Unitary operator in the spectral theorem is unique?

Question

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.

Answer 1

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.

Answer 2

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).