This question follows from this one and especially from Willie Wong's answer: link.
In Reed & Simon's book Methods of modern mathematical physics, vol. I, pag.277, the form domain of a self-adjoint operator $(A, D(A))$ on a Hilbert space $H$ is defined by passing to a spectral representation, that is by taking a unitary isomorphism
$$U \colon H \to \bigoplus_{j=1}^N L^2(\mathbb{R}, d\mu_j), $$
(where $N\in \{1, 2 \ldots +\infty\}$ and $\mu_j$ are finite Borel measures) such that $(UA)\varphi=(x\psi_j(x))_{j=1}^N$ [$A$ is unitarily equivalent to multiplication by $x$]. The sought domain is then said to be
$$Q(q)=\left\{ (\psi_j)_{j=1}^N \ :\ \sum_{j=1}^N \int_{-\infty}^\infty \lvert x \rvert \lvert \psi_j(x)\rvert^2\, d\mu_j <+\infty \right\}.$$
Question. Let
$$D(\lvert A\rvert^{1/2})=\left\{ \varphi \in H\ :\ \int_{-\infty}^\infty \lvert \lambda \rvert\, d\big(E_A(\lambda)\varphi, \varphi\big)<+\infty\right\},$$
where $\{E_A(\lambda)\}_{\lambda \in \mathbb{R}}$ is the spectral family of $A$ (cfr. Reed & Simon vol. I Theorem VIII.6).
Is it true that $Q(q)=D(\lvert A\rvert^{1/2})$?
I believe the answer to be affirmative. This should make for a characterization of the form domain a bit more transparent than the one based on spectral representations. For example, it is not immediately clear that the latter is independent on the particular representation chosen.
Yes, it is true. First, note that $Q(q) = D(|A|^{1/2})$ holds in the case that $A$ is "multiplication by $x$". Then observe that the spectral family is preseved by unitary transformations.