Posted on Categories:Quantum mechanics, 物理代写, 量子力学

# 物理代写|量子力学代写Quantum mechanics代考|Unitary and Isometric Channels

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 物理代写|量子力学代写Quantum mechanics代考|Unitary and Isometric Channels

Unitary evolution is a special kind of quantum channel in which there is a single Kraus operator $U \in \mathcal{L}(\mathcal{H})$, satisfying $U U^{\dagger}=U^{\dagger} U=I_{\mathcal{H}}$. Unitary channels are thus completely positive, trace-preserving, and unital. Let $\rho \in \mathcal{D}(\mathcal{H})$. Under the action of a unitary channel $\mathcal{U}$, this state evolves as
$$\mathcal{U}(\rho)=U \rho U^{\dagger}$$
where $\mathcal{U}(\rho) \in \mathcal{D}(\mathcal{H})$. Our convention henceforth is to denote a unitary channel by $\mathcal{U}$ and a unitary operator by $U$.

There is a related, but more general kind of quantum channel called an isometric quantum channel. Before defining it, we need to define the notion of a linear isometry:

DEFINition 4.6.3 (Isometry) Let $\mathcal{H}$ and $\mathcal{H}^{\prime}$ be Hilbert spaces such that $\operatorname{dim}(\mathcal{H}) \leq \operatorname{dim}\left(\mathcal{H}^{\prime}\right)$. An isometry $V$ is a linear map from $\mathcal{H}$ to $\mathcal{H}^{\prime}$ such that $V^{\dagger} V=I_{\mathcal{H}}$. Equivalently, an isometry $V$ is a linear, norm-preserving operator, in the sense that $||\psi\rangle\left|_2=\right| V|\psi\rangle |_2$ for all $|\psi\rangle \in \mathcal{H}$.

An isometry is a generalization of a unitary, because it maps between spaces of different dimensions and is thus generally rectangular and need not satisfy $V V^{\dagger}=I_{\mathcal{H}^{\prime}}$. Rather, it satisfies $V V^{\dagger}=\Pi_{\mathcal{H}^{\prime}}$, where $\Pi_{\mathcal{H}^{\prime}}$ is some projection onto $\mathcal{H}^{\prime}$, because
$$\left(V V^{\dagger}\right)\left(V V^{\dagger}\right)=V\left(V^{\dagger} V\right) V^{\dagger}=V I_{\mathcal{H}} V^{\dagger}=V V^{\dagger}$$
In later chapters, we repeatedly use the notion of an isometry.

## 物理代写|量子力学代写Quantum mechanics代考|Reversing Unitary and Isometric Channels

Suppose that we would like to reverse the action of a unitary channel $\mathcal{U}$. It is easy to do so: the adjoint map $\mathcal{U}^{\dagger}$ is a unitary channel, and by performing it after $\mathcal{U}$, we get
$$\left(\mathcal{U}^{\dagger} \circ \mathcal{U}\right)(X)=U^{\dagger} U X U^{\dagger} U=X$$
for $X \in \mathcal{L}(\mathcal{H})$.
If we would like to reverse the action of an isometric channel $\mathcal{V}$, we need to be a bit more careful. In this case, the adjoint map $\mathcal{V}^{\dagger}$ is not a channel, because it is not trace-preserving. Consider that
\begin{aligned} \operatorname{Tr}\left{\mathcal{V}^{\dagger}(Y)\right} & =\operatorname{Tr}\left{V^{\dagger} Y V\right}=\operatorname{Tr}\left{V V^{\dagger} Y\right} \ & =\operatorname{Tr}\left{\Pi_{\mathcal{H}^{\prime}} Y\right} \leq \operatorname{Tr}{Y}, \end{aligned}
for $Y \in \mathcal{L}\left(\mathcal{H}^{\prime}\right)$ and where the projection $\Pi_{\mathcal{H}^{\prime}} \equiv V V^{\dagger}$.
However, it is possible to construct a reversal channel $\mathcal{R}$ for any isometric channel $\mathcal{V}$ in the following way:
$$\mathcal{R}(Y) \equiv \mathcal{V}^{\dagger}(Y)+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \sigma,$$
where $\sigma \in \mathcal{D}(\mathcal{H})$. One can verify that the map $\mathcal{R}$ is completely positive, and it is trace-preserving because
\begin{aligned} \operatorname{Tr}{\mathcal{R}(Y)} & =\operatorname{Tr}\left{\left[\mathcal{V}^{\dagger}(Y)+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \sigma\right]\right} \ & =\operatorname{Tr}\left{\mathcal{V}^{\dagger}(Y)\right}+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \operatorname{Tr}{\sigma} \ & =\operatorname{Tr}\left{\Pi_{\mathcal{H}^{\prime}} Y\right}+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \ & =\operatorname{Tr}{Y} . \end{aligned}
Furthermore, it perfectly reverses the action of the isometric channel $\mathcal{V}$ because
\begin{aligned} (\mathcal{R} \circ \mathcal{V})(X) & =\mathcal{V}^{\dagger}(\mathcal{V}(X))+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) \mathcal{V}(X)\right} \sigma \ & =V^{\dagger} V X V^{\dagger} V+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-V V^{\dagger}\right) V X V^{\dagger}\right} \sigma \ & =X+\left[\operatorname{Tr}\left{V X V^{\dagger}\right}-\operatorname{Tr}\left{V V^{\dagger} V X V^{\dagger}\right}\right] \sigma \ & =X+\left[\operatorname{Tr}\left{V^{\dagger} V X\right}-\operatorname{Tr}\left{V^{\dagger} V V^{\dagger} V X\right}\right] \sigma \end{aligned}
\begin{aligned} & =X+[\operatorname{Tr}{X}-\operatorname{Tr}{X}] \sigma \ & =X, \end{aligned}
for $X \in \mathcal{L}(\mathcal{H})$.

## 物理代写|量子力学代写Quantum mechanics代考|Unitary and Isometric Channels

$$\mathcal{U}(\rho)=U \rho U^{\dagger}$$

$$\left(V V^{\dagger}\right)\left(V V^{\dagger}\right)=V\left(V^{\dagger} V\right) V^{\dagger}=V I_{\mathcal{H}} V^{\dagger}=V V^{\dagger}$$

## 物理代写|量子力学代写Quantum mechanics代考|Reversing Unitary and Isometric Channels

$$\left(\mathcal{U}^{\dagger} \circ \mathcal{U}\right)(X)=U^{\dagger} U X U^{\dagger} U=X$$

\begin{aligned} \operatorname{Tr}\left{\mathcal{V}^{\dagger}(Y)\right} & =\operatorname{Tr}\left{V^{\dagger} Y V\right}=\operatorname{Tr}\left{V V^{\dagger} Y\right} \ & =\operatorname{Tr}\left{\Pi_{\mathcal{H}^{\prime}} Y\right} \leq \operatorname{Tr}{Y}, \end{aligned}

$$\mathcal{R}(Y) \equiv \mathcal{V}^{\dagger}(Y)+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \sigma,$$

\begin{aligned} \operatorname{Tr}{\mathcal{R}(Y)} & =\operatorname{Tr}\left{\left[\mathcal{V}^{\dagger}(Y)+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \sigma\right]\right} \ & =\operatorname{Tr}\left{\mathcal{V}^{\dagger}(Y)\right}+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \operatorname{Tr}{\sigma} \ & =\operatorname{Tr}\left{\Pi_{\mathcal{H}^{\prime}} Y\right}+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) Y\right} \ & =\operatorname{Tr}{Y} . \end{aligned}

\begin{aligned} (\mathcal{R} \circ \mathcal{V})(X) & =\mathcal{V}^{\dagger}(\mathcal{V}(X))+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-\Pi_{\mathcal{H}^{\prime}}\right) \mathcal{V}(X)\right} \sigma \ & =V^{\dagger} V X V^{\dagger} V+\operatorname{Tr}\left{\left(I_{\mathcal{H}^{\prime}}-V V^{\dagger}\right) V X V^{\dagger}\right} \sigma \ & =X+\left[\operatorname{Tr}\left{V X V^{\dagger}\right}-\operatorname{Tr}\left{V V^{\dagger} V X V^{\dagger}\right}\right] \sigma \ & =X+\left[\operatorname{Tr}\left{V^{\dagger} V X\right}-\operatorname{Tr}\left{V^{\dagger} V V^{\dagger} V X\right}\right] \sigma \end{aligned}
\begin{aligned} & =X+[\operatorname{Tr}{X}-\operatorname{Tr}{X}] \sigma \ & =X, \end{aligned}

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。