Physlib.QuantumMechanics.Operators.Unbounded

Let $H$ and $H'$ be Hilbert spaces over $\mathbb{C}$. For a linear partial map $T$ from $H$ to $H'$, this property holds if the domain of $T$, denoted as $\text{dom}(T)$, is a dense subspace of $H$.

Let $H$ and $H'$ be Hilbert spaces over $\mathbb{C}$. For any linear partial map $U: H \to H'$, the property $U.\text{HasDenseDomain}$ holds if and only if the domain of $U$ is a dense subset of $H$.

Let $H$ and $H'$ be complex Hilbert spaces. For a partially defined linear map $T$ from $H$ to $H'$, the predicate `LinearPMap.IsUnbounded` holds if the domain of $T$ is dense in $H$ and $T$ is a closable operator.

Let $H$ and $H'$ be complex Hilbert spaces. For a partially defined linear operator $U$ from $H$ to $H'$, $U$ is defined as an "unbounded operator" if and only if $U$ is densely defined (i.e., its domain is dense in $H$) and $U$ is closable.

Let $\mathcal{H}$ be a complex Hilbert space. A partially defined linear operator $T$ on $\mathcal{H}$ (represented by a `LinearPMap`) is said to be **symmetric** if for every pair of vectors $x, y$ in the domain of $T$, the inner product satisfies the condition: $$\langle Tx, y \rangle = \langle x, Ty \rangle$$ where $\langle \cdot, \cdot \rangle$ denotes the inner product on $\mathcal{H}$.

Let $H$ be a complex inner product space. For any partially defined linear operator $T$ on $H$ (represented as a `LinearPMap`), $T$ is symmetric if and only if it is formally adjoint to itself. In this context, $T$ is symmetric if $\langle Tx, y \rangle = \langle x, Ty \rangle$ for all $x, y \in \text{dom}(T)$, and $T$ is formally adjoint to $S$ if $\langle Tx, y \rangle = \langle x, Sy \rangle$ for all $x \in \text{dom}(T)$ and $y \in \text{dom}(S)$.

Let $H$ be a complex Hilbert space. A partially defined linear map $T: D(T) \subseteq H \to H$ (of type `LinearPMap`) is **essentially self-adjoint** if it is densely defined and its closure $\overline{T}$ is a self-adjoint operator, such that $\overline{T} = (\overline{T})^*$.

Let $H$ be a complex Hilbert space. For a partially defined linear operator $T$ on $H$, $T$ is essentially self-adjoint if and only if its closure $\bar{T}$ is self-adjoint.

Let $H$ be a complex Hilbert space and $T$ be a partial linear map on $H$. Then $T$ is star-normal if and only if $T^* T = T T^*$, where $T^*$ denotes the adjoint of $T$ and the product refers to the restricted composition of partial linear maps. The restricted composition $g \circ_r f$ is defined on the domain $\{x \in \text{dom}(f) \mid f(x) \in \text{dom}(g)\}$.

Let $H$ and $H'$ be complex Hilbert spaces. Let $U$ be a linear operator from a subspace of $H$ to $H'$ (a `LinearPMap`). If the domain of $U$ is dense in $H$, then $U$ satisfies the definition of an unbounded operator if and only if $U$ is closable. Here, an "unbounded operator" is defined as a linear operator that is both densely defined and closable.

Let $H$ and $H'$ be complex Hilbert spaces. Let $U: \mathcal{D}(U) \subseteq H \to H'$ be a partially defined linear operator. If $U$ is densely defined (i.e., its domain $\mathcal{D}(U)$ is dense in $H$), then its closure $\overline{U}$ is also densely defined.

Let $H$ and $H'$ be complex Hilbert spaces. For any partially defined linear operator $U$ from $H$ to $H'$ with domain $D(U)$, let $\overline{U}$ denote the closure of the operator $U$. Then the domain of the closure is a subset of the closure of the domain, i.e., $D(\overline{U}) \subseteq \overline{D(U)}$.

Let $H$ and $H'$ be Hilbert spaces. For any partially defined linear operator $U: H \to H'$, the domain of $U$ is dense in $H$ if and only if the domain of its closure $\overline{U}$ is dense in $H$.

Let $H$ and $H'$ be complex Hilbert spaces. For any partially defined linear map $U$ from $H$ to $H'$, if $U$ has a dense domain, then its negation $-U$ also has a dense domain.

Let $H$ and $H'$ be complex Hilbert spaces. For any partially defined linear map $U: H \to H'$, if the domain of $U$ is dense in $H$, then for any scalar $c \in \mathbb{C}$, the domain of the scalar multiple $c U$ is also dense in $H$.

Let $H$ and $H'$ be complex Hilbert spaces. Let $U_1$ and $U_2$ be linear partially defined maps from $H$ to $H'$. If $U_1$ has a dense domain and the domain of $U_1$ is contained within the domain of $U_2$ (i.e., $\text{dom}(U_1) \subseteq \text{dom}(U_2)$), then the sum operator $U_1 + U_2$ also has a dense domain.

Let $H$ and $H'$ be complex Hilbert spaces. Let $U_1$ and $U_2$ be linear partial maps from $H$ to $H'$. If $U_1$ has a dense domain $\mathcal{D}(U_1)$ and $\mathcal{D}(U_1) \subseteq \mathcal{D}(U_2)$, then the domain of the difference $U_1 - U_2$ is also dense in $H$.

Let $H$ and $H'$ be Hilbert spaces over $\mathbb{C}$, and let $\{W_a\}_{a \in \alpha}$ be a finite family of partial linear maps from $H$ to $H'$. If there exists a submodule $E$ of $H$ that is dense in $H$ and is contained in the domain of each $W_a$ (i.e., $E \subseteq \text{dom}(W_a)$ for all $a \in \alpha$), then the sum of these partial linear maps $\sum_{a \in \alpha} W_a$ has a dense domain.

Let $H$ be a Hilbert space over $\mathbb{C}$ and $T$ be a partial linear map on $H$ with domain $D(T) \subseteq H$. If $T$ is densely defined (i.e., $D(T)$ is dense in $H$) and the domain is invariant under $T$ (i.e., for all $x \in D(T)$, $T(x) \in D(T)$), then for every natural number $n$, the $n$-th power of $T$ is also densely defined.

Let $H$ be a complex Hilbert space and $T$ be a linear partial map from $H$ to itself. For any natural number $n$, if the domain of the $n$-th power $T^n$ (defined via restricted composition) is dense in $H$, then for any natural number $k$ such that $k \leq n$, the domain of $T^k$ is also dense in $H$.

Let $H$ and $H'$ be Hilbert spaces over $\mathbb{C}$, where $H$ is a complete space. Let $U: \mathcal{D}(U) \to H'$ be a linear operator with a dense domain $\mathcal{D}(U) \subseteq H$. Then the orthogonal complement of the range of $U$ is equal to the kernel of its adjoint operator $U^*$: $$(\text{ran } U)^\perp = \ker U^*$$ Here, $(\text{ran } U)^\perp$ denotes the set of vectors in $H'$ orthogonal to the image of $U$, and $\ker U^*$ denotes the subspace of vectors $y \in \mathcal{D}(U^*) \subseteq H'$ such that $U^* y = 0$.

Let $H$ and $H'$ be complex Hilbert spaces. For any densely defined linear operator $U: D(U) \subseteq H \to H'$, the orthogonal complement of the kernel of its adjoint operator $U^*$ is equal to the closure of the range of $U$. That is, $$(\ker U^*)^\perp = \overline{\text{ran } U},$$ where $\ker U^*$ denotes the kernel of the adjoint operator, $(\cdot)^\perp$ denotes the orthogonal complement, and $\overline{\text{ran } U}$ denotes the closure of the range of $U$.

Let $H$ and $H'$ be complex Hilbert spaces. For any partially defined linear map $U$ from $H$ to $H'$, if $U$ is a closed operator, then its closure is equal to $U$.

Let $H$ and $H'$ be Hilbert spaces over $\mathbb{C}$. For any partially defined linear operator $U: H \to H'$, if $U$ is closable, then $U$ is closed if and only if $U = \overline{U}$, where $\overline{U}$ denotes the closure of $U$.

Let $H$ and $H'$ be complex Hilbert spaces. Let $U$ be a linear operator defined on a dense subspace $\text{dom}(U) \subseteq H$ taking values in $H'$. If there exists a densely defined linear operator $U'$ from $\text{dom}(U') \subseteq H'$ to $H$ that is a formal adjoint of $U$ (i.e., $\langle U' \phi, \psi \rangle_H = \langle \phi, U \psi \rangle_{H'}$ for all $\phi \in \text{dom}(U')$ and $\psi \in \text{dom}(U)$), then $U$ is closable.

Let $\mathcal{H}$ and $\mathcal{H}'$ be Hilbert spaces over $\mathbb{C}$. For any partially defined linear operator $U: D(U) \subseteq \mathcal{H} \to \mathcal{H}'$ (represented as a `LinearPMap`), if $U$ is the zero operator on its domain, then $U$ is closable.

Let $H$ and $H'$ be complex Hilbert spaces, and let $U$ be a partially defined linear operator from $H$ to $H'$. If $U$ is closable, then for any complex scalar $c \in \mathbb{C}$, the scalar multiple $c U$ is also closable.

Let $H$ and $H'$ be complex Hilbert spaces, and let $U$ be a linear map defined on a subspace of $H$ mapping into $H'$ (a `LinearPMap`). For any non-zero complex scalar $c \in \mathbb{C}$ ($c \neq 0$), the operator $cU$ is closable if and only if $U$ is closable.

Let $H$ and $H'$ be inner product spaces over $\mathbb{C}$. For any linear partial map $U$ from $H$ to $H'$, the negation of the operator is equal to the scalar multiplication of the operator by $-1$, that is, $-U = -1 \cdot U$.

Let $H$ and $H'$ be complex Hilbert spaces. If a partially defined linear operator $U: \mathcal{D}(U) \subseteq H \to H'$ is closable, then its negation $-U$ is also closable.

Let $H$ and $H'$ be Hilbert spaces over the complex numbers $\mathbb{C}$. For any partially defined linear operator $U$ from $H$ to $H'$ and any non-zero scalar $c \in \mathbb{C}$, the closure of the scalar multiple $c \cdot U$ is equal to the scalar multiple of the closure of $U$: $$(c \cdot U)^{\text{closure}} = c \cdot U^{\text{closure}}$$

Let $H$ be a complex Hilbert space. Let $1$ denote the identity operator on $H$ defined on the entire space (the identity element of the monoid of partial linear maps). Then the adjoint of this identity operator is equal to itself, i.e., $1^\dagger = 1$.

Let $H$ be a Hilbert space and $H'$ be an inner product space over $\mathbb{C}$. For any partially defined linear map $U$ from $H$ to $H'$, if $U$ is the zero operator, then its adjoint operator $U^\dagger$ is also the zero operator.

Let $H$ be a complex Hilbert space and $H'$ be a complex inner product space. The adjoint of the zero operator $0: H \to H'$ is the zero operator $0: H' \to H$.

Let $H$ and $H'$ be complex Hilbert spaces, and let $U: H \to H'$ be a partially defined linear operator. For any non-zero complex scalar $c \in \mathbb{C}$, the adjoint of the scalar multiple $cU$ is given by the product of the complex conjugate $\bar{c}$ and the adjoint $U^*$, i.e., $(cU)^* = \bar{c}U^*$.

Let $H$ and $H'$ be complex Hilbert spaces, with $H$ being a complete space. Let $U_1$ and $U_2$ be partially defined linear operators (linear pre-maps) from $H$ to $H'$. If the sum $U_1 + U_2$ has a dense domain in $H$, then the sum of the adjoints of $U_1$ and $U_2$ is a restriction of the adjoint of their sum, denoted as $U_1^\dagger + U_2^\dagger \leq (U_1 + U_2)^\dagger$.

Let $H$ be a Hilbert space and $H'$ be an inner product space. Let $U_1, U_2: H \to H'$ be partially defined linear operators (linear PMaps). If the operator $U_1 - U_2$ has a dense domain, then the difference of their adjoints is extended by the adjoint of their difference: $$U_1^\dagger - U_2^\dagger \leq (U_1 - U_2)^\dagger$$ where $U^\dagger$ denotes the adjoint of $U$ and $\leq$ denotes the extension relation for partially defined operators.

Let $H$, $H'$, and $H''$ be complex Hilbert spaces. Let $U: H \to H'$ and $V: H' \to H''$ be partial linear maps. If $V$ is densely defined and the restricted composition $V \circ_r U$ is also densely defined, then the restricted composition of the adjoints is extended by the adjoint of the restricted composition: $$U^\dagger \circ_r V^\dagger \subseteq (V \circ_r U)^\dagger$$ where $\dagger$ denotes the operator adjoint and $A \subseteq B$ means that the partial linear map $B$ extends $A$ (i.e., $\text{dom}(A) \subseteq \text{dom}(B)$ and $A(x) = B(x)$ for all $x \in \text{dom}(A)$).

Let $H$ be a complex Hilbert space and $T$ be a partial linear operator on $H$. For any natural number $n$, if the $n$-th power of the operator $T^n$ (defined via restricted composition) has a dense domain, then the $n$-th power of the adjoint of $T$ is a restriction of the adjoint of the $n$-th power of $T$. That is, $$(T^*)^n \le (T^n)^*$$ where $T^*$ denotes the adjoint of $T$, and the relation $A \le B$ indicates that the operator $B$ is an extension of the operator $A$.

Let $\mathbb{k}$ be a nontrivially normed field, and let $E$ and $F$ be seminormed spaces over $\mathbb{k}$. A linear map $f: E \to F$ is continuous if and only if it is bounded, i.e., there exists a constant $M > 0$ such that for all $x \in E$, $\|f(x)\| \le M \|x\|$.

Let $H$ and $H'$ be complex inner product spaces. For any partially defined linear map $U: D(U) \subseteq H \to H'$, if $U$ is continuous on its domain $D(U)$, then $U$ is a closable operator.

Let $H$ and $H'$ be complex Hilbert spaces, and let $U$ be a partially defined linear operator from $H$ to $H'$ with domain $D(U) \subseteq H$. If $U$ is continuous on its domain, then the domain of its operator closure $\overline{U}$ is equal to the topological closure of its domain, denoted as $D(\overline{U}) = \overline{D(U)}$.

Let $H$ and $H'$ be complex inner product spaces, and assume $H'$ is a complete space (a Hilbert space). Let $U: \mathcal{D}(U) \to H'$ be a linear operator with domain $\mathcal{D}(U) \subseteq H$. If $U$ is continuous on its domain, then $U$ is a closed operator if and only if its domain $\mathcal{D}(U)$ is a closed subspace of $H$.

Let $H$ and $H'$ be Hilbert spaces over $\mathbb{C}$, and let $U$ be a partially defined linear operator from $H$ to $H'$. If $U$ is a closed operator, then its graph, $\{(x, Ux) \mid x \in \operatorname{dom}(U)\}$, is a closed subset of $H \times H'$.

Let $H$ and $H'$ be complex Hilbert spaces. Let $U: D(U) \subseteq H \to H'$ be a partially defined linear operator. If $U$ is a closed operator and its domain $D(U)$ is a closed subspace of $H$, then $U$ is continuous on its domain.

Let $H$ and $H'$ be Hilbert spaces over $\mathbb{C}$. Let $U_1$ and $U_2$ be partially defined linear operators from $H$ to $H'$. If $U_1$ is closable, $U_2$ is continuous, and the domain of $U_1$ is contained within the domain of $U_2$ ($\text{dom}(U_1) \subseteq \text{dom}(U_2)$), then the sum operator $U_1 + U_2$ is closable.

Let $H$ and $H'$ be complex Hilbert spaces. Let $U_1$ and $U_2$ be partially defined linear operators from $H$ to $H'$ with domains $D(U_1)$ and $D(U_2)$, respectively. If $U_1$ is closable, $U_2$ is continuous, and $D(U_1) \subseteq D(U_2)$, then the operator $U_1 - U_2$ (defined on $D(U_1)$) is also closable.

Let $H$ and $H'$ be complex inner product spaces, where $H'$ is a complete space (a Hilbert space). Let $U_1$ and $U_2$ be partially defined linear maps from $H$ to $H'$. If $U_1$ is a closed operator, $U_2$ is continuous on its domain, and the domain of $U_1$ is contained within the domain of $U_2$ ($\text{dom}(U_1) \subseteq \text{dom}(U_2)$), then the sum $U_1 + U_2$ is also a closed operator.

Let $H$ and $H'$ be complex Hilbert spaces, with $H'$ being a complete space. Let $U_1$ and $U_2$ be partially defined linear operators from $H$ to $H'$. If $U_1$ is a closed operator, $U_2$ is a continuous operator, and the domain of $U_1$ is contained in the domain of $U_2$ ($\operatorname{dom}(U_1) \subseteq \operatorname{dom}(U_2)$), then the operator $U_1 - U_2$ is also a closed operator.

Let $H$ and $H'$ be complex Hilbert spaces. Given a partially defined linear operator $A$ on $H$ and a unitary operator $u : H \to H'$, the unitary conjugation $u A u^{-1}$ is a partially defined linear operator on $H'$ with domain $$\text{dom}(u A u^{-1}) = u(\text{dom}(A)) = \{ y \in H' \mid u^{-1} y \in \text{dom}(A) \}$$ and the action defined by $$(u A u^{-1}) y = u(A(u^{-1} y))$$ for all $y \in \text{dom}(u A u^{-1})$. Since $u$ and $u^{-1}$ are $\mathbb{C}$-linear, the resulting operator is also $\mathbb{C}$-linear.

Let $H$ and $H'$ be complex Hilbert spaces. Let $A$ be a partially defined linear operator on $H$ and $u : H \to H'$ be a unitary operator. For any vector $x \in H'$, $x$ is in the domain of the unitarily conjugated operator $u A u^{-1}$ if and only if $u^{-1} x$ is in the domain of $A$. That is, $x \in \operatorname{dom}(u A u^{-1}) \iff u^{-1} x \in \operatorname{dom}(A)$.

Let $H$ and $H'$ be complex Hilbert spaces. Let $A$ be a partially defined linear operator on $H$ and $u : H \to H'$ be a unitary operator. For any vector $x$ in the domain of the unitarily conjugated operator $u A u^{-1}$, the value of the operator at $x$ is given by $$(u A u^{-1}) x = u(A(u^{-1} x)).$$ In this expression, the vector $u^{-1} x$ is an element of the domain of $A$.

Let $H$ and $H'$ be complex Hilbert spaces. Let $A$ be a partially defined linear operator on $H$ and $u: H \to H'$ be a unitary operator. For any vector $y$ in the domain of $A$, its image $u y$ is contained in the domain of the unitary conjugation $u A u^{-1}$.

Let $H$ and $H'$ be complex Hilbert spaces. Let $A$ be a partially defined linear operator on $H$ with domain $\text{dom}(A)$, and let $u: H \to H'$ be a unitary operator. For any vector $y \in \text{dom}(A)$, the action of the unitary conjugation $u A u^{-1}$ on the image vector $u y$ is given by $$(u A u^{-1})(u y) = u(A y).$$

Let $H$ and $H'$ be complex Hilbert spaces and $u : H \to H'$ be a unitary operator. Let $A$ and $B$ be partially defined linear operators on $H$ (of type $H \to_l.[{\mathbb{C}}] H$). If $A$ is a formal adjoint of $B$, then the unitary conjugation $u A u^{-1}$ is a formal adjoint of $u B u^{-1}$ as partially defined linear operators on $H'$.