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|PHYS402 Formal Development of Perturbation Expansion

avatest.org™量子力学Quantum mechanics代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.org™， 最高质量的量子力学Quantum mechanics作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此量子力学Quantum mechanics作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

avatest™ 为您的留学生涯保驾护航 在网课代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的网课代写服务。我们的专家在量子力学Quantum mechanics代写方面经验极为丰富，各种量子力学Quantum mechanics相关的作业也就用不着 说。

## 物理代写|量子力学代写Quantum mechanics代考|Formal Development of Perturbation Expansion

We now state in more precise terms the basic problem we wish to solve. Suppose we know completely and exactly the energy eigenkets and energy eigenvalues of the unperturbed Hamiltonian $H_{0}$, that is
$$H_{0}\left|n^{(0)}\right\rangle=E_{n}^{(0)}\left|n^{(0)}\right\rangle .$$
The set $\left{\left|n^{(0)}\right\rangle\right}$ is complete in the sense that the closure relation $1=\sum_{n}\left|n^{(0)}\right\rangle\left\langle n^{(0)}\right|$ holds. Furthermore, we assume here that the energy spectrum is nondegenerate; in the next section we will relax this assumption. We are interested in obtaining the energy eigenvalues and eigenkets for the problem defined by (5.4). To be consistent with (5.18) we should write $(5.4)$ as
$$\left(H_{0}+\lambda V\right)|n\rangle_{\lambda}=E_{n}^{(\lambda)}|n\rangle_{\lambda}$$
to denote the fact that the energy eigenvalues $E_{n}^{(\lambda)}$ and energy eigenkets $|n\rangle_{\lambda}$ are functions of the continuous parameter $\lambda$; however, we will usually dispense with this correct but more cumbersome notation.

As the continuous parameter $\lambda$ is increased from zero, we expect the energy eigenvalue $E_{n}$ for the $n$th eigenket to depart from its unperturbed value $E_{n}^{(0)}$, so we define the energy shift for the $n$th level as follows:
$$\Delta_{n} \equiv E_{n}-E_{n}^{(0)} .$$
The basic Schrödinger equation to be solved (approximately) is
$$\left(E_{n}^{(0)}-H_{0}\right)|n\rangle=\left(\lambda V-\Delta_{n}\right)|n\rangle .$$
We may be tempted to invert the operator $E_{n}^{(0)}-H_{0}$; however, in general, the inverse operator $1 /\left(E_{n}^{(0)}-H_{0}\right)$ is ill defined because it may act on $\left|n^{(0)}\right\rangle$. Fortunately in our case $\left(\lambda V-\Delta_{n}\right)|n\rangle$ has no component along $\left|n^{(0)}\right\rangle$, as can easily be seen by multiplying both sides of $(5.21)$ by $\left\langle n^{(0)}\right|$ on the left:
$$\left\langle n^{(0)}\left|\left(\lambda V-\Delta_{n}\right)\right| n\right\rangle=0 .$$

## 物理代写|量子力学代写Quantum mechanics代考|Wave Function Renormalization

We are in a position to look at the normalization of the perturbed ket. Recalling the normalization convention we use, (5.31), we see that the perturbed ket $|n\rangle$ is not normalized in the usual manner. We can renormalize the perturbed ket by defining
$$|n\rangle_{N}=Z_{n}^{1 / 2}|n\rangle,$$
where $Z_{n}$ is simply a constant with ${ }{N}\langle n \mid n\rangle{N}=1$. Multiplying $\left\langle n^{(0)}\right|$ on the left we obtain [because of (5.31)]
$$Z_{n}^{1 / 2}=\left\langle n^{0} \mid n\right\rangle_{N} .$$
What is the physical meaning of $Z_{n}$ ? Because $|n\rangle_{N}$ satisfies the usual normalization requirement (5.30), $Z_{n}$ can be regarded as the probability for the perturbed energy eigenstate to be found in the corresponding unperturbed energy eigenstate. Noting
$${ }{N}\langle n \mid n\rangle{N}=Z_{n}\langle n \mid n\rangle=1,$$
we have
\begin{aligned} Z_{n}^{-1}=&\langle n \mid n\rangle=\left(\left\langle n^{(0)}\right|+\lambda\left\langle n^{(1)}\right|+\lambda^{2}\left\langle n^{(2)}\right|+\cdots\right) \ & \times\left(\left|n^{(0)}\right\rangle+\lambda\left|n^{(1)}\right\rangle+\lambda^{2}\left|n^{(2)}\right\rangle+\cdots\right) \ =& 1+\lambda^{2}\left\langle n^{(1)} \mid n^{(1)}\right\rangle+0\left(\lambda^{3}\right) \ =& 1+\lambda^{2} \sum_{k \neq n} \frac{\left|V_{k n}\right|^{2}}{\left(E_{n}^{(0)}-E_{k}^{(0)}\right)^{2}}+0\left(\lambda^{3}\right), \end{aligned}
so up to order $\lambda^{2}$, we get for the probability of the perturbed state to be found in the corresponding unperturbed state
$$Z_{n} \simeq 1-\lambda^{2} \sum_{k \neq n} \frac{\left|V_{k n}\right|^{2}}{\left(E_{n}^{0}-E_{k}^{0}\right)^{2}} .$$
The second term in (5.48b) is to be understood as the probability for “leakage” to states other than $\left|n^{(0)}\right\rangle$. Notice that $Z_{n}$ is less than 1, as expected on the basis of the probability interpretation for $Z$.

## 物理代写|量子力学代写Quantum mechanics代考|Formal Development of Perturbation Expansion

$$H_{0}\left|n^{(0)}\right\rangle=E_{n}^{(0)}\left|n^{(0)}\right\rangle .$$

$$\left(H_{0}+\lambda V\right)|n\rangle_{\lambda}=E_{n}^{(\lambda)}|n\rangle_{\lambda}$$

$$\Delta_{n} \equiv E_{n}-E_{n}^{(0)} .$$

$$\left(E_{n}^{(0)}-H_{0}\right)|n\rangle=\left(\lambda V-\Delta_{n}\right)|n\rangle .$$

$$\left\langle n^{(0)}\left|\left(\lambda V-\Delta_{n}\right)\right| n\right\rangle=0 .$$

## 物理代写|量子力学代写Quantum mechanics代考|Wave Function Renormalization

$$|n\rangle_{N}=Z_{n}^{1 / 2}|n\rangle,$$

$$Z_{n}^{1 / 2}=\left\langle n^{0} \mid n\right\rangle_{N} .$$

$$N\langle n \mid n\rangle N=Z_{n}\langle n \mid n\rangle=1,$$

$$Z_{n}^{-1}=\langle n \mid n\rangle=\left(\left\langlen ^ { ( 0 ) } \left|+\lambda\left\langle n^{(1)}\left|+\lambda^{2}\left\langle n^{(2)}\right|+\cdots\right) \quad \times\left(\left|n^{(0)}\right\rangle+\lambda\left|n^{(1)}\right\rangle+\lambda^{2}\left|n^{(2)}\right\rangle+\cdots\right)=1+\lambda^{2}\left\langle n^{(1)} \mid n^{(1)}\right\rangle+0\left(\lambda^{3}\right)=\quad 1+\lambda^{2} \sum_{k \neq n}\right.\right.\right.\right.$$

$$Z_{n} \simeq 1-\lambda^{2} \sum_{k \neq n} \frac{\left|V_{k n}\right|^{2}}{\left(E_{n}^{0}-E_{k}^{0}\right)^{2}} .$$
(5.48b) 中的第二项应理解为“泄漏”到除 $\left|n^{(0)}\right\rangle$. 请注意 $Z_{n}$ 小于 1 ，正如在概率解释的基础上所预期的那样 $Z$.

## 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

## 物理代写|量子力学代写Quantum mechanics代考|PHYSICS3544 Time-Independent Perturbation Theory: Nondegenerate Case

avatest.org™量子力学Quantum mechanics代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.org™， 最高质量的量子力学Quantum mechanics作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此量子力学Quantum mechanics作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

avatest™ 为您的留学生涯保驾护航 在网课代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的网课代写服务。我们的专家在量子力学Quantum mechanics代写方面经验极为丰富，各种量子力学Quantum mechanics相关的作业也就用不着 说。

## 物理代写|量子力学代写Quantum mechanics代考|Statement of the Problem

The approximation method we consider here is time-independent perturbation theory, sometimes known as the Rayleigh-Schrödinger perturbation theory. We consider a timeindependent Hamiltonian $H$ such that it can be split into two parts, namely,
$$H=H_{0}+V,$$
where the $V=0$ problem is assumed to have been solved in the sense that both the exact energy eigenkets $\left|n^{(0)}\right\rangle$ and the exact energy eigenvalues $E_{n}^{(0)}$ are known:
$$H_{0}\left|n^{(0)}\right\rangle=E_{n}^{(0)}\left|n^{(0)}\right\rangle .$$
We are required to find approximate eigenkets and eigenvalues for the full Hamiltonian problem
$$\left(H_{0}+V\right)|n\rangle=E_{n}|n\rangle,$$
where $V$ is known as the perturbation; it is not, in general, the full-potential operator. For example, suppose we consider the hydrogen atom in an external electric or magnetic field. The unperturbed Hamiltonian $H_{0}$ is taken to be the kinetic energy $\mathbf{p}^{2} / 2 \mathrm{~m}$ and the Coulomb potential due to the presence of the proton nucleus $-e^{2} / r$. Only that part of the potential due to the interaction with the external $\mathbf{E}$ or $\mathbf{B}$ field is represented by the perturbation $V$.

## 物理代写|量子力学代写Quantum mechanics代考|The Two-State Problem

Before we embark on a systematic presentation of the basic method, let us see how the expansion in $\lambda$ might indeed be valid in the exactly soluble two-state problem we have encountered many times already. Suppose we have a Hamiltonian that can be written as
$$H=E_{1}^{(0)}\left|1^{(0)}\right\rangle\left\langle 1^{(0)}\left|+E_{2}^{(0)}\right| 2^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{12}\right| 1^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{21}\right| 2^{(0)}\right\rangle\left\langle 1^{(0)}\right|,$$
where $\left|1^{(0)}\right\rangle$ and $\left|2^{(0)}\right\rangle$ are the energy eigenkets for the $\lambda=0$ problem, and we consider the case $V_{11}=V_{22}=0$. In this representation the $H$ may be represented by a square matrix as follows:
$$H=\left(\begin{array}{ll} E_{1}^{(0)} & \lambda V_{12} \ \lambda V_{21} & E_{2}^{(0)} \end{array}\right),$$
where we have used the basis formed by the unperturbed energy eigenkets. The $V$ matrix must, of course, be Hermitian; let us solve the case when $V_{12}$ and $V_{21}$ are real:
$$V_{12}=V_{12}^{}, \quad V_{21}=V_{21}^{} ;$$
hence, by Hermiticity
$$V_{12}=V_{21} \text {. }$$

## 物理代写|量子力学代写Quantum mechanics代考|Statement of the Problem

$$H=H_{0}+V,$$

$$H_{0}\left|n^{(0)}\right\rangle=E_{n}^{(0)}\left|n^{(0)}\right\rangle .$$

$$\left(H_{0}+V\right)|n\rangle=E_{n}|n\rangle,$$

## 物理代写|量子力学代写Quantum mechanics代考|The Two-State Problem

$$H=E_{1}^{(0)}\left|1^{(0)}\right\rangle\left\langle 1^{(0)}\left|+E_{2}^{(0)}\right| 2^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{12}\right| 1^{(0)}\right\rangle\left\langle 2^{(0)}\left|+\lambda V_{21}\right| 2^{(0)}\right\rangle\left\langle 1^{(0)}\right|,$$

$$H=\left(\begin{array}{lll} E_{1}^{(0)} & \lambda V_{12} \lambda V_{21} & E_{2}^{(0)} \end{array}\right)$$

$$V_{12}=V_{12}, \quad V_{21}=V_{21}$$

$$V_{12}=V_{21} .$$

## 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

## 物理代写|量子力学代考Quantum mechanics代考|PHYS519 The Schrödinger Versus the Heisenberg Picture

avatest.org™量子力学Quantum mechanics代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.org™， 最高质量的量子力学Quantum mechanics作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此量子力学Quantum mechanics作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

my-assignmentexpert™ 为您的留学生涯保驾护航 在网课代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的网课代写服务。我们的专家在量子力学Quantum mechanics代写方面经验极为丰富，各种量子力学Quantum mechanics相关的作业也就用不着 说。

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Unitary Operators

In the previous section we introduced the concept of time development by considering the time-evolution operator that affects state kets; that approach to quantum dynamics is known as the Schrödinger picture. There is another formulation of quantum dynamics where observables, rather than state kets, vary with time; this second approach is known as the Heisenberg picture. Before discussing the differences between the two approaches in detail, we digress to make some general comments on unitary operators.

Unitary operators are used for many different purposes in quantum mechanics. In this book we introduced (Section 1.5) an operator satisfying the unitarity property. In that section we were concerned with the question of how the base kets in one representation are related to those in some other representations. The state kets themselves are assumed not to change as we switch to a different set of base kets even though the numerical values of the expansion coefficients for $|\alpha\rangle$ are, of course, different in different representations.

Subsequently we introduced two unitary operators that actually change the state kets, the translation operator of Section $1.6$ and the time-evolution operator of Section 2.1. We have
$$|\alpha\rangle \rightarrow U|\alpha\rangle,$$
where $U$ may stand for $\mathscr{T}(d \mathbf{x})$ or $\mathscr{U}\left(t, t_{0}\right)$. Here $U|\alpha\rangle$ is the state ket corresponding to a physical system that actually has undergone translation or time evolution.

It is important to keep in mind that under a unitary transformation that changes the state kets, the inner product of a state bra and a state ket remains unchanged:
$$\langle\beta \mid \alpha\rangle \rightarrow\left\langle\beta\left|U^{\dagger} U\right| \alpha\right\rangle=\langle\beta \mid \alpha\rangle .$$
Using the fact that these transformations affect the state kets but not operators, we can infer how $\langle\beta|X| \alpha\rangle$ must change:
$$\langle\beta|X| \alpha\rangle \rightarrow\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|U^{\dagger} X U\right| \alpha\right\rangle$$
We now make a very simple mathematical observation that follows from the associative axiom of multiplication.
$$\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|\cdot\left(U^{\dagger} X U\right) \cdot\right| \alpha\right\rangle$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|State Kets and Observables in the Schrödinger and the Heisenberg Pictures

We now return to the time-evolution operator $\mathscr{U}\left(t, t_{0}\right)$. In the previous section we examined how state kets evolve with time. This means that we were following approach 1 , known as the Schrödinger picture when applied to time evolution. Alternatively we may follow approach 2, known as the Heisenberg picture when applied to time evolution.

In the Schrödinger picture the operators corresponding to observables like $x, p_{y}$, and $S_{z}$ are fixed in time, while state kets vary with time, as indicated in the previous section. In contrast, in the Heisenberg picture the operators corresponding to observables vary with time; the state kets are fixed, frozen so to speak, at what they were at $t_{0}$. It is convenient to set $t_{0}$ in $\mathscr{U}\left(t, t_{0}\right)$ to zero for simplicity and work with $\mathscr{U}(t)$, which is defined by
$$\mathscr{U}\left(t, t_{0}=0\right) \equiv \mathscr{U}(t)=\exp \left(\frac{-i H t}{\hbar}\right) .$$
Motivated by (2.79b) of approach 2, we define the Heisenberg picture observable by
$$A^{(H)}(t) \equiv \mathscr{U}^{\dagger}(t) A^{(S)} \mathscr{U}(t)$$
where the superscripts $H$ and $S$ stand for Heisenberg and Schrödinger, respectively. At $t=0$, the Heisenberg picture observable and the corresponding Schrödinger picture observable coincide:
$$A^{(H)}(0)=A^{(S)}$$
The state kets also coincide between the two pictures at $t=0$; at later $t$ the Heisenberg picture state ket is frozen to what it was at $t=0$ :
$$\left|\alpha, t_{0}=0 ; t\right\rangle_{H}=\left|\alpha, t_{0}=0\right\rangle$$
independent of $t$. This is in dramatic contrast with the Schrödinger picture state ket,
$$\left|\alpha, t_{0}=0 ; t\right\rangle_{S}=\mathscr{U}(t)\left|\alpha, t_{0}=0\right\rangle$$
The expectation value $\langle A\rangle$ is obviously the same in both pictures:
$s\left\langle\alpha, t_{0}=0 ; t\left|A^{(S)}\right| \alpha, t_{0}=0 ; t\right\rangle_{S}=\left\langle\alpha, t_{0}=0\left|\mathscr{U}^{\dagger} A^{(S)} \mathscr{U}\right| \alpha, t_{0}=0\right\rangle$
$={ }{H}\left\langle\alpha, t{0}=0 ; t\left|A^{(H)}(t)\right| \alpha, t_{0}=0 ; t\right\rangle_{H} .$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Unitary Operators

$$|\alpha\rangle \rightarrow U|\alpha\rangle$$

$$\langle\beta \mid \alpha\rangle \rightarrow\left\langle\beta\left|U^{\dagger} U\right| \alpha\right\rangle=\langle\beta \mid \alpha\rangle .$$

$$\langle\beta|X| \alpha\rangle \rightarrow\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|U^{\dagger} X U\right| \alpha\right\rangle$$

$$\left(\langle\beta| U^{\dagger}\right) \cdot X \cdot(U|\alpha\rangle)=\left\langle\beta\left|\cdot\left(U^{\dagger} X U\right) \cdot\right| \alpha\right\rangle$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|State Kets and Observables in the Schrödinger and the Heisenberg Pictures

$$\mathscr{U}\left(t, t_{0}=0\right) \equiv \mathscr{U}(t)=\exp \left(\frac{-i H t}{\hbar}\right) .$$

$$A^{(H)}(t) \equiv \mathscr{U}^{\dagger}(t) A^{(S)} \mathscr{U}(t)$$

$$A^{(H)}(0)=A^{(S)}$$

$$\left|\alpha, t_{0}=0 ; t\right\rangle_{H}=\left|\alpha, t_{0}=0\right\rangle$$

$$\left|\alpha, t_{0}=0 ; t\right\rangle_{S}=\mathscr{U}(t)\left|\alpha, t_{0}=0\right\rangle$$

\begin{aligned} &s\left\langle\alpha, t_{0}=0 ; t\left|A^{(S)}\right| \alpha, t_{0}=0 ; t\right\rangle_{S}=\left\langle\alpha, t_{0}=0\left|\mathscr{U}^{\dagger} A^{(S)} \mathscr{U}\right| \alpha, t_{0}=0\right\rangle \ &=H\left\langle\alpha, t 0=0 ; t\left|A^{(H)}(t)\right| \alpha, t_{0}=0 ; t\right\rangle_{H} . \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。

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

## 物理代写|量子力学代考Quantum mechanics代考|PHY402 Neutrino Oscillations

avatest.org™量子力学Quantum mechanics代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.org™， 最高质量的量子力学Quantum mechanics作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此量子力学Quantum mechanics作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

my-assignmentexpert™ 为您的留学生涯保驾护航 在网课代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的网课代写服务。我们的专家在量子力学Quantum mechanics代写方面经验极为丰富，各种量子力学Quantum mechanics相关的作业也就用不着 说。

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Neutrino Oscillations

A lovely example of quantum-mechanical dynamics leading to interference in a two-state system, based on current physics research, is provided by the phenomenon known as neutrino oscillations. ${ }^{3}$

Neutrinos are elementary particles with no charge, and very small mass, much smaller than that of an electron. They are known to occur in nature in three distinct “flavors,”although for this discussion it suffices to only consider two of them. These two flavors are identified by their interactions which may be either with electrons, in which case we write $v_{e}$, or with muons, that is $v_{\mu}$. These are in fact eigenstates of a Hamiltonian which controls those interactions.

On the other hand, it is possible (and, in fact, now known to be true) that neutrinos may have some other interactions, in which case their energy eigenvalues correspond to states that have a well-defined mass. These “mass eigenstates” would have eigenvalues $E_{1}$ and $E_{2}$, say, corresponding to masses $m_{1}$ and $m_{2}$, and might be denoted as $\left|v_{1}\right\rangle$ and $\left|v_{2}\right\rangle$. The “flavor eigenstates” are related to these through a simple unitary transformation, specified by some mixing angle $\theta$, as follows:
\begin{aligned} &\left|v_{e}\right\rangle=\cos \theta\left|v_{1}\right\rangle-\sin \theta\left|v_{2}\right\rangle \ &\left|v_{\mu}\right\rangle=\sin \theta\left|v_{1}\right\rangle+\cos \theta\left|v_{2}\right\rangle . \end{aligned}
If the mixing angle were zero, then $\left|v_{e}\right\rangle$ and $\left|v_{\mu}\right\rangle$ would respectively be the same as $\left|v_{1}\right\rangle$ and $\left|v_{2}\right\rangle$. However, we know of no reason why this should be the case. Indeed, there is no strong theoretical bias for any particular value of $\theta$, and it is a free parameter which, today, can only be determined through experiment.

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Correlation Amplitude and the Energy-Time Uncertainty Relation

We conclude this section by asking how state kets at different times are correlated with each other. Suppose the initial state ket at $t=0$ of a physical system is given by $|\alpha\rangle$. With time it changes into $\left|\alpha, t_{0}=0 ; t\right\rangle$, which we obtain by applying the time-evolution operator. We are concerned with the extent to which the state ket at a later time $t$ is similar to the state ket at $t=0$; we therefore construct the inner product between the two state kets at different times:
\begin{aligned} C(t) & \equiv\left\langle\alpha \mid \alpha, t_{0}=0 ; t\right\rangle \ &=\langle\alpha|\mathscr{U}(t, 0)| \alpha\rangle \end{aligned}
which is known as the correlation amplitude. The modulus of $C(t)$ provides a quantitative measure of the “resemblance” between the state kets at different times.

As an extreme example, consider the very special case where the initial ket $|\alpha\rangle$ is an eigenket of $H$; we then have
$$C(t)=\left\langle a^{\prime} \mid a^{\prime}, t_{0}=0 ; t\right\rangle=\exp \left(\frac{-i E_{\alpha^{\prime}} t}{\hbar}\right),$$
so the modulus of the correlation amplitude is unity at all times, which is not surprising for a stationary state. In the more general situation where the initial ket is represented by a superposition of $\left{\left|a^{\prime}\right\rangle\right}$, as in (2.37), we have
\begin{aligned} C(t) &=\left(\sum_{a^{\prime}} c_{a^{\prime}}^{*}\left\langle a^{\prime}\right|\right)\left[\sum_{a^{\prime \prime}} c_{a^{\prime \prime}} \exp \left(\frac{-i E_{a^{\prime \prime}} t}{\hbar}\right)\left|a^{\prime \prime}\right\rangle\right] \ &=\sum_{a^{\prime}}\left|c_{a^{\prime}}\right|^{2} \exp \left(\frac{-i E_{a^{\prime}} t}{\hbar}\right) . \end{aligned}
As we sum over many terms with oscillating time dependence of different frequencies, a strong cancellation is possible for moderately large values of $t$. We expect the correlation amplitude that starts with unity at $t=0$ to decrease in magnitude with time.

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Neutrino Oscillations

$$\left|v_{e}\right\rangle=\cos \theta\left|v_{1}\right\rangle-\sin \theta\left|v_{2}\right\rangle \quad\left|v_{\mu}\right\rangle=\sin \theta\left|v_{1}\right\rangle+\cos \theta\left|v_{2}\right\rangle .$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Correlation Amplitude and the Energy-Time Uncertainty Relation

$$C(t) \equiv\left\langle\alpha \mid \alpha, t_{0}=0 ; t\right\rangle \quad=\langle\alpha|\mathscr{U}(t, 0)| \alpha\rangle$$

$$C(t)=\left\langle a^{\prime} \mid a^{\prime}, t_{0}=0 ; t\right\rangle=\exp \left(\frac{-i E_{\alpha^{\prime}} t}{\hbar}\right)$$

$$C(t)=\left(\sum_{a^{\prime}} c_{a^{\prime}}^{*}\left\langle a^{\prime}\right|\right)\left[\sum_{a^{\prime \prime}} c_{a^{\prime \prime}} \exp \left(\frac{-i E_{a^{\prime \prime}} t}{\hbar}\right)\left|a^{\prime \prime}\right\rangle\right] \quad=\sum_{a^{\prime}}\left|c_{a^{\prime}}\right|^{2} \exp \left(\frac{-i E_{a^{\prime}} t}{\hbar}\right)$$

## 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

## 物理代写|量子力学代考Quantum mechanics代考|PHY401 Time Dependence of Expectation Values

avatest.org™量子力学Quantum mechanics代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。avatest.org™， 最高质量的量子力学Quantum mechanics作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于统计Statistics作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此量子力学Quantum mechanics作业代写的价格不固定。通常在经济学专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。

my-assignmentexpert™ 为您的留学生涯保驾护航 在网课代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的网课代写服务。我们的专家在量子力学Quantum mechanics代写方面经验极为丰富，各种量子力学Quantum mechanics相关的作业也就用不着 说。

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Time Dependence of Expectation Values

It is instructive to study how the expectation value of an observable changes as a function of time. Suppose that at $t=0$ the initial state is one of the eigenstates of anservable $A$ that commutes with $H$, as in (2.40). We now look at the expectation value of some other observable $B$, which need not commute with $A$ nor with $H$. Because at a later time we have
$$\left|a^{\prime}, t_{0}=0 ; t\right\rangle=\mathscr{U}(t, 0)\left|a^{\prime}\right\rangle$$
for the state ket, $\langle B\rangle$ is given by
\begin{aligned} \langle B\rangle &=\left(\left\langle a^{\prime}\right| \mathscr{U}^{\dagger}(t, 0)\right) \cdot B \cdot\left(\mathscr{U}(t, 0)\left|a^{\prime}\right\rangle\right) \ &=\left\langle a^{\prime}\left|\exp \left(\frac{i E_{a^{\prime}} t}{\hbar}\right) B \exp \left(\frac{-i E_{a^{\prime}} t}{\hbar}\right)\right| a^{\prime}\right\rangle \ &=\left\langle a^{\prime}|B| a^{\prime}\right\rangle \end{aligned}
which is independent of $t$. So the expectation value of an observable taken with respect to an energy eigenstate does not change with time. For this reason an energy eigenstate is often referred to as a stationary state.

The situation is more interesting when the expectation value is taken with respect to a superposition of energy eigenstates, or a nonstationary state. Suppose that initially we have
$$\left|\alpha, t_{0}=0\right\rangle=\sum_{a^{\prime}} c_{\alpha^{\prime}}\left|a^{\prime}\right\rangle .$$
We easily compute the expectation value of $B$ to be
\begin{aligned} \langle B\rangle &=\left[\sum_{a^{\prime}} c_{\alpha^{\prime}}^{}\left\langle a^{\prime}\right| \exp \left(\frac{i E_{a^{\prime}} t}{\hbar}\right)\right] \cdot B \cdot\left[\sum_{a^{\prime \prime}} c_{d^{\prime \prime}} \exp \left(\frac{-i E_{a^{\prime \prime}} t}{\hbar}\right)\left|a^{\prime \prime}\right\rangle\right] \ &=\sum_{d^{\prime}} \sum_{d^{\prime \prime}} c_{d^{\prime}}^{} c_{d^{\prime \prime}}\left\langle a^{\prime}|B| a^{\prime \prime}\right\rangle \exp \left[\frac{-i\left(E_{d^{\prime \prime}}-E_{d^{\prime}}\right) t}{\hbar}\right] \end{aligned}
So this time the expectation value consists of oscillating terms whose angular frequencies are determined by N. Bohr’s frequency condition
$$\omega_{d^{\prime \prime} \alpha^{\prime}}=\frac{\left(E_{a^{\prime \prime}}-E_{\alpha^{\prime}}\right)}{\hbar} .$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Spin Precession

It is appropriate to treat an example here. We consider an extremely simple system which, however, illustrates the basic formalism we have developed.

We start with a Hamiltonian of a spin $\frac{1}{2}$ system with magnetic moment $e \hbar / 2 m_{e} c$ subjected to an external magnetic field $\mathbf{B}$ :
$$H=-\left(\frac{e}{m_{e} c}\right) \mathbf{S} \cdot \mathbf{B}$$
( $e<0$ for the electron). Furthermore, we take $\mathbf{B}$ to be a static, uniform magnetic field in the $z$-direction. We can then write $H$ as
$$H=-\left(\frac{e B}{m_{e} c}\right) S_{z} .$$
Because $S_{z}$ and $H$ differ just by a multiplicative constant, they obviously commute. The $S_{z}$ eigenstates are also energy eigenstates, and the corresponding energy eigenvalues are
$$E_{\pm}=\mp \frac{e \hbar B}{2 m_{e} c}, \text { for } S_{z} \pm .$$
It is convenient to define $\omega$ in such a way that the difference in the two energy eigenvalues is $\hbar \omega$ :
$$\omega \equiv \frac{|e| B}{m_{e} c}$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Time Dependence of Expectation Values

$$\left|a^{\prime}, t_{0}=0 ; t\right\rangle=\mathscr{U}(t, 0)\left|a^{\prime}\right\rangle$$

$$\langle B\rangle=\left(\left\langle a^{\prime}\right| \mathscr{U}^{\dagger}(t, 0)\right) \cdot B \cdot\left(\mathscr{U}(t, 0)\left|a^{\prime}\right\rangle\right) \quad=\left\langle a^{\prime}\left|\exp \left(\frac{i E_{a^{\prime}} t}{\hbar}\right) B \exp \left(\frac{-i E_{a^{\prime}} t}{\hbar}\right)\right| a^{\prime}\right\rangle=\left\langle a^{\prime}|B| a^{\prime}\right\rangle$$

$$\left|\alpha, t_{0}=0\right\rangle=\sum_{a^{\prime}} c_{\alpha^{\prime}}\left|a^{\prime}\right\rangle$$

$$\langle B\rangle=\left[\sum_{a^{\prime}} c_{\alpha^{\prime}}\left\langle a^{\prime}\right| \exp \left(\frac{i E_{a^{\prime}} t}{\hbar}\right)\right] \cdot B \cdot\left[\sum_{a^{\prime \prime}} c_{d^{\prime \prime}} \exp \left(\frac{-i E_{a^{\prime \prime}} t}{\hbar}\right)\left|a^{\prime \prime}\right\rangle\right] \quad=\sum_{d^{\prime}} \sum_{d^{\prime \prime}} c_{d^{\prime}} c_{d^{\prime \prime}}\left\langle a^{\prime}|B| a^{\prime \prime}\right\rangle \exp \left[\frac{-i\left(E_{d^{\prime \prime}}-E_{d^{\prime}}\right) t}{\hbar}\right]$$

$$\omega_{d^{\prime \prime} \alpha^{\prime}}=\frac{\left(E_{a^{\prime \prime}}-E_{\alpha^{\prime}}\right)}{\hbar}$$

## 物理代写|量子力学代考QUANTUM MECHANICS代考|Spin Precession

$$H=-\left(\frac{e}{m_{e} c}\right) \mathbf{S} \cdot \mathbf{B}$$
$(e<0$ 为电子) 。此外，我们取 $\mathbf{B}$ 是一个静态的、均匀的磁场 $z$-方向。然后我们可以写 $H$ 作为
$$H=-\left(\frac{e B}{m_{e} c}\right) S_{z} .$$

$$E_{\pm}=\mp \frac{e \hbar B}{2 m_{e} c}, \text { for } S_{z} \pm$$

$$\omega \equiv \frac{|e| B}{m_{e} c}$$

## 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。