# 物理代写|费曼图代写Feynman diagram代考|PHY610

## 物理代写|费曼图代写Feynman diagram代考|The electron gas

A metallic crystal has a large number of mobile electrons, of the order of Avogadro’s number, and a correspondingly large number of ions. If our interest is in the bulk properties of a crystal, we may take the volume $V$ of the crystal to be infinite, and the number of electrons $N$ to be infinite, while keeping $N / V$, the number density of electrons. finite; this is called the thermodynamic limit. The ions incessantly vibrate about their equilibrium positions, but due to their large mass, they move very slowly in comparison with the electrons, so that the electrons quickly adjust their state to reflect whatever positions the ions occupy at any given time. Consequently, to a good approximation, one may solve the Schrödinger equation for electrons by assuming that the ions are fixed; this is the Born-Oppenheimer approximation. The influence of the ionic vibrations on the electronic states, described through the electron-phonon interaction, may be treated by perturbation theory: this is discussed in Chapter $11 .$

A more drastic approximation in the description of a metal is to replace the mesh of positive ions with a uniform positive background, which results in the socalled jellium model. In a model such as this. any results obtained are necessarily qualitative in nature. In this chapter. we study the jellium model. One of our goals in this study is to show that the divergent term in the Coulomb interaction. corresponding to $\mathbf{q}=\mathbf{0}$ (see Eq. [3.29]), is cancelled by contributions to the total energy from the positive background. This cancellation is a consequence of the charge neutrality of the crystal. and it holds true even if the approximation of a uniform positive background is relaxed. Another goal of this chapter is to show the neccssity of performing perturbation expansions to higher. generally infinite, orders. We will later proceed to study Green’s functions, whereby such a program may be carried out more easily.

## 物理代写|费曼图代写Feynman diagram代考|The Hamiltonian in the jellium model

Let us consider the jellium model in thermodynamic limit: $N \rightarrow \infty, V \rightarrow \infty$, while $N / V$ remains constant. The Hamiltonian consists of three terms.
$$H=H_e+H_b+H_{c-b} .$$
The first term is the sum of the kinetic energies of the electrons and their Coulomb interactions. From Eqs (3.19) and (3.29).
$$H_c=\sum_{\mathrm{k} \sigma} \frac{\hbar^2 k^2}{2 m} c_{k \sigma}^{\dagger} c_{k \sigma}+\lim {\mu \rightarrow 0} \frac{1}{2 V} \sum_q \sum{k \sigma} \sum_{k^{\prime} \sigma^{\prime}} \frac{4 \pi e^2}{q^2+\mu^2} c_{k+q \sigma}^{\dagger} c_{k^{\prime}-q \sigma^{\prime}}^{\dagger} c_{k^{\prime} \sigma^{\prime}} c_k \sigma .$$
The second term. $H_h$, represents the Coulomb energy of the uniform positive background. To find the correct expression for $\mathrm{H}b$, consider a collection of point charges $q_1, q_2 \ldots$ at positions $\mathbf{r}_1, \mathbf{r}_2, \ldots$ Their Coulomb energy is $$E{\text {Coul }}=\frac{1}{2} \sum_{i \neq j} \frac{q_i q i}{\left|\mathbf{r}i-\mathbf{r}_j\right|} \quad \text { (cgs units). }$$ The factor 1/2 ensures that pairs of point charges are counted only once. For a continuous charge distribution, $q_i$ is replaced by $\rho\left(\mathbf{r}_i\right) d^3 r_i$, where $\rho(\mathbf{r})$ is the chatge density, and the summation is replaced by integration. Therefore, $$H_b=\lim {\mu \rightarrow 0^{+}} \frac{1}{2} \int \frac{\rho(\mathbf{r}) \rho\left(\mathbf{r}^{\prime}\right) e^{-\mu\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d^3 r d^3 r^{\prime}=\lim _{\mu \rightarrow 0^{+}} \frac{N^2 e^2}{2 V^2} \int \frac{e^{-\mu\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}}{\mid \mathbf{r}-\overline{\mathbf{r}^{\prime} \mid}} d^3 r d^3 r^{\prime} .$$

## 物理代写|费曼图代写Feynman diagram代考|The Hamiltonian in the jellium model

$$H=H_e+H_b+H_{c-b} .$$

$$H_c=\sum_{\mathrm{k} \sigma} \frac{\hbar^2 k^2}{2 m} c_{k \sigma}^{\dagger} c_{k \sigma}+\lim \mu \rightarrow 0 \frac{1}{2 V} \sum_q \sum k \sigma \sum_{k^{\prime} \sigma^{\prime}} \frac{4 \pi e^2}{q^2+\mu^2} c_{k+q \sigma}^{\dagger} c_{k^{\prime}-q \sigma^{\prime}}^{\dagger} c_{k^{\prime} \sigma^{\prime}} c_k \sigma .$$

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: