# 算符

（重定向自算符 (物理學)

## 經典力學

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

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

### 經典力學算符表格

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

### 生成元概念

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

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

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

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

### 指數映射

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

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

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

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

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

## 量子力學

### 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

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

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

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

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

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

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

### 對易算符

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

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

### 厄米算符

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

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

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

### 矩陣力學

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

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

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

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

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

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

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

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

### 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}}} ${\displaystyle \mathbf {\hat {d}} =q\mathbf {r} }$

### 範例

#### 位置算符

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

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

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

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

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

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

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

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

#### 動量算符

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

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

{\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}}}

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

${\displaystyle |\psi \rangle }$改寫為本徵值為${\displaystyle p}$的本徵態${\displaystyle |p\rangle }$，方程式改寫為

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

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

## 參考文獻

1. ^ Kittel charles著,洪連輝等譯，固態物理學導論，第681頁。
2. Sakurai, J. J.; Napolitano, Jim, Modern Quantum Mechanics 2nd, Addison-Wesley, 2010, ISBN 978-0805382914
3. Griffiths, David J., Introduction to Quantum Mechanics (2nd ed.), Prentice Hall, 2004, ISBN 0-13-111892-7
4. ^ Ballentine, L. E., The Statistical Interpretation of Quantum Mechanics, Reviews of Modern Physics, 1970, 42: 358–381, doi:10.1103/RevModPhys.42.358
5. ^ 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
6. ^ 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