## 物理代写|统计力学代写Statistical mechanics代考|PHYC30017

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|High-temperature form

As $\beta \rightarrow 0$ there are contributions to Eq. (5.33) from large values of the quantum number $l$, which suggests we approximate the sum in Eq. (5.33) with an integral, using the form of $Z$ in Eq. (4.15). That route requires the density-of-states function, $\Omega(E)$, the derivative with respect to energy of the total number of energy states up to and including $E$. Energy at a specified value $E$ implies a maximum value of $l$ determined by $E=\hbar^2 l_{\max }\left(l_{\max }+1\right) /(2 I) \approx \hbar^2 l_{\max }^2 /(2 I)$ because $l_{\max } \gg 1$. How many states are there for $0 \leq l \leq l_{\max }$ ? It can be shown that
$$\sum_{l=0}^{l_{\max }}(2 l+1)=\left(l_{\max }+1\right)^2 \approx l_{\max }^2 \approx \frac{2 I}{\hbar^2} E .$$
The density of states is therefore $\Omega(E)=2 I / \hbar^2$. Thus, we can approximate Eq. (5.33),
$$Z_{1, \text { rol }}(T)=\frac{2 I}{\hbar^2} \int_0^{\infty} \mathrm{e}^{-\beta E} \mathrm{~d} E=\frac{2 I}{\beta \hbar^2} \equiv \frac{T}{\Theta_r}, \quad\left(T \gg \Theta_r\right)$$
where $\Theta_r=\hbar^2 /(2 I k)$ sets a characteristic temperature for rotational motions. ${ }^{23}$ Using equations that we’ve now used several times (Eqs. (4.40) and (P4.1)), with $Z=\left(Z_1\right)^N$,
\begin{aligned} \langle E\rangle_{\mathrm{rot}} &=N k T \ \left(C_V\right){\mathrm{rot}} &=N k, \quad(T \rightarrow \infty) \end{aligned} the same as what we obtain from the equipartition theorem. A more accurate high-temperature form can be obtained using the result of Exercise 5.11: $$Z{1, \mathrm{rot}}(T)=\frac{T}{\Theta_r}+\frac{1}{3}+\frac{1}{15} \frac{\Theta_r}{T}+\frac{4}{315}\left(\frac{\Theta_r}{T}\right)^2+\cdots . \quad\left(T \gg \Theta_r\right)$$
From Eq. (5.37) we obtain an expression for the heat capacity more general than Eq. (5.36) (see Exercise 5.12),
$$\left(C_V(T)\right){\mathrm{rot}}=N k\left[1+\frac{1}{45}\left(\frac{\Theta_r}{T}\right)^2+\frac{16}{945}\left(\frac{\Theta_r}{T}\right)^3+\cdots\right] .$$ We see that $\left(C_V(T)\right){\text {rot }}$ exceeds the classical value $N k$, a value that it tends to as $T \rightarrow \infty$.

## 物理代写|统计力学代写Statistical mechanics代考|Low-temperature form

In the low-temperature regime, $T \ll \Theta_r$, we have, from Eq. (5.33),
$$Z(T){1, \mathrm{rot}}=1+3 \mathrm{e}^{-2 \Theta_r / T}+5 \mathrm{e}^{-6 \Theta_r / T}+\cdots .$$ In this case, the variable $\mathrm{e}^{-\Theta_r / T}$ is exponentially small as $T \rightarrow 0$. From Eq. (5.39), we find to lowest order $$\left(C_V(T)\right){\mathrm{rot}} \approx 12 N k\left(\frac{\Theta_r}{T}\right)^2 \mathrm{e}^{-2 \Theta_r / T} . \quad\left(T \ll \Theta_r\right)$$

As $T \rightarrow 0,\left(C_V(T)\right)_{\text {rot }}$ drops to zero exponentially fast; rotational degrees of freedom can’t be excited at sufficiently low temperature – they become “frozen out.”

The two equations, (5.38) and (5.40), are limiting forms of $\left(C_V(T)\right){\mathrm{rot}}$ in the high- and lowtemperature regimes. They each show that the heat capacity is temperature dependent. To obtain the complete temperature dependence of $\left(C_V(T)\right){\text {rot }}$ requires the use of a computer to evaluate the sum in Eq. (5.33) at each temperature. A detailed analysis shows there is a maximum value of $\left(C_V(T)\right)_{\mathrm{rot}} \approx 1.1 \mathrm{Nk}$ at $T \approx 0.81 \Theta_r$. Given that $\Theta_r \approx 10 \mathrm{~K}$, measurements of $C_V$ on diatomic gases at room temperature are consistent with the prediction of the equipartition theorem.

## 物理代写|统计力学代写Statistical mechanics代考|High-temperature form

$$\sum_{l=0}^{l_{\max }}(2 l+1)=\left(l_{\max }+1\right)^2 \approx l_{\max }^2 \approx \frac{2 I}{\hbar^2} E$$

$$Z_{1, \text { rol }}(T)=\frac{2 I}{\hbar^2} \int_0^{\infty} \mathrm{e}^{-\beta E} \mathrm{~d} E=\frac{2 I}{\beta \hbar^2} \equiv \frac{T}{\Theta_r}, \quad\left(T \gg \Theta_r\right)$$

$$\langle E\rangle_{\mathrm{rot}}=N k T\left(C_V\right) \text { rot }=N k, \quad(T \rightarrow \infty)$$

$$Z 1, \operatorname{rot}(T)=\frac{T}{\Theta_r}+\frac{1}{3}+\frac{1}{15} \frac{\Theta_r}{T}+\frac{4}{315}\left(\frac{\Theta_r}{T}\right)^2+\cdots . \quad\left(T \gg \Theta_r\right)$$

$$\left(C_V(T)\right) \operatorname{rot}=N k\left[1+\frac{1}{45}\left(\frac{\Theta_r}{T}\right)^2+\frac{16}{945}\left(\frac{\Theta_r}{T}\right)^3+\cdots\right] .$$

## 物理代写|统计力学代写Statistical mechanics代考|Low-temperature form

$$Z(T) 1, \text { rot }=1+3 \mathrm{e}^{-2 \Theta_r / T}+5 \mathrm{e}^{-6 \Theta_r / T}+\cdots .$$

$$\left(C_V(T)\right) \operatorname{rot} \approx 12 N k\left(\frac{\Theta_r}{T}\right)^2 \mathrm{e}^{-2 \Theta_r / T} . \quad\left(T \ll \Theta_r\right)$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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)， ${ }^{22}$ $$Z{1, \mathrm{rol}}(T)=\sum_{l=0}^{\infty}(2 l+1) \mathrm{e}^{-\beta E_l}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 物理代写|统计力学代写Statistical mechanics代考|PHYS3034

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|THE MAXWELL SPEED DISTRIBUTION

The Hamiltonian of a gas of $N$ noninteracting particles is $H=\sum_{i=1}^N \boldsymbol{p}i^2 /(2 m)$. The partition function for this system (volume $V$, temperature $T$ ) is found from Eqs. (4.47) and (4.53), $$Z{\operatorname{can}}(N, V, T)=\frac{1}{N !}\left(\frac{V}{\lambda_T^3}\right)^N \equiv \frac{1}{N !} Z(N, V, T),$$
where $\lambda_T$ is the thermal wavelength, Eq. (1.65), which results from integrating over the momentum variables. With $Z_{\mathrm{can}}$ one can calculate the equation of state and the entropy using Eq. (4.58) (Exercise 5.1). The phase-space probability density is, from Eq. (4.54),
$$\rho(p, q)=\frac{1}{Z} \exp \left(-\beta \sum_{i=1}^N \boldsymbol{p}i^2 /(2 m)\right)=\prod{i=1}^N\left(\frac{\lambda_T^3}{V} \mathrm{e}^{-\beta \boldsymbol{p}i^2 /(2 m)}\right) \equiv \prod{i=1}^N \rho_i,$$
where $\rho_i$ is a one-particle distribution function. Because the Hamiltonian is separable, the $N$ particle distribution occurs as the product of $N$, single-particle distributions, i.e., the particles are independently distributed. ${ }^2$ Note that $\rho_i$ is normalized on a one-particle phase space:
$$\int \rho_i \mathrm{~d} \Gamma_i \equiv \frac{\lambda_T^3}{h^3 V} \int_V \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \int_{-\infty}^{\infty} \mathrm{d} p_x \mathrm{~d} p_y \mathrm{~d} p_z \mathrm{e}^{-\beta\left(p_x^2+p_y^2+p_z^3\right) /(2 m)}=1 .$$
Another way to calculate the entropy is through the distribution function, Eq. (4.60). One can show that Eq. (4.60) yields the Sackur-Tetrode formula when combined with Eq. (5.2) (see Exercise 5.3).

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

Some of the most successful applications of statistical mechanics involve the magnetic properties of materials. Under the general banner of magnetism there are different types of magnetic phenomena: ferromagnetism, antiferromagnetism, paramagnetism, diamagnetism, and others. In the limited space of this book we can only offer a cursory treatment of the subject. Ferro- and antiferromagnetism are cooperative effects produced by interactions among the magnetic dipoles of the atoms in a solid. Paramagnetism is the “ideal gas” of magnetism, in which magnetic moments interact only with an applied magnetic field and not with each other.

For a collection of magnetic moments $\left{\boldsymbol{\mu}i\right}$ that interact only with the external field, we need treat only the statistical mechanics of a single magnetic moment. The partition function for $N$ identical, noninteracting particles $Z_N=\left(Z_1\right)^N$, where $Z_1$ is the single-particle partition function. The energy of interaction between a magnetic dipole moment $\mu$ and a magnetic field ${ }^9 \boldsymbol{B}$ is $E=-\boldsymbol{\mu} \cdot \boldsymbol{B}$. Should we adopt a classical or a quantum treatment of this problem? It turns out that a quantum treatment leads to excellent agreement with experimental results. Thus, we consider the energy of interaction between $\mu$ and $B$ as the Hamiltonian operator, $$\hat{H}=-\boldsymbol{B} \cdot \hat{\boldsymbol{\mu}}=\frac{g \mu_B}{\hbar} \boldsymbol{B} \cdot \hat{\boldsymbol{J}}=\frac{g B \mu_B}{\hbar} \hat{J}_z,$$ where we’ve used Eq. (E.4), $\boldsymbol{\mu}=-g \mu_B \boldsymbol{J} / \hbar$, where $\mu_B \equiv e \hbar /(2 m)$ is the Bohr magneton, $g$ is the Landé g-factor (see Appendix E), and the operator $\hat{J}_z$ is the $z$-component of the total angular momentum (the $B$-field defines the $z$-direction). To use Eqs. (4.123) or (4.125) (quantum statistical mechanics in the canonical ensemble), we require the eigenfunctions and eigenvalues of the Hamiltonian operator, which in this case is proportional to $\hat{J}_z$ (Eq. (5.9)). As is well known, $\hat{J}^2$ and $\hat{J}_z$ have a common set of eigenfunctions $|J, m\rangle$ (a complete orthonormal set), such that \begin{aligned} &\hat{J}^2|J, m\rangle=J(J+1) \hbar^2|J, m\rangle \ &\hat{J}_z|J, m\rangle=m \hbar|J, m\rangle \end{aligned} where the quantum number $J$ has the values $J=0,1,2, \cdots$ or $J=\frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \cdots$, and $m=$ $-J,-J+1, \cdots, J-1, J$ so that there are $(2 J+1)$ values of $m$. The energy eigenvalues are therefore $E_m=g \mu_B m B$. From Eq. (4.123), ${ }^{10}$ $$Z_1=\sum{m=-J}^J \mathrm{e}^{-\beta m \mu_B g B}=\frac{\sinh \left(y\left(J+\frac{1}{2}\right)\right)}{\sinh (y / 2)},$$
where $y \equiv \beta \mu_B g B$. The summation in Eq. (5.10) is simple because it’s a finite geometric series.

## 物理代写|统计力学代写Statistical mechanics代考|THE MAXWELL SPEED DISTRIBUTION

$$Z \operatorname{can}(N, V, T)=\frac{1}{N !}\left(\frac{V}{\lambda_T^3}\right)^N \equiv \frac{1}{N !} Z(N, V, T),$$

$$\rho(p, q)=\frac{1}{Z} \exp \left(-\beta \sum_{i=1}^N \boldsymbol{p} i^2 /(2 m)\right)=\prod i=1^N\left(\frac{\lambda_T^3}{V} \mathrm{e}^{-\beta p i^2 /(2 m)}\right) \equiv \prod i=1^N \rho_i,$$

$$\int \rho_i \mathrm{~d} \Gamma_i \equiv \frac{\lambda_T^3}{h^3 V} \int_V \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z \int_{-\infty}^{\infty} \mathrm{d} p_x \mathrm{~d} p_y \mathrm{~d} p_z \mathrm{e}^{-\beta\left(p_x^2+p_y^2+p_z^3\right) /(2 m)}=1 .$$

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

$$\hat{H}=-\boldsymbol{B} \cdot \hat{\boldsymbol{\mu}}=\frac{g \mu_B}{\hbar} \boldsymbol{B} \cdot \hat{\boldsymbol{J}}=\frac{g B \mu_B}{\hbar} \hat{J}_z,$$

$$\hat{J}^2|J, m\rangle=J(J+1) \hbar^2|J, m\rangle \quad \hat{J}_z|J, m\rangle=m \hbar|J, m\rangle$$

$$Z_1=\sum m=-J^J \mathrm{e}^{-\beta m \mu_B g B}=\frac{\sinh \left(y\left(J+\frac{1}{2}\right)\right)}{\sinh (y / 2)}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 物理代写|统计力学代写Statistical mechanics代考|PHYC90010

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|Binomial distribution

Probability thrives on the repeatability of experiments. Much can be learned about random processes realized through repeated measurements of a quantity that produces only a few, perhaps just two, outcomes. Consider a pair of coins that’s tossed 200 times. What is the probability that $x$ of the 200 tosses shows two heads ( $x$ is an integer)? Let $S$ denote the probability of “success” in obtaining two heads in a given trial, with $F$ the probability of “failure.” Referring to the sample space of Fig. 3.1, $S=1 / 4$ and $F=3 / 4$. The tosses are independent and thus the probability of any realization of $x$ successes and $(200-x)$ failures is the same: $S^{x} F^{200-x}$. There are $\left(\begin{array}{c}200 \ x\end{array}\right)$ ways that $x$ successes can occur among the 200 outcomes. Thus, we have the probability distribution ( $x$ is a random variable)
$$f(x)=\frac{200 !}{x !(200-x) !}\left(\frac{1}{4}\right)^{x}\left(\frac{3}{4}\right)^{200-x} .$$
Equation (3.38) readily lends itself to generalization. Let the probability of success in an individual trial be $p$, with the probability of failure $q=1-p$, and let there be $N$ trials. ${ }^{12}$ The probability distribution $f(x)$ of $x$ successes (whatever “success” refers to) in $N$ trials is
$$f(x)=\left(\begin{array}{c} N \ x \end{array}\right) p^{x} q^{N-x}=\frac{N !}{x !(N-x) !} p^{x} q^{N-x}$$

Equation (3.39) is the binomial distribution; it applies to many problems involving a discrete variable $x$ where the probability $p$ is known. Is it normalized-is $\sum_{x=0}^{N} f(x)=1$ ? That is indeed the case, as can be seen by applying the binomial theorem, Eq. (3.7):
$$1=(p+q)^{N}=\sum_{x=0}^{N}\left(\begin{array}{l} N \ x \end{array}\right) p^{x} q^{N-x} .$$

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

When $N$ becomes large, direct calculations using Eq. (3.39) become unwieldy. In that case having approximate expressions is quite useful. We develop the Poisson distribution,
$$\lim {\substack{N \rightarrow \infty \ N p=\mu}} f(x=k)=\frac{\mu^{k}}{k !} \mathrm{e}^{-\mu},$$ which holds for $p \ll 1$, such that $N p \equiv \mu$ is fixed. The Poisson distribution is normalized; $\sum{k=0}^{\infty} f(k)=1$. We can let $k \rightarrow \infty$ because we’ve already let $N \rightarrow \infty$. A formula like Eq. (3.43) is known as a limit theorem or as an asymptotic theorem; see Section 3.6.
To derive Eq. (3.43), first note that for fixed $x$, (see Exercise 3.22)
$$\left(\begin{array}{l} N \ x \end{array}\right) \stackrel{N \rightarrow \infty}{\sim} \frac{N^{x}}{x !} .$$

From Eq. (3.39),
$$f(x) \sim \frac{N^{x}}{x !} p^{x} q^{N-x}=\frac{\mu^{x}}{x !}(1-p)^{N-x}=\frac{\mu^{x}}{x !}\left(1-\frac{\mu}{N}\right)^{N-x},$$
where we’ve used $\mu=N p$. Equation (3.43) follows in the limit $N \rightarrow \infty$ when we make use of the Euler form of the exponential, $\mathrm{e}^{y}=\lim _{N \rightarrow \infty}(1+y / N)^{N}$.

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

$$f(x)=\frac{200 !}{x !(200-x) !}\left(\frac{1}{4}\right)^{x}\left(\frac{3}{4}\right)^{200-x} .$$

$$f(x)=(N x) p^{x} q^{N-x}=\frac{N !}{x !(N-x) !} p^{x} q^{N-x}$$

$$1=(p+q)^{N}=\sum_{x=0}^{N}(N x) p^{x} q^{N-x} .$$

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

$\lim {N \rightarrow \infty} N p=\mu f(x=k)=\frac{\mu^{k}}{k !} \mathrm{e}^{-\mu}$ 这适用于 $p \ll 1$ ，这样 $N p \equiv \mu$ 是固定的。泊松分布被归一化; $\sum k=0^{\infty} f(k)=1$. 我们可以让 $k \rightarrow \infty$ 因为我 们已经让 $N \rightarrow \infty$. 像方程式这样的公式。(3.43) 被称为极限定理或渐近定理；见第 $3.6$ 节。 推导出方程。(3.43)，首先注意对于固定 $x$ ，(见习题 3.22) $$(N x)^{N \rightarrow \infty} \frac{N^{x}}{x !} .$$ 从方程式。(3.39), $$f(x) \sim \frac{N^{x}}{x !} p^{x} q^{N-x}=\frac{\mu^{x}}{x !}(1-p)^{N-x}=\frac{\mu^{x}}{x !}\left(1-\frac{\mu}{N}\right)^{N-x}$$ 我们用过的地方 $\mu=N p$. 方程 (3.43) 遵循极限 $N \rightarrow \infty$ 当我们利用指数的欧拉形式时， $\mathrm{e}^{y}=\lim {N \rightarrow \infty}(1+y / N)^{N}$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 物理代写|统计力学代写Statistical mechanics代考|PHYC30017

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|EXAMPLES INVOLVING DISCRETE PROBABILITIES

• Two cards are drawn from a 52-card deck, with the first being replaced before the second is drawn. What is the probability that both cards are spades? Let $A$ be the event of drawing a spade, with $B$ the event of drawing another spade after the first has been replaced in the deck. This is an “and” kind of problem: What is the probability of a spade being drawn and another spade being drawn. $P(A)=P(B)=13 / 52=1 / 4$. The two events are independent, and thus from Eq. (3.5), $P(A \cap B)=P(A) P(B)=1 / 16$.
• What is the probability of at least one spade in drawing two cards, when the first is replaced? The slick way to work this problem is to calculate the probability of not drawing a spade-the probability of at least one spade is the complement of the probability of no spades in two draws. The probability of no spades (not drawing a spade and not drawing another one) is $(39 / 52)^{2}=9 / 16$ (independent events). The probability of at least one spade is then $1-P$ (no spades) $=7 / 16$. The direct approach is to treat this as an “or” problem: What is the probability of drawing one or two spades? Let $A$ be the event of drawing a spade and not drawing a spade on the other draw, with $B$ the event of drawing two spades. The probability of at least one spade is $P(A$ or $B)=P(A)+P(B)$ (mutually exclusive). $P(A)=P$ (spade on one draw and not a spade on the other $)=(1 / 4)(3 / 4)=3 / 16$ (independent). There are two ways to realize the first experiment, however, draw a spade and then not, or not draw a spade and then a spade, so we add the probabilities: The probability of one spade is $2 \times(3 / 16)$. The probability of two spades, $P(B)=(1 / 4)^{2}=1 / 16$. The probability of at least one spade is $2 \times(3 / 16)+(1 / 16)=7 / 16$, in agreement with the first answer.
• Two cards are drawn from a deck, but now suppose the first is not put back. What is the probability that both are spades? This is an “and” problem, the probability of drawing a spade and drawing another one. The events are independent. Thus, $P=(13 / 52) \times(12 / 51)=1 / 17$.
• What is the probability that the second card is a spade, when it’s not known what the first card was? Let $B$ be the event of drawing a spade on the second draw. All we know about the first event is that a card was drawn and not replaced. There are two mutually exclusive possibilities: The first card was a spade or not, call these events $A$ and $\bar{A}$. Then, $P(A \cap B)+P(\bar{A} \cap B)=$ $P(A) P(B)+P(\bar{A}) P(B)=(P(A)+P(\bar{A})) P(B)=P(B)$. Thus, $P(B)=1 / 4$. The probability of a spade on the second draw, when the result of the first draw is unknown, is the probability of a spade on the first draw.

## 物理代写|统计力学代写Statistical mechanics代考|Probability distributions on discrete sample spaces

The collection of probabilities associated with the range of values of a random variable is known as a probability distribution. ${ }^{10}$ For each value $x_{j}$ of a random variable $x$, the aggregate of sample points associated with $x_{j}$ form the event for which $x-x_{j}$; its probability is denoted $P\left(x-x_{j}\right)$. From Fig. 3.4, for example, $f(1)=1 / 2$ is associated with the event $T H$ or $H T$.
Definition. A function $f(x)$ such that $f\left(x_{j}\right)=P\left(x=x_{j}\right)$ is the probability distribution of $x$.
For the range of values $\left{x_{j}\right}$ of $x, f\left(x_{j}\right) \geq 0$ and $\sum_{j} f\left(x_{j}\right)=1$; see Figs. $3.4$ and 3.5.
There can be more than one random variable defined on the same sample space. Consider random variables $x$ and $y$ that take on the values $x_{1}, x_{2}, \ldots$ and $y_{1}, y_{2}, \ldots$, and let the corresponding probability distributions be $f\left(x_{j}\right)$ and $g\left(y_{k}\right)$. The aggregate of events for which the two conditions $x=x_{j}$ and $y=y_{k}$ are satisfied forms the event having probability denoted $P\left(x=x_{j}, y=y_{k}\right)$.
Definition. A function $p(x, y)$ for which $p\left(x_{j}, y_{k}\right)=P\left(x=x_{j}, y=y_{k}\right)$ is called the joint probability distribution of $x$ and $y$.

Clearly, $p\left(x_{j}, y_{k}\right) \geq 0$ and $\sum_{j k} p\left(x_{j}, y_{k}\right)=1$. Moreover, for fixed $x_{j}$,
$$\sum_{k} p\left(x_{j}, y_{k}\right)=f\left(x_{j}\right),$$
while for fixed $y_{k}$
$$\sum_{j} p\left(x_{j}, y_{k}\right)=g\left(y_{k}\right) .$$
That is, adding the probabilities for all events $y_{k}$ for fixed $x_{j}$ produces the probability distribution for $x_{j}$, and adding the probabilities for all events $x_{j}$ produces the probability distribution for $y_{k}$.

## 物理代写|统计力学代写Statistical mechanics代考|EXAMPLES INVOLVING DISCRETE PROBABILITIES

• 从一副 52 张牌中抽出两张牌，在抽出第二张牌之前先替换第一张牌。两张牌都是黑桃的概率是多少? 让 $A$ 是 绘制黑桃的事件，与 $B$ 在第一个黑桃被替换在甲板上之后绘制另一个黑桃的事件。这是一个”和”类型的问题: 一个铲子被抽出和另一个铲子被抽出的概率是多少。 $P(A)=P(B)=13 / 52=1 / 4$. 这两个事件是独立 的，因此来自方程式。(3.5), $P(A \cap B)=P(A) P(B)=1 / 16$.
• 当第一张被替换时，至少有一张黑桃抽两张牌的概率是多少? 解决这个问题的巧妙方法是计算没有抽到黑桃的 概率一一至少有一个黑桃的概率是两次抽到没有黑桃的概率的补数。没有黑桃的概率 (没有画黑桃，也没有画 另一个) 是 $(39 / 52)^{2}=9 / 16$ (独立事件) 。那么至少有一把铁锹的概率是 $1-P$ (没有黑桃) $=7 / 16$. 直 接的方法是将其视为一个”或”问题: 抽到一两个黑桃的概率是多少? 让 $A$ 是画黑挑而不是在另一张画上画黑桃 的事件，与 $B$ 绘制两个黑桃的事件。至少有一把铲子的概率是 $P(A$ 或者 $B)=P(A)+P(B)$ (互斥) 。 $P(A)=P($ (一平局是铁锹，另一个不是铁锹 $)=(1 / 4)(3 / 4)=3 / 16$ (独立的)。实现第一个实验有两 种方法，但是，画一个黑桃然后不画，或者不画一个黑桃然后一个黑桃，所以我们添加概率: 一个黑桃的概率 是 $2 \times(3 / 16)$. 两个黑桃的概率， $P(B)=(1 / 4)^{2}=1 / 16$. 至少有一把铲子的概率是 $2 \times(3 / 16)+(1 / 16)=7 / 16$ ，与第一个答案一致。
• -从一副牌中抽出两张牌，但现在假设第一张牌没有放回。两者都是黑桃的概率是多少? 这是一个”与”问题，即 画出一把铁锹再画另一个的概率。事件是独立的。因此， $P=(13 / 52) \times(12 / 51)=1 / 17$.
• 当不知道第一张牌是什么时，第二张牌是黑桃的概率是多少? 让 $B$ 是在第二次抽奖时抽到黑桃的事件。关于第 一个事件，我们所知道的只是一张牌被抽出来而不是被替换。有两种相互排斥的可能性: 第一张牌是否是黑 桃，调用这些事件 $A$ 和 $\bar{A}$. 然后， $P(A \cap B)+P(\bar{A} \cap B)=$ $P(A) P(B)+P(\bar{A}) P(B)=(P(A)+P(\bar{A})) P(B)=P(B)$. 因此， $P(B)=1 / 4$. 当第一次平局的 结果末知时，第二次平局出现黑桃的概率是第一次平局出现黑桃的概率。

## 物理代写|统计力学代写Statistical mechanics代考|Probability distributions on discrete sample spaces

$$\sum_{k} p\left(x_{j}, y_{k}\right)=f\left(x_{j}\right),$$

$$\sum_{j} p\left(x_{j}, y_{k}\right)=g\left(y_{k}\right) .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|Multiplying probabilities

How is $P(A \cap B)$ in Eq. (3.2) calculated? To answer that, it’s necessary to introduce another kind of probability, the conditional probability, denoted $P(A \mid B)$, the probability of $A$ occurring, given that $B$ has occurred. Referring to Fig. 3.3, we’re interested in the probability that $A$ occurs given that $B$ has definitely occurred, a type of problem where the sample space has changed-in this case the certain event is $B$. The probability we want is the ratio of the number of sample points in the intersection, $N_{A \cap B}$, to that in $B$ :
$$P(A \mid B)=\frac{\bar{N}{A \cap B}}{N{B}}=\frac{\bar{N}{A \cap B}}{N{\Omega}} \frac{\bar{N}{\Omega}}{N{B}}=\frac{\bar{P}(A \cap \bar{B})}{P(B)},$$
or
$$P(A \wedge B)=P(A \mid B) P(B) .$$
In words, Eq. (3.4) indicates that the probability of $A$ and $B$ is the probability of $A$ given that $B$ has occurred, multiplied by the probability that $B$ occurs. This relation is symmetrical between $A$ and $B: P(A \cap B)=P(B \mid A) P(A)$, implying $P(B \mid A)=P(A \mid B) P(B) / P(A)$

Suppose $A$ and $B$ are such that $P(A \mid B)=P(A)$. In that case $A$ is said to be independent of $B$-the probability of $A$ occurring is independent of the condition that $B$ has occurred. For independent events, Eq. (3.4) reduces to
$$P(A \cap B)=P(A) P(B) . \quad \text { (independent events) }$$
For independent events, the probability of $A$ and $B$ is the product of the probabilities. Many problems in physics implicitly assume independent events; many problems implicitly ask for the probability of “this and that and that.” Be on the lookout for how statements are worded; there may be implied “ands.” Thus, for mutually exclusive events, probabilities are added, Eq. (3.3), whereas for independent events, probabilities are multiplied, Eq. (3.5). In Section 3.4, we give examples of how to calculate probabilities using these rules. First we must learn to count.

## 物理代写|统计力学代写Statistical mechanics代考|Stirling’s approximation

In its simplest form, Stirling’s approximation is, for $n \gg 1$,
$$\ln n !=n(\ln n-1)+O(\ln n),$$
where $O(\ln n)$ indicates that terms of order $\ln n$ have been neglected (which are negligible compared to $n$ for large $n$ ). Equation (3.14) is one of those results that should work only for $n \rightarrow \infty$, but which is accurate for relatively small values of $n(n \approx 10)$; see Exercise 3.8. Equation (3.14) is surprisingly easy to derive: $\ln n !=\sum_{k=1}^{n} \ln k \approx \int_{1}^{n} \ln x \mathrm{~d} x=\left.(x \ln x-x)\right|{1} ^{n} \approx n \ln -n$. The $O(\ln n)$ remainder is evaluated below. A more accurate version of Stirling’s approximation is $$n !^{n \rightarrow \infty} \underset{\sim}{2 \pi n}\left(\frac{n}{\mathrm{e}}\right)^{n},$$ where the notation $\sim$ indicates asymptotic equivalence. ${ }^{8}$ Equation (3.15) can be derived from $\Gamma(x)$ (see Eq. (B.1)): $$\Gamma(n+1)=n !=\int{0}^{\infty} x^{n} \mathrm{e}^{-x} \mathrm{~d} x \stackrel{x=n y}{=} n n^{n} \int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y .$$
The integral on the right of Eq. (3.16) can be approximated using the method of steepest descent[16. p233] for large $n$ :
$$\int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y \sim \sqrt{\frac{2 \pi}{n}} \mathrm{e}^{-n} .$$
Combining Eqs. (3.17) and (3.16) yields Eq. (3.15). By taking the logarithm of Eq. (3.15), we see that the remainder term in Eq. (3.14) is $\frac{1}{2} \ln (2 \pi n)$.

Sometimes we require the logarithm of the gamma function (the log-gamma function), $\ln \Gamma(x)$. From the recursion relation, Eq. (B.3), $\ln \Gamma(x+1)=\ln x+\ln \Gamma(x)$, and thus
$$\ln \Gamma(x)=\ln \Gamma(x+1)-\ln x .$$
Use Stirling’s approximation,
$$\Gamma(x+1) \sim \sqrt{2 \pi x}\left(\frac{x}{\mathrm{e}}\right)^{x} .$$

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

$$P(A \mid B)=\frac{\bar{N} A \cap B}{N B}=\frac{\bar{N} A \cap B}{N \Omega} \frac{\bar{N} \Omega}{N B}=\frac{\bar{P}(A \cap \bar{B})}{P(B)},$$

$$P(A \wedge B)=P(A \mid B) P(B) .$$

$$P(A \cap B)=P(A) P(B) . \quad \text { (independent events) }$$

## 物理代写|统计力学代写Statistical mechanics代考|Stirling’s approximation

$$\ln n !=n(\ln n-1)+O(\ln n),$$

$\ln n !=\sum_{k=1}^{n} \ln k \approx \int_{1}^{n} \ln x \mathrm{~d} x=(x \ln x-x) \mid 1^{n} \approx n \ln -n$. 这 $O(\ln n)$ 余数在下面评估。斯特林近似的 更准确版本是
$$n !^{n \rightarrow \infty} \underset{\sim}{2 \pi n}\left(\frac{n}{\mathrm{e}}\right)^{n},$$

$$\Gamma(n+1)=n !=\int 0^{\infty} x^{n} \mathrm{e}^{-x} \mathrm{~d} x \stackrel{x=n y}{=} n n^{n} \int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y .$$

$$\int_{0}^{\infty} \mathrm{e}^{n(\ln y-y)} \mathrm{d} y \sim \sqrt{\frac{2 \pi}{n}} \mathrm{e}^{-n} .$$

$$\ln \Gamma(x)=\ln \Gamma(x+1)-\ln x .$$

$$\Gamma(x+1) \sim \sqrt{2 \pi x}\left(\frac{x}{\mathrm{e}}\right)^{x} .$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 物理代写|统计力学代写Statistical mechanics代考|PHYS3034

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|EVENTS, SAMPLE SPACE, AND PROBABILITY

The sample space of two coins tossed is $\Omega={H H, H T, T H, T T}$. One way to represent these outcomes would be to assign $H$ the number 1 and $T$ the number 0 , so that they’re given by the points $(1,1),(1,0),(0,1),(0,0)$ in the $x y$ plane; see Fig. 3.1. One doesn’t have to depict the sample space as in Fig. 3.1-one could mark off any four points on the $x$-axis for example. For three coins tosised there are eight outcomes; one could depict the sample space using a three-dimensional Cartesian space, or, again, mark off any eight points on the $x$-axis. Sample space is a useful mathematical concept for discussing probability. How we display these sets is a matter of convenience. It’s often simpler to display the sample space in an abstract manner. The right part of Fig. $3.1$ shows the 36 elements of $\Omega$ for the roll of two dice simply as points in a box.

Experiments that produce a finite number of outcomes, such as the roll of a die, have discrete sample spaces where the events can be represented as isolated points, as in Fig. 3.1. Probabilities defined on discrete sample spaces are referred to as discrete probabilities. Not every sample space is discrete. Continuous sample spaces are associated with experiments that produce a continuous range of possibilities, such as the heights of individuals in a certain population. Probabilities defined on continuous sample spaces are referred to as probability densities.

The individual elements of $\Omega$ are elementary events. ${ }^{1}$ The word event (not elementary event) is reserved for subsets of $\Omega$, aggregates of sample points. A subset $A$ of $\Omega$ is a set such that every element of $A$ is an element of $\Omega$, a relationship indicated $A \subset \Omega$. In tossing two coins, the event $A$ might be the occurrence of $T T$ or $H H ; A \subset \Omega$ is then the set of elementary events $A={T T, H H}$, where $\Omega={T T, H H, T H, H T}$. The terms “sample point” and “event” have an intuitive appeal, that, once specified for a given experiment, can be treated using the mathematics of point sets.

## 物理代写|统计力学代写Statistical mechanics代考|Adding probabilities—“or” statements

Consider events $A$ and $B$ (such as in Fig. 3.3), which have $N_{A}$ and $N_{B}$ sample points (elementary events). In $A \cup \perp$ there are $N_{A \cup B}=N_{A}+N_{B}-N_{A \cap B}$ elements, where $N_{A \cap B}$ is the number of elements of the intersection $A \cap B$, which must be subtracted to prevent overcounting. ${ }^{6}$ We then have using Eq. (3.1) the analogous formula for probabilities,
$$P(A \cup B)=P(A)+P(B)-P(A \cap B) .$$
If $A$ and $B$ have no sample points in common (mutually exclusive), $A \cap B=\emptyset$. In that case,
$$P(A \cup B)=P(A)+P(B) . \quad(A, B \text { mutually exclusive })$$
Equation (3.3) is used frequently in applications-it tells us that the probability of $A$ or $B$ is the sum of the probabilities when $A, B$ are mutually exclusive. It pays to get in the habit of noticing how many calculations stem from questions of the form “what is the probability of the occurrence of this or that or that?” There’s often an implicit “or” statement underlying calculations in physics. Equation (3.3) easily generalizes to more than two mutually exclusive events.

## 物理代写|统计力学代写Statistical mechanics代考|Adding probabilities—“or” statements

$N_{A \cup B}=N_{A}+N_{B}-N_{A \cap B}$ 元素，其中 $N_{A \cap B}$ 是交点的元素个数 $A \cap B$, 必须减去以防止多算。 ${ }^{6}$ 然后我们使用 方程式。(3.1) 概率的类似公式，
$$P(A \cup B)=P(A)+P(B)-P(A \cap B) .$$

$$P(A \cup B)=P(A)+P(B) . \quad(A, B \text { mutually exclusive })$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 物理代写|统计力学代写Statistical mechanics代考|PHYS3020

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|Symmetrization and occupancy of single-particle states

Particle statistics, or state occupancy, for bosons and for fermions arise from the rules for wave function symmetrization. The purpose of this section is to show that the symmetrization factor defined above from the normalization of the symmetrized wave function, equation (2.13), is exactly the correct weight factor that is required to replace an arbitrary sum over occupied states by a sum over all states whether the particles be bosons or fermions. This is one example of the utility of the symmetrization factor. Two other advantages are that it generalizes the occupancy rules to multi-particle states, and to overlapping states (i.e., those with nonorthogonal wave functions).

The usual texts on quantum mechanics treat the subject of wave function symmetrization by invoking quantum states that comprise single, identical particle, states (Messiah 1961, Merzbacher 1970, Pathria 1972). Indeed the familiar concept that an arbitrary number of bosons, but at most one fermion, can occupy the same state is predicated upon, and only makes sense, if the state referred to is a singleparticle state. Since the present basis consists of wave packets, whose states are single-particle, we need to show that the symmetrization factor is related to these usual rules of particle occupancy. In section 3.3, the analysis is generalized to systems in which the concept of occupancy is inapplicable because the relevant states are not single-particle.

The position-momentum states are discrete so that $\Gamma_{j}$ is equivalent to $\ell_{j}$; it is the single-particle state occupied by particle $j$. The state of the system is $\boldsymbol{\Gamma}=\left{\mathbf{\Gamma}{1}, \mathbf{\Gamma}{2}, \ldots, \mathbf{\Gamma}{N}\right}$, which is equivalently $\ell=\left{\ell{1}, \ell_{2}, \ldots, \ell_{N}\right}$. We may order the possible single-particle states $a=1,2, \ldots, A$, and say equivalently that particle $j$ is in the state $\ell_{j}$, or else that it is in the $a$ th state, with $a=a\left(\ell_{j}\right)$. Let $m_{a}(\ell)=\sum_{j} \delta_{a\left(\ell_{j}\right), a}$ be the number of particles in the single-particle state $a$ when the system is in state $\ell$. We may regard $m_{a}$ as a component of the $A$-dimensional vector $\mathbf{m}(\ell)$, which tells the occupancy numbers of the possible single-particle states when the system is in the labeled particle state $\ell$. Clearly there is a many to one mapping from the system state $\ell$ to the occupancy state $\mathbf{m}$, since the latter doesn’t distinguish which particle or particles are in the given single-particle state.

A function of the state of the system may be written $f(\ell)$, or as $f(\mathbf{m})$, the latter being short-hand for the more precise $f_{\mathrm{s}}(\mathbf{m}(\ell))=f(\ell)$. Since the particles are identical, and since the labels are arbitrary, either description should suffice, provided that the rules are properly accounted for.

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

We consider a canonical equilibrium system, where the subsystem has number of particles $N$ and volume $V$, and the reservoir has temperature $T$. Instead of the latter we usually exhibit the inverse temperature, $\beta \equiv 1 / k_{\mathrm{B}} T$, where $k_{\mathrm{B}}$ is Boltzmann’s constant. The partition function is (Messiah 1961, Merzbacher 1970)
$$Z^{\pm}(N, V, T)=\mathrm{TR}^{\prime} e^{-\beta \hat{H}} .$$
This is derived in chapter 12, and it is used in chapter 7 as the starting point for a formally exact transformation of quantum statistical mechanics to classical phase space, valid in all regions of the phase diagram. Here that transformation is performed using wave packets, and it is strictly valid only in the classical limit.
The prime on the trace in the above formula is quite important as it signifies that only allowed quantum states should be included, and that these should be distinct and each counted once only. Messiah (1961, chapter XIV, sections 6 and 7) makes a similar point that the trace should be performed on a restricted subspace containing only allowed distinct states. The symmetrized wave function normalization factor given by him in the case of one-particle, orthogonal states is equivalent to the symmetrization factor given here, at least for the same case.

Unfortunately not all workers avert to the need to restrict the trace. Pathria (1972, equation (9.6.2)), gives the partition function as the Boltzmann weighted sum over energy states, with the implication being that these are all states (the issue of distinct states is not raised), but no symmetrization correction is exhibited.

As mentioned in the previous section it is usually most convenient to invoke an unrestricted sum, together with a weight factor that is zero for forbidden states and in inverse proportion to the number of times an allowed distinct state is counted. It was shown that the symmetrization factor had these properties.

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

$$Z^{\pm}(N, V, T)=\mathrm{TR}^{\prime} e^{-\beta \hat{H}}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 物理代写|统计力学代写Statistical mechanics代考|PHYSICS7546

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|Wave packet symmetrization and overlap

A fundamental axiom of quantum mechanics is that the wave function must be either fully symmetric (bosons) or fully anti-symmetric (fermions) with respect to interchange of identical particles (Messiah 1961, Merzbacher 1970). For the present wave packets, the symmetrized form is

\begin{aligned} \zeta_{\Gamma}^{\pm}(\mathbf{r}) & \equiv \frac{1}{\sqrt{N ! \chi_{\Gamma}^{\pm}}} \sum_{\hat{\mathrm{P}}}(\pm 1)^{p} \zeta_{\Gamma}(\hat{\mathrm{P}} \mathbf{r}) \ & \equiv \frac{1}{\sqrt{N ! \chi_{\Gamma}^{\pm}}} \sum_{\hat{\mathrm{P}}}(\pm 1)^{p} \zeta_{\hat{\mathrm{P}} \mathrm{\Gamma}}(\mathbf{r}) . \end{aligned}
The meaning of particle permutation can be seen from the fact that a vector is an ordered set. In the present case the first element of the vector of coordinates, or of the vector of position-momentum labels, is associated with the first particle, the second element with the second particle, etc. Hence the particle permutator $\hat{\mathrm{P}}$ can be applied to one vector relative to the other.

In the above expression for the symmetrized wave packet, the sum is over all $N$ ! permutations of the $N$ particles. Recall that we sometimes write the combined position and momentum labels as $\boldsymbol{\Gamma} \equiv{\mathbf{q}, \mathbf{p}}$. Here and below $p(\hat{\mathrm{P}})$ is the number of pair transpositions that comprise the permutation operator $\hat{\mathrm{P}}$ (or simply their parity). The plus sign is for bosons and the minus sign is for fermions.

By inspection one sees that the symmetrized wave functions show the mandated behavior for bosons and fermions under particle interchange,
$$\zeta_{\Gamma}^{\pm}(\hat{P} \mathbf{r})=(\pm 1)^{p} \zeta_{\Gamma}^{\pm}(\mathbf{r}) .$$

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

Before exploring the general properties of the symmetrization factor, it may be worth illustrating the nature of particle permutations with a simple example.

Because the wave function is the product of individual particle wave packets, it is simplest to focus on a single pair transposition, since all permutations can be decomposed into consecutive pair transpositions. Let $\hat{P}{j k}$ transpose particles $j$ and $k$, so that \begin{aligned} \zeta{\Gamma}\left(\hat{\mathbf{P}}{j k} \mathbf{r}\right) &=\zeta{\Gamma_{1} \ldots \Gamma_{j} \ldots \Gamma_{k} \ldots}\left(\mathbf{r}{1} \ldots \mathbf{r}{k} \ldots \mathbf{r}{j \ldots} \ldots\right) \ &=\zeta{\Gamma_{1}}^{(1)}\left(\mathbf{r}{1}\right) \ldots \zeta{\Gamma_{j}}^{(1)}\left(\mathbf{r}{k}\right) \ldots \zeta{\mathbf{r}{k}}^{(1)}\left(\mathbf{r}{j}\right) \ldots \end{aligned}
The so-called dimer symmetrization or overlap factor consists of the inner product of the original and the transposed wave function,
$\chi_{j k}^{\pm,(2)} \equiv \pm\left\langle\zeta_{\boldsymbol{\Gamma}}\left(\hat{\mathbf{P}}{j k} \mathbf{r}\right) \mid \zeta{\boldsymbol{\Gamma}}(\mathbf{r})\right\rangle$
$=\pm\left\langle\zeta_{\Gamma_{j}}^{(1)}\left(\mathbf{r}{k}\right) \zeta{\mathbf{\Gamma}{k}}^{(1)}\left(\mathbf{r}{j}\right) \mid \zeta_{\Gamma_{j}}^{(1)}\left(\mathbf{r}{j}\right) \zeta{\mathbf{r}{k}}^{(1)}\left(\mathbf{r}{k}\right)\right\rangle$
$=\pm\left|\left\langle\zeta_{\Gamma_{j}}^{(1)} \mid \zeta_{\Gamma_{k}}^{(1)}\right\rangle\right|^{2}$
$=\frac{\pm 1}{\left(2 \pi \xi^{2}\right)^{3}} \mid \int \mathrm{d} \mathbf{r}{j} e^{-\left(\mathbf{r}{j}-\mathbf{q}{k}\right)^{2} / 4 \xi^{2}} e^{\mathbf{p}{k}-\left(\mathbf{r} /-\mathbf{q}{k}\right) / / \mathrm{in}}$ $\times\left. e^{-\left(\mathbf{r}{j}-\mathbf{q}{j}\right)^{2} / 4 \xi^{2}} e^{-\mathbf{p}{j}-\left(\mathbf{r}{j}-\mathbf{q}{j}\right) / i^{\prime}}\right|^{2}$
$=\pm \exp \left{\frac{-1}{4 \xi^{2}}\left(\mathbf{q}{k}-\mathbf{q}{j}\right)^{2}-\frac{\xi^{2}}{\hbar^{2}}\left(\mathbf{p}{k}-\mathbf{p}{j}\right)^{2}\right} .$
The second equality follows because the unpermuted single-particle wave packets are normalized and so their respective inner product each gives a factor of unity. The third equality follows because what remains in the product of two integrals over the dummy variables $\mathbf{r}{j}$ and $\mathbf{r}{k}$, respectively, and these integrals are the complex conjugate of each other. The final equality shows that for the general wave packet the dimer overlap factor is evidently an un-normalized Gaussian in position and momentum that ties the two particles together.

## 物理代写|统计力学代写Statistical mechanics代考|Wave packet symmetrization and overlap

$$\zeta_{\Gamma}^{\pm}(\mathbf{r}) \equiv \frac{1}{\sqrt{N ! \chi_{\Gamma}^{\pm}}} \sum_{\hat{\mathrm{P}}}(\pm 1)^{p} \zeta_{\Gamma}(\hat{\mathrm{P}} \mathbf{r}) \equiv \frac{1}{\sqrt{N ! \chi_{\Gamma}^{\pm}}} \sum_{\hat{\mathrm{P}}}(\pm 1)^{p} \zeta_{\hat{\mathrm{P}}}(\mathbf{r}) .$$

$$\zeta_{\Gamma}^{\pm}(\hat{P} \mathbf{r})=(\pm 1)^{p} \zeta_{\Gamma}^{\pm}(\mathbf{r}) .$$

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

$$\zeta \Gamma\left(\hat{\mathbf{P}}{j k \mathbf{r}}\right)=\zeta \Gamma{1} \ldots \Gamma_{j} \ldots \Gamma_{k} \ldots(\mathbf{r} 1 \ldots \mathbf{r} k \ldots \mathbf{r} j \ldots \ldots)=\zeta \Gamma_{1}^{(1)}(\mathbf{r} 1) \ldots \zeta \Gamma_{j}^{(1)}(\mathbf{r} k)^{\prime}$$

$\chi_{j k}^{\pm,(2)} \equiv \pm\left\langle\zeta_{\boldsymbol{\Gamma}}(\hat{\mathbf{P}} j k \mathbf{r}) \mid \zeta \boldsymbol{\Gamma}(\mathbf{r})\right\rangle$
$=\pm\left\langle\zeta_{\Gamma_{j}}^{(1)}(\mathbf{r} k) \zeta \mathbf{\Gamma} k^{(1)}(\mathbf{r} j) \mid \zeta_{\Gamma_{j}}^{(1)}(\mathbf{r} j) \zeta \mathbf{r} k^{(1)}(\mathbf{r} k)\right\rangle$
$=\pm\left|\left\langle\zeta_{\Gamma_{j}}^{(1)} \mid \zeta_{\Gamma_{k}}^{(1)}\right\rangle\right|^{2}$
$=\frac{\pm 1}{\left(2 \pi \xi^{2}\right)^{3}}\left|\int \mathrm{d} \mathbf{r} j e^{-(\mathbf{r} j-\mathbf{q} k)^{2} / 4 \xi^{2}} e^{\mathbf{p} k-(\mathbf{r} /-\mathbf{q} k) / / \mathrm{in}} \times e^{-(\mathbf{r} j-\mathbf{q} j)^{2} / 4 \xi^{2}} e^{-\mathbf{p} j-(\mathbf{r} j-\mathbf{q} j) / i^{t}}\right|^{2}$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 物理代写|统计力学代写Statistical mechanics代考|PHYS3034

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
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|Wave packets as eigenfunctions in the classical limit

The primary goal here is to show how to get from the quantum picture to classical statistical mechanics, which is of course formulated in the classical phase space of the particles’ positions and momenta. To this end we use wave packets because they are localized simultaneously in position and momentum. Some care must be taken with using these as a basis, as there are issues with their orthogonality and completeness. Also the meaning of simultaneous localization must be reconciled with the Heisenberg uncertainty principle.

Consider a system consisting of $N$ particles in three dimensions, with position representation coordinates $\mathbf{r}=\left{\mathbf{r}{1}, \mathbf{r}{2}, \ldots, \mathbf{r}{N}\right}, \mathbf{r}{j}=\left{r_{j x}, r_{j y}, r_{j z}\right}, j=1,2, \ldots, N$. Define analogous vectors for the position label $\mathbf{q}$, and the momentum label $\mathbf{p}$. It will sometimes be convenient to write the combined labels as $\mathbf{\Gamma} \equiv{\mathbf{q}, \mathbf{p}}$. The representation coordinates $\mathbf{r}$ belong to the real continuum. The labels are discretized, for example $q_{j \alpha}=\ell_{\mathrm{q}, j \alpha} \Delta_{\mathrm{q}}$ and $p_{j a}=\ell_{\mathrm{p}, j \alpha} \Delta_{\mathrm{p}}$, with the $\ell_{\mathrm{q}}$ and $\ell_{\mathrm{p}}$ being integers. Below $\Gamma$ and $\ell$ are used interchangeably to label the system state. The meaning of position and momentum labels emerges from the following analysis.

At this stage we do not need to be more precise about the grid spacing or the system size. Messiah (1961, chapter V, section 11) insists that in order for the momentum operator to be Hermitian periodic boundary conditions must be imposed on all wave functions, and discrete momentum eigenvalues must have spacing $\Delta_{\mathrm{p}}=2 \pi \hbar / L$ where $I$. is the system edge length. This is true if the wave function does not go to zero at the boundaries. In any case, here we do not explicitly use this (but we consistently do so in later chapters), but instead observe that for a macroscopic system, $L \rightarrow \infty$, the present grid spacing could be made an integer multiple of this.

Spin could be included in the set of commuting dynamical variables, with $\mathbf{x}{j}=\left{\mathbf{r}{j}, \sigma_{j}\right}$, where $\sigma_{j} \in{-S,-S+1, \ldots, S}$ is the $z$-component of the spin of particle j. See Messiah (1961, section 14.1), or Merzbacher (1970, section 20.5), or section $6.4$ below. However, we do not include spin in the present analysis.

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

The probability amplitude in the position representation is
$$\zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r})^{*} \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r})=\frac{1}{C^{2}} e^{-(\mathbf{r}-\mathbf{q})(\mathbf{r}-\mathbf{q}) / 2 \xi^{2}} .$$
This is a Gaussian of width $\xi$ (per particle, per direction), peaked at the position label $\mathbf{q}$. Similarly in the momentum representation, the probability amplitude has width $\hbar / 2 \xi$ and is peaked at the momentum label p. To within an error of these magnitudes, the minimum uncertainty wave function is approximately a simultaneous eigenfunction of the position and momentum operators,
$$\hat{\mathbf{q}} \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r})=\mathbf{r} \zeta_{\mathbf{q}, \mathrm{p}}(\mathbf{r}) \approx \mathbf{q} \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}),$$
and
$$\hat{\mathbf{p}} \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r})=\left[\frac{\mathrm{i} \hbar}{2 \xi^{2}}(\mathbf{r}-\mathbf{q})+\mathbf{p}\right] \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r}) \approx \mathbf{p} \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r}) .$$
(Recall that the momentum operator is $\hat{\mathbf{p}}=-i \hbar \nabla_{\mathbf{r}}$.) Since the probability amplitude is sharply peaked about ${\mathbf{q}, \mathbf{p}}$, the approximation in these is to neglect contributions $\mathcal{O}(\xi)$ and $\mathcal{O}(\hbar / 2 \xi)$, respectively.

Because the wave packet is an approximate simultaneous position and momentum eigenfunction, for any sufficiently slowly varying phase function one can write $\int(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}) \approx f(\mathbf{q}, \mathbf{p}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r})$. In particular, the wave packet is an approximate energy eigenfunction,
$$\hat{H}(\mathbf{r}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}) \equiv H(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}) \approx H(\mathbf{q}, \mathbf{p}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}) .$$
The eigenvalue is the classical Hamiltonian energy function of phase space, which is usually split into kinetic and potential energies,
$$H(\mathbf{q}, \mathbf{p})=\mathcal{K}(\mathbf{p})+U(\mathbf{q}), \quad \mathcal{K}(\mathbf{p})=\frac{p^{2}}{2 m}=\frac{1}{2 m} \sum_{j, \alpha} p_{j \alpha}^{2} .$$

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

$$\zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r})^{*} \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r})=\frac{1}{C^{2}} e^{-(\mathbf{r}-\mathbf{q})(\mathbf{r}-\mathbf{q}) / 2 \xi^{2}}$$

$$\hat{\mathbf{q}} \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r})=\mathbf{r} \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}) \approx \mathbf{q} \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r})$$

$$\hat{\mathbf{p}} \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r})=\left[\frac{\mathrm{i} \hbar}{2 \xi^{2}}(\mathbf{r}-\mathbf{q})+\mathbf{p}\right] \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r}) \approx \mathbf{p} \zeta_{\mathrm{q}, \mathrm{p}}(\mathbf{r})$$
(回想一下，动量算子是 $\hat{\mathbf{p}}=-i \hbar \nabla_{\mathbf{r}}$.) 由于概率幅度在大约 $\mathbf{q}, \mathbf{p}$ ，其中的近似值是忽略贡献 $\mathcal{O}(\xi)$ 和 $\mathcal{O}(\hbar / 2 \xi)$ ，分别。

$$\hat{H}(\mathbf{r}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}) \equiv H(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r}) \approx H(\mathbf{q}, \mathbf{p}) \zeta_{\mathbf{q}, \mathbf{p}}(\mathbf{r})$$

$$H(\mathbf{q}, \mathbf{p})=\mathcal{K}(\mathbf{p})+U(\mathbf{q}), \quad \mathcal{K}(\mathbf{p})=\frac{p^{2}}{2 m}=\frac{1}{2 m} \sum_{j, \alpha} p_{j \alpha}^{2}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。