---
title: "Lecture Note 03"
author: "sin-iu ho"
date: "9 March 2020"
header-includes:
output:
pdf_document
---
# Lecture Note 03
###### tags: `Quantum Physics II`
###### date: `9 March 2020`
---
## Additional Topics
1. [Feynman diagram講義](http://www.fisica.uniud.it/~cobal/Site/PIF_7_Feynman.pdf)
2. [natural unit](https://en.wikipedia.org/wiki/Natural_units): Let $\hbar=c=1$
3. [21cm Line](https://en.wikipedia.org/wiki/Hydrogen_line) $\leftrightarrow$ 1/million precision
4. 把一個介子中的b夸克換成c夸克，外面的反u夸克不會感覺到
$\left<b|\hat{H'}|c\right>\propto\left<b|c\right>$

## §7.1.2: First-Order Theory 一階理論 ([Review](https://hackmd.io/@ulynx/HJwNZ1ANU))
- Reference
- **[Textbook §7.1](https://hackmd.io/@ulynx/rym_o5_8I) (editing)**
- Wikipedia: [Perturbation theory (quantum mechanics)#First order corrections](https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#First_order_corrections)

- Comments
- ==有關在課本6.13式後面加一項 $c\psi^0_n$ 以及產生之相 $e^{i\lambda\tilde{c}}$ 可以消掉的問題==
- Conclusion
- $\displaystyle \boxed{E^1_n=\left<\psi^0_n\middle|\hat{H'}\middle|\psi^0_n\right>}\tag{7.9}$
- $\displaystyle \boxed{\psi^1_n=\sum_{m\neq n}\dfrac{\left<\psi^0_m\middle|\hat{H'}\middle|\psi^0_n\right>}{E^0_n-E^0_m}\psi^0_m}\tag{7.13}$

## §7.1.3: Second-Order Energies 二階能量
- Reference
- **[Textbook §7.1](https://hackmd.io/@ulynx/rym_o5_8I) (editing)**
- Wikipedia: [Perturbation theory (quantum mechanics)#Second-order and higher corrections](https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#Second-order_and_higher_corrections)
- 基礎量子力學(二)/姚珩
- 7-3. 無退化狀態二級能量的微擾修正・[Youtube](http://ocw.lib.ntnu.edu.tw/mod/resource/view.php?id=9513)・[講義](http://ocw.lib.ntnu.edu.tw/mod/resource/view.php?id=9514)
- Conclusion
$\displaystyle \boxed{E^2_n=\sum_{m\neq n}\dfrac{\left|\left<\psi^0_m\middle|\hat{H'}\middle|\psi^0_n\right>\right|^2}{E^0_n-E^0_m}}\tag{7.15}$

## §7.2 Degenerate Perturbation Theory 簡併微擾理論
- Reference
- **[Textbook §7.2](https://hackmd.io/nY58RMEyQceIaEoIe-_a_A?view#%C2%A7721-Two-fold-degenracy-%E4%BA%8C%E9%87%8D%E7%B0%A1%E4%BD%B5) (editing)**
- 基礎量子力學(二)/姚珩
- 7-5. 有退化狀態能量與波函數的微擾修正・[Youtube](http://ocw.lib.ntnu.edu.tw/mod/resource/view.php?id=9620)・[講義](http://ocw.lib.ntnu.edu.tw/mod/resource/view.php?id=9621)
- Idea
代表7.9與7.15式分母的兩個eigenvalue相等，相減為零
分母為零，分子也要為零才「比較合理」
也就是 $\left<\psi^0_m|\hat{H'}|\psi^0_n\right>$ 為零
==用矩陣表示，這代表 $\left\{\left<\psi^0_m|\hat{H'}|\psi^0_n\right>\right\}$ 是一個反對角線為零的 $2\times 2$ 矩陣==？？

### §7.2.1 Two-fold degenracy 二重簡併
- Reference
- **[Textbook §7.2](https://hackmd.io/nY58RMEyQceIaEoIe-_a_A?view#%C2%A7721-Two-fold-degenracy-%E4%BA%8C%E9%87%8D%E7%B0%A1%E4%BD%B5) (editing)**
- Wikipedia: [Perturbation theory (quantum mechanics)#Effects of degeneracy](https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#Effects_of_degeneracy)
- Goal: 要找到一對簡併態 $\psi^0_a$ 和 $\psi^0_b$ 的兩個線性組合，也許是
$\beta\psi^0_a-\alpha\psi^0_b$ 和$\alpha\psi^0_a+\beta\psi^0_b$，使得這兩個線性組合前後夾住 $\hat{H'}$ 的矩陣表示是對角化。

#### Textbook Example 7.2
- Reference :book:
- **[Textbook Example 7.2 - 7.3](https://hackmd.io/1wOFvGKyTiatvhUpAqCFoQ?both) ✅**</big>
- [北京大學講義](http://www.phy.pku.edu.cn/~qhcao/resources/class/QMA15/note_pert.pdf)，頁11-14
- [蘇黎世大學講義](https://www.uzh.ch/cmsssl/physik/dam/jcr:2b07e3c6-ea07-4791-bfaa-96e853d2555c/Results1.pdf)，Exercise 3

~~~
點擊Details以展開速記內容，詳細分析請參閱課後筆記7.2.1。
~~~
::: spoiler
考慮一個二維量子諧振子
$\hat{H_0}=\dfrac{\hat{p}^2}{2m}+\dfrac{1}{2}m\omega^2(\hat{x}^2+\hat{y}^2)$
$\hat{H'}=\epsilon m\omega^2xy$
有簡併態
$\psi_a^0=\psi_0(x)\psi_1(y)=\sqrt{\dfrac{2}{\pi}}\dfrac{m\omega}{\hbar}ye^{-\frac{m\omega}{2\hbar}(x^2+y^2)}$ 是Gassian
$\psi_b^0=\psi_1(x)\psi_0(y)=\sqrt{\dfrac{2}{\pi}}\dfrac{m\omega}{\hbar}xe^{-\frac{m\omega}{2\hbar}(x^2+y^2)}$
簡併的能量為
$E^0_a=E^0_b=\left(\dfrac{1}{2}+\dfrac{3}{2}\right)\hbar\omega=2\hbar\omega$。

選一組好用的{算符|}
令 $\hat{x'}=\dfrac{\hat{x}+\hat{y}}{\sqrt{2}}$、$\hat{y'}=\dfrac{\hat{x}-\hat{y}}{\sqrt{2}}$，使得
$\hat{H}=\dfrac{-\hbar^2}{2m}\left(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}\right)+\dfrac{1}{2}m(1+\epsilon)\omega^2\hat{x'}^2+\dfrac{1}{2}m(1-\epsilon)\omega^2\hat{y'}^2$，
且令$\omega^2_{+}\equiv(1+\epsilon)\omega^2$ 和 $\omega^2_{-}\equiv(1-\epsilon)\omega^2$，
$\psi_{mn}=\psi^+_m(x')\psi^-_n(y')$
$E_{mn}=\left(m+\dfrac{1}{2}\right)\hbar\omega_{+}+\left(n+\dfrac{1}{2}\right)\hbar\omega_{-} = (m+n+1)\hbar\omega$

$\epsilon\to 0$ 時，
$\displaystyle \lim_{\epsilon\to 0}\psi_{01}=\lim_{\epsilon\to 0}\psi_0^+(x')\psi_1^-(y')=\psi_0\left(\dfrac{x+y}{\sqrt{2}}\right)\psi_1^-\left(\dfrac{x-y}{\sqrt{2}}\right)=\sqrt{\dfrac{2}{\pi}}\dfrac{m\omega}{\hbar}\dfrac{x-y}{\sqrt{2}}e^{-\frac{m\omega}{\hbar}(x^2+y^2)}=\dfrac{-\psi_a^0+\psi_b^0}{\sqrt{2}}$
同樣的，
$\displaystyle \lim_{\epsilon\to 0}\psi_{10}=\dfrac{\psi_a^0+\psi_b^0}{\sqrt{2}}$
找到兩個線性組合了
:::