# 算符

## 经典力学

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

${\displaystyle T_{a}f(q_{i})=f(q_{i}-a)}$

### 经典力学算符表格

• ${\displaystyle R({\hat {\mathbf {n} )),\theta )}$旋转矩阵${\displaystyle {\hat {\mathbf {n} ))}$是旋转轴矢量，${\displaystyle \theta }$是旋转角弧。

### 生成元概念

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

${\displaystyle T_{\epsilon }f(x)=f(x-\epsilon )}$

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

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

### 指数映射

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

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

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

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

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

## 量子力学

### 量子算符

${\displaystyle \langle e_{i}|e_{j}\rangle =\delta _{ij))$

${\displaystyle |\psi \rangle =\sum _{i}\ c_{i}|e_{i}\rangle }$

### 期望值

${\displaystyle \langle O\rangle \ {\stackrel {def}{=))\ \langle \psi |{\hat {O))|\psi \rangle }$

${\displaystyle |\phi \rangle ={\hat {O))|\psi \rangle =\sum _{i}\ c_{i}{\hat {O))|e_{i}\rangle =\sum _{i}\ c_{i}O_{i}|e_{i}\rangle }$

${\displaystyle \langle \psi |\phi \rangle =\langle \psi |{\hat {O))|\psi \rangle =\sum _{i}\ c_{i}O_{i}\langle \psi |e_{i}\rangle =\sum _{i}\ |c_{i}|^{2}O_{i}=\sum _{i}\ p_{i}O_{i))$

${\displaystyle \langle O\rangle =\sum _{i}\ p_{i}O_{i))$

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

${\displaystyle \langle O^{2}\rangle =\langle \psi \vert {\hat {O))^{2}\vert \psi \rangle }$

### 对易算符

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

${\displaystyle [{\hat {A)),{\hat {B))]|\psi \rangle ={\hat {A)){\hat {B))|\psi \rangle -{\hat {B)){\hat {A))|\psi \rangle }$

### 厄米算符

${\displaystyle \langle O\rangle =\langle O\rangle ^{*))$

${\displaystyle \langle \psi |{\hat {O))|\psi \rangle =\langle \psi |{\hat {O))|\psi \rangle ^{*))$

${\displaystyle {\hat {O))={\hat {O))^{\dagger ))$

### 矩阵力学

${\displaystyle {\hat {O))=\sum _{i,j}|e_{i}\rangle \langle e_{i}|{\hat {O))|e_{j}\rangle \langle e_{j}|=\sum _{ij}O_{i,j}|e_{i}\rangle \langle e_{j}|}$

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

${\displaystyle \langle e_{i}|{\hat {O))|e_{j}\rangle =\langle e_{j}|{\hat {O))^{\dagger }|e_{i}\rangle ^{*))$

${\displaystyle |\phi \rangle ={\hat {O))|\psi \rangle }$

${\displaystyle \langle e_{i}|\phi \rangle =\langle e_{i}|{\hat {O))|\psi \rangle =\sum _{j}\langle e_{i}|{\hat {O))|e_{j}\rangle \langle e_{j}|\psi \rangle =\sum _{ij}O_{ij}\langle e_{j}|\psi \rangle }$

${\displaystyle |\phi \rangle \ {\stackrel {rep}{=))\ {\begin{pmatrix}\langle e_{1}|\phi \rangle \\\langle e_{2}|\phi \rangle \\\vdots \\\langle e_{n}|\phi \rangle \\\end{pmatrix))}$ 　　　　${\displaystyle |\psi \rangle \ {\stackrel {rep}{=))\ {\begin{pmatrix}\langle e_{1}|\psi \rangle \\\langle e_{2}|\psi \rangle \\\vdots \\\langle e_{n}|\psi \rangle \\\end{pmatrix))}$

${\displaystyle {\begin{pmatrix}\langle e_{1}|\phi \rangle \\\langle e_{2}|\phi \rangle \\\vdots \\\langle e_{n}|\phi \rangle \\\end{pmatrix))={\begin{pmatrix}O_{11}&O_{12}&\cdots &O_{1n}\\O_{21}&O_{22}&\cdots &O_{2n}\\\vdots &\vdots &\ddots &\vdots \\O_{n1}&O_{n2}&\cdots &O_{nn}\\\end{pmatrix)){\begin{pmatrix}\langle e_{1}|\psi \rangle \\\langle e_{2}|\psi \rangle \\\vdots \\\langle e_{n}|\psi \rangle \\\end{pmatrix))}$

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

### 量子算符表格

{\displaystyle {\begin{aligned}{\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{aligned))}

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

{\displaystyle {\begin{aligned}{\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{aligned))}

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

{\displaystyle {\begin{aligned}{\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{aligned))}

{\displaystyle {\begin{aligned}{\hat {T))&={\hat {T))_{x}+{\hat {T))_{y}+{\hat {T))_{z}\\&={\frac {-\hbar ^{2)){2m))\nabla ^{2}\\\end{aligned))}

{\displaystyle {\begin{aligned}{\hat {T))_{x}&={\frac {1}{2m))\left(-i\hbar {\frac {\partial }{\partial x))-qA_{x}\right)^{2}\\{\hat {T))_{y}&={\frac {1}{2m))\left(-i\hbar {\frac {\partial }{\partial y))-qA_{y}\right)^{2}\\{\hat {T))_{z}&={\frac {1}{2m))\left(-i\hbar {\frac {\partial }{\partial z))-qA_{z}\right)^{2}\end{aligned))}

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

{\displaystyle {\begin{aligned}{\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))_{z}^{2)){2I_{zz))}\\\end{aligned))}

${\displaystyle {\hat {T))={\frac {\mathbf {\hat {J)) \cdot \mathbf {\hat {J)) }{2I))}$

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

${\displaystyle {\hat {E))=E}$

${\displaystyle \sigma _{x}={\begin{pmatrix}0&1\\1&0\end{pmatrix))}$

${\displaystyle \sigma _{y}={\begin{pmatrix}0&-i\\i&0\end{pmatrix))}$

${\displaystyle \sigma _{z}={\begin{pmatrix}1&0\\0&-1\end{pmatrix))}$

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

（transition moment）
{\displaystyle {\begin{aligned}{\hat {d))_{x}&=qx\\{\hat {d))_{y}&=qy\\{\hat {d))_{z}&=qz\end{aligned))} ${\displaystyle \mathbf {\hat {d)) =q\mathbf {r} }$

### 范例

#### 位置算符

${\displaystyle {\hat {x))|x\rangle =x|x\rangle }$

${\displaystyle |\psi \rangle =\int _{-\infty }^{\infty }\ |x\rangle \langle x|\psi \rangle \mathrm {d} x}$

${\displaystyle {\hat {x))|\psi \rangle ={\hat {x))\int _{-\infty }^{\infty }\ |x\rangle \langle x|\psi \rangle \mathrm {d} x=\int _{-\infty }^{\infty }\ {\hat {x))|x\rangle \langle x|\psi \rangle \mathrm {d} x=\int _{-\infty }^{\infty }\ x|x\rangle \langle x|\psi \rangle \mathrm {d} x}$

${\displaystyle \langle \psi |{\hat {x))|\psi \rangle =\int _{-\infty }^{\infty }\ x\langle \psi |x\rangle \langle x|\psi \rangle \mathrm {d} x}$

${\displaystyle \langle \psi |\alpha \rangle =\int _{-\infty }^{\infty }\ \langle \psi |x\rangle \langle x|\alpha \rangle \mathrm {d} x=\int _{-\infty }^{\infty }\ \langle \psi |x\rangle \langle x|{\hat {x))|\psi \rangle \mathrm {d} x}$

${\displaystyle \langle x|{\hat {x))|\psi \rangle =x\langle x|\psi \rangle }$

${\displaystyle \Psi (x)\ {\stackrel {def}{=))\ \langle x|\Psi \rangle }$
${\displaystyle \psi (x)\ {\stackrel {def}{=))\ \langle x|\psi \rangle }$

${\displaystyle \Psi (x)=x\psi (x)}$

#### 动量算符

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

${\displaystyle \langle x|{\hat {p))|\psi \rangle =-i\hbar {\frac {\partial }{\partial x))\langle x|\psi \rangle }$

{\displaystyle {\begin{aligned}\langle \phi |{\hat {p))|\psi \rangle &=\int _{-\infty }^{\infty }\ \langle \phi |x\rangle \langle x|{\hat {p))|\psi \rangle \mathrm {d} x\\&=\int _{-\infty }^{\infty }\ \langle \phi |x\rangle \left(-i\hbar {\frac {\partial }{\partial x))\right)\langle x|\psi \rangle \mathrm {d} x\\&=\int _{-\infty }^{\infty }\ \phi ^{*}(x)\left(-i\hbar {\frac {\partial }{\partial x))\right)\psi (x)\mathrm {d} x\\\end{aligned))}

${\displaystyle \langle x|{\hat {p))|\psi \rangle =p\langle x|\psi \rangle =-i\hbar {\frac {\partial }{\partial x))\langle x|\psi \rangle }$

${\displaystyle |\psi \rangle }$改写为本征值为${\displaystyle p}$的本征态${\displaystyle |p\rangle }$，方程改写为

${\displaystyle -i\hbar {\frac {\partial }{\partial x))\langle x|p\rangle =p\langle x|p\rangle }$

${\displaystyle \langle x|p\rangle ={\frac {1}{\sqrt {2\pi ))}e^{ipx/\hbar ))$

## 参考文献

1. Sakurai, J. J.; Napolitano, Jim, Modern Quantum Mechanics 2nd, Addison-Wesley, 2010, ISBN 978-0805382914
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. ^ 费曼, 理查; 雷顿, 罗伯; 山德士, 马修, 費曼物理學講義III量子力學(3)薛丁格方程式, 台湾: 天下文化书: pp. 205–237, 2006, ISBN 986-417-672-2