Physlib.QuantumMechanics.DDimensions.Operators.Multiplication

Given a complex-valued function $f : \text{Space } d \to \mathbb{C}$, the multiplication operator $\mathcal{M}_f$ is the partially defined linear operator on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ that maps a vector $\psi$ to the product $f \cdot \psi$. The domain of this operator is the maximal subspace consisting of all $\psi \in L^2(\text{Space } d, \mathbb{C})$ such that the resulting product $f \cdot \psi$ remains square-integrable, i.e., $f \cdot \psi \in L^2(\text{Space } d, \mathbb{C})$.

This definition introduces the notation $\mathcal{M}$ for the multiplication operator. Given a function $f : \text{Space } d \to \mathbb{C}$, the expression $\mathcal{M} f$ represents the unbounded linear operator on the Hilbert space $\text{SpaceDHilbertSpace } d$ that maps a vector $\psi$ to the product $f \cdot \psi$, defined on the maximal domain where such a product remains in the Hilbert space.

For a complex-valued function $f: \text{Space } d \to \mathbb{C}$ and a vector $\psi$ in the Hilbert space $L^2(\text{Space } d, \mathbb{C})$, $\psi$ belongs to the domain of the multiplication operator $\mathcal{M}_f$ if and only if the pointwise product $f \cdot \psi$ is square-integrable, i.e., $f \cdot \psi \in L^2(\text{Space } d, \mathbb{C})$.

For any complex-valued function $f : \text{Space } d \to \mathbb{C}$ and any vector $\psi$ in the domain of the multiplication operator $\mathcal{M}_f$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$, the application of the operator $(\mathcal{M}_f) \psi$ is equal to the pointwise product $f \cdot \psi$ almost everywhere with respect to the volume measure.

For any almost everywhere strongly measurable function $f : \text{Space } d \to \mathbb{C}$, the multiplication operator $\mathcal{M}_f$ has a dense domain in the Hilbert space $L^2(\text{Space } d, \mathbb{C})$.

Let $d$ be a natural number and $f : \text{Space } d \to \mathbb{C}$ be a function. If $f$ has temperate growth, then the Schwartz space $\mathcal{S}(\text{Space } d)$ is contained within the domain of the multiplication operator $\mathcal{M}_f$ acting on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$.

Let $f : \text{Space } d \to \mathbb{C}$ be an almost everywhere strongly measurable function. Let $\mathcal{M}_f$ be the multiplication operator on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ and let $\bar{f} = \text{conj} \circ f$ be the complex conjugate of $f$. Then the domain of the multiplication operator $\mathcal{M}_{\bar{f}}$ is equal to the domain of the multiplication operator $\mathcal{M}_f$.

For any almost everywhere strongly measurable function $f : \text{Space } d \to \mathbb{C}$, the domain of the adjoint of the multiplication operator $\mathcal{M}_f$ on $L^2(\text{Space } d, \mathbb{C})$ is a subspace of the domain of the multiplication operator $\mathcal{M}_{\bar{f}}$, where $\bar{f} = \text{conj} \circ f$ is the complex conjugate of $f$. That is, $\text{dom}(\mathcal{M}_f^\dagger) \subseteq \text{dom}(\mathcal{M}_{\bar{f}})$.

Let $d$ be a natural number and let $f : \text{Space } d \to \mathbb{C}$ be an almost everywhere strongly measurable function. The adjoint $(\mathcal{M}_f)^\dagger$ of the multiplication operator $\mathcal{M}_f$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ is equal to the multiplication operator $\mathcal{M}_{\bar{f}}$ defined by the complex conjugate function $\bar{f} = \text{conj} \circ f$.

Let $f : \text{Space } d \to \mathbb{C}$ be an almost everywhere strongly measurable function. If $f$ is real-valued, meaning that the composition of the complex conjugate with $f$ satisfies $\text{conj} \circ f = f$, then the multiplication operator $\mathcal{M}_f$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ is self-adjoint.

For any almost everywhere strongly measurable function $f : \text{Space } d \to \mathbb{C}$, the multiplication operator $\mathcal{M}_f$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ is a closable operator.

Let $d$ be a natural number and $f : \text{Space } d \to \mathbb{C}$ be a function that is almost everywhere strongly measurable. Then the multiplication operator $\mathcal{M}_f$ (defined as $\psi \mapsto f \cdot \psi$ with its maximal domain) is an unbounded operator on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$.

For any complex-valued functions $f, g : \text{Space } d \to \mathbb{C}$, the restricted composition of the multiplication operators $\mathcal{M}_f$ and $\mathcal{M}_g$ is a sub-operator of the multiplication operator $\mathcal{M}_{f \cdot g}$ associated with the pointwise product $f \cdot g$, denoted by: $$\mathcal{M}_f \circ_r \mathcal{M}_g \le \mathcal{M}_{f \cdot g}$$ This implies that the domain of the composed operator $\mathcal{M}_f \circ_r \mathcal{M}_g$ is a subset of the domain of $\mathcal{M}_{f \cdot g}$, and both operators agree on the domain of the composition.

For any complex-valued functions $f, g : \text{Space } d \to \mathbb{C}$, if the domain of the multiplication operator $\mathcal{M}_g$ is the entire Hilbert space $L^2(\text{Space } d, \mathbb{C})$ (i.e., $\text{dom}(\mathcal{M}_g) = \top$), then the restricted composition of the multiplication operators $\mathcal{M}_f$ and $\mathcal{M}_g$ is equal to the multiplication operator of their pointwise product: $$\mathcal{M}_f \circ_r \mathcal{M}_g = \mathcal{M}_{f \cdot g}$$ where $\circ_r$ denotes the restricted composition of partial linear maps and $f \cdot g$ is the pointwise product of the functions.