# Unitary operator in the spectral theorem is unique?

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.

• Yes, I think it should be unique, cf. for example the following paper: arxiv.org/pdf/2212.13953.pdf (Thm. II.2) Commented Aug 15, 2023 at 13:44
• Thank you very much. The author seem to make further assumtptions on $U$. I wonder if this holds also with no further assumptions. Commented Aug 15, 2023 at 14:26
• 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. Commented Aug 15, 2023 at 17:19

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.
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).