## 数学代写|信息论作业代写information theory代考|Stinespring representation

Let $\mathcal{A}$ be a $C^$-algebra of bounded linear operators (but not necessarily on any specifically known complex Hilbert space). Recall from GNS representation of $\mathcal{A}$ (see Definition 2.5.1 and Proposition 2.5.2) that if there exists a ${ }^$-homomorphism $\pi: \mathcal{A} \rightarrow \mathbb{K}$ and a vector $\zeta \in \mathbb{K}$ such that $\pi(\mathcal{A}) \zeta$ spans $\mathbb{K}$ (i. e., $\operatorname{span}{\pi(\mathbf{a}) \zeta \mid \mathbf{a} \in \mathcal{A}}=\mathbb{K}$ ), then the pair ( $\pi, \mathbb{K}$ ) is called a $G N S$ representation of $\mathcal{A}$ on $\mathbb{K}$ and the vector $\zeta \in \mathbb{K}$ is called a cyclic vector.

The following result due to Stinespring [167] gives a representation of completely positive maps. Although the result holds true for any $C^*$-algebra of bounded linear operators $\mathcal{A}$ on $\mathbb{H}$, however, for representation simplicity of the material, we assume that $\mathcal{A}=\mathfrak{B}(\mathbb{H})$ below.

Theorem 4.3.1 (Stinespring [167]). A linear map $\mathrm{Y}: \mathfrak{B}(\mathbb{H}) \rightarrow \mathfrak{B}(\mathbb{H})$ is completely positive if and only if it has the form

$$Y(\mathbf{a})=\mathbf{V}^* \pi(\mathbf{a}) \mathbf{V}, \quad \forall \mathbf{a} \in \mathcal{A}$$
where $(\pi, \mathbb{K})$ is a GNS representation of $\mathfrak{B}(\mathbb{H})$ on $\mathbb{K}$ for some Hilbert space $\mathbb{K}$, and $\mathbf{V}$ is a bounded linear operator from $\mathbb{H}$ to $\mathbb{K}$. If the map $\mathrm{Y}$ is normal, then the representation $\pi$ can be chosen to be normal.

Proof. $(\Leftarrow)$ We first assume that $Y$ be a linear map of the form (4.7) and let $\left[\mathbf{a}{i j}\right]{i, j=1}^n$ be a positive matrix in $\mathfrak{B}(\mathbb{H}) \otimes \mathcal{M}n$. For all vectors $\left(u_j\right){j=1}^n$ in $\mathbb{H}$, we have then
\begin{aligned} \sum_{1 \leq i, j \leq n}\left\langle u_i, \mathrm{Y}\left(\mathbf{a}{i j}\right) u_j\right\rangle{\mathbf{H}} & =\sum_{1 \leq i, j \leq n}\left\langle u_i, \mathbf{V}^* \pi\left(\mathbf{a}{i j}\right) \mathbf{V} u_j\right\rangle{\mathrm{H}} \ & =\sum_{1 \leq i, j \leq n}\left\langle\mathbf{V} u_i, \pi\left(\mathbf{a}{i j}\right) \mathbf{V} u_j\right\rangle{\mathrm{H}} \geq 0 \end{aligned}
because $\pi: \mathfrak{B}(\mathbb{H}) \rightarrow \mathfrak{B}(\mathbb{K})$ is a *-homomorphism and, therefore, is completely positive by Proposition 4.1.2. This shows that the map $Y: \mathfrak{B}(\mathbb{H}) \rightarrow \mathfrak{B}(\mathbb{H})$ is completely positive by Proposition 4.1.6.

## 数学代写|信息论作业代写information theory代考|Kraus representation

In this section, we explore and prove the Kraus version of the Stinespring theorem (see Kraus [98]). Kraus representation provides a characterization of $\sigma$-weakly continuous (i. e., normal) completely positive maps.

The presentation of Kraus representation in this section is largely based on the lecture note by Attal [3].

Recall that a map $Y$ on a von Neumann algebra $\mathcal{A}$ is said to be normal if $\Upsilon\left(\vee_a \mathbf{a}\alpha\right)=$ $\vee\alpha Y\left(\mathbf{a}\alpha\right)$ for all increasing net $\left(\mathbf{a}\alpha\right){\alpha \in L}$ that is bounded above where $\vee\alpha(\cdot)$ denotes the least upper bound.
We first need the following preliminary result.
Lemma 4.4.1. Let $\pi$ be a normal representation of $\mathfrak{B}(\mathbb{H})$ on a separable Hilbert space $\mathbb{K}$. Then there exists a direct sum decomposition of $\mathbb{K}$,
$$\mathbb{K}=\bigoplus_{n=1}^{+\infty} \mathbb{K}n,$$ where the subspaces $\mathbb{K}_n$ are invariant under $\pi$ (i.e., $\pi\left(\mathbb{K}_n\right) \subset \mathbb{K}_n$ ) and $\left.\pi\right|{\mathrm{K}n}$, the restriction of $\pi$ to each $\mathbb{K}_n$, is unitarily equivalent to the standard representation of $\mathfrak{B}(\mathbb{H})$. Proof. If $\phi$ is a unit vector in $\mathbb{H}$, then the projection $\mathbf{P}=\pi\left(|\phi\rangle{\mathrm{H}}\langle\phi|\right)$ is nonzero. This is because if $\mathbf{U}n$ are unitary operators in $\mathfrak{B}(\mathbb{H})$ such that $\phi_n=\mathbf{U}_n \phi$ form a maximal orthonormal set in $\mathbb{H}$ and if $\mathbf{P}=\mathbf{0}$ then $$\pi\left(\left|\phi_n\right\rangle{\mathrm{H}}\left\langle\phi_n\right|\right)=\pi\left(\mathbf{U}_n\right) \mathbf{P} \pi\left(\mathbf{U}_n\right)=\mathbf{0}$$
and, by normality of $\pi$,$$\pi(\mathbf{I})=\sum_n \pi\left(\left|\phi_n\right\rangle_{\mathrm{H}}\left\langle\phi_n\right|\right)=\mathbf{0}$$
instead of being equal to $\mathrm{I}$ as it should be. Therefore, $\mathbf{P} \neq \mathbf{0}$. If $\psi$ is a unit vector in $\mathbb{K}$ such that $\mathbf{P} \psi=\psi$, then
\begin{aligned} \langle\psi, \pi(\mathbf{X}) \psi\rangle_{\mathrm{K}} & =\langle\mathbf{P} \psi, \pi(\mathbf{X}) \mathbf{P} \psi\rangle_{\mathrm{K}}=\left\langle\psi, \pi\left(|\phi\rangle_{\mathbf{H}}\langle\phi|\mathbf{X}| \phi\rangle_{\mathrm{H}}\langle\phi|\right) \psi\right\rangle_{\mathbb{K}} \ & =\langle\phi, \mathbf{X} \phi\rangle_{\mathbb{H}},\left\langle\psi, \pi\left(|\phi\rangle_{\mathbb{H}}\langle\phi|\right) \psi\right\rangle_{\mathrm{K}}=\langle\phi, \mathbf{X} \phi\rangle_{\mathbb{H}} . \end{aligned}

# 信息论代考

## 数学代写|信息论作业代写information theory代考|Stinespring representation

$$\sum_{1 \leq i, j \leq n}\left\langle u_i, \mathrm{Y}(\mathbf{a} i j) u_j\right\rangle \mathbf{H}=\sum_{1 \leq i, j \leq n}\left\langle u_i, \mathbf{V}^* \pi(\mathbf{a} i j) \mathbf{V} u_j\right\rangle \mathbf{H} \quad=\sum_{1 \leq i, j \leq n}\left\langle\mathbf{V} u_i, \pi(\mathbf{a} i j)\right.$$

## 数学代写|信息论作业代写information theory代考|Kraus representation

$$\mathbb{K}=\bigoplus_{n=1}^{+\infty} \mathbb{K} n,$$

$$\langle\psi, \pi(\mathbf{X}) \psi\rangle_{\mathrm{K}}=\langle\mathbf{P} \psi, \pi(\mathbf{X}) \mathbf{P} \psi\rangle_{\mathrm{K}}=\left\langle\psi, \pi\left(|\phi\rangle_{\mathbf{H}}\langle\phi|\mathbf{X}| \phi\rangle_{\mathrm{H}}\langle\phi|\right) \psi\right\rangle_{\mathbb{K}} \quad=\langle\phi, \mathbf{X} \phi\rangle_{\mathbb{H}}$$

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