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• 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|Partition function

We might assume that the canonical partition function can be written in the form
$$Z_N=\sum_{m_1} \cdots \sum_{m_N} \exp \left(-\beta \sum_{n=1}^N E^{(n)}\right)$$
(wrong!)

Equation (5.43) is incorrect because it overcounts the allowed states of identical particles. The occupation number $n_k$ is the number of particles in the system having the eigenstate associated with eigenvalue $E_k$ (see Appendix D). The energy of the system can therefore be written not as a sum over particles, as in $E=\sum_{n=1}^N E^{(n)}$, but as a sum over energy levels,
$$E=\sum_m n_m E_m .$$
Equation (5.44) is an important step in setting up the statistical mechanics of identical particles. The occupation numbers must satisfy the constraint
$$\sum_m n_m=N .$$
Equations (5.44) and (5.45) are unrestricted sums over all possible energy states. ${ }^{24}$
How many ways can the energy $E$ be partitioned over the particles of the system? For a given set of occupation numbers $\left{n_k\right}$ satisfying Eq. (5.45), there are, using Eq. (3.12), $N ! / \prod_k\left(n_{k} !\right)$ ways of permuting the particles, which, by the indistinguishability of identical particles, are equivalent and have to be treated as a single state. To correct for overcounting, Eq. (5.43) should be written
$$Z_N=\frac{1}{N !} \sum_{m_1} \cdots \sum_{m_N} \prod_k\left(n_{k} !\right) \exp \left(-\beta \sum_{n=1}^N E^{(n)}\right) .$$
Equation (5.46) presents a challenging combinatorial problem because it connects the energy levels of individual particles, $E^{(n)}$, to the occupation numbers $n_k$, which apply to the entire system. To apply Eq. (5.46), we must know the occupation numbers associated with a system in thermal equilibrium, which is what we’re trying to solve for! At high temperature $(\beta \rightarrow 0)$ the number of energy levels that can make a significant contribution to the sum in Eq. (5.46) becomes quite large. One would expect, for fixed values of $N$ and $E$, that in this limit the occupation numbers will be predominately 0 or 1 and thus $\left(n_k\right) !=1$ for most configurations. If we set $\left(n_k\right) !=1$ in Eq. (5.46), we have an approximate expression for the partition function (which factorizes)
$$Z \approx \frac{1}{N !} \sum_{m_1} \cdots \sum_{m_N} \mathrm{e}^{-\beta \sum_{n=1}^N E^{(n)}}=\frac{1}{N !} \prod_{k=1}^N\left(\sum_{m_k} \mathrm{e}^{-\beta E^{(k)}}\right)=\frac{1}{N !}\left(Z_1\right)^N$$
where in the last step we’ve used that all particles are identical, where $Z_1$ is the partition function for a single particle. Equation (5.47) is the high-temperature limit of the partition function for bosons or fermions.

## 物理代写|统计力学代写Statistical mechanics代考|Equation of state, fugacity expansions

From Eq. (4.76), $\Phi=-k T \ln Z_G$. For a system with $V$ as the only relevant external parameter, $\Phi=-P V$, Eq. (4.69). For such a system, $P V=k T \ln Z_G$. Using Eq. (5.55) for $Z_G$, we have the equation of state:
$$P V=-\theta g k T \sum_k \ln \left(1-\theta \mathrm{e}^{\beta\left(\mu-E_k\right)}\right) .$$
The sum in Eq. (5.64) can be converted to an integral, $\sum_k \longrightarrow\left(V /\left(8 \pi^3\right)\right) \int \mathrm{d}^3 k$ (see Section 2.1.5). Thus,
$$P V=-\theta g k T \frac{V}{8 \pi^3} \int \mathrm{d}^3 k \ln \left(1-\theta \mathrm{e}^{\beta\left(\mu-E_k\right)}\right) .$$
Because $E_k$ is isotropic in $k$-space, work with spherical coordinates:
$$P V=-g \theta k T \frac{V}{8 \pi^3} \int_0^{\infty} 4 \pi k^2 \mathrm{~d} k \ln \left(1-\theta \mathrm{e}^{\beta\left[\mu-\hbar^2 k^2 /(2 m)\right]}\right) \text {. }$$
Change variables: Let $x \equiv \beta \hbar^2 k^2 /(2 m)$, a dimensionless variable. Then,
$$P=-\frac{g \theta k T}{\lambda_T^3} \frac{2}{\sqrt{\pi}} \int_0^{\infty} \mathrm{d} x \sqrt{x} \ln \left(1-\theta \mathrm{e}^{\beta \mu-x}\right) .$$
Integrate by parts:
$$P(T, \mu)=g \frac{k T}{\lambda_T^3} \frac{1}{\Gamma(5 / 2)} \int_0^{\infty} \frac{x^{3 / 2}}{\mathrm{e}^{x-\beta \mu}-\theta} \mathrm{d} x .$$
Note that Eq. (5.66) provides an expression for $P$ that’s intensive in character, $P=P(T, \mu)$ is independent of $V$. Pressure $P$, which has the dimension of energy density, is equal to $k T$, an energy, divided by $\lambda_T^3$, a volume, multiplied by a dimensionless function of $\mathrm{e}^{\beta \mu}$ (the integral in Eq. (5.66)).
The two equations implied by Eq. (5.66) (one for each of $\theta=\pm 1$ ) involve a class of integrals known as Bose-Einstein integrals for $\theta=1, G_n(z)$, defined in Eq. (B.18), and Fermi-Dirac integrals for $\theta=-1, F_n(z)$, defined in Eq. (B.29), where $z=\mathrm{e}^{\beta \mu}$. Written in terms of these functions, we have from Eq. (5.66),
$$P(T, \mu)=g \frac{k T}{\lambda_T^3} \begin{cases}G_{5 / 2}(z) & \theta=+1 \ F_{5 / 2}(z) & \theta=-1 .\end{cases}$$

# 统计力学代考

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

$$Z_N=\sum_{m_1} \cdots \sum_{m_N} \exp \left(-\beta \sum_{n=1}^N E^{(n)}\right)$$
(错误的! )

$$E=\sum_m n_m E_m .$$

$$\sum_m n_m=N .$$

$$Z_N=\frac{1}{N !} \sum_{m_1} \cdots \sum_{m_N} \prod_k\left(n_{k} !\right) \exp \left(-\beta \sum_{n=1}^N E^{(n)}\right) .$$

$$Z \approx \frac{1}{N !} \sum_{m_1} \cdots \sum_{m_N} \mathrm{e}^{-\beta \sum_{n 1}^N E^{(n)}}=\frac{1}{N !} \prod_{k=1}^N\left(\sum_{m_k} \mathrm{e}^{-\beta E^{(k)}}\right)=\frac{1}{N !}\left(Z_1\right)^N$$

## 物理代写|统计力学代写Statistical mechanics代考|Equation of state, fugacity expansions

$$P V=-\theta g k T \sum_k \ln \left(1-\theta \mathrm{e}^{\beta\left(\mu-E_k\right)}\right) .$$

$$P V=-\theta g k T \frac{V}{8 \pi^3} \int \mathrm{d}^3 k \ln \left(1-\theta \mathrm{e}^{\beta\left(\mu-E_k\right)}\right) .$$

$$P V=-g \theta k T \frac{V}{8 \pi^3} \int_0^{\infty} 4 \pi k^2 \mathrm{~d} k \ln \left(1-\theta \mathrm{e}^{\beta\left[\mu-\hbar^2 k^2 /(2 m)\right]}\right) .$$

$$P=-\frac{g \theta k T}{\lambda_T^3} \frac{2}{\sqrt{\pi}} \int_0^{\infty} \mathrm{d} x \sqrt{x} \ln \left(1-\theta \mathrm{e}^{\beta \mu-x}\right) .$$

$$P(T, \mu)=g \frac{k T}{\lambda_T^3} \frac{1}{\Gamma(5 / 2)} \int_0^{\infty} \frac{x^{3 / 2}}{\mathrm{e}^{x-\beta \mu}-\theta} \mathrm{d} x .$$

$$P(T, \mu)=g \frac{k T}{\lambda_T^3}\left{G_{5 / 2}(z) \quad \theta=+1 F_{5 / 2}(z) \quad \theta=-1 .\right.$$

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