Physlib.QuantumMechanics.DDimensions.Operators.Unbounded

Let $H$ and $H'$ be Hilbert spaces, and let $U_1$ and $U_2$ be unbounded operators between them. If the underlying linear partial maps of $U_1$ and $U_2$ are equal, then $U_1 = U_2$.

Let $U$ be an unbounded operator between Hilbert spaces $H$ and $H'$. This instance allows $U$ to be treated as a function mapping an element $x$ of its domain $\mathcal{D}(U)$ to its image $U(x)$ in $H'$.

Let $H$ be a complex Hilbert space and $T$ be an unbounded operator on $H$. For any vectors $x$ and $y$ in the domain of $T$ (denoted $\mathcal{D}(T)$), the following polarization identity holds: $$\langle Tx, y \rangle = \frac{1}{4} \left( \langle T(x + y), x + y \rangle - \langle T(x - y), x - y \rangle - i \langle T(x + iy), x + iy \rangle + i \langle T(x - iy), x - iy \rangle \right)$$ where $\langle \cdot, \cdot \rangle$ denotes the inner product on $H$ and $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be an unbounded operator on $H$. For any vectors $x$ and $y$ in the domain of $T$, the inner product of $x$ and $Ty$ can be expressed via the polarization identity: $$\langle x, Ty \rangle = \frac{1}{4} \left( \langle x+y, T(x+y) \rangle - \langle x-y, T(x-y) \rangle - i \langle x+iy, T(x+iy) \rangle + i \langle x-iy, T(x-iy) \rangle \right)$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product and $i$ is the imaginary unit.

The type of unbounded operators between Hilbert spaces $H$ and $H'$ is equipped with a partial order structure. For two unbounded operators $U_1$ and $U_2$, the relation $U_1 \le U_2$ holds if $U_2$ is an extension of $U_1$. Specifically, this means that the domain of $U_1$ is a subspace of the domain of $U_2$ ($\mathcal{D}(U_1) \subseteq \mathcal{D}(U_2)$) and the operators agree on the smaller domain ($U_1 x = U_2 x$ for all $x \in \mathcal{D}(U_1)$).

Let $E$ and $F$ be complex inner product spaces. If $f$ is a partially defined linear map from $E$ to $F$ (represented as a `LinearPMap`) such that $f(x) = 0$ for all $x$ in its domain, then $f$ is closable.

The zero element $0$ of the type of unbounded operators between Hilbert spaces $H$ and $H'$ represents the operator whose domain is the entire space $H$ and which maps every vector in $H$ to the zero vector in $H'$.

Let $H$ and $H'$ be Hilbert spaces. The linear partial map underlying the zero unbounded operator $0$ between $H$ and $H'$ is equal to the zero linear partial map.

The type of unbounded operators between Hilbert spaces $H$ and $H'$ is inhabited, meaning it contains at least one element. This instance identifies the zero unbounded operator $0$ as the default or canonical element of the type.

The addition of two unbounded operators $U_1, U_2$ between Hilbert spaces $H$ and $H'$ is defined based on the density of the intersection of their domains $D(U_1) \cap D(U_2)$ and the closability of their sum. Let $U_1 +_{p} U_2$ denote the sum of the underlying linear partial maps. The sum $U_1 + U_2$ is defined as follows: - If $D(U_1) \cap D(U_2)$ is dense in $H$: - If $U_1 +_{p} U_2$ is a closable operator, then $U_1 + U_2$ is the unbounded operator with domain $D(U_1) \cap D(U_2)$ mapping $x$ to $U_1 x + U_2 x$. - If $U_1 +_{p} U_2$ is not closable, $U_1 + U_2$ is the zero operator restricted to the domain $D(U_1) \cap D(U_2)$. - If $D(U_1) \cap D(U_2)$ is not dense in $H$, $U_1 + U_2$ is the zero operator $0$ with domain $H$.

Let $H$ and $H'$ be Hilbert spaces, and let $U_1, U_2$ be unbounded operators from $H$ to $H'$ with domains $D(U_1)$ and $D(U_2)$, respectively. If the intersection of their domains $D(U_1) \cap D(U_2)$ is dense in $H$, then the domain of the sum $U_1 + U_2$ is precisely that intersection, i.e., $D(U_1 + U_2) = D(U_1) \cap D(U_2)$.

Let $U_1$ and $U_2$ be unbounded operators between Hilbert spaces $H$ and $H'$ with domains $D(U_1)$ and $D(U_2)$, respectively. If the intersection of their domains $D(U_1) \cap D(U_2)$ is not dense in $H$, then the domain of their sum $D(U_1 + U_2)$ is equal to the entire space $H$.

Let $H$ and $H'$ be Hilbert spaces, and let $U_1$ and $U_2$ be unbounded operators from $H$ to $H'$ with domains $D(U_1)$ and $D(U_2)$, respectively. If the intersection of their domains $D(U_1) \cap D(U_2)$ is dense in $H$, then any vector $\psi$ in the domain of the sum $D(U_1 + U_2)$ must belong to both $D(U_1)$ and $D(U_2)$.

Let $H$ and $H'$ be Hilbert spaces, and let $U_1$ and $U_2$ be unbounded operators from $H$ to $H'$. If the intersection of their domains $D(U_1) \cap D(U_2)$ is dense in $H$, and if the sum of their underlying linear partial maps $U_1 + U_2$ is closable, then the linear partial map associated with the sum of the unbounded operators $U_1 + U_2$ is equal to the sum of the linear partial maps of $U_1$ and $U_2$.

Let $H$ and $H'$ be Hilbert spaces and $U_1, U_2$ be unbounded operators between them. If the intersection of their domains $D(U_1) \cap D(U_2)$ is dense in $H$ and the sum of their underlying linear partial maps is not closable, then the linear partial map of the operator sum $U_1 + U_2$ is defined to be the zero map on the domain $D(U_1) \cap D(U_2)$.

Let $H$ and $H'$ be Hilbert spaces, and let $U_1$ and $U_2$ be unbounded operators between them. If the intersection of their domains $D(U_1) \cap D(U_2)$ is not dense in $H$, then the sum $U_1 + U_2$ is defined as the zero operator $0$ (which has the entire space $H$ as its domain).

Let $H$ and $H'$ be Hilbert spaces and let $U_1, U_2$ be unbounded operators from $H$ to $H'$ with domains $D(U_1)$ and $D(U_2)$ respectively. Suppose that the intersection of their domains $D(U_1) \cap D(U_2)$ is dense in $H$ and that the sum of their underlying linear partial maps is closable. Then, for any vector $\psi$ in the domain of the operator sum $D(U_1 + U_2)$, the application of the sum to $\psi$ satisfies $(U_1 + U_2)\psi = U_1\psi + U_2\psi$.

The type of unbounded operators between Hilbert spaces $H$ and $H'$ forms an additive structure with a zero element (an `AddZeroClass`). This means that the zero operator $0$ (defined on the whole space $H$) acts as the identity for the addition of unbounded operators, such that for any unbounded operator $U$, we have $0 + U = U$ and $U + 0 = U$.

Let $H$ and $H'$ be Hilbert spaces, and let $U_1, U_2,$ and $U_3$ be unbounded operators from $H$ to $H'$. Suppose the following conditions are satisfied: 1. The intersection of the domains $D(U_1) \cap D(U_2)$ is dense in $H$, and the sum of the underlying linear partial maps of $U_1$ and $U_2$ is closable. 2. The intersection of the domains $D(U_2) \cap D(U_3)$ is dense in $H$, and the sum of the underlying linear partial maps of $U_2$ and $U_3$ is closable. Under these conditions, the addition of unbounded operators is associative, meaning $(U_1 + U_2) + U_3 = U_1 + (U_2 + U_3)$.

Let $E$ and $F$ be complex inner product spaces and let $f$ be a partially defined linear map from $E$ to $F$. For any non-zero complex scalar $c \in \mathbb{C}$ and any pair $(x, y) \in E \times F$, if $(x, y)$ is an element of the topological closure of the graph of the scaled map $c \cdot f$, then $(x, c^{-1} y)$ is an element of the topological closure of the graph of $f$.

Let $E$ and $F$ be complex inner product spaces and let $f$ be a partially defined linear map from $E$ to $F$. If $f$ is closable, then for any complex scalar $c \in \mathbb{C}$, the scaled map $c f$ is also closable.

Let $H$ and $H'$ be complex Hilbert spaces. For any complex scalar $c \in \mathbb{C}$ and any unbounded operator $U$ (which is defined as a densely defined and closable linear map from $H$ to $H'$), the scalar multiplication $c \cdot U$ is an unbounded operator. This operator is defined by scaling the underlying partially defined linear map of $U$ by $c$, while preserving the property of having a dense domain and being closable.

For any complex scalar $c \in \mathbb{C}$ and any unbounded operator $U$ between Hilbert spaces, the domain of the scalar-multiplied operator $c \cdot U$ is equal to the domain of $U$: $\text{dom}(c \cdot U) = \text{dom}(U)$.

For any complex scalar $c \in \mathbb{C}$ and any unbounded operator $U$ between complex Hilbert spaces, the underlying linear partial map of the scaled operator $c \cdot U$ is equal to the scalar $c$ multiplied by the underlying linear partial map of $U$, expressed as $(c \cdot U).\text{toLinearPMap} = c \cdot U.\text{toLinearPMap}$.

Let $U$ be an unbounded operator between Hilbert spaces $H$ and $H'$. Then the scalar multiplication of $U$ by the complex number $0$ is less than or equal to the zero operator $0$ in the partial order of unbounded operators, written as $(0 : \mathbb{C}) \cdot U \le 0$. This inequality states that the zero operator (which is defined on the entire space $H$) is an extension of the operator $(0 : \mathbb{C}) \cdot U$ (which is defined on the domain of $U$).

Let $H$ and $H'$ be complex Hilbert spaces. The type of unbounded operators between $H$ and $H'$ admits a distributive scalar multiplication by complex numbers $c \in \mathbb{C}$. Specifically, for any $c \in \mathbb{C}$ and any unbounded operators $U_1, U_2$, the distributive law $c \cdot (U_1 + U_2) = c \cdot U_1 + c \cdot U_2$ holds, and the action preserves the zero operator such that $c \cdot 0 = 0$.

For an unbounded operator $U$ between Hilbert spaces $H$ and $H'$, its closure $\overline{U}$ is the operator whose graph is the closure of the graph of $U$ in $H \times H'$. The resulting operator $\overline{U}$ is itself a densely defined and closed (and thus closable) unbounded operator.

For an unbounded operator $U$ between Hilbert spaces, let $\overline{U}$ denote its closure. The underlying linear partial map of the closure $\overline{U}$ is equal to the closure of the linear partial map of $U$.

Let $U$ be an unbounded operator between Hilbert spaces. Then $U \le \overline{U}$, where $\overline{U}$ denotes the closure of $U$. This relation indicates that the closure $\overline{U}$ is an extension of $U$, meaning that the domain of $U$ is contained in the domain of $\overline{U}$ ($\mathcal{D}(U) \subseteq \mathcal{D}(\overline{U})$) and the operators agree on $\mathcal{D}(U)$.

An unbounded operator $U$ is **closed** if the graph of its underlying linear partial map $T$, defined as $\{ (x, Tx) \mid x \in \text{dom}(T) \}$, is a closed subspace of the product of the Hilbert spaces.

Let $U$ be an unbounded operator between Hilbert spaces. Its closure $\overline{U}$ is a closed operator.

Let $U$ be an unbounded operator between Hilbert spaces. $U$ is closed if and only if it equals its closure, denoted as $U = \overline{U}$.

Let $H$ and $H'$ be Hilbert spaces and $f$ be a linear partially defined map from $H$ to $H'$. If the domain of $f$ is dense in $H$ and $f$ is closable, then the domain of its adjoint operator $f^\dagger$ is dense in $H'$.

Given an unbounded operator $U$ from a Hilbert space $H$ to a Hilbert space $H'$, the adjoint operator $U^\dagger$ is an unbounded operator from $H'$ to $H$. It is constructed from the adjoint of the underlying linear partial map of $U$. This definition ensures that $U^\dagger$ is an `UnboundedOperator` by proving that its domain is dense (which follows from $U$ being closable and densely defined) and that it is a closed (and therefore closable) operator.

This definition introduces the postfix notation $\dagger$ to represent the adjoint of an unbounded operator. For an unbounded operator $T$, $T^\dagger$ denotes its adjoint operator between Hilbert spaces.

Let $H$ and $H'$ be Hilbert spaces and let $U$ be an unbounded operator from $H$ to $H'$. The underlying linear partial map of the adjoint operator $U^\dagger$ is equal to the adjoint of the underlying linear partial map of $U$, denoted as $(U.\text{toLinearPMap})^\dagger$.

Let $H$ and $H'$ be Hilbert spaces and let $U$ be an unbounded operator from $H$ to $H'$. The adjoint operator $U^\dagger$ from $H'$ to $H$ is a closed operator.

Let $H$ and $H'$ be Hilbert spaces and let $U$ be an unbounded operator from $H$ to $H'$. The closure of the adjoint operator $U^\dagger$ is equal to the adjoint operator itself, that is, $\overline{U^\dagger} = U^\dagger$.

Let $H$ and $H'$ be Hilbert spaces and let $U$ be an unbounded operator from $H$ to $H'$. Let $\overline{U}$ denote the closure of $U$ and $U^\dagger$ denote the adjoint operator of $U$. Then the adjoint of the closure of $U$ is equal to the adjoint of $U$, that is, $(\overline{U})^\dagger = U^\dagger$.

Let $H$ and $H'$ be Hilbert spaces and $U$ be an unbounded operator from $H$ to $H'$. The double adjoint of $U$, denoted by $U^{\dagger\dagger}$, is equal to the closure of $U$, denoted by $\overline{U}$.

Let $H$ and $H'$ be Hilbert spaces and let $U$ be a densely defined, closable unbounded operator from $H$ to $H'$. Then $U \le U^{\dagger\dagger}$, where $U^{\dagger\dagger}$ denotes the double adjoint of $U$. This relation indicates that $U^{\dagger\dagger}$ is an extension of $U$, meaning that the domain of $U$ is a subspace of the domain of $U^{\dagger\dagger}$ ($\mathcal{D}(U) \subseteq \mathcal{D}(U^{\dagger\dagger})$) and both operators agree on $\mathcal{D}(U)$.

Let $H$ and $H'$ be Hilbert spaces and $U$ be an unbounded operator from $H$ to $H'$. $U$ is closed if and only if it is equal to its double adjoint, $U = U^{\dagger\dagger}$.

Let $H$ and $H'$ be Hilbert spaces. For any two unbounded operators $U_1, U_2$ from $H$ to $H'$, if $U_1 \le U_2$ (meaning $U_2$ is an extension of $U_1$), then their adjoints satisfy $U_2^\dagger \le U_1^\dagger$ (meaning $U_1^\dagger$ is an extension of $U_2^\dagger$). Here, $U^\dagger$ denotes the adjoint operator of $U$.

Let $H$ and $H'$ be Hilbert spaces. For any two unbounded operators $U_1$ and $U_2$ from $H$ to $H'$, if $U_1 \le U_2$ (meaning $U_2$ is an extension of $U_1$), then their closures satisfy $\overline{U_1} \le \overline{U_2}$ (meaning the closure of $U_2$ is an extension of the closure of $U_1$). Here, $\overline{U}$ denotes the closure of the operator $U$.

Given a dense submodule $E$ of a Hilbert space $H$ and a symmetric linear map $f: E \to E$ (satisfying $\langle f(x), y \rangle = \langle x, f(y) \rangle$ for all $x, y \in E$), this definition constructs an unbounded operator on $H$. The resulting operator has domain $E$ and maps $\psi \in E$ to $f(\psi) \in H$. It is guaranteed to be closable because it is symmetric and densely defined.

Let $H$ be a Hilbert space and $E$ be a dense subspace of $H$. Suppose $f: E \to E$ is a symmetric linear map (i.e., $\langle f(x), y \rangle = \langle x, f(y) \rangle$ for all $x, y \in E$). If $T$ is the unbounded operator on $H$ constructed from $f$ (via `ofSymmetric`), then for any vector $\psi \in E$, the application of the operator $T$ to $\psi$ is given by $T\psi = f(\psi)$.

An unbounded operator $T$ on a Hilbert space $H$ is symmetric if for all $x, y$ in the domain of $T$, the inner product satisfies the condition $\langle Tx, y \rangle = \langle x, Ty \rangle$.

Let $H$ be a complex Hilbert space and $T$ be an unbounded operator on $H$. The operator $T$ is symmetric if and only if for every $x$ in the domain of $T$, denoted as $\mathcal{D}(T)$, the inner product $\langle Tx, x \rangle$ is a real number, i.e., $\overline{\langle Tx, x \rangle} = \langle Tx, x \rangle$.

Let $H$ be a Hilbert space and $T$ be an unbounded operator on $H$. The operator $T$ is symmetric if and only if $T \le T^\dagger$, where $T^\dagger$ denotes the adjoint of $T$ and the relation $\le$ signifies that $T^\dagger$ is an extension of $T$ (i.e., the domain of $T$ is a subspace of the domain of $T^\dagger$, and they agree on the domain of $T$).

The star operation on the set of unbounded operators mapping a Hilbert space $H$ to itself is defined by the adjoint operator, mapping $T$ to $T^\dagger$.

Let $T$ be an unbounded operator on a Hilbert space $H$. $T$ is self-adjoint if and only if its adjoint $T^\dagger$ is equal to $T$.

Let $H$ be a Hilbert space and $T$ be an unbounded operator on $H$. Then $T$ is self-adjoint if and only if its underlying linear partial map, $T.\text{toLinearPMap}$, is self-adjoint.

Let $H$ be a Hilbert space and let $T$ be an unbounded operator on $H$. If $T$ is self-adjoint, then $T$ is closed.

Let $H$ be a Hilbert space and $T$ be an unbounded operator on $H$. If $T$ is self-adjoint, then $T$ is symmetric.

An unbounded operator $T$ on a Hilbert space $H$ is essentially self-adjoint if its closure $\overline{T}$ is a self-adjoint operator, satisfying $\overline{T} = \overline{T}^\dagger$.

Let $T$ be an unbounded operator on a Hilbert space $H$. $T$ is essentially self-adjoint if and only if its adjoint operator $T^\dagger$ is equal to its closure $\overline{T}$.

Let $H$ be a Hilbert space and $T$ be an unbounded operator on $H$. If $T$ is self-adjoint, then $T$ is essentially self-adjoint (meaning its closure $\overline{T}$ is self-adjoint).

Let $T$ be an unbounded operator with domain $D(T)$. A continuous linear functional $F : D(T) \to \mathbb{C}$ is a **generalized eigenvector** of $T$ with eigenvalue $c \in \mathbb{C}$ if for all vectors $\psi \in D(T)$, the relation $F(T \psi) = c F(\psi)$ holds.

Let $H$ be a Hilbert space and $E$ be a dense submodule of $H$. Let $f: E \to E$ be a symmetric linear map, and let $T$ be the unbounded operator on $H$ constructed from $f$ with domain $D(T) = E$ (such that $T\psi = f(\psi)$ for all $\psi \in E$). For a continuous linear functional $F : E \to \mathbb{C}$ and a scalar $c \in \mathbb{C}$, $F$ is a generalized eigenvector of $T$ with eigenvalue $c$ if and only if the relation $F(f(\psi)) = c F(\psi)$ holds for all $\psi \in E$.