Physlib.QuantumMechanics.HilbertSpaces.SpaceD.Basic

For a natural number $d$, let $\mathcal{H}_d = L^2(\text{Space } d, \mathbb{C})$ be the Hilbert space of square-integrable functions. For any two state vectors $\psi, \varphi \in \mathcal{H}_d$, the action of the dual vector (bra) $\text{toBra}(\psi)$ on the vector (ket) $\varphi$ is equal to the inner product $\langle \psi, \varphi \rangle$ in $\mathcal{H}_d$.

Let $\mathcal{H}_d = L^2(\text{Space } d, \mathbb{C})$ be the Hilbert space of square-integrable functions for a given dimension $d$. Let $\text{toBra} : \mathcal{H}_d \to \mathcal{H}_d^*$ be the antilinear isomorphism that maps a vector $\phi \in \mathcal{H}_d$ to its corresponding dual functional in the strong dual space $\mathcal{H}_d^*$, defined by the inner product $\langle \phi, \cdot \rangle$. For any linear functional $f \in \mathcal{H}_d^*$ and any vector $\psi \in \mathcal{H}_d$, the inner product of the vector $\text{toBra}^{-1}(f)$ with $\psi$ is equal to the result of the functional $f$ applied to $\psi$: $$\langle \text{toBra}^{-1}(f), \psi \rangle = f(\psi)$$

For any dimension $d \in \mathbb{N}$, if $\psi$ is an element of the Hilbert space $L^2(\text{Space } d, \mathbb{C})$, then its underlying function satisfies the square-integrability property $\text{MemHS}$. That is, the function is almost everywhere strongly measurable and satisfies $\int_{\text{Space } d} \| \psi(x) \|^2 \, dx < \infty$.

For any dimension $d \in \mathbb{N}$, the constant zero function $0: \text{Space } d \to \mathbb{C}$ is a member of the Hilbert space $L^2(\text{Space } d)$.

For any natural number $d$ and any complex-valued function $f: \text{Space } d \to \mathbb{C}$, if $f$ is a member of the square-integrable Hilbert space $L^2(\text{Space } d)$, then its pointwise negation $-f$ is also a member of $L^2(\text{Space } d)$.

For any dimension $d \in \mathbb{N}$ and complex-valued functions $f, g: \text{Space } d \to \mathbb{C}$, if $f$ and $g$ are square-integrable (i.e., they satisfy the $\text{MemHS}$ condition), then their sum $f + g$ is also square-integrable.

For any dimension $d \in \mathbb{N}$ and any complex-valued functions $f, g: \text{Space } d \to \mathbb{C}$, if both $f$ and $g$ are square-integrable (i.e., they satisfy the condition $\text{MemHS}$, meaning $\int \|f(x)\|^2 \, dx < \infty$ and $\int \|g(x)\|^2 \, dx < \infty$), then their difference $f - g$ is also square-integrable.

For any $d \in \mathbb{N}$ and any complex-valued function $f: \text{Space } d \to \mathbb{C}$, if $f$ is a member of the Hilbert space $L^2(\text{Space } d)$ (i.e., $f$ is square-integrable), then for any scalar $c \in \mathbb{C}$, the function $c f$ is also a member of $L^2(\text{Space } d)$.

For any dimension $d \in \mathbb{N}$ and complex-valued functions $f, g: \text{Space } d \to \mathbb{C}$, if $f$ and $g$ are equal almost everywhere with respect to the volume measure ($f = g$ a.e.), then $f \in L^2(\text{Space } d)$ implies $g \in L^2(\text{Space } d)$. Here, the condition $f \in L^2(\text{Space } d)$ (denoted as `MemHS f` in the formal text) means that $f$ is almost everywhere strongly measurable and square-integrable, satisfying $\int \|f(x)\|^2 \, dx < \infty$.

Let $\mu$ and $\mu'$ be measures on $\text{Space } d$, and let $f: \text{Space } d \to \mathbb{C}$ be a complex-valued function. If $\mu' \leq \mu$ and $f$ is a member of the Hilbert space $L^2(\text{Space } d, \mu)$, then $f$ is also a member of the Hilbert space $L^2(\text{Space } d, \mu')$.

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$. For any complex-valued function $f: \text{Space } d \to \mathbb{C}$, if $f$ belongs to the $L^2$ Hilbert space with respect to the measure $\mu$, then for any subset $\Omega \subseteq \text{Space } d$, $f$ also belongs to the $L^2$ Hilbert space with respect to the restricted measure $\mu|_\Omega$.

Let $d$ be a natural number and $\mu$ be a measure on $\text{Space } d$. For any complex-valued function $f: \text{Space } d \to \mathbb{C}$ and any two subsets $\Omega, \Omega' \subseteq \text{Space } d$, if $\Omega' \subseteq \Omega$ and $f$ is square-integrable on $\Omega$ with respect to $\mu$ (i.e., $f \in L^2(\Omega, \mu)$), then $f$ is also square-integrable on $\Omega'$ with respect to $\mu$ (i.e., $f \in L^2(\Omega', \mu)$). Here, the square-integrability condition $f \in L^2(S, \mu)$ corresponds to the property that $f$ is almost everywhere strongly measurable and $\int_S |f|^2 \, d\mu < \infty$.

For any natural number $d$, let $\mu$ be a measure on $\text{Space } d$. For any complex-valued function $f: \text{Space } d \to \mathbb{C}$ and any subset $\Omega \subseteq \text{Space } d$, if $\Omega$ is a measurable set and $f$ is a member of the Hilbert space $L^2(\text{Space } d, \mu)$, then the indicator function $\mathbf{1}_\Omega f$ (the function equal to $f$ on $\Omega$ and $0$ elsewhere) is also a member of the Hilbert space $L^2(\text{Space } d, \mu)$.

For any natural number $d$ and any measure $\mu$ on $\text{Space } d$, let $f: \text{Space } d \to \mathbb{C}$ be a complex-valued function and $\Omega \subseteq \text{Space } d$ be a measurable set. If $f$ is square-integrable with respect to the restricted measure $\mu|_{\Omega}$ (i.e., $f \in L^2(\text{Space } d, \mu|_{\Omega})$), then the indicator function $\mathbb{1}_{\Omega} f$ (defined as $f(x)$ for $x \in \Omega$ and $0$ otherwise) is square-integrable with respect to the original measure $\mu$ (i.e., $\mathbb{1}_{\Omega} f \in L^2(\text{Space } d, \mu)$).

For any natural number $d$ and any measure $\mu$ on $\text{Space } d$, let $f, g: \text{Space } d \to \mathbb{C}$ be functions. If $f$ is a member of the Hilbert space $L^2(\text{Space } d, \mu)$, $g$ is $\mu$-almost everywhere strongly measurable, and $\|g(x)\| \leq \|f(x)\|$ for $\mu$-almost every $x$, then $g$ is also a member of the Hilbert space $L^2(\text{Space } d, \mu)$.

For any natural number $d$ and any complex-valued function $f: \text{Space } d \to \mathbb{C}$ that is square-integrable (i.e., $f$ satisfies the property $\text{MemHS}$), let $[f]$ denote the corresponding element in the Hilbert space $L^2(\text{Space } d, \mathbb{C})$. Then, the element in the Hilbert space represented by the function $-f$ is equal to the additive inverse of the element represented by $f$: $$[-f] = -[f]$$

For any dimension $d \in \mathbb{N}$ and complex-valued functions $f, g: \text{Space } d \to \mathbb{C}$, if $f$ and $g$ are square-integrable (i.e., they satisfy the property $\text{MemHS}$), then the element in the Hilbert space $L^2(\text{Space } d, \mathbb{C})$ corresponding to the function $f - g$ is equal to the difference of the elements corresponding to $f$ and $g$. That is, $$\text{mk}(f - g) = \text{mk}(f) - \text{mk}(g)$$ where $\text{mk}$ denotes the map from a square-integrable function to its equivalence class in the Hilbert space.

For any natural number $d$ and any complex-valued function $f: \text{Space } d \to \mathbb{C}$, if $f$ satisfies the condition for being in the Hilbert space (i.e., $\text{MemHS}(f)$ holds), then the function representative of the Hilbert space element constructed from $f$ is equal to $f$ almost everywhere with respect to the volume measure.

For any natural number $d$, let $L^2(\text{Space } d, \mathbb{C})$ be the Hilbert space of square-integrable functions from $\text{Space } d$ to $\mathbb{C}$. For any element $\psi$ in this Hilbert space, the function representative of the additive inverse $-\psi$ is equal to the negative of the function representative of $\psi$ almost everywhere with respect to the volume measure on $\text{Space } d$. That is, $(-\psi)(x) = -\psi(x)$ for almost every $x \in \text{Space } d$.

This linear map projects the Hilbert space $L^2(\text{Space } d, \mu)$ onto $L^2(\text{Space } d, \mu|_\Omega)$, which is the Hilbert space of square-integrable functions defined on $\text{Space } d$ with respect to the restricted measure $\mu|_\Omega$ for a given set $\Omega$. It maps an equivalence class of functions $\psi$ in the global space to its corresponding equivalence class in the restricted space, effectively disregarding the behavior of the functions on the complement of $\Omega$.

For any element $\psi$ in the Hilbert space $L^2(\text{Space } d, \mu)$, its projection into the Hilbert space $L^2(\text{Space } d, \mu|_\Omega)$ is equal to $\psi$ almost everywhere with respect to the restricted measure $\mu|_\Omega$. That is, $$\text{subspaceProjection}(\Omega, \mu, \psi) = \psi \pmod{\mu|_\Omega\text{-a.e.}}$$ where $\mu|_\Omega$ denotes the restriction of the measure $\mu$ to the set $\Omega$.

For any vector $\psi$ in the Hilbert space $L^2(\text{Space } d, \mu)$, the norm of its projection onto the Hilbert space $L^2(\text{Space } d, \mu|_\Omega)$ (where $\mu|_\Omega$ is the restriction of the measure $\mu$ to the set $\Omega$) is less than or equal to the norm of $\psi$. That is, $$\|\text{subspaceProjection}(\psi)\| \leq \|\psi\|.$$

This definition defines the complex linear isometry $$\text{subspaceIncl} : L^2(\Omega, \mu|_{\Omega}) \to L^2(\text{Space } d, \mu)$$ which includes the Hilbert space of square-integrable functions on a measurable subset $\Omega \subseteq \text{Space } d$ into the Hilbert space on the entire space. The map sends an equivalence class of functions $\psi \in L^2(\Omega, \mu|_{\Omega})$ to its extension by zero outside $\Omega$, represented by the indicator function $\mathbb{1}_{\Omega} \psi$. Here $\mu|_{\Omega}$ denotes the restriction of the measure $\mu$ to the set $\Omega$.

For a measurable subset $\Omega \subseteq \text{Space } d$ and a measure $\mu$ on $\text{Space } d$, let $\text{subspaceIncl} : L^2(\Omega, \mu|_\Omega) \to L^2(\text{Space } d, \mu)$ be the linear isometry that includes the Hilbert space on $\Omega$ into the Hilbert space on the whole space. For any $\varphi \in L^2(\Omega, \mu|_\Omega)$, the result of this inclusion is equal $\mu$-almost everywhere to the extension of $\varphi$ by zero outside $\Omega$, expressed using the indicator function as: $$\text{subspaceIncl}(\varphi) = \mathbb{1}_\Omega \varphi$$

For a measurable subset $\Omega \subseteq \text{Space } d$ and a measure $\mu$, let $\text{subspaceIncl} : L^2(\Omega, \mu|_{\Omega}) \to L^2(\text{Space } d, \mu)$ be the linear isometry that includes the Hilbert space on $\Omega$ into the global Hilbert space by extending functions as zero outside $\Omega$. Let $\text{subspaceProjection} : L^2(\text{Space } d, \mu) \to L^2(\Omega, \mu|_{\Omega})$ be the linear map that projects functions onto the restricted space. Then, $\text{subspaceProjection}$ is a left inverse of $\text{subspaceIncl}$, meaning that for any $\psi \in L^2(\Omega, \mu|_{\Omega})$, it holds that $\text{subspaceProjection}(\text{subspaceIncl}(\psi)) = \psi$.

Let $d$ be a natural number and $\mu$ a measure on $\text{Space } d$. Let $\Omega \subseteq \text{Space } d$ be a measurable subset. Let $\text{subspaceIncl} : L^2(\Omega, \mu|_\Omega) \to L^2(\text{Space } d, \mu)$ be the linear isometry that includes the Hilbert space on $\Omega$ into the global Hilbert space, and let $\text{subspaceProjection} : L^2(\text{Space } d, \mu) \to L^2(\Omega, \mu|_\Omega)$ be the linear projection onto the Hilbert space on $\Omega$. Then for any $\phi \in L^2(\Omega, \mu|_\Omega)$, it holds that $$\text{subspaceProjection}(\text{subspaceIncl}(\phi)) = \phi.$$

For a natural number $d$, a measure $\mu$ on $\text{Space } d$, and a set $\Omega \subseteq \text{Space } d$, the linear projection map $\text{subspaceProjection } \Omega \ \mu: L^2(\text{Space } d, \mu) \to L^2(\text{Space } d, \mu|_{\Omega})$ is surjective. This means that every square-integrable function in the Hilbert space defined with respect to the restricted measure $\mu|_{\Omega}$ corresponds to the projection of at least one square-integrable function from the global Hilbert space $L^2(\text{Space } d, \mu)$.