PhyslibAlpha.ProbabilisticTheory.WStarAlgebra.TracePairingSurjectivity

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

Let $H$ be a complex Hilbert space. For any vectors $x, y \in H$, let $|x\rangle\langle y|$ be the rank-one operator defined by $z \mapsto \langle y, z \rangle x$. The trace norm $\|\cdot\|_1$ of this operator satisfies the inequality $\| |x\rangle\langle y| \|_1 \le \|x\| \|y\|$.

Let $H$ be a complex Hilbert space and $\mathcal{S}_1(H)$ denote the space of trace-class operators on $H$. The `rankOneSpan` is the $\mathbb{C}$-linear subspace of $\mathcal{S}_1(H)$ spanned by the set of all rank-one operators $|x\rangle\langle y|$, where for $x, y \in H$, the operator $|x\rangle\langle y|$ is defined by $z \mapsto \langle y, z \rangle x$.

Let $H$ be a complex Hilbert space and $\mathcal{S}_1(H)$ denote the space of trace-class operators on $H$ equipped with the trace norm. Given a continuous linear functional $\phi \in \mathcal{S}_1(H)^*$, this definition constructs a bounded sesquilinear form $B: H \times H \to \mathbb{C}$ defined by $$B(x, y) = \overline{\phi(|x\rangle\langle y|)}$$ for all $x, y \in H$. Here, $|x\rangle\langle y|$ denotes the rank-one operator in $\mathcal{S}_1(H)$ that maps a vector $z \in H$ to $\langle y, z \rangle x$, and the overline denotes complex conjugation. The form is shown to be bounded by the operator norm of the functional, $\|\phi\|$.

Let $H$ be a complex Hilbert space and $\mathcal{S}_1(H)$ denote the space of trace-class operators on $H$ equipped with the trace norm. For any continuous linear functional $\phi \in \mathcal{S}_1(H)^*$ and any vectors $x, y \in H$, the value of the associated bounded sesquilinear form $B$ at $(x, y)$ is given by $$B(x, y) = \overline{\phi(|x\rangle\langle y|)}$$ where $|x\rangle\langle y|$ is the rank-one operator $z \mapsto \langle y, z \rangle x$ and the overline denotes complex conjugation.

Let $H$ be a complex Hilbert space and $\mathcal{S}_1(H)$ be the space of trace-class operators on $H$ equipped with the trace norm. Given a continuous linear functional $\phi \in \mathcal{S}_1(H)^*$, let $A$ be the bounded linear operator on $H$ associated with the sesquilinear form $B(u, v) = \overline{\phi(|u\rangle\langle v|)}$ such that $\langle v, Au \rangle = \overline{B(u, v)}$ for all $u, v \in H$. For any vectors $x, y \in H$, the trace pairing of $A$ with the rank-one operator $|x\rangle\langle y|$ (defined by $z \mapsto \langle y, z \rangle x$) is equal to the value of the functional $\phi$ evaluated at $|x\rangle\langle y|$: $$\text{Tr}(A |x\rangle\langle y|) = \phi(|x\rangle\langle y|)$$

Let $H$ be a complex Hilbert space. If the $\mathbb{C}$-linear span of the rank-one operators $\{ |x\rangle\langle y| : x, y \in H \}$ is dense in the space of trace-class operators $\mathcal{T}(H)$ with respect to the trace norm, then the trace pairing map from the space of bounded linear operators $\mathcal{B}(H)$ to the continuous dual space $\mathcal{T}(H)^*$, defined by $A \mapsto \operatorname{Tr}(A \cdot)$, is surjective. That is, every continuous linear functional $\phi$ on the trace class can be represented as $\phi(B) = \operatorname{Tr}(AB)$ for some $A \in \mathcal{B}(H)$.

Let $H$ be a complex Hilbert space and $\mathcal{B}(H)$ be the space of bounded linear operators on $H$. Given an operator $S \in \mathcal{B}(H)$, a Hilbert basis $\{b_i\}_{i \in w}$ for $H$, and a finite subset of indices $F \subseteq w$, the Hilbert-Schmidt partial sum is the bounded linear operator defined by the sum of rank-one operators: $$\sum_{i \in F} |S b_i\rangle\langle b_i|$$ Specifically, its action on a vector $x \in H$ is given by $\sum_{i \in F} \langle b_i, x \rangle S(b_i)$, where $\langle \cdot, \cdot \rangle$ denotes the inner product on $H$.

Let $H$ be a complex Hilbert space. For a bounded linear operator $S \in \mathcal{B}(H)$, a Hilbert basis $\{b_i\}_{i \in w}$ of $H$, a finite subset of indices $F \subseteq w$, and any vector $x \in H$, the action of the operator `hilbertSchmidtPartial` is given by the formula: $$\text{hilbertSchmidtPartial}(S, b, F)x = \sum_{i \in F} \langle b_i, x \rangle S(b_i)$$ where $\langle \cdot, \cdot \rangle$ denotes the inner product on $H$ (linear in the second argument).

Let $H$ be a complex 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 $F \subseteq w$, the Hilbert-Schmidt partial sum operator $P_F = \sum_{i \in F} |S b_i\rangle\langle b_i|$ satisfies the following property for any basis vector $b_j$ ($j \in w$): $$P_F(b_j) = \begin{cases} S(b_j) & \text{if } j \in F \\ 0 & \text{if } j \notin F \end{cases}$$

Let $H$ be a complex 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: $$(S - S_F)(b_j) = \begin{cases} 0 & \text{if } j \in F \\ S(b_j) & \text{if } j \notin F \end{cases}$$ where the partial sum $S_F$ acts on a vector $x$ as $\sum_{i \in F} \langle b_i, x \rangle S(b_i)$.

Let $H$ be a complex Hilbert space and $\{b_i\}_{i \in w}$ be a Hilbert basis for $H$. Let $S: H \to H$ be a bounded linear operator. If $S$ is a Hilbert-Schmidt operator, then for any finite subset of indices $F \subseteq w$, the operator $$S - \sum_{i \in F} |S b_i\rangle\langle b_i|$$ is also a Hilbert-Schmidt operator, where $|S b_i\rangle\langle b_i|$ denotes the rank-one operator $x \mapsto \langle b_i, x \rangle S(b_i)$.

Let $H$ be a Hilbert space over $\mathbb{C}$ and $S: H \to H$ be a Hilbert-Schmidt operator. Let $\{b_i\}_{i \in w}$ be an orthonormal basis for $H$. For any finite subset of indices $F \subseteq w$, let $S_F$ denote the Hilbert-Schmidt partial sum $\sum_{i \in F} |S b_i\rangle\langle b_i|$. Then the sum over the squared norms of the operator $S - S_F$ applied to the basis vectors is equal to the sum of the squared norms of $S$ applied to the basis vectors whose indices are 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 Hilbert space over $\mathbb{C}$. Given continuous linear operators $R, S: H \to H$, a Hilbert basis $\{b_i\}_{i \in w}$ of $H$, and a finite subset of indices $F \subseteq w$, the trace-class partial sum is defined as the operator: $$\sum_{i \in F} |R(S b_i)\rangle\langle b_i|$$ where $|u\rangle\langle v|$ denotes the rank-one operator defined by $z \mapsto \langle v, z \rangle u$. This construction results in an element of the trace-class operators $\mathcal{T}(H)$.

Let $H$ be a Hilbert space over $\mathbb{C}$. For any continuous linear operator $R: H \to H$ and any vectors $x, y \in H$, let $|x\rangle\langle y|$ denote the rank-one operator defined by the map $z \mapsto \langle y, z \rangle x$ for all $z \in H$. Then, the composition of $R$ with this rank-one operator satisfies: $$R \circ (|x\rangle\langle y|) = |Rx\rangle\langle y|$$

Let $H$ be a complex Hilbert space. For any bounded linear operators $R, S: H \to H$, a Hilbert basis $\{b_i\}_{i \in w}$ of $H$, and a finite subset $F \subseteq w$, the operator underlying the trace-class partial sum satisfies: $$\sum_{i \in F} |R(S b_i)\rangle\langle b_i| = R \circ \left( \sum_{i \in F} |S b_i\rangle\langle b_i| \right)$$ where $|u\rangle\langle v|$ denotes the rank-one operator defined by $z \mapsto \langle v, z \rangle u$, and the right-hand side involves the composition of $R$ with the Hilbert-Schmidt partial sum.

Let $H$ be a complex Hilbert space. For any continuous linear operators $R, S : H \to H$, any Hilbert basis $\{b_i\}_{i \in w}$ of $H$, and any finite subset of indices $F \subseteq w$, the trace-class 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 space of trace-class operators $\mathcal{T}(H)$, where $|u\rangle\langle v|$ denotes the rank-one operator $z \mapsto \langle v, z \rangle u$.

Let $H$ be a complex Hilbert space. Let $T, R, S$ be bounded linear operators on $H$ such that $T = RS$, where $R$ and $S$ are Hilbert-Schmidt operators. Let $\{b_i\}_{i \in w}$ be a Hilbert basis of $H$, and let $F \subseteq w$ be a finite subset of indices. Define the trace-class partial sum $T_F = \sum_{i \in F} |R(S b_i)\rangle\langle b_i|$ and the Hilbert-Schmidt partial sum $S_F = \sum_{i \in F} |S b_i\rangle\langle b_i|$. Then the trace norm $\|\cdot\|_1$ of the approximation error $T - T_F$ is bounded by the product of the Hilbert-Schmidt norms of $R$ and the tail of $S$: $$\|T - T_F\|_1 \leq \sqrt{\sum_{i \in w} \|R b_i\|^2} \cdot \sqrt{\sum_{i \in w} \|(S - S_F) b_i\|^2}$$ where the sums are taken over the index set $w$ of the Hilbert basis.

Let $H$ be a complex Hilbert space and let $\mathcal{S}_1(H)$ denote the space of trace-class operators on $H$ equipped with the trace norm $\|\cdot\|_1$. The subspace of $\mathcal{S}_1(H)$ formed by the linear span of rank-one operators of the form $|x\rangle\langle y|$ (where $x, y \in H$ and $|x\rangle\langle y|(z) = \langle y, z \rangle x$) is dense in $\mathcal{S}_1(H)$.

Let $H$ be a complex Hilbert space. Let $\mathcal{B}(H)$ be the space of bounded linear operators on $H$, and let $\mathcal{T}(H)$ be the space of trace-class operators on $H$. The trace pairing map $\Phi: \mathcal{B}(H) \to \mathcal{T}(H)^*$, which sends a bounded operator $A$ to the continuous linear functional $f_A(B) = \operatorname{Tr}(AB)$ for all $B \in \mathcal{T}(H)$, is surjective. In other words, every continuous linear functional on the trace-class operators can be represented as a trace pairing with some bounded operator.