $L^2(\sigma(A), E_x)$ acting as pointwise multiplication by $u$ then $$(U M_u)^*A(U M_u)x(t) = M_u^*(U^* A U) M_u x(t) = \bar{u}(t)(U^*AU)(u(t)x(t)) = \bar{u}(t)tu(t)x(t)= tx(t)$$ So $UM_u$ is another such unitary.

Instead of $M_u$ you can use any unitary of $L^2(\sigma(A), E_x)$ that commutes with $M_t$ where $M_t$ is multiplication by $t$ - ($M_t(x(t)) = tx(t)$). I think that unitaries of $L^2(\sigma(A), E_x)$ that commute with $M_t$ all have the form $M_u$ as above, but there may be some $\sigma(A)$ or $E_x$ of special form where there are more such unitaries. I'm not sure about the last part.

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To complement John Doe's answer: If $U,V$ are unitary operators such that $A=U M_t U^\ast=V M_t V^\ast$, then $V^\ast U$ is a unitary on $L^2(\sigma(A),E_x)$ such that $$ V^\ast UM_t=M_t V^\ast U. $$ Iterating this identity, we get $V^\ast UM_{t^k}=M_{t^k}V^\ast U$ for all $k\in\mathbb N$, and then by linearity and continuity also $V^\ast U M_f=M_f V^\ast U$ for all $f\in C(\sigma(A))$.

Thus, with $u=V^\ast U(1)$ we get $$ V^\ast U f=V^\ast U M_f (1)=M_f V^\ast U(1)=uf $$ for $f\in C(\sigma(A))$. By density, the same identity holds for all $f\in L^2(\sigma(A),E_x)$. Hence $V^\ast U=M_u$. From the fact that $V^\ast U$ is unitary, it follows easily that $u=1$ a.e.

Thus, given one unitary operator $U$ such that $A=U M_t U^\ast$, every other unitary operator $V$ that satisfies the same identity is of the form $V=U M_{u}$ with $u\colon \sigma(A)\to\mathbb T$ measurable (if you do the computations, you'll see that this $u$ is actually $\bar u$ from above).

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