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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|凝聚态物理代写condensed matter physics代考|Probing the Structure

In order to measure the detailed microscopic structure of condensed matter systems we need a probe with a wavelength comparable to the typical distance between nuclei. For photons of energy $E$, the wavelength is $\lambda=h c / E$ so that
$$\lambda \sim \frac{12.4 \AA}{E / \mathrm{keV}},$$
where keV is kilo-electron-volts. For neutrons, the de Broglie wavelength is $\lambda=h / \sqrt{2 M E}$, thus
$$\lambda \sim \frac{0.28}{(E / \mathrm{eV})^{1 / 2}} \AA,$$
where $M$ is the neutron mass. ${ }^1$ For electrons the wavelength is longer by a factor of the square root of the neutron-to-electron mass ratio:
$$\lambda \sim \frac{12 \AA}{(E / \mathrm{eV})^{1 / 2}}$$

It turns out that for typical condensed matter systems with an average distance between nuclei of $\sim$ 1-10 Å we would need to use X-ray photons in the keV energy range, electrons in the eV range, or thermal neutrons, like those that come out of the thermal moderator of a nuclear reactor. Lowenergy electrons have a very short mean free path in solids since the electron charge interacts so strongly with the charges in the solid. It turns out that it is possible to use high-energy electrons, provided that one can study their deflections through small angles (small momentum transfers). X-rays are weakly scattering and penetrate samples easily. The Born approximation for weak scattering is generally applicable, which is convenient both mathematically and for physical interpretation of the scattering intensity. X-rays are conveniently generated by accelerating electrons into a metal target. The wavelength can be tuned by adjusting the accelerating voltage and by choice of target material. Extremely intense X-ray beams of widely tunable energy are also generated at synchrotron radiation facilities.

Neutrons are very weakly interacting (which means that samples smaller than tens of grams in mass are difficult to work with), and can more easily provide information on magnetic structures of the system than X-rays (although the latter is also possible). X-ray diffraction is the most common technique, so we shall discuss it first, and then discuss neutron scattering at length a little later. It turns out that neutrons have advantages in inelastic scattering experiments that can tell us something about the excitations of condensed matter systems. This will be discussed in detail in Chapter 4. In the remainder of this chapter we will focus on X-ray scattering [17].

## 物理代写|凝聚态物理代写condensed matter physics代考|Semiclassical Theory of X-Ray Scattering

Let us begin by considering classical X-ray scattering by a free, classical, non-relativistic electron (Thomson scattering). We can ignore scattering by the nuclei since they are too massive to be much perturbed (i.e. accelerated) by the electromagnetic field. The electron is assumed to have speed $v / c \ll 1$, so it couples primarily to the electric field $\vec{\epsilon}$ (rather than the magnetic field) of the X-rays (recall that in the Bohr model of hydrogen $v / c=\alpha \sim 1 / 137$ in the ground state). Hence we neglect the Lorentz force and Newton’s law gives ${ }^2$
$$\delta \ddot{\vec{r}}=\frac{-e}{m_{\mathrm{e}}} \vec{\epsilon}(\vec{r}+\delta \vec{r}, t),$$
where $\vec{r}$ is the equilibrium position of the electron, and $\delta \vec{r}$ is the deviation caused by external force. Now consider a plane wave incident on the electron,
$$\vec{\epsilon}(\vec{r}, t)=\vec{E}{\mathrm{in}} e^{i \vec{k} \cdot \vec{r}} e^{-i \omega t} .$$ The oscillatory electric force induces a harmonically oscillating dipole: $$\vec{p}(t)=-e \delta \vec{r}(t)=-\frac{e^2}{m{\mathrm{e}} \omega^2} \vec{E}{\mathrm{in}} e^{i \vec{k} \cdot \vec{r}} e^{-i \omega t},$$ where we have assumed that $\vec{k} \cdot \delta \vec{r} \ll 1$ (or small oscillation), so the position dependence of the phase of electric field is negligible. The electric field radiated by this oscillating dipole at position $\vec{R}$ is ${ }^3$ $$\vec{\epsilon}{\mathrm{a}}=\frac{e^2}{m_{\mathrm{e}} c^2}\left[\hat{n} \times\left(\hat{n} \times \vec{E}_{\mathrm{in}}\right)\right] e^{i \vec{k} \cdot \vec{r}} e^{-i \omega t} \frac{e^{i k|\vec{R}-\vec{r}|}}{|\vec{R}-\vec{r}|},$$ where $\hat{n}=(\vec{R}-\vec{r}) /|\vec{R}-\vec{r}| \approx \vec{R} /|\vec{R}|$ is the unit vector along the direction of radiation, $e^2 / m_{\mathrm{e}} c^2 \equiv$ $r_{\mathrm{c}}=\alpha^2 a_{\mathrm{B}}$ is the “classical radius of the electron” and $k \equiv \omega / c$ (we assume that the frequency of the radiated wave is the same as that of the incident wave, i.e. we neglect the small inelasticity of the scattering that is present quantum mechanically due to the Compton effect). The factor of $|\vec{R}-\vec{r}|^{-1}$ gives the usual fall off of amplitude for spherical waves due to energy conservation. The factor of $r_{\mathrm{c}}$ cancels out the length units on the right hand side of Eq. (2.7), and it will turn out that the scattering cross section for photons is proportional to $r_{\mathrm{c}}^2$.

# 凝聚态物理代考

## 物理代写|凝聚态物理代写condensed matter physics代考|Probing the Structure

$$\lambda \sim \frac{12.4 \backslash \mathrm{AA}}{E / \mathrm{keV}},$$

$$\lambda \sim \frac{0.28}{(E / \mathrm{eV})^{1 / 2}} \backslash \mathrm{AA},$$

$$\lambda \sim \frac{12 \backslash \mathrm{AA}}{(E / \mathrm{eV})^{1 / 2}}$$

## 物理代写|凝聚态物理代写condensed matter physics代考|Semiclassical Theory of X-Ray Scattering

$$\ddot{\vec{r}}=\frac{-e}{m_{\mathrm{e}}} \vec{\epsilon}(\vec{r}+\delta \vec{r}, t),$$

$$\vec{\epsilon}(\vec{r}, t)=\vec{E} \text { in } e^{i \dot{k} \cdot r} e^{-i \omega t} .$$

$$\vec{p}(t)=-e \delta \vec{r}(t)=-\frac{e^2}{m \mathrm{e} \omega^2} \vec{E} \text { ine }^{i i \cdot \cdot r} e^{-i \omega t},$$

$$\vec{\epsilon} \mathrm{a}=\frac{e^2}{m_{\mathrm{e}} c^2}\left[\hat{n} \times\left(\hat{n} \times \vec{E}{\mathrm{in}}\right)\right] e^{i \dot{k} \cdot \vec{r}} e^{-i \omega t} \frac{e^{i k|\hat{R}-\vec{r}|}}{|\vec{R}-\vec{r}|},$$ 在哪里 $\hat{n}=(\vec{R}-\vec{r}) /|\vec{R}-\vec{r}| \approx \vec{R} /|\vec{R}|$ 是沿辐射方向的单位矢量， $e^2 / m{\mathrm{e}} c^2 \equiv r_{\mathrm{c}}=\alpha^2 a_{\mathrm{B}}$ 是“电子的经典半径”和 $k \equiv \omega / c$ (我们假设辐射波的频率与入射波的频 率相同，即我们忽略了由于康普顿效应而在量子力学上存在的散射的小非弹性)。的因素 $|\vec{R}-\vec{r}|^{-1}$ 由于能量守恒，给出了球面波振幅的通常下降. 的因素 $r_{\mathrm{c}}$ 取消等 式右侧的长度单位。(2.7)，结果表明光子的散射截面与 $r_{\mathrm{c}}^2$.

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