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

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## 物理代写|量子力学代写Quantum mechanics代考|Generalized Dephasing Channels

A generalized dephasing channel is one that preserves states diagonal in some preferred orthonormal basis ${|x\rangle}$, but it can add arbitrary phases to the offdiagonal elements of a density operator represented in this basis. An isometric extension of a generalized dephasing channel acts as follows on the basis ${|x\rangle}$ :
$$U_{A \rightarrow B E}^{\mathcal{N}_{\mathrm{D}}}|x\rangle_A=|x\rangle_B\left|\varphi_x\right\rangle_E$$

where $\left|\varphi_x\right\rangle_E$ is some state for the environment (these states need not be mutually orthogonal). Thus, we can represent the isometry as follows:
$$U_{A \rightarrow B E}^{\mathcal{N}{\mathrm{D}}} \equiv \sum_x|x\rangle_B\left|\varphi_x\right\rangle_E\left\langle\left. x\right|_A\right.$$ and its action on a density operator $\rho$ is $$U^{\mathcal{N}{\mathrm{D}}} \rho\left(U^{\mathcal{N}{\mathrm{D}}}\right)^{\dagger}=\sum{x, x^{\prime}}\left\langle x|\rho| x^{\prime}\right\rangle \quad|x\rangle\left\langle\left. x^{\prime}\right|B \otimes \mid \varphi_x\right\rangle\left\langle\left.\varphi{x^{\prime}}\right|E .\right.$$ Tracing out the environment gives the action of the channel $\mathcal{N}{\mathrm{D}}$ to the receiver
$$\mathcal{N}{\mathrm{D}}(\rho)=\sum{x, x^{\prime}}\left\langle x|\rho| x^{\prime}\right\rangle\left\langle\varphi_{x^{\prime}} \mid \varphi_x\right\rangle \quad|x\rangle\left\langle\left. x^{\prime}\right|B\right.$$ where we observe that this channel preserves the diagonal components ${|x\rangle\langle x|}$ of $\rho$, but it multiplies the $d(d-1)$ off-diagonal elements of $\rho$ by arbitrary phases, depending on the $d(d-1)$ overlaps $\left\langle\varphi{x^{\prime}} \mid \varphi_x\right\rangle$ of the environment states (where $\left.x \neq x^{\prime}\right)$. Tracing out the receiver gives the action of the complementary channel $\mathcal{N}{\mathrm{D}}^c$ to the environment $$\mathcal{N}{\mathrm{D}}^c(\rho)=\sum_x\langle x|\rho| x\rangle\left|\varphi_x\right\rangle\left\langle\left.\varphi_x\right|_E .\right.$$

Quantum Hadamard channels are those whose complements are entanglementbreaking, and so generalized dephasing channels are a subclass of quantum Hadamard channels. We can write the output of a quantum Hadamard channel as the Hadamard product (element-wise multiplication) of a representation of the input density operator with another operator. To discuss how this comes about, suppose that the complementary channel $\mathcal{N}{A \rightarrow E}^c$ of a channel $\mathcal{N}{A \rightarrow B}$ is entanglement-breaking. Then, using the fact that its Kraus operators $\left|\xi_i\right\rangle_E\left\langle\left.\zeta_i\right|_A\right.$ are unit rank (see Theorem 4.6.1) and the construction in (5.36) for an isometric extension, we can write an isometric extension $U^{\mathcal{N}^c}$ for $\mathcal{N}^c$ as

\begin{aligned} U^{\mathcal{N}^c} \rho_A\left(U^{\mathcal{N}^c}\right)^{\dagger} & =\sum_{i, j}\left|\xi_i\right\rangle_E\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A\left\langle\left.\xi_j\right|_E \otimes \mid i\right\rangle_B\left\langle\left. j\right|_B\right. \ & =\sum{i, j}\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A\left|\xi_i\right\rangle_E\left\langle\left.\xi_j\right|_E \otimes \mid i\right\rangle_B\left\langle\left. j\right|_B .\right. \end{aligned} The sets $\left{\left|\xi_i\right\rangle_E\right}$ and $\left{\left|\zeta_i\right\rangle_A\right}$ each do not necessarily consist of orthonormal states, but the set $\left{|i\rangle_B\right}$ does because it is the environment of the complementary channel. Tracing over the system $E$ gives the original channel from system $A$ to $B:$ $$\mathcal{N}{A \rightarrow B}^{\mathrm{H}}\left(\rho_A\right)=\sum_{i, j}\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A\left\langle\xi_j \mid \xi_i\right\rangle_E|i\rangle_B\left\langle\left. j\right|_B .\right.$$ Let $\Sigma$ denote the matrix with elements $[\Sigma]{i, j}=\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A$, a representation of the input state $\rho$, and let $\Gamma$ denote the matrix with elements $[\Gamma]{i, j}=\left\langle\xi_i \mid \xi_j\right\rangle_E$. Then, from (5.62), it is clear that the output of the channel is the Hadamard product $*$ of $\Sigma$ and $\Gamma^{\dagger}$ with respect to the basis $\left{|i\rangle_B\right}$ :
$$\mathcal{N}{A \rightarrow B}^{\mathrm{H}}(\rho)=\Sigma * \Gamma^{\dagger} .$$ For this reason, such a channel is known as a Hadamard channel. Hadamard channels are degradable, as introduced in the following definition: DEfinition 5.2.3 (Degradable Channel) Let $\mathcal{N}{A \rightarrow B}$ be a quantum channel, and let $\mathcal{N}{A \rightarrow E}^c$ denote a complementary channel for $\mathcal{N}{A \rightarrow B}$. The channel $\mathcal{N}{A \rightarrow B}$ is degradable if there exists a degrading channel $\mathcal{D}{B \rightarrow E}$ such that
$$\mathcal{D}{B \rightarrow E}\left(\mathcal{N}{A \rightarrow B}\left(X_A\right)\right)=\mathcal{N}_{A \rightarrow E}^c\left(X_A\right),$$
for all $X_A \in \mathcal{L}\left(\mathcal{H}_A\right)$.

## 物理代写|量子力学代写Quantum mechanics代考|Generalized Dephasing Channels

$$U_{A \rightarrow B E}^{\mathcal{N}_{\mathrm{D}}}|x\rangle_A=|x\rangle_B\left|\varphi_x\right\rangle_E$$

$$U_{A \rightarrow B E}^{\mathcal{N}{\mathrm{D}}} \equiv \sum_x|x\rangle_B\left|\varphi_x\right\rangle_E\left\langle\left. x\right|A\right.$$及其对密度算子$\rho$的作用是$$U^{\mathcal{N}{\mathrm{D}}} \rho\left(U^{\mathcal{N}{\mathrm{D}}}\right)^{\dagger}=\sum{x, x^{\prime}}\left\langle x|\rho| x^{\prime}\right\rangle \quad|x\rangle\left\langle\left. x^{\prime}\right|B \otimes \mid \varphi_x\right\rangle\left\langle\left.\varphi{x^{\prime}}\right|E .\right.$$跟踪环境将通道$\mathcal{N}{\mathrm{D}}$的作用提供给接收器 $$\mathcal{N}{\mathrm{D}}(\rho)=\sum{x, x^{\prime}}\left\langle x|\rho| x^{\prime}\right\rangle\left\langle\varphi{x^{\prime}} \mid \varphi_x\right\rangle \quad|x\rangle\left\langle\left. x^{\prime}\right|B\right.$$，我们观察到该通道保留了$\rho$的对角分量${|x\rangle\langle x|}$，但它将$\rho$的$d(d-1)$非对角元素乘以任意相位，这取决于环境状态的$d(d-1)$重叠$\left\langle\varphi{x^{\prime}} \mid \varphi_x\right\rangle$(其中$\left.x \neq x^{\prime}\right)$。跟踪接收器给出了互补信道$\mathcal{N}{\mathrm{D}}^c$对环境的作用 $$\mathcal{N}{\mathrm{D}}^c(\rho)=\sum_x\langle x|\rho| x\rangle\left|\varphi_x\right\rangle\left\langle\left.\varphi_x\right|_E .\right.$$

\begin{aligned} U^{\mathcal{N}^c} \rho_A\left(U^{\mathcal{N}^c}\right)^{\dagger} & =\sum_{i, j}\left|\xi_i\right\rangle_E\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A\left\langle\left.\xi_j\right|E \otimes \mid i\right\rangle_B\left\langle\left. j\right|_B\right. \ & =\sum{i, j}\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A\left|\xi_i\right\rangle_E\left\langle\left.\xi_j\right|_E \otimes \mid i\right\rangle_B\left\langle\left. j\right|_B .\right. \end{aligned} 布景 $\left{\left|\xi_i\right\rangle_E\right}$ 和 $\left{\left|\zeta_i\right\rangle_A\right}$ 每个都不一定由正交态组成，而是集合 $\left{|i\rangle_B\right}$ 因为它是环境的互补渠道。系统跟踪 $E$ 给出来自系统的原始通道 $A$ 到 $B:$ $$\mathcal{N}{A \rightarrow B}^{\mathrm{H}}\left(\rho_A\right)=\sum{i, j}\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A\left\langle\xi_j \mid \xi_i\right\rangle_E|i\rangle_B\left\langle\left. j\right|B .\right.$$ 让 $\Sigma$ 用元素表示矩阵 $[\Sigma]{i, j}=\left\langle\left.\zeta_i\right|A \rho_A \mid \zeta_j\right\rangle_A$表示输入状态 $\rho$，让 $\Gamma$ 用元素表示矩阵 $[\Gamma]{i, j}=\left\langle\xi_i \mid \xi_j\right\rangle_E$． 然后，由式(5.62)可以清楚地看出通道的输出是Hadamard积 $*$ 的 $\Sigma$ 和 $\Gamma^{\dagger}$ 关于基底 $\left{|i\rangle_B\right}$ ： $$\mathcal{N}{A \rightarrow B}^{\mathrm{H}}(\rho)=\Sigma * \Gamma^{\dagger} .$$ 由于这个原因，这样的通道被称为Hadamard通道。Hadamard通道是可降解的，定义如下:定义5.2.3(可降解通道 $\mathcal{N}{A \rightarrow B}$ 做一个量子通道，让 $\mathcal{N}{A \rightarrow E}^c$ 表示为的互补通道 $\mathcal{N}{A \rightarrow B}$． 频道 $\mathcal{N}{A \rightarrow B}$ 如果存在降解通道，是否可降解 $\mathcal{D}{B \rightarrow E}$ 这样 $$\mathcal{D}{B \rightarrow E}\left(\mathcal{N}{A \rightarrow B}\left(X_A\right)\right)=\mathcal{N}{A \rightarrow E}^c\left(X_A\right),$$

## 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。

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

## avatest™帮您通过考试

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## 物理代写|量子力学代写Quantum mechanics代考|Depolarizing Channels

The depolarizing channel is a “worst-case scenario” channel. It assumes that we completely lose the input qubit with some probability, i.e., it replaces the lost qubit with the maximally mixed state. The map for the depolarizing channel is
$$\rho \rightarrow(1-p) \rho+p \pi$$
where $\pi$ is the maximally mixed state: $\pi=I / 2$.
Most of the time, this channel is too pessimistic. Usually, we can learn something about the physical nature of the channel by some estimation process. We should only consider using the depolarizing channel as a model if we have little to no information about the actual physical channel.

EXercise 4.7.3 (Pauli Twirl) Show that randomly applying the Pauli operators $I, X, Y, Z$ with uniform probability to any density operator gives the maximally mixed state:
$$\frac{1}{4} \rho+\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z=\pi$$
(Hint: Represent the density operator as $\rho=\left(I+r_x X+r_y Y+r_z Z\right) / 2$ and apply the commutation rules of the Pauli operators.) This is known as the “twirling” operation.

EXERCISE 4.7.4 Show that we can rewrite the depolarizing channel as the following Pauli channel:
$$\rho \rightarrow(1-3 p / 4) \rho+p\left(\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z\right) .$$
EXERCISE 4.7.5 Show that the action of a depolarizing channel on the Bloch vector is
$$\left(r_x, r_y, r_z\right) \rightarrow\left((1-p) r_x,(1-p) r_y,(1-p) r_z\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|Amplitude Damping Channels

The amplitude damping channel is an approximation to a noisy evolution that occurs in many physical systems ranging from optical systems to chains of spin$1 / 2$ particles to spontaneous emission of a photon from an atom.

In order to motivate this channel, we give a physical interpretation to our computational basis states. Let us think of the $|0\rangle$ state as the ground state of a two-level atom and let us think of the state $|1\rangle$ as the excited state of the atom. Spontaneous emission is a process that tends to decay the atom from its excited state to its ground state, even if the atom is in a superposition of the ground and excited states. Let the parameter $\gamma$ denote the probability of decay so that $0 \leq \gamma \leq 1$. One Kraus operator that captures the decaying behavior is
$$A_0=\sqrt{\gamma}|0\rangle\langle 1|$$
The operator $A_0$ annihilates the ground state:
$$A_0|0\rangle\langle 0| A_0^{\dagger}=0$$
and it decays the excited state to the ground state:
$$A_0|1\rangle\left\langle 1\left|A_0^{\dagger}=\gamma\right| 0\right\rangle\langle 0| .$$
The Kraus operator $A_0$ alone does not specify a physical map because $A_0^{\dagger} A_0=$ $\gamma|1\rangle\langle 1|$ (recall that the Kraus operators of any channel should satisfy the condition $\left.\sum_k A_k^{\dagger} A_k=I\right)$. We can satisfy this condition by choosing another operator $A_1$ such that
$$A_1^{\dagger} A_1=I-A_0^{\dagger} A_0=|0\rangle\langle 0|+(1-\gamma)| 1\rangle\langle 1|$$

## 物理代写|量子力学代写Quantum mechanics代考|Depolarizing Channels

$$\rho \rightarrow(1-p) \rho+p \pi$$

$$\frac{1}{4} \rho+\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z=\pi$$
(提示:将密度算子表示为$\rho=\left(I+r_x X+r_y Y+r_z Z\right) / 2$，并应用泡利算子的交换规则。)这就是所谓的“旋转”操作。

$$\rho \rightarrow(1-3 p / 4) \rho+p\left(\frac{1}{4} X \rho X+\frac{1}{4} Y \rho Y+\frac{1}{4} Z \rho Z\right) .$$

$$\left(r_x, r_y, r_z\right) \rightarrow\left((1-p) r_x,(1-p) r_y,(1-p) r_z\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|Amplitude Damping Channels

$$A_0=\sqrt{\gamma}|0\rangle\langle 1|$$

$$A_0|0\rangle\langle 0| A_0^{\dagger}=0$$

$$A_0|1\rangle\left\langle 1\left|A_0^{\dagger}=\gamma\right| 0\right\rangle\langle 0| .$$

## 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。