# 物理代写|费曼图代写Feynman diagram代考|PHYS501

## 物理代写|费曼图代写Feynman diagram代考|High density limit

Under what conditions could the Coulomb interaction between electrons be treated as a small perturbation? To answer this question. we define a dimensionless parameter $r_{\mathrm{y}}$ by
$$(4 \pi / 3) r_s^3 a_0^2=V / N .$$
where $a_0=\hbar^2 /\left(m e^2\right.$ ) is the Bohr radius (in SI units. $\left.e^2 \rightarrow e^2 / 4 \pi \varepsilon_0\right) . r_s a_0$ is the radius of a sphere whose volume is equal to the average volume occupied by one electron. Defining the dimensionless quantities
$$V^{\prime}=V /\left(r_s a_0\right)^3 . \quad \mathbf{K}=r_s a_0 \mathbf{k} . \quad \mathbf{Q}=r_s a_0 \mathbf{q} .$$
we may recast the Hamiltonian in Eq. $(4.11)$ into the following form:
$$H=\frac{e^2}{r_s^2 a_0}\left[\sum_{\mathrm{K} / r} K^2 c_{\mathrm{k} \sigma}^{\dagger} c_{\mathrm{k} \sigma}+\frac{r_{\mathrm{s}}}{V^{\prime}} \sum_Q^{\prime} \sum_{\mathrm{k} \sigma} \sum_{\mathrm{k}^{\prime} \sigma^{\prime}} \frac{2 \pi}{Q^2} c_{\mathrm{k}+\mathrm{q} \sigma}^{\dagger} c_{\mathrm{k}^{\prime}-\mathbf{q} \sigma^{\prime}}^{\dagger} c_{\mathbf{k}^{\prime} \cdot \sigma^{\prime} c_{\mathrm{k}}}\right]{(4.13)}$$ This expression for $H$ is very telling: compared to the kinetic energy of electrons, the Coulomb interaction is negligible in the high density limit, $r_s \rightarrow 0$. This conclusion appears to be counterintuitive, but a moment’s reflection reveals its validity. Coulomb repulsion scales as $1 / r{\mathrm{s}}$, and from Heisenberg’s uncertainty principle, the electron’s momentum also scales as $1 / r_{\text {s }}$. Therefore, the kinetic energy scales as $1 / r^2$. Thus, as $r_s \rightarrow 0$, even though the Coulomb energy grows larger. the kinetic energy of the electrons grows larger at a faster rate. We conclude that in the high-density limit. the Coulomb repulsion is weak in comparison with the kinetic energy, and it is permissible to treat it within the framework of perturbation theory. In real metals, $r_s=2-6$, which is neither too small nor too large. Nevertheless, in most metals, the single-particle approximation explains many of their low energy properties. This is because the Coulomb interaction, even when it is strong, is not very effective at changing the momentum distribution of the electrons: most of the states into which they could scatter are already occupied.

## 物理代写|费曼图代写Feynman diagram代考|First order perturbation

Treating $V_C$ as a perturbation, the energy per electron in the ground state is written as a perturbation series
$$E / N=E_0 / N+E_1 / N+E_2 / N+\cdots$$
$E_1$ is given by
$$E_1=\frac{1}{2 V} \sum_q^{\prime} \sum_{\mathrm{k} \sigma} \sum_{\mathrm{k}^{\prime} \sigma^{\prime}} \frac{4 \pi e^2}{q^2}\left\langle F\left|c_{\mathrm{k}+\mathrm{q} \sigma}^{\dagger} c_{\mathrm{k}^{\prime}-\mathrm{q}^{\prime}}^{\dagger} \cdot c_{\mathrm{k}^{\prime} \sigma^{\prime} \cdot c_{\mathrm{k} \sigma}}\right| F\right\rangle .$$
The action of $c_{\mathbf{k}^{\prime} \sigma} \cdot c_{\mathrm{k} \sigma}$ on $|F\rangle$, for $k, k^{\prime}k_F$. Hence
$$c_{\mathrm{k} \pi}^{\dagger} c_{\mathrm{k} \sigma}|F\rangle=\theta\left(k_F-k\right)|F\rangle$$

## 物理代写|费曼图代写Feynman diagram代考|High density limit

$$(4 \pi / 3) r_s^3 a_0^2=V / N .$$

$$V^{\prime}=V /\left(r_s a_0\right)^3 . \quad \mathbf{K}=r_s a_0 \mathbf{k} . \quad \mathbf{Q}=r_s a_0 \mathbf{q} .$$

## 物理代写|费曼图代写Feynman diagram代考|First order perturbation

$$E / N=E_0 / N+E_1 / N+E_2 / N+\cdots$$
$E_1$ 是 (谁) 给的
$$E_1=\frac{1}{2 V} \sum_q^{\prime} \sum_{\mathrm{k} \sigma} \sum_{\mathbf{k}^{\prime} \sigma^{\prime}} \frac{4 \pi e^2}{q^2}\left\langle F\left|c_{\mathrm{k}+4 \sigma}^{\dagger} c_{\mathrm{k}^{\prime}-\mathrm{q}^{\prime}}^{\dagger} \cdot c_{\mathrm{k}^{\prime} \sigma^{\prime} \cdot c_{\mathrm{k} \sigma}}\right| F\right\rangle .$$

$$c_{\mathrm{k} \pi}^{\dagger} c_{\mathrm{k} \sigma}|F\rangle=\theta\left(k_F-k\right)|F\rangle$$

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