PhyslibAlpha.AlgebraicFramework.WStarAlgebra.TracePairingSurjectivity

Let $H$ be a complex Hilbert space. For any vectors $x, y \in H$, the rank-one operator $x \otimes y^*: H \to H$ (also denoted as $|x\rangle\langle y|$), defined by $z \mapsto \langle y, z \rangle x$, is a trace-class operator.

Let $H$ be a complex Hilbert space. For any vectors $x, y \in H$, this definition constructs the rank-one operator $|x\rangle\langle y|$ (defined by $z \mapsto \langle y, z \rangle x$) as an element of the trace-class Banach space $\mathcal{S}_1(H)$.

Let $H$ be a complex Hilbert space. For any bounded linear operator $A$ on $H$ and vectors $x, y \in H$, the trace pairing of $A$ with the rank-one operator $|x\rangle\langle y|$ (defined as the operator $z \mapsto \langle y, z \rangle x$) is equal to the matrix coefficient $\langle y, Ax \rangle$: $$\text{Tr}(A(|x\rangle\langle y|)) = \langle y, Ax \rangle$$

Let $H$ be a complex Hilbert space and $\mathcal{S}_1(H)$ be the Banach space of trace-class operators on $H$. The `rankOneSpan` is the $\mathbb{C}$-submodule of $\mathcal{S}_1(H)$ generated by the set of all rank-one operators. Specifically, it is the set of all finite $\mathbb{C}$-linear combinations of operators of the form $u \mapsto \langle y, u \rangle x$ for $x, y \in H$.

Let $H$ be a complex Hilbert space. Let $\mathcal{S}_1(H)$ denote the space of trace-class operators on $H$ equipped with the trace norm $\|\cdot\|_1$, and let $\mathcal{L}(H)$ denote the space of bounded linear operators on $H$. If the $\mathbb{C}$-linear span of rank-one operators is dense in $\mathcal{S}_1(H)$, then the trace pairing map $\Phi : \mathcal{L}(H) \to \mathcal{S}_1(H)^*$, which sends an operator $A \in \mathcal{L}(H)$ to the continuous linear functional $T \mapsto \text{Tr}(AT)$, is surjective. That is, every continuous linear functional $\varphi \in \mathcal{S}_1(H)^*$ can be represented as $\varphi(T) = \text{Tr}(AT)$ for some $A \in \mathcal{L}(H)$.

Given a bounded linear operator $S: H \to H$ on a Hilbert space $H$, a Hilbert basis $\{b_i\}_{i \in w}$ of $H$, and a finite subset of indices $F \subseteq w$, the Hilbert-Schmidt partial sum is the bounded linear operator defined by $$\sum_{i \in F} |S(b_i) \rangle \langle b_i |$$ which maps a vector $x \in H$ to $\sum_{i \in F} \langle b_i, x \rangle S(b_i)$.

Let $H$ be a complex Hilbert space, $S: H \to H$ be a bounded linear operator, and $\{b_i\}_{i \in w}$ be a Hilbert basis for $H$. For any finite subset of indices $F \subseteq w$ and any vector $x \in H$, the Hilbert–Schmidt partial sum operator, defined as $\sum_{i \in F} |S(b_i) \rangle \langle b_i |$, maps the vector $x$ to the sum $$\text{hilbertSchmidtPartial}(S, b, F)x = \sum_{i \in F} \langle b_i, x \rangle S(b_i)$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product.

Let $H$ be a Hilbert space and $S: H \to H$ be a bounded linear operator. Let $\{b_i\}_{i \in w}$ be a Hilbert basis for $H$, and let $F \subseteq w$ be a finite subset of indices. The Hilbert-Schmidt partial sum operator $P_F$, defined by $$P_F = \sum_{i \in F} |S b_i \rangle \langle b_i |$$ acts on a basis vector $b_j$ (where $j \in w$) as follows: $$P_F(b_j) = \begin{cases} S(b_j) & \text{if } j \in F \\ 0 & \text{if } j \notin F \end{cases}$$

Let $H$ be a Hilbert space and $S: H \to H$ be a bounded linear operator. Let $\{b_i\}_{i \in w}$ be a Hilbert basis for $H$. For any finite subset of indices $F \subseteq w$, let $S_F$ be the Hilbert–Schmidt partial sum operator defined by $S_F = \sum_{i \in F} |S b_i \rangle \langle b_i |$. Then for any index $j \in w$, the action of the operator $S - S_F$ on the basis vector $b_j$ is given by: $$(S - S_F) b_j = \begin{cases} 0 & \text{if } j \in F \\ S b_j & \text{if } j \notin F \end{cases}$$

Let $H$ be a complex Hilbert space and $S: H \to H$ be a Hilbert-Schmidt operator. Given a Hilbert basis $\{b_i\}_{i \in w}$ of $H$ and a finite subset of indices $F \subseteq w$, let $S_F$ be the Hilbert-Schmidt partial sum operator defined by $S_F = \sum_{i \in F} |S b_i \rangle \langle b_i |$. Then the operator $S - S_F$ is also a Hilbert-Schmidt operator.

Let $H$ be a complex Hilbert space with a Hilbert basis $\{b_i\}_{i \in w}$. Let $S : H \to H$ be a Hilbert–Schmidt operator and $F \subseteq w$ be a finite subset of indices. Let $S_F$ be the Hilbert–Schmidt partial sum operator defined by $S_F = \sum_{i \in F} | S b_i \rangle \langle b_i |$ (which maps $x \in H$ to $\sum_{i \in F} \langle b_i, x \rangle S b_i$). Then the sum over the basis indices of the squared norms of $(S - S_F) b_i$ is equal to the sum of the squared norms of $S b_j$ for all indices $j$ not in $F$: $$\sum_{i \in w} \|(S - S_F) b_i\|^2 = \sum_{j \in w \setminus F} \|S b_j\|^2$$

Let $H$ be a complex Hilbert space with a Hilbert basis $\{b_i\}_{i \in w}$. Given two bounded linear operators $R, S \in \mathcal{L}(H)$ and a finite subset of indices $F \subseteq w$, this definition constructs a trace-class operator in $\mathcal{S}_1(H)$ defined by the partial sum: $$\sum_{i \in F} |R(S b_i)\rangle\langle b_i|$$ where $|u\rangle\langle v|$ denotes the rank-one operator $x \mapsto \langle v, x \rangle u$.

Let $H$ be a complex Hilbert space. For any bounded linear operators $R, S \in \mathcal{L}(H)$, any Hilbert basis $\{b_i\}_{i \in w}$ of $H$, and any finite set of indices $F \subseteq w$, the operator corresponding to the trace-class partial sum $\sum_{i \in F} |R(S b_i)\rangle\langle b_i|$ is equal to the composition of $R$ with the Hilbert-Schmidt partial sum $\sum_{i \in F} |S b_i \rangle \langle b_i |$. That is, $$\sum_{i \in F} |R(S b_i)\rangle\langle b_i| = R \left( \sum_{i \in F} |S b_i \rangle \langle b_i | \right).$$

Let $H$ be a complex Hilbert space with a Hilbert basis $\{b_i\}_{i \in w}$. For any bounded linear operators $R, S \in \mathcal{L}(H)$ and any finite subset $F \subseteq w$, the trace-class operator defined by the partial sum $$\sum_{i \in F} |R(S b_i)\rangle\langle b_i|$$ belongs to the $\mathbb{C}$-linear span of rank-one operators in the trace-class space $\mathcal{S}_1(H)$. Here, $|u\rangle\langle v|$ denotes the rank-one operator mapping $x \mapsto \langle v, x \rangle u$.

Let $H$ be a complex Hilbert space. Let $T, R, S$ be bounded linear operators on $H$ such that $T$ is a trace-class operator, $R$ and $S$ are Hilbert–Schmidt operators, and $T = RS$. Let $\{b_i\}_{i \in w}$ be a Hilbert basis of $H$ and $F \subseteq w$ be a finite subset of indices. Let $T_F$ be the trace-class partial sum $\sum_{i \in F} |R(S b_i)\rangle\langle b_i|$, where $|u\rangle\langle v|$ denotes the rank-one operator $x \mapsto \langle v, x \rangle u$. Let $S_F$ be the Hilbert-Schmidt partial sum $\sum_{i \in F} |S b_i \rangle \langle b_i |$. Then the trace norm $\|\cdot\|_1$ of the difference between $T$ and $T_F$ satisfies: $$\|T - T_F\|_1 \leq \sqrt{\sum_{i \in w} \|R(b_i)\|^2} \cdot \sqrt{\sum_{i \in w} \|(S - S_F)(b_i)\|^2}$$

Let $H$ be a complex Hilbert space and $\mathcal{S}_1(H)$ be the Banach space of trace-class operators on $H$ equipped with the trace norm $\|\cdot\|_1$. The $\mathbb{C}$-linear span of rank-one operators is dense in $\mathcal{S}_1(H)$.

Let $H$ be a complex Hilbert space. Let $\mathcal{S}_1(H)$ denote the Banach space of trace-class operators on $H$ equipped with the trace norm $\|\cdot\|_1$, and let $\mathcal{L}(H)$ denote the space of bounded linear operators on $H$. The trace pairing map $\Phi: \mathcal{L}(H) \to \mathcal{S}_1(H)^*$, which maps each operator $A \in \mathcal{L}(H)$ to the continuous linear functional $\varphi_A(T) = \text{Tr}(AT)$ for $T \in \mathcal{S}_1(H)$, is surjective. That is, for every continuous linear functional $\ell \in \mathcal{S}_1(H)^*$, there exists a bounded linear operator $A \in \mathcal{L}(H)$ such that $\ell(T) = \text{Tr}(AT)$ for all $T \in \mathcal{S}_1(H)$.