PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.SelfAdjointSpectralTheorem

Let $T$ be an essentially self-adjoint operator. Then its closure, denoted by $\bar{T}$, is a self-adjoint operator.

Let $T$ be a linear operator such that its closure $\bar{T}$ is self-adjoint (i.e., $T$ is essentially self-adjoint). Then the closure $\bar{T}$ is a closed operator.

Let $T$ be a linear operator such that its closure $\bar{T}$ is self-adjoint. Then $T \le \bar{T}$, which means that the domain of $T$ is contained in the domain of its closure $\bar{T}$, and the two operators coincide on the domain of $T$.

Let $H$ be a complex Hilbert space and $T$ be a linear operator on $H$ such that its closure $\bar{T}$ is self-adjoint (i.e., $T$ is essentially self-adjoint). If $S$ is any self-adjoint extension of $T$ (denoted $T \leq S$, meaning $\text{dom}(T) \subseteq \text{dom}(S)$ and $S$ restricted to $\text{dom}(T)$ equals $T$), then $S = \bar{T}$.

Let $H$ be a complex Hilbert space and $T$ be a densely defined symmetric linear operator on $H$. If the defect numbers of $T$ at $i$ and $-i$ are both zero—that is, $\dim(\text{Ran}(T - iI)^\perp) = 0$ and $\dim(\text{Ran}(T + iI)^\perp) = 0$—then $T$ is essentially self-adjoint, meaning its closure $\bar{T}$ is a self-adjoint operator. This result represents the von Neumann defect-number criterion for essential self-adjointness.

Let $H$ be a complex Hilbert space and $T$ be a linear operator on $H$. If there exists a defect index certificate $C$ for $T$, then $T$ possesses self-adjoint closure data, which implies that the closure of $T$ is a self-adjoint operator.

For a linear operator $T$ on a complex Hilbert space $H$ and a spectral measure $\mu$ on $\mathbb{R}$, this property states that $\mu$ is a weak spectral resolution of $T$. Specifically, for every $x$ in the domain of $T$, the identity function $f(\lambda) = \lambda$ must be integrable with respect to the complex measure $\mu_{x,y}$ (defined as $\mu_{x,y}(A) = \langle y, \mu(A)x \rangle$) for all $y \in H$, and the following relation must hold: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu(\lambda)x \rangle$$

Given a spectral measure $\mu$ on $\mathbb{R}$ acting on a Hilbert space $H$, the set consists of all vectors $x \in H$ for which the second moment with respect to the spectral measure is finite. Specifically, a vector $x$ is in this set if the function $\lambda \mapsto \lambda^2$ is integrable with respect to the scalar measure $\mu_x$ defined by $\mu_x(A) = \langle x, \mu(A)x \rangle$. In integral notation, this is the set of $x \in H$ such that $$\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty.$$ This set defines the natural domain of the self-adjoint operator $T = \int \lambda \, d\mu(\lambda)$ associated with the spectral measure.

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ acting on $H$. For any vector $x \in H$, $x$ belongs to the domain of finite second moment of $\mu$ (denoted as `spectralSquareMomentDomain μ`) if and only if the function $\lambda \mapsto \lambda^2$ is integrable with respect to the diagonal measure $\mu_x$ (defined as $\mu_x(A) = \langle x, \mu(A)x \rangle$). In other words, $x$ is in the domain if and only if $$\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty.$$

Given a spectral measure $\mu$ on $\mathbb{R}$ acting on a Hilbert space $H$ and a real number $C$, this property states that the support of the spectral measure is contained within the closed interval $[-C, C]$. Specifically, for every measurable set $S \subseteq \mathbb{R}$, if $S$ is disjoint from $[-C, C]$, then the spectral measure of $S$ is zero, i.e., $\mu(S) = 0$.

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ acting on $H$. If the spectral measure $\mu$ has bounded support contained within the interval $[-C, C]$ for some $C \in \mathbb{R}$, then its domain of finite second moment (the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty$) is the entire Hilbert space $H$.

Let $H$ be a Hilbert space and $T$ be a linear operator on $H$ with domain $\mathcal{D}(T)$. Suppose $\mu$ is a spectral measure on $\mathbb{R}$ that resolves $T$, meaning that for all $x \in \mathcal{D}(T)$ and $y \in H$, the identity $\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu(\lambda)x \rangle$ holds. If the domain of $T$ is the entire Hilbert space ($\mathcal{D}(T) = H$) and the support of $\mu$ is bounded (contained in some interval $[-C, C]$ for $C \in \mathbb{R}$), then $\mu$ provides a domain-aware spectral resolution for $T$. This implies that the domain of $T$ is exactly the set of vectors with finite second moment, i.e., $\mathcal{D}(T) = \{x \in H \mid \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty \}$.

Let $H$ be a Hilbert space and $T$ be an operator acting on $H$. If $\mu_S$ is a spectral measure that provides a domain-aware spectral resolution for $T$ (meaning $T$ is represented by the integral $\int \lambda \, d\mu_S$ and its domain coincides with the set of vectors of finite second moment), then $T$ is a self-adjoint operator.

If $T$ is a linear operator on a complex Hilbert space $H$ and $\mu$ is a spectral measure on $\mathbb{R}$ that provides a domain-aware spectral resolution for $T$, then $\mu$ is a weak spectral resolution of $T$. Specifically, for every vector $x$ in the domain of $T$ and every vector $y \in H$, the identity function $f(\lambda) = \lambda$ is integrable with respect to the complex measure $\mu_{x,y}$ (defined by $\mu_{x,y}(A) = \langle y, \mu(A)x \rangle$), and the following identity holds: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu(\lambda)x \rangle$$

Let $H$ be a Hilbert space, $T$ be an operator on $H$, and $\mu$ be a spectral measure on $\mathbb{R}$ that resolves $T$. For a spectral resolution that accounts for the operator's domain (represented by `DomainAwareSelfAdjointSpectralTheorem`), a vector $x \in H$ belongs to the domain of $T$ if and only if the second spectral moment of $x$ with respect to $\mu$ is finite. That is, $$x \in \text{dom}(T) \iff \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty.$$

Given a domain-aware spectral resolution for a self-adjoint operator $T$ with associated spectral measure $\mu$ on a Hilbert space $H$, this definition constructs the strongly continuous one-parameter unitary group $U: \mathbb{R} \to \mathcal{B}(H)$. For each $t \in \mathbb{R}$, the operator $U(t)$ is defined via the spectral integral: $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda).$$ This construction provides the unitary group $U(t) = e^{itT}$ associated with the operator $T$ as described by Stone's theorem, satisfying the group properties $U(0) = I$ and $U(t+s) = U(t)U(s)$.

Let $T$ be a self-adjoint operator on a Hilbert space $H$ with an associated spectral measure $\mu$ provided by a domain-aware spectral resolution. Let $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda)$ be the strongly continuous one-parameter unitary group associated with $T$ (often denoted $e^{itT}$). Then, at $t = 0$, the operator is the identity operator, i.e., $U(0) = I$.

Given a domain-aware spectral resolution for a self-adjoint operator $T$ with associated spectral measure $\mu$ on a Hilbert space $H$, let $U(t) = e^{itT}$ be the strongly continuous one-parameter unitary group defined by the spectral integral $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda).$$ Then for any $t, s \in \mathbb{R}$, the group law holds: $$U(t+s) = U(t)U(s).$$

Let $H$ be a complex Hilbert space and $T$ a self-adjoint operator with a domain-aware spectral resolution provided by the spectral measure $\mu$ on $\mathbb{R}$. Let $U(t) = e^{itT}$ be the one-parameter unitary group defined by the spectral integral $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu(\lambda).$$ For any vector $x \in H$, the map $t \mapsto U(t)x$ is continuous from $\mathbb{R}$ to $H$.

Let $H$ and $H'$ be complex Hilbert spaces. Suppose $T$ is a self-adjoint operator on $H$ with a spectral resolution given by a spectral measure $\mu_S$ on $\mathbb{R}$. Let $u : H \to H'$ be a unitary operator (a linear isometric equivalence). Then, the unitarily conjugated operator $u T u^{-1}$ on $H'$ is resolved by the unitarily conjugated spectral measure $u \mu_S u^{-1}$. This theorem allows for transporting a spectral resolution from one operator to any unitarily equivalent operator.

Let $H$ and $H'$ be complex Hilbert spaces. Let $T$ be a partially defined linear operator on $H$ and $\mu$ be a spectral measure on $\mathbb{R}$. Suppose that $\mu$ provides a domain-aware spectral resolution of $T$, meaning that for all $x$ in the domain of $T$ and all $y \in H$, the relation $\langle y, T x \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu(\lambda) x \rangle$ holds, and the domain of $T$ is precisely the set of vectors $x$ for which the second moment is finite: $\int_{\mathbb{R}} \lambda^2 \, d\mu_{x,x} < \infty$. Given a unitary operator $u : H \to H'$, then the unitarily conjugated spectral measure $u \mu u^{-1}$ provides a domain-aware spectral resolution for the unitarily conjugated operator $u T u^{-1}$. This implies that the conjugation preserves both the weak spectral integral representation and the characterization of the operator's domain via the square-moment of the spectral measure.