PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.SelfAdjointSpectralTheorem

For a core operator $T$ that is essentially self-adjoint, its canonical graph closure $T_{\text{cl}}$ is a self-adjoint operator.

For a core operator $T$, its canonical graph closure $T_{\text{cl}}$ is a closed operator.

For a core operator $T$, the operator $T$ is a sub-operator of its canonical self-adjoint closure $T_{\text{cl}}$. Mathematically, this is expressed as $T \leq T_{\text{cl}}$, meaning that the domain of $T$ is contained within the domain of $T_{\text{cl}}$, and for every vector $x$ in the domain of $T$, the equality $Tx = T_{\text{cl}}x$ holds.

Let $H$ be a complex Hilbert space and $T$ be a core operator such that its closure $T_{\text{cl}}$ is self-adjoint (i.e., $T$ is essentially self-adjoint). If $S$ is a self-adjoint linear operator on $H$ that extends $T$ (denoted $T \leq S$), then $S$ must be equal to the canonical closure of $T$, written as $S = T_{\text{cl}}$.

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, i.e., $$n_+(T) = \text{defectNumber}(T, i) = 0$$ and $$n_-(T) = \text{defectNumber}(T, -i) = 0$$ then $T$ is essentially self-adjoint, meaning its graph closure is a self-adjoint operator.

Let $H$ be a complex Hilbert space and $T$ be a linear operator on $H$. If there exists a defect index certificate for $T$ (providing evidence that the defect indices of $T$ are zero), then $T$ is essentially self-adjoint, meaning its graph closure is a self-adjoint operator.

For a linear operator $T$ on a Hilbert space $H$ and a weak operator topology (WOT) spectral measure $\mu_S$ on $\mathbb{R}$, the **weak spectral resolution** property holds if for every $x$ in the domain of $T$, and for all $y \in H$: 1. The identity function $f(\lambda) = \lambda$ is integrable with respect to the complex scalar measure $S \mapsto \langle y, \mu_S(S) x \rangle$. 2. The inner product $\langle y, Tx \rangle$ satisfies the weak matrix-element reconstruction law: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu_S(\lambda) x \rangle$$

Given a weak spectral measure $\mu_S$ on $\mathbb{R}$ acting on a Hilbert space $H$, the **square-moment domain** is the set of vectors $x \in H$ for which the function $\lambda \mapsto \lambda^2$ is integrable with respect to the diagonal measure $\mu_x$ associated with $x$. Mathematically, $x$ belongs to this domain if and only if: $$\int_{\mathbb{R}} \lambda^2 d\mu_x(\lambda) < \infty$$ where $\mu_x$ is the positive diagonal measure defined by $\mu_x(S) = \langle x, \mu_S(S) x \rangle$ for measurable sets $S \subseteq \mathbb{R}$. In the context of the spectral theorem, this set identifies the domain of the unbounded self-adjoint operator associated with the spectral measure.

Let $H$ be a Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ acting on $H$. For any vector $x \in H$, $x$ belongs to the square-moment domain of $\mu_S$ if and only if the function $f(\lambda) = \lambda^2$ is integrable with respect to the diagonal measure $\mu_x$. This is equivalent to the condition: $$\int_{\mathbb{R}} \lambda^2 \, d\mu_x(\lambda) < \infty$$ where $\mu_x$ is the positive measure defined by $\mu_x(S) = \langle x, \mu_S(S) x \rangle$ for measurable sets $S \subseteq \mathbb{R}$.

A weak spectral measure $\mu_S$ defined on $\mathbb{R}$ and acting on a Hilbert space $H$ has **bounded spectral support** with respect to a constant $C \in \mathbb{R}$ if $C \geq 0$ and, for every measurable set $S \subseteq \mathbb{R}$, if $S$ is disjoint from the closed interval $[-C, C]$, then the spectral measure of $S$ is zero ($\mu_S(S) = 0$). This condition essentially states that the support of the spectral measure is contained within the compact interval $[-C, C]$, which implies that the associated self-adjoint operator is bounded.

Let $H$ be a Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ acting on $H$. If $\mu_S$ has bounded spectral support contained in the interval $[-C, C]$ for some $C \in \mathbb{R}$, then the square-moment domain of $\mu_S$ (the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 d\mu_x(\lambda) < \infty$) is the entire Hilbert space $H$.

Let $H$ be a Hilbert space, $T$ an operator on $H$, and $\mu_S$ a spectral measure on $\mathbb{R}$. Suppose that $T$ and $\mu_S$ satisfy the weak spectral resolution (i.e., the matrix elements of $T$ are given by $\langle y, T x \rangle = \int_{\mathbb{R}} \lambda \, d\mu_{y,x}(\lambda)$ for $x \in \text{dom}(T)$). If the domain of $T$ is the entire Hilbert space ($\text{dom}(T) = H$) and $\mu_S$ has bounded spectral support in some interval $[-C, C]$, then $T$ and $\mu_S$ satisfy the domain-aware self-adjoint spectral theorem. This implies that the domain of $T$ is exactly the square-moment domain of the spectral measure $\mu_S$.

Let $T$ be an operator on a Hilbert space $H$ and $\mu_S$ be a spectral measure on $\mathbb{R}$. If $T$ and $\mu_S$ satisfy the domain-aware self-adjoint spectral theorem, then $T$ is a self-adjoint operator.

Suppose a linear operator $T$ on a Hilbert space $H$ and a spectral measure $\mu_S$ satisfy the domain-aware self-adjoint spectral theorem. Then $T$ and $\mu_S$ satisfy the weak spectral resolution property. That is, for every $x$ in the domain of $T$ and for all $y \in H$, the inner product $\langle y, Tx \rangle$ is given by the integral of the identity function with respect to the complex measure $\langle y, \mu_S(\cdot) x \rangle$: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu_S(\lambda) x \rangle$$

Let $H$ be a Hilbert space, $T$ be an operator on $H$, and $\mu_S$ be a weak spectral measure on $\mathbb{R}$. If $T$ and $\mu_S$ satisfy the conditions of the domain-aware self-adjoint spectral theorem, then a vector $x \in H$ belongs to the domain of $T$ if and only if $x$ is in the square-moment domain of $\mu_S$. This condition is equivalent to the finiteness of the second spectral moment: $$x \in \text{dom}(T) \iff \int_{\mathbb{R}} \lambda^2 d\mu_x(\lambda) < \infty$$ where $\mu_x$ is the positive diagonal measure defined by $\mu_x(B) = \langle x, \mu_S(B) x \rangle$ for measurable sets $B \subseteq \mathbb{R}$.

Given a domain-aware self-adjoint spectral theorem $D$ for an operator $T$ and its spectral measure $\mu_S$ 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 unitary operator $U(t)$ is defined via the spectral integral: $$U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu_S(\lambda)$$ This group represents the operator exponential $e^{itT}$, where $T$ is the self-adjoint operator whose domain and action are characterized by $\mu_S$.

Let $H$ be a Hilbert space and $T$ be a self-adjoint operator with spectral measure $\mu_S$ as characterized by a domain-aware spectral theorem. Let $U(t) = e^{itT}$ be the strongly continuous one-parameter unitary group defined via the spectral integral $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu_S(\lambda)$. Then, at $t = 0$, the unitary operator is the identity operator, i.e., $U(0) = I$.

Let $T$ be a self-adjoint operator on a Hilbert space $H$ with an associated spectral measure $\mu_S$. Let $U(t) = e^{itT}$ be the strongly continuous one-parameter unitary group defined via the spectral theorem as $U(t) = \int_{\mathbb{R}} e^{it\lambda} \, d\mu_S(\lambda)$. For any real numbers $t$ and $s$, the group satisfies the composition law: $$U(t + s) = U(t) U(s)$$

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

Let $H$ and $H'$ be complex Hilbert spaces. Suppose $T$ is a self-adjoint operator on $H$ with a weak spectral resolution $\mu_S$, such that they satisfy the properties recorded in a `SelfAdjointSpectralTheorem` (i.e., $T$ is self-adjoint and $\langle y, T x \rangle = \int \lambda \, d\langle y, \mu_S(\lambda) x \rangle$ for $x \in \text{dom}(T)$). Given a unitary operator $u : H \to H'$, the unitarily conjugated operator $u T u^{-1}$ on $H'$ and the unitarily conjugated spectral measure $u \mu_S u^{-1}$ also satisfy the conditions of the self-adjoint spectral theorem.

Let $H$ and $H'$ be complex Hilbert spaces and $u: H \to H'$ be a unitary operator (an isometric isomorphism). Let $T$ be a partially defined linear operator on $H$ and $\mu$ be a spectral measure. If $T$ and $\mu$ satisfy the domain-aware self-adjoint spectral theorem—which entails that $T$ is self-adjoint, the weak integral identity $\langle y, Tx \rangle = \int \lambda \, d\mu_{y,x}(\lambda)$ holds for all $x \in \text{dom}(T)$, and the domain of $T$ coincides exactly with the square-moment domain of $\mu$—then the unitarily conjugated operator $u T u^{-1}$ and the unitarily conjugated spectral measure $u \mu u^{-1}$ also satisfy the domain-aware self-adjoint spectral theorem on $H'$.