---
# System prepended metadata

title: 固態物理第二次作業

---

# 固態物理第二次作業
    王榮椿109022201
[TOC]
## 問題一：
- 已知一維材料個子點分佈為$\rho(r)=\sum_n\delta(r-an)$ ，且有一入射光為$e^{ikr}$ ，k表示為wavevector。利用一維繞射的方式證明下列關係式，並進而推廣到三維空間。
$$F[\rho(r)]=\dfrac{|a|}{2\pi}\int_{0}^{\dfrac{2\pi}{|a|}} \rho(r)e^{ikr}\, dr =\dfrac{|a|}{2\pi}\int_{0}^{\dfrac{2\pi}{|a|}}\sum_n\delta(r-an) e^{ikr}\, dr \\
=\sum_G\delta(k-G)$$ 

- How to solve?
1.已知入射光為$e^{ikr}$ ，根據Tight Binding Chain model，以及Bragg Law我們知道，經過晶格後的繞射光線$e^{ik'r}$ ,k 與k' 的關係。
2.利用光線相位相干來計算繞射結果。有了1.的觀念才會知道為何入射光線的k會最終與晶格中wavevector space的G有關，但這裡我們就不多做說明。
$$F = \sum e^{ik'r}$$其中$k'=k-ma \Delta k$ ，並且所有的k'都有一個overall number k 且$|e^{ikr}|=1$，因此我們將F除以$e^{ikr}$整理成下面的樣子
$$F = \sum_{m=0}^{M-1} e^{-ima \Delta kr}=\dfrac{1-e^{-iMa \Delta kr}}{1-e^{-ia \Delta kr}}$$
光強度大小為$|F|^2$
$$|F|^2=\dfrac{1-e^{-iMa \Delta kr}}{1-e^{-ia \Delta kr}}*\dfrac{1-e^{-iMa \Delta kr}}{1-e^{-ia \Delta kr}}\\
=\dfrac{1-Re[e^{-iMa \Delta kr}]}{1-Re[e^{-ia \Delta kr}]}=\dfrac{\sin^2(\dfrac{M}{2}a \Delta k)}{\sin^2(\dfrac{1}{2}a \Delta k)}$$
3. 接著我們讓chatgpt來幫忙：

```python =
import numpy as np
import matplotlib.pyplot as plt

# Define constants
#m = 10
a = 0.001
M = 100

# Define function for |F|^2
def f(x):
    return np.sin(M/2 * a * x)**2 / np.sin(1/2 * a * x)**2

# Define x-axis values
x = np.linspace(0, np.pi/a*10, 1000)

# Compute y-axis values
y = f(x)

# Plot figure
fig, ax = plt.subplots()
ax.plot(a*x/np.pi, y)
ax.set_xlabel('a*Δk(π)')
ax.set_ylabel('|F|^2 (Intensity)')
ax.set_title('Plot of |F|^2 vs. a*Δk for M=100')
plt.show()
```
![](https://i.imgur.com/TudFzmf.png)
因此我們發現$a \Delta k = 2n\pi$時 $|F|^2$才有值。且一維的$G=\dfrac{ 2n\pi}{a}$ ,因此可以證明
$$F[\rho(r)]=\sum_G\delta(\Delta k-G) \Rightarrow \Delta k=G$$


## 問題二：
- 已知繞射前強度為$F$,則$F = NS_\vec{G}$，證明下列關係：
$$S_\vec{G} = \sum_{j} f_je^{-i\vec{G}\cdot\vec{r_j}} $$
where $\vec{r_j}$ is reletive position in atoms and$f_j$ is:
$$f_j =\int_{cell} \rho(\vec{\rho})e^{-i\vec{G}\cdot\vec{\rho}}\, dV$$ where $\vec{\rho} =$mass center position

- How to solve?
1.as we know $F[\rho(\vec{r})]=\int \rho(\vec{r})e^{i\vec{k}\cdot\vec{r}}\, d\vec{r}$ and $\vec{r} =\vec{\rho}+\vec{r_j}$
2.And we know $\rho(\vec{r}) =\sum\delta(\vec{r_j})$ 
3.Thus we have the formula $F[\rho(\vec{r})]=N\sum_\vec{r_j}\int_{unit-cell} \rho(\vec{\rho}+\vec{r_j})e^{i(\vec{\rho}+\vec{r_j})\cdot\vec{k}}\, d\vec{r}$
4.From qeusion No.1, we can let $F$ divide by overall factor $e^{i\vec{k}\cdot\vec{r}}$ and get $\Delta \vec{k}=\vec{G}$ 

$$F[\rho(\vec{r})]=N\sum_\vec{r_j}e^{i\vec{r_j}\cdot\vec{k'}}\int_{unit-cell} \rho(\vec{\rho})e^{i\vec{k'}\cdot\vec{\rho}}\, d\vec{\rho}\\
\Rightarrow F[\rho(\vec{r})]=N\sum_\vec{r_j}e^{i\vec{r_j}\cdot(-\Delta \vec{k})}\int_{unit-cell} \rho(\vec{\rho})e^{i(-\Delta \vec{k})\cdot\vec{\rho}}\, d\vec{\rho}\\
\Rightarrow F[\rho(\vec{r})]=N\sum_\vec{r_j}e^{-i\vec{r_j}\cdot\vec{G}}\int_{unit-cell} \rho(\vec{\rho})e^{-i\vec{G}\cdot\vec{\rho}}\, d\vec{\rho}\\
\Rightarrow F[\rho(\vec{r})]=NS_\vec{G}=N\sum_{j} f_je^{-i\vec{G}\cdot\vec{r_j}}$$
................prove end.......