# 算符

（重定向自算符 (物理學)

## 經典力學

$S\mathcal{H}(\mathbf{q},\ \mathbf{p})=\mathcal{H}(\mathbf{q}',\ \mathbf{p}')=\mathcal{H}(\mathbf{q},\ \mathbf{p})$

$T_a f(q_i)=f(q_i-a)$

### 經典力學算符表格

• $R(\hat{\mathbf{n}},\theta)$旋轉矩陣$\hat{\mathbf{n}}$是旋轉軸向量，$\theta$是旋轉角弧。

### 生成元概念

$T_{\epsilon}\approx I+\epsilon A$

$T_\epsilon f(x)=f(x - \epsilon)$

$T_\epsilon f(x)=f(x-\epsilon)\approx f(x) - \epsilon f'(x)$

$T_\epsilon f(x) = (I-\epsilon \mathrm{D}) f(x)$

### 指數映射

$T_a f(x)=T_{a/N} \cdots T_{a/N}\ f(x)$

$T_a f(x)=\lim_{N\to\infty} T_{a/N} \cdots T_{a/N} f(x)= \lim_{N\to\infty} (I -(a/N)\mathrm{D})^N f(x)$

$T_a f(x)= e^{-a\mathrm{D}} f(x)$

$T_a f(x) = \left(I - a\mathrm{D} + {a^2\mathrm{D}^2\over 2!} - {a^3\mathrm{D}^3\over 3!} + \cdots \right) f(x)$

$f(x) - a f'(x) + {a^2\over 2!} f''(x) - {a^3\over 3!} f'''(x) + \cdots$

## 量子力學

### 量子算符

$\lang e_i |e_j\rang=\delta_{ij}$

$|\psi\rang=\sum_i \ c_i|e_i\rang$

### 期望值

$\lang O \rang\ \stackrel{def}{=}\ \lang \psi |\hat{O} | \psi \rang$

$|\phi\rang=\hat{O}|\psi\rang=\sum_i \ c_i\hat{O}| e_i\rang=\sum_i \ c_i O_i| e_i\rang$

$\lang\psi|\phi\rang =\lang\psi|\hat{O}|\psi\rang=\sum_i \ c_i O_i\lang\psi| e_i\rang=\sum_i\ |c_i|^2O_i =\sum_i\ p_iO_i$

$\lang O\rang=\sum_i\ p_iO_i$

$\langle F( O ) \rangle = \lang \psi | F( \hat{O} ) | \psi \rang$

$\lang O^2 \rang= \lang\psi \vert \hat{O}^2 \vert \psi \rang$

### 對易算符

$[\hat{A},\hat{B}]\ \stackrel{def}{=}\ \hat{A}\hat{B}-\hat{B}\hat{A}$

$[\hat{A},\hat{B}]|\psi\rang=\hat{A}\hat{B}|\psi\rang-\hat{B}\hat{A}|\psi\rang$

### 厄米算符

$\lang O\rang=\lang O\rang^*$

$\lang \psi|\hat{O}|\psi\rang=\lang \psi|\hat{O}|\psi\rang^*$

$\hat{O}=\hat{O}^{\dagger}$

### 矩陣力學

$\hat{O}=\sum_{i,j}|e_i\rang\lang e_i|\hat{O}|e_j\rang\lang e_j|=\sum_{ij}O_{i,j}|e_i\rang\lang e_j|$

$\hat{O}\ \stackrel{rep}{=}\ \begin{pmatrix} O_{11} & O_{12} & \cdots & O_{1n} \\ O_{21} & O_{22} & \cdots & O_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ O_{n1} & O_{n2} & \cdots & O_{nn} \\ \end{pmatrix}$

$\lang e_i|\hat{O}|e_j\rang=\lang e_j|\hat{O}^{\dagger}|e_i\rang^*$

$|\phi\rang=\hat{O}|\psi\rang$

$\lang e_i|\phi\rang= \lang e_i|\hat{O}|\psi\rang=\sum_j \lang e_i|\hat{O}|e_j\rang\lang e_j|\psi\rang=\sum_{ij}O_{ij}\lang e_j|\psi\rang$

$|\phi\rang\ \stackrel{rep}{=}\ \begin{pmatrix} \lang e_1|\phi\rang \\ \lang e_2|\phi\rang \\ \vdots \\ \lang e_n|\phi\rang \\ \end{pmatrix}$ 　　　　$|\psi\rang\ \stackrel{rep}{=}\ \begin{pmatrix} \lang e_1|\psi\rang \\ \lang e_2|\psi\rang \\ \vdots \\ \lang e_n|\psi\rang \\ \end{pmatrix}$

$\begin{pmatrix} \lang e_1|\phi\rang \\ \lang e_2|\phi\rang \\ \vdots \\ \lang e_n|\phi\rang \\ \end{pmatrix} = \begin{pmatrix} O_{11} & O_{12} & \cdots & O_{1n} \\ O_{21} & O_{22} & \cdots & O_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ O_{n1} & O_{n2} & \cdots & O_{nn} \\ \end{pmatrix} \begin{pmatrix} \lang e_1|\psi\rang \\ \lang e_2|\psi\rang \\ \vdots \\ \lang e_n|\psi\rang \\ \end{pmatrix}$

$\det\left ( \hat{O} - \lambda\hat{I} \right ) = 0$

### 量子算符表格

\begin{align} \hat{p}_x & = -i \hbar \frac{\partial }{\partial x} \\ \hat{p}_y & = -i \hbar \frac{\partial }{\partial y} \\ \hat{p}_z & = -i \hbar \frac{\partial }{\partial z} \end{align}

$\mathbf{\hat{p}} = -i \hbar \nabla$

\begin{align} \hat{p}_x = -i \hbar \frac{\partial }{\partial x} - qA_x \\ \hat{p}_y = -i \hbar \frac{\partial }{\partial y} - qA_y \\ \hat{p}_z = -i \hbar \frac{\partial }{\partial z} - qA_z \end{align}

$\mathbf{\hat{p}} = -i \hbar \nabla - q\bold{A}$

\begin{align} \hat{T}_x & = -\frac{\hbar^2}{2m}\frac{\partial^2 }{\partial x^2} \\ \hat{T}_y & = -\frac{\hbar^2}{2m}\frac{\partial^2 }{\partial y^2} \\ \hat{T}_z & = -\frac{\hbar^2}{2m}\frac{\partial^2 }{\partial z^2} \\ \end{align}

\begin{align} \hat{T} & = \hat{T}_x+ \hat{T}_y+ \hat{T}_z \\ & = \frac{-\hbar^2 }{2m}\nabla^2 \\ \end{align}

\begin{align} \hat{T}_x & = \frac{1}{2m}\left(-i \hbar \frac{\partial }{\partial x } - q A_x \right)^2 \\ \hat{T}_y & = \frac{1}{2m}\left(-i \hbar \frac{\partial }{\partial y} - q A_y \right)^2 \\ \hat{T}_z & = \frac{1}{2m}\left(-i \hbar \frac{\partial }{\partial z} - q A_z \right)^2 \end{align}

\begin{align} \hat{T} & = \frac{\mathbf{\hat{p}}\cdot\mathbf{\hat{p}}}{2m} \\ & = \frac{1}{2m}(-i \hbar \nabla - q\bold{A})\cdot(-i \hbar \nabla - q\bold{A}) \\ & = \frac{1}{2m}(-i \hbar \nabla - q\bold{A})^2 \end{align}

\begin{align} \hat{T}_{xx} & = \frac{\hat{J}_x^2}{2I_{xx}} \\ \hat{T}_{yy} & = \frac{\hat{J}_y^2}{2I_{yy}} \\ \hat{T}_{zz} & = \frac{\hat{J}_y^2}{2I_{zz}} \\ \end{align}

$\hat{T} = \frac{\bold{\hat{J}}\cdot\bold{\hat{J}}}{2I}$

$\hat{E} = i \hbar \frac{\partial }{\partial t}$

$\hat{E} = E$

$\sigma_x = \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$

$\sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}$

$\sigma_z = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}$

$\mathbf{\hat{S}} = {\hbar \over 2} \boldsymbol{\sigma}$

（transition moment）
\begin{align} \hat{d}_x & = qx\\ \hat{d}_y & = qy\\ \hat{d}_z & = qz \end{align} $\mathbf{\hat{d}} = q \mathbf{r}$

### 範例

#### 位置算符

$\hat{x}|x\rang=x|x\rang$

$|\psi\rang=\int_{ - \infty}^{\infty}\ |x\rang\lang x|\psi\rang \mathrm{d}x$

$\hat{x}|\psi\rang=\hat{x}\int_{ - \infty}^{\infty}\ |x\rang\lang x|\psi\rang\mathrm{d}x =\int_{ - \infty}^{\infty}\ \hat{x}|x\rang\lang x|\psi\rang \mathrm{d}x =\int_{ - \infty}^{\infty}\ x|x\rang\lang x|\psi\rang \mathrm{d}x$

$\lang\psi|\hat{x}|\psi\rang =\int_{ - \infty}^{\infty}\ x\lang\psi|x\rang\lang x|\psi\rang \mathrm{d}x$

$\lang\psi|\alpha\rang=\int_{ - \infty}^{\infty}\ \lang \psi| x\rang\lang x|\alpha\rang\mathrm{d}x=\int_{ - \infty}^{\infty}\ \lang \psi| x\rang\lang x|\hat{x}|\psi\rang\mathrm{d}x$

$\lang x|\hat{x}|\psi\rang=x\lang x|\psi\rang$

$\Psi(x)\ \stackrel{def}{=}\ \lang x|\Psi\rang$
$\psi(x)\ \stackrel{def}{=}\ \lang x|\psi\rang$

$\Psi(x)=x\psi(x)$

#### 動量算符

$\hat{p} = -i\hbar\frac{\partial }{\partial x}$

$\lang x|\hat{p}|\psi\rang= -i\hbar\frac{\partial }{\partial x}\lang x|\psi\rang$

\begin{align}\lang \phi|\hat{p}|\psi\rang & =\int_{ - \infty}^{\infty}\ \lang \phi| x\rang\lang x|\hat{p}|\psi\rang\mathrm{d}x \\ & =\int_{ - \infty}^{\infty}\ \lang \phi| x\rang\left( -i\hbar\frac{\partial }{\partial x}\right)\lang x|\psi\rang\mathrm{d}x \\ & =\int_{ - \infty}^{\infty}\ \phi^*(x)\left( -i\hbar\frac{\partial }{\partial x}\right)\psi(x)\mathrm{d}x \\ \end{align}

$\lang x|\hat{p}|\psi\rang=p\lang x|\psi\rang = -i\hbar\frac{\partial }{\partial x}\lang x|\psi\rang$

$|\psi\rang$改寫為本徵值為$p$的本徵態$|p\rang$，方程式改寫為

$-i\hbar\frac{\partial }{\partial x}\lang x|p\rang =p\lang x|p\rang$

$\lang x|p\rang=\frac{1}{\sqrt{2\pi}} e^{ipx/\hbar}$

## 參考文獻

1. ^ 1.0 1.1 1.2 1.3 1.4 Sakurai, J. J.; Napolitano, Jim, Modern Quantum Mechanics 2nd, Addison-Wesley, 2010, ISBN 978-0805382914
2. ^ 2.0 2.1 2.2 Griffiths, David J., Introduction to Quantum Mechanics (2nd ed.), Prentice Hall, 2004, ISBN 0-13-111892-7
3. ^ Ballentine, L. E., The Statistical Interpretation of Quantum Mechanics, Reviews of Modern Physics, 1970, 42: 358–381, doi:10.1103/RevModPhys.42.358
4. ^ Molecular Quantum Mechanics Parts I and II: An Introduction to QUANTUM CHEMISRTY (Volume 1), P.W. Atkins, Oxford University Press, 1977, ISBN 0-19-855129-0
5. ^ 費曼, 理查; 雷頓, 羅伯; 山德士, 馬修, 費曼物理學講義III量子力學(3)薛丁格方程式, 台灣: 天下文化書, pp. 205–237, 2006, ISBN 986-417-672-2