### 物理代写|统计力学代写Statistical mechanics代考|PHYS3006

statistics-lab™ 为您的留学生涯保驾护航 在代写统计力学Statistical mechanics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写统计力学Statistical mechanics代写方面经验极为丰富，各种代写统计力学Statistical mechanics相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|统计力学代写Statistical mechanics代考|Quantum treatment

Harmonic oscillators have quantized energy levels ${ }^{14} E_n=\left(n+\frac{1}{2}\right) \hbar \omega, n=0,1,2, \cdots$. The energy associated with $n=0, \frac{1}{2} \hbar \omega$, is the zero-point energy, the lowest possible energy that a quantum system may have (which, we note, is not zero). ${ }^{15}$ The canonical partition function for a single oscillator is, from Eq. (4.123), ${ }^{16}$
$$Z_1(\beta)=\sum_{n=0}^{\infty} \mathrm{e}^{-\beta\left(n+\frac{1}{2}\right) \hbar \omega}=\frac{1}{2 \sinh (\beta \hbar \omega / 2)} .$$
The partition function specifies the number of states a system has available to it at temperature $T$. As $\beta \rightarrow 0$ (high temperature), we have from Eq. (5.20),
$$Z_1(\beta) \stackrel{\beta \rightarrow 0}{\sim} \frac{1}{\beta \hbar \omega},$$
that all of the infinite number of energy states of the harmonic oscillator become thermally accessible, that $Z$ diverges as we (formally) allow $T \rightarrow \infty$. Compare with the $\beta \rightarrow 0$ limit of the partition function for a paramagnetic ion, Eq. (5.17), $Z(\beta \rightarrow 0)=2$. In that case there are only two states available to the system: aligned or antialigned with the direction of the magnetic field. Consider the other limit of Eq. (5.20),
$$Z_1(\beta) \stackrel{\beta \rightarrow \infty}{\sim} \mathrm{e}^{-\beta \hbar \omega / 2} .$$

For temperatures such that $k T \leqslant \hbar \omega / 2, Z_1 \leqslant 1$; the number of states available to the system is exponentially smaller than unity. As $T \rightarrow 0$ there are no states available to the system: $Z \rightarrow 0$.
Applying Eq. $(5.20)$ to Eq. (4.40), we have the average energy of the oscillator,
$$\langle E\rangle=\frac{\hbar \omega}{2} \operatorname{coth}\left(\frac{1}{2} \beta \hbar \omega\right)=\hbar \omega\left(\frac{1}{\mathrm{e}^{\beta \hbar \omega}-1}+\frac{1}{2}\right) \equiv \hbar \omega\left(\langle n\rangle+\frac{1}{2}\right) \text {. }$$
Let’s look at the limiting forms of Eq. (5.23):
$$\begin{array}{ll} \langle E\rangle=\frac{\hbar \omega}{2} & (T \rightarrow 0) \ \langle E\rangle=k T . & (T \rightarrow \infty) \end{array}$$

## 物理代写|统计力学代写Statistical mechanics代考|Rotatonal motion

The rigid rotor problem treats the two atoms of a diatomic molecule as having a fixed separation distance $r_0$. The allowed rotational energies depend on the moment of inertia $I=\mu r_0^2$, where $\mu$ is the reduced mass of the two atomic masses, $\mu=m_1 m_2 /\left(m_1+m_2\right)$. The rotational state is determined by the angular momentum operator, $\hat{L} . \hat{L}^2$ and $\hat{L}z$ have a common set of eigenfunctions, \begin{aligned} &\hat{L}^2|l, m\rangle=l(l+1) \hbar^2|l, m\rangle \ &\hat{L}_z|l, m\rangle=m \hbar|l, m\rangle, \end{aligned} where $l=0,1,2, \cdots$ and $m=-l,-l+1, \cdots, l-1, l$ so that there are $2 l+1$ values of $m$. The Hamiltonian for rotational motion about the center of mass is $\hat{H}{\mathrm{rot}}=L^2 /(2 I)$, and thus the rotational energy eigenvalues are $E_l=\hbar^2 l(l+1) /(2 I)$. Because $E_l$ is independent of the quantum number $m$, each state is ( $2 l+1)$-fold degenerate. The partition function is, using Eq. (4.123), ${ }^{22}$
$$Z_{1, \mathrm{rol}}(T)=\sum_{l=0}^{\infty}(2 l+1) \mathrm{e}^{-\beta E_l} .$$ The sum in Eq. (5.33) cannot be evaluated in closed analytic form, and we must introduce approximations. We examine the high and low-temperature limits.

## 物理代写|统计力学代写Statistical mechanics代考|Quantum treatment

$$Z_1(\beta)=\sum_{n=0}^{\infty} \mathrm{e}^{-\beta\left(n+\frac{1}{2}\right) \hbar \omega}=\frac{1}{2 \sinh (\beta \hbar \omega / 2)}$$

$$Z_1(\beta) \stackrel{\beta \rightarrow 0}{\sim} \frac{1}{\beta \hbar \omega},$$

$$Z_1(\beta) \stackrel{\beta \rightarrow \infty}{\sim} \mathrm{e}^{-\beta \hbar \omega / 2} .$$

$$\langle E\rangle=\frac{\hbar \omega}{2} \operatorname{coth}\left(\frac{1}{2} \beta \hbar \omega\right)=\hbar \omega\left(\frac{1}{\mathrm{e}^{\beta \hbar \omega}-1}+\frac{1}{2}\right) \equiv \hbar \omega\left(\langle n\rangle+\frac{1}{2}\right) .$$

$$\langle E\rangle=\frac{\hbar \omega}{2} \quad(T \rightarrow 0)\langle E\rangle=k T . \quad(T \rightarrow \infty)$$

## 物理代写|统计力学代写Statistical mechanics代考|Rotatonal motion

$$\hat{L}^2|l, m\rangle=l(l+1) \hbar^2|l, m\rangle \quad \hat{L}z|l, m\rangle=m \hbar|l, m\rangle$$ 在哪里 $l=0,1,2, \cdots$ 和 $m=-l,-l+1, \cdots, l-1, l$ 所以有 $2 l+1$ 的值 $m$. 绕质心旋转运动的哈 密顿量是 $\hat{H} \operatorname{rot}=L^2 /(2 I)$ ，因此旋转能量特征值为 $E_l=\hbar^2 l(l+1) /(2 I)$. 因为 $E_l$ 与量子数无关 $m$ ，每个状态是 $(2 l+1)$-折㿿退化。分区函数是，使用方程。(4.123)， ${ }^{22}$ $$Z{1, \mathrm{rol}}(T)=\sum_{l=0}^{\infty}(2 l+1) \mathrm{e}^{-\beta E_l}$$

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## 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。