Physlib.QuantumMechanics.Operators.Multiplication

Let $\mu$ be a measure on $\text{Space } d$. For any $\mu$-a.e. strongly measurable function $f : \text{Space } d \to \mathbb{C}$, if there exists a constant $c \in \mathbb{R}$ such that $\|f(x)\| \le c$ for $\mu$-almost every $x \in \text{Space } d$, then the domain of the multiplication operator $\mathcal{M}_f$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ is the entire space (denoted by $\top$).

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$. Let $f, g : \text{Space } d \to \mathbb{C}$ be complex-valued functions, and suppose $g$ is $\mu$-almost everywhere strongly measurable. If $\|g(x)\| \leq \|f(x)\|$ for $\mu$-almost every $x$, then the domain of the multiplication operator $\mathcal{M}_f$ is a subset of the domain of the multiplication operator $\mathcal{M}_g$, that is, $\text{dom}(\mathcal{M}_f) \subseteq \text{dom}(\mathcal{M}_g)$.

Let $\mu$ be a measure on $\text{Space } d$. For any two $\mu$-almost everywhere strongly measurable functions $f, g : \text{Space } d \to \mathbb{C}$ such that their norms are equal $\mu$-almost everywhere (i.e., $\|f(x)\| = \|g(x)\|$ for $\mu$-almost every $x$), the domain of the multiplication operator $\mathcal{M}_f$ is equal to the domain of the multiplication operator $\mathcal{M}_g$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$.

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$ that is finite on compact sets. If $f : \text{Space } d \to \mathbb{C}$ is a function that is $\mu$-almost everywhere strongly measurable, then the multiplication operator $\mathcal{M}_f$ acting on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ is a closed operator.

For any measure $\mu$ on $\text{Space } d$, the multiplication operator $\mathcal{M}_0$ associated with the constant zero function $f(x) = 0$ on the Hilbert space $L^2(\text{Space } d, \mu)$ is equal to the zero operator. The domain of this operator is the entire Hilbert space.

Let $\mu$ be a measure on $\text{Space } d$. For any two functions $f, g : \text{Space } d \to \mathbb{C}$, if $f = g$ almost everywhere with respect to $\mu$, then the corresponding multiplication operators $\mathcal{M}_f$ and $\mathcal{M}_g$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ are equal.

Let $\mu$ be a measure on $\text{Space } d$, $f : \text{Space } d \to \mathbb{C}$ be a complex-valued function, and $c \in \mathbb{C}$ be a scalar. Let $\mathcal{M}_f$ denote the multiplication operator on the Hilbert space $L^2(\text{Space } d, \mu)$. The operator $c \cdot \mathcal{M}_f$ is a restriction of the multiplication operator $\mathcal{M}_{cf}$, i.e., $c \cdot \mathcal{M}_f \le \mathcal{M}_{cf}$. Equality holds when $c \neq 0$, but for $c = 0$, the domain of $0 \cdot \mathcal{M}_f$ (which is the domain of $\mathcal{M}_f$) may be smaller than the domain of $\mathcal{M}_0$ (which is the entire space).

Let $\mu$ be a measure on $\text{Space } d$, $f : \text{Space } d \to \mathbb{C}$ be a function, and $c \in \mathbb{C}$ be a non-zero complex scalar ($c \neq 0$). Then the multiplication operator $\mathcal{M}_{cf}$ (defined by the function $x \mapsto c \cdot f(x)$) is equal to the scalar multiplication of the operator $\mathcal{M}_f$ by $c$, satisfying $\mathcal{M}_{cf} = c \mathcal{M}_f$.

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$. For any complex-valued function $f : \text{Space } d \to \mathbb{C}$, let $\mathcal{M}_f$ denote the multiplication operator on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$. Then, negation commutes with the multiplication operator, such that the multiplication operator by $-f$ is equal to the negation of the multiplication operator by $f$: $$\mathcal{M}_{-f} = -\mathcal{M}_f$$

For a measure $\mu$ on $\text{Space } d$ and functions $f, g : \text{Space } d \to \mathbb{C}$, the multiplication operator associated with the sum $f + g$, denoted $\mathcal{M}_{f+g}$, is an extension of the sum of the individual multiplication operators $\mathcal{M}_f + \mathcal{M}_g$ on the Hilbert space $L^2(\text{Space } d, \mu)$. This means that the domain of the sum $\text{dom}(\mathcal{M}_f + \mathcal{M}_g)$, which consists of vectors $\psi$ such that both $f \cdot \psi$ and $g \cdot \psi$ are square-integrable, is contained within the domain $\text{dom}(\mathcal{M}_{f+g})$, which only requires the weaker condition that $(f+g) \cdot \psi$ be square-integrable. On the shared domain, the operators agree: $(\mathcal{M}_f + \mathcal{M}_g)\psi = \mathcal{M}_{f+g}\psi$.

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$. Let $f, g : \text{Space } d \to \mathbb{C}$ be complex-valued functions, and let $\mathcal{M}_f$ and $\mathcal{M}_g$ be the corresponding multiplication operators on the Hilbert space $L^2(\text{Space } d, \mu)$. If the domain of $\mathcal{M}_g$ is the entire Hilbert space (i.e., $\text{dom}(\mathcal{M}_g) = \top$), then the multiplication operator of the sum of the functions is equal to the sum of the individual multiplication operators: $$\mathcal{M}_{f+g} = \mathcal{M}_f + \mathcal{M}_g$$

Consider the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ with measure $\mu$. Let $f, g: \text{Space } d \to \mathbb{C}$ be $\mu$-almost everywhere strongly measurable functions. Suppose there exist constants $c, c' \in \mathbb{R}$ such that the inequalities $\|f(x)\| \le c \|f(x) + g(x)\|$ and $\|g(x)\| \le c' \|f(x) + g(x)\|$ hold for $\mu$-almost every $x \in \text{Space } d$. Then the multiplication operator associated with the sum of the functions is equal to the sum of the individual multiplication operators, i.e., $\mathcal{M}_{f+g} = \mathcal{M}_f + \mathcal{M}_g$.

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$. For any complex-valued functions $f, g : \text{Space } d \to \mathbb{C}$, the multiplication operator $\mathcal{M}_{f-g}$ is an extension of the difference of operators $\mathcal{M}_f - \mathcal{M}_g$. That is, $\text{dom}(\mathcal{M}_f - \mathcal{M}_g) \subseteq \text{dom}(\mathcal{M}_{f-g})$ and for all $\psi \in \text{dom}(\mathcal{M}_f - \mathcal{M}_g)$, $(\mathcal{M}_f - \mathcal{M}_g)\psi = \mathcal{M}_{f-g}\psi$. The domain of the difference $\mathcal{M}_f - \mathcal{M}_g$ is the intersection of the individual domains, $\text{dom}(\mathcal{M}_f) \cap \text{dom}(\mathcal{M}_g)$, requiring both $f \cdot \psi$ and $g \cdot \psi$ to be in $L^2(\text{Space } d, \mu)$. In contrast, the domain of $\mathcal{M}_{f-g}$ only requires the difference $(f - g) \cdot \psi$ to be in $L^2(\text{Space } d, \mu)$, which is a weaker condition.

Let $\mu$ be a measure on $\text{Space } d$ and let $L^2(\text{Space } d, \mathbb{C})$ be the Hilbert space of square-integrable complex-valued functions. For any two functions $f, g : \text{Space } d \to \mathbb{C}$, if the domain of the multiplication operator $\mathcal{M}_g$ is the entire Hilbert space (i.e., $\text{dom}(\mathcal{M}_g) = L^2(\text{Space } d, \mathbb{C})$), then the multiplication operator associated with the difference $f - g$ is equal to the difference of the individual multiplication operators: $$\mathcal{M}_{f-g} = \mathcal{M}_f - \mathcal{M}_g$$

Let $\mu$ be a measure on the space $\text{Space } d$. Let $f, g: \text{Space } d \to \mathbb{C}$ be almost everywhere strongly measurable functions. Suppose there exist constants $c, c' \in \mathbb{R}$ such that for $\mu$-almost every $x$, the following conditions hold: $$\|f(x)\| \leq c \|f(x) - g(x)\| \quad \text{and} \quad \|g(x)\| \leq c' \|f(x) - g(x)\|$$ Then the multiplication operator $\mathcal{M}_{f-g}$ is equal to the difference of the multiplication operators $\mathcal{M}_f$ and $\mathcal{M}_g$ on the Hilbert space $L^2(\text{Space } d, \mathbb{C})$. This equality implies that the domain of $\mathcal{M}_{f-g}$ coincides with the intersection of the domains of $\mathcal{M}_f$ and $\mathcal{M}_g$.

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$. For any complex-valued functions $f, g : \text{Space } d \to \mathbb{C}$, let $\mathcal{M}_f$ and $\mathcal{M}_g$ be the corresponding multiplication operators on the Hilbert space $L^2(\text{Space } d, \mu)$. Then the multiplication operator associated with the pointwise product $f \cdot g$, denoted $\mathcal{M}_{f \cdot g}$, is an extension of the restricted composition of the individual operators $\mathcal{M}_f \circ_r \mathcal{M}_g$. Specifically, the domain of the composition satisfies $\text{dom}(\mathcal{M}_f \circ_r \mathcal{M}_g) \subseteq \text{dom}(\mathcal{M}_{f \cdot g})$, and for any $\psi$ in the domain of the composition, $(\mathcal{M}_f \circ_r \mathcal{M}_g)\psi = \mathcal{M}_{f \cdot g}\psi$.

Let $\mu$ be a measure on $\text{Space } d$, and let $f, g: \text{Space } d \to \mathbb{C}$ be functions. Let $\mathcal{M}_f$ and $\mathcal{M}_g$ be the multiplication operators on the Hilbert space $L^2(\text{Space } d, \mu)$ associated with $f$ and $g$ respectively. If the domain of $\mathcal{M}_g$ is the entire Hilbert space (i.e., $\text{dom}(\mathcal{M}_g) = \top$), then the restricted composition $\mathcal{M}_f \circ_r \mathcal{M}_g$ is equal to the multiplication operator $\mathcal{M}_{f \cdot g}$ of the pointwise product $f \cdot g$: $$\mathcal{M}_f \circ_r \mathcal{M}_g = \mathcal{M}_{f \cdot g}$$ This theorem provides a sufficient condition (the totality of the domain of the second operator) to turn the operator inclusion $\mathcal{M}_f \circ_r \mathcal{M}_g \subseteq \mathcal{M}_{f \cdot g}$ into an equality.