PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.ScalarMeasure

For fixed vectors $x, y$ in a Hilbert space $H$, this is the additive homomorphism from the space of continuous linear operators on $H$ (equipped with the weak operator topology) to the complex numbers $\mathbb{C}$ that maps an operator $A$ to the inner product $\langle y, Ax \rangle$.

Let $H$ be a Hilbert space and let $x, y \in H$ be fixed vectors. The map from the space of bounded linear operators on $H$ (equipped with the weak operator topology) to the complex numbers $\mathbb{C}$ given by $A \mapsto \langle y, Ax \rangle$ is continuous.

Given a weak operator topology (WOT) spectral measure $\mu$ on a measurable space $\alpha$ acting on a Hilbert space $H$, and two vectors $x, y \in H$, the **scalar measure** is the complex-valued measure on $\alpha$ defined by the mapping $S \mapsto \langle y, \mu(S) x \rangle$ for every measurable set $S \subseteq \alpha$. This measure represents the result of "testing" the spectral measure against the pair of vectors $x$ and $y$.

Let $\mu$ be a weak operator topology (WOT) spectral measure on a measurable space $\alpha$ acting on a Hilbert space $H$. For any vectors $x, y \in H$ and any measurable set $S \subseteq \alpha$, the value of the scalar measure $\mu_{x,y}$ at $S$ is given by the complex inner product $\langle y, \mu(S) x \rangle$.

Let $\alpha$ and $\beta$ be measurable spaces and $f : \alpha \to \beta$ be a measurable function. Let $\mu$ be a weak operator topology (WOT) spectral measure on $\alpha$ acting on a Hilbert space $H$. For any vectors $x, y \in H$, the scalar measure associated with the push-forward spectral measure $f_* \mu$ and the vectors $x, y$ is equal to the push-forward of the scalar measure of $\mu$ under $f$. That is, $$(f_* \mu)_{x,y} = f_* (\mu_{x,y}),$$ where $(f_* \mu)_{x,y}$ is the scalar measure $B \mapsto \langle y, (f_* \mu)(B) x \rangle$ for measurable sets $B \subseteq \beta$, and $f_* (\mu_{x,y})$ is the push-forward of the complex measure $\mu_{x,y} : S \mapsto \langle y, \mu(S) x \rangle$.

Let $H$ be a Hilbert space and $\alpha$ be a measurable space. Suppose $\mu$ and $\nu$ are two weak operator topology (WOT) spectral measures on $\alpha$ acting on $H$. If for every pair of vectors $x, y \in H$, their scalar measures are equal—that is, for every measurable set $S \subseteq \alpha$, the complex inner products satisfy $\langle y, \mu(S) x \rangle = \langle y, \nu(S) x \rangle$—then the spectral measures themselves are equal, $\mu = \nu$.

Let $H$ be a Hilbert space and $\alpha$ be a measurable space. For a weak spectral measure $\mu_S$ on $\alpha$ taking values in the bounded operators on $H$, for any measurable set $S \subseteq \alpha$ and any vector $x \in H$, the real part of the complex inner product $\langle x, \mu_S(S) x \rangle$ is non-negative: $$0 \leq \operatorname{Re}(\langle x, \mu_S(S) x \rangle).$$

Let $H$ be a Hilbert space and $\alpha$ be a measurable space. For a weak spectral measure $\mu_S$ defined on $\alpha$ with values in the operators on $H$, any set $S \subseteq \alpha$, and any vector $x \in H$, the real part of the inner product of $x$ and $\mu_S(S)x$ is equal to the squared norm of the vector $\mu_S(S)x$: $$\operatorname{Re} \langle x, \mu_S(S) x \rangle = \|\mu_S(S) x\|^2$$

Let $H$ be a Hilbert space and $\mu_S$ be a weak spectral measure on a space $\alpha$. For any set $S \subseteq \alpha$ and any vector $x \in H$, the inner product of $x$ with the image of $x$ under the measure of $S$ is equal to the inner product of $\mu_S(S) x$ with itself: $$\langle x, \mu_S(S) x \rangle = \langle \mu_S(S) x, \mu_S(S) x \rangle$$

Let $H$ be a Hilbert space and $\mu_S$ be a weak spectral measure on a space $\alpha$. For any two disjoint measurable sets $A, B \subseteq \alpha$ and any vector $x \in H$, the inner product of the vectors $\mu_S(A)x$ and $\mu_S(B)x$ is zero: $$\langle \mu_S(A)x, \mu_S(B)x \rangle = 0$$ where $\mu_S(A)$ and $\mu_S(B)$ are the operators associated with the sets $A$ and $B$ by the spectral measure.

Let $\mu_S$ be a weak spectral measure on a measurable space $\alpha$ with values in operators on a Hilbert space $H$. For any vector $x \in H$ and any sequence of pairwise disjoint measurable sets $\{f_i\}_{i \in \mathbb{N}} \subseteq \alpha$, the real part of the diagonal matrix element is countably additive: $$\text{Re} \langle x, \mu_S(\bigcup_{i \in \mathbb{N}} f_i) x \rangle = \sum_{i=0}^\infty \text{Re} \langle x, \mu_S(f_i) x \rangle$$ where $\langle \cdot, \cdot \rangle$ denotes the inner product on $H$.

Given a weak spectral measure $\mu_S$ on a measurable space $\alpha$ taking values in the operators on a Hilbert space $H$, the diagonal measure associated with a vector $x \in H$ is a positive measure on $\alpha$. For any measurable set $S \subseteq \alpha$, its value is defined as the real part of the inner product $\text{Re} \langle x, \mu_S(S) x \rangle$, which is also equal to the squared norm $\|\mu_S(S) x\|^2$.

Let $H$ be a Hilbert space and $\mu_S$ be a weak spectral measure on a measurable space $\alpha$. For any vector $x \in H$ and any measurable set $S \subseteq \alpha$, the value of the diagonal measure $\mu_x$ evaluated at $S$ is equal to the real part of the inner product of $x$ and $\mu_S(S)x$, written as: $$\mu_x(S) = \text{Re} \langle x, \mu_S(S) x \rangle$$ where the real part is mapped into the extended non-negative real numbers $[0, \infty]$.

Let $\mu_S$ be a weak spectral measure on a measurable space $\alpha$ taking values in the bounded operators on a Hilbert space $H$. For any vector $x \in H$ and any measurable set $S \subseteq \alpha$, the value of the diagonal measure $\mu_x$ at $S$ is equal to the squared norm of the vector $\mu_S(S)x$: $$\mu_x(S) = \|\mu_S(S) x\|^2$$

Let $\mu_S$ be a weak spectral measure on a measurable space $\alpha$ with values in the operators on a Hilbert space $H$. For any vector $x \in H$, let $\mu_x$ be the diagonal measure associated with $x$, which is defined for measurable sets $S \subseteq \alpha$ as $\mu_x(S) = \|\mu_S(S) x\|^2$. Then the value of this measure on the entire space $\alpha$ is equal to the squared norm of the vector $x$: $$\mu_x(\alpha) = \|x\|^2.$$

Let $\mu_S$ be a weak spectral measure on a measurable space $\alpha$ taking values in the bounded operators on a Hilbert space $H$. Let $f: \alpha \to \beta$ be a measurable function into another measurable space $\beta$. For any vector $x \in H$, let $\mu_x$ denote the diagonal measure of $x$ with respect to $\mu_S$, which satisfies $\mu_x(S) = \|\mu_S(S)x\|^2$ for measurable sets $S$. Then the diagonal measure of $x$ with respect to the pushforward spectral measure $f_* \mu_S$ is equal to the pushforward of the diagonal measure $\mu_x$ under $f$. That is, $$(f_* \mu_S)_x = f_* \mu_x$$

Let $H$ be a Hilbert space and $\mu_S$ be a weak spectral measure on a measurable space $\alpha$. For any vector $x \in H$, the diagonal measure $\mu_x$ (defined by $\mu_x(S) = \|\mu_S(S)x\|^2$ for measurable sets $S \subseteq \alpha$) is a finite measure.

Let $\mu_S$ be a weak spectral measure on a measurable space $\alpha$ taking values in the operators on a Hilbert space $H$. For any vector $x \in H$, the diagonal measure associated with the negation $-x$ is equal to the diagonal measure associated with $x$, denoted as $\mu_{-x} = \mu_x$. (The diagonal measure $\mu_x$ is the positive measure defined by $\mu_x(S) = \|\mu_S(S)x\|^2$ for measurable sets $S \subseteq \alpha$).

Let $H$ be a Hilbert space and $\mu_S$ be a weak spectral measure. For any vector $x \in H$, the diagonal measure associated with $ix$ is equal to the diagonal measure associated with $x$, where $i$ is the imaginary unit. That is, $\mu_{ix} = \mu_x$.

Let $\mu_S$ be a weak spectral measure on a measurable space $\alpha$ taking values in the operators on a Hilbert space $H$. For any vector $x \in H$, let $\mu_x$ denote the diagonal measure of $x$, defined such that for any measurable set $S \subseteq \alpha$, $\mu_x(S) = \|\mu_S(S)x\|^2$. Then for any complex scalar $c \in \mathbb{C}$ and vector $x \in H$, the diagonal measure of the scaled vector $cx$ satisfies: $$\mu_{cx} = |c|^2 \mu_x$$ where $|c|$ denotes the absolute value of the complex number $c$.

Let $\mu_S$ be a weak spectral measure on a measurable space $\alpha$ taking values in the operators on a Hilbert space $H$. For any vector $x \in H$, let $\mu_x$ denote the diagonal measure associated with $x$, which is defined for any measurable set $S \subseteq \alpha$ as $\mu_x(S) = \|\mu_S(S) x\|^2 = \text{Re} \langle x, \mu_S(S) x \rangle$. For any two vectors $x, y \in H$, the diagonal measures satisfy the parallelogram identity: $$\mu_{x+y} + \mu_{x-y} = 2\mu_x + 2\mu_y$$