## 物理代写|凝聚态物理代写condensed matter physics代考|X-Ray Scattering from Crystals

We saw previously that, in the Born approximation, the X-ray scattering amplitude is proportional to the Fourier transform of the electron density at a wave vector $\vec{q}$ determined by the scattering angle: $\vec{k}{\text {final }}=\vec{k}{\text {initial }}+\vec{q}$. With our new knowledge about the reciprocal lattice we can see that it will play a central role in X-ray scattering structural determinations. If we assume zero temperature and classical atoms so that quantum (zero-point) and thermal fluctuations can be neglected, the electron density clearly has the periodicity of the lattice. Suppose we write it in the form
$$\rho(\vec{r})=\sum_i \rho_{\mathrm{a}}\left(\vec{r}-\vec{R}_i\right),$$

where the sum is over sites of the underlying Bravais lattice and $\rho_{\mathrm{a}}$ is the atomic electron density (in the crystalline environment). Let $F(\vec{q})$ be the X-ray form factor for the crystal,
$$F(\vec{q})=\int d^3 \vec{r} e^{-i \vec{q} \cdot \vec{r}} \sum_i \rho_{\mathrm{a}}\left(\vec{r}-\vec{R}i\right) .$$ It is easily seen that $$F(\vec{q})=W(\vec{q}) f(\vec{q}),$$ where $$W(\vec{q}) \equiv \sum_i e^{-i \vec{q} \cdot \vec{R}_i}$$ is the lattice form factor introduced in Chapter 2. As we show below, in a periodic crystal $W$ has the property that, if $\vec{q}$ is not an element of the reciprocal lattice, then $W$ vanishes. If $\vec{q}=\vec{G}$ then $e^{-i \vec{G} \cdot \vec{R}_i}=1$, which implies that $W$ diverges. We can see this by writing the lattice sum explicitly, $$W(\vec{q})=\sum{\ell, m, n} e^{-i \vec{q} \cdot\left(\ell \vec{a}1+m \vec{a}_2+n \vec{a}_3\right)}=N \sum{{\vec{G}]} \delta_{\vec{q}, \vec{G}},$$
or using the Poisson summation formula (see Appendix B) to obtain
\begin{aligned} W(\vec{q}) &=\sum_{\text {K,L,M }}(2 \pi)^3 \delta\left(2 \pi K-\vec{q} \cdot \vec{a}1\right) \delta\left(2 \pi L-\vec{q} \cdot \vec{a}_2\right) \delta\left(2 \pi M-\vec{q} \cdot \vec{a}_3\right) \ &=\frac{1}{|\omega|} \sum{{\vec{G}}}(2 \pi)^3 \delta^3(\vec{q}-\vec{G}) \end{aligned}

## 物理代写|凝聚态物理代写condensed matter physics代考|Effects of Lattice Fluctuations on X-Ray Scattering

Up until this point we have been assuming that the atoms are precisely located on the points of a perfect lattice. Using this assumption we found infinitely sharp diffraction beams characterized by $\vec{q}=\vec{G}$. Even neglecting lattice defects, we have been ignoring the effects of:
(1) finite temperature;
(2) quantum zero-point motion;
(3) dynamies (time-dependence).
There was a great debate early in the twentieth century about whether or not X-ray diffraction could ever be observed at finite temperatures (quantum fluctuations were unknown at the time). It was argued that at room temperature all the atoms would be randomly displaced by at least a small amount from their nominal lattice positions and this would totally destroy the diffraction pattern. This argument turns out to be incorrect, however, as we show in the following classical argument. (The full quantum derivation will be given later.)
Let us assume that thermal fluctuations cause
$$\vec{r}_j=\vec{R}_j+\vec{u}_j,$$

where $\vec{u}j$ is the displacement of the $j$ th atom from its nominal lattice position $\vec{R}_j$. Let us further assume that the time-dependence of $\vec{u}_j$ can be neglected. This is typically a good approximation for X-ray scattering, where the energy resolution $\Delta \epsilon$ is so poor that the atomic vibration periods are much longer than the corresponding temporal resolution given by the Heisenberg uncertainty relation, $\hbar / \Delta \epsilon$ In the presence of fluctuations, the lattice structure factor becomes $$W(\vec{q})=\sum_i e^{-i \vec{q} \cdot\left(\vec{R}_i+\vec{u}_i\right)} .$$ The scattered intensity is proportional to the square of this quantity: $$\left.I_L(\vec{q})=N S(\vec{q})=\left\langle|W(\vec{q})|^2\right\rangle\right\rangle=\sum{i, j} e^{i \vec{q} \cdot\left(\vec{R}_i-\vec{R}_j\right)}\left\langle\left\langle e^{i \vec{q} \cdot\left(\vec{u}_i-\vec{u}_j\right)}\right\rangle\right\rangle,$$
where $\langle\langle\cdot\rangle\rangle$ indicates the thermal expectation value. ${ }^{22}$
Thermal averaged quantities are perfectly translation-invariant (i.e. have the symmetry of the lattice), so we can take $\vec{R}_i=\overrightarrow{0}$ and multiply by a factor of $N$ :
$$I_L(\vec{q})=N \sum_j e^{-i \vec{q} \cdot \vec{R}_j}\left\langle\left\langle e^{i \vec{q} \cdot\left(\vec{u}_0-\vec{u}_j\right)}\right\rangle .\right.$$

## 物理代写|凝聚态物理代写condensed matter physics代考|X-Ray Scattering from Crystals

$$\rho(\vec{r})=\sum_i \rho_{\mathrm{a}}\left(\vec{r}-\vec{R}i\right),$$ 其中总和超过了底层 Bravais 晶格的位点，并且 $\rho{\mathrm{a}}$ 是原子电子密度（在结晶环境中）。让 $F(\vec{q})$ 是晶体的 $\mathrm{X}$ 射线形状因子，
$$F(\vec{q})=\int d^3 \vec{r} e^{-i \vec{q} \cdot \vec{r}} \sum_i \rho_{\mathrm{a}}(\vec{r}-\vec{R} i)$$

$$F(\vec{q})=W(\vec{q}) f(\vec{q}),$$

$$W(\vec{q}) \equiv \sum_i e^{-i \vec{q} \cdot \vec{R}i}$$ 是第 2 章中介绍的晶格形状因子。如下所示，在周期性晶体中 $W$ 有这样的性质，如果 $\vec{q}$ 不是倒易格的元 素，则 $W$ 消失。如果 $\vec{q}=\vec{G}$ 然后 $e^{-i \vec{G} \cdot \vec{R}_i}=1$ ，这意味着 $W$ 分歧。我们可以通过明确写出格和来看到这一 orusingthePoissonsummation formula(seeAppendix $B$ ) toobtain $$W(\vec{q})=\sum{K, \mathrm{~L}, \mathrm{M}}(2 \pi)^3 \delta(2 \pi K-\vec{q} \cdot \vec{a} 1) \delta\left(2 \pi L-\vec{q} \cdot \vec{a}_2\right) \delta\left(2 \pi M-\vec{q} \cdot \vec{a}_3\right) \quad=\frac{1}{|\omega|} \sum \vec{G}(2 \pi)^3 \delta^3(\vec{q}-\vec{G})$$

## 物理代写|凝聚态物理代写condensed matter physics代考|Effects of Lattice Fluctuations on X-Ray Scattering

(1) 有限温度；
(2) 量子零点运动;
(3) 动态 (时间(依赖性)。

$$\vec{r}_j=\vec{R}_j+\vec{u}_j,$$

$$W(\vec{q})=\sum_i e^{-i \vec{q} \cdot\left(\vec{R}_i+\vec{u}_i\right)} .$$

$$\left.\left.I_L(\vec{q})=N S(\vec{q})=\left\langle|W(\vec{q})|^2\right\rangle\right\rangle=\sum i, j e^{i \cdot \cdot\left(\vec{R}_i-\vec{R}_j\right.}\right)\left\langle\left\langle e^{i \vec{q} \cdot\left(\vec{u}_i-\vec{u}_j\right)}\right\rangle\right\rangle,$$

$$I_L(\vec{q})=N \sum_j e^{-i \vec{q} \cdot \vec{R}_j}\left\langle\left\langle e^{i \vec{q} \cdot\left(\vec{u}_0-\vec{u}_j\right)}\right\rangle .\right.$$

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