# 正规序

## 记号

${\displaystyle {\hat {O}}}$ 為任意創生和湮灭算符之乘積，則我們將${\displaystyle {\hat {O}}}$ 按照正规序重新排列之后得到的算符用${\displaystyle {\mathcal {N}}({\hat {O}})}$  或 ${\displaystyle {\mathopen {:}}{\hat {O}}{\mathclose {:}}}$ 表示。注意正規序只對算符乘積有意義，因為正規序不是線性關係，將正規序用在算符和並無太大作用。

## 玻色子

### 单个玻色子

• ${\displaystyle {\hat {b}}^{\dagger }}$ ：玻色子的产生算符
• ${\displaystyle {\hat {b}}}$ ：玻色子的湮灭算符

${\displaystyle \left[{\hat {b}}^{\dagger },{\hat {b}}^{\dagger }\right]_{-}=0}$
${\displaystyle \left[{\hat {b}},{\hat {b}}\right]_{-}=0}$
${\displaystyle \left[{\hat {b}},{\hat {b}}^{\dagger }\right]_{-}=1}$

#### 例子

1. 最简单的例子是 ${\displaystyle {\hat {b}}^{\dagger }{\hat {b}}}$  的正规序，根据正规序的定义，可见这里的算符已经按照正规序排列，所以${\displaystyle {\hat {b}}^{\dagger }{\hat {b}}}$ 的正规序就是它自身：

${\displaystyle {:\,}{\hat {b}}^{\dagger }\,{\hat {b}}{\,:}={\hat {b}}^{\dagger }\,{\hat {b}}.}$

2. 第二个例子是 ${\displaystyle {\hat {b}}\,{\hat {b}}^{\dagger }}$  的正规序，

${\displaystyle {:\,}{\hat {b}}\,{\hat {b}}^{\dagger }{\,:}={\hat {b}}^{\dagger }\,{\hat {b}}.}$

${\displaystyle {\hat {b}}\,{\hat {b}}^{\dagger }-{:\,}{\hat {b}}\,{\hat {b}}^{\dagger }{\,:}=1.}$

3. 一个多算符的例子：

${\displaystyle {:\,}{\hat {b}}^{\dagger }\,{\hat {b}}\,{\hat {b}}\,{\hat {b}}^{\dagger }\,{\hat {b}}\,{\hat {b}}^{\dagger }\,{\hat {b}}{\,:}={\hat {b}}^{\dagger }\,{\hat {b}}^{\dagger }\,{\hat {b}}^{\dagger }\,{\hat {b}}\,{\hat {b}}\,{\hat {b}}\,{\hat {b}}=({\hat {b}}^{\dagger })^{3}\,{\hat {b}}^{4}.}$

### 多个玻色子

• ${\displaystyle {\hat {b}}_{i}^{\dagger }}$ ：第 ${\displaystyle i}$  个玻色子的产生算符
• ${\displaystyle {\hat {b}}_{i}}$ ：第 ${\displaystyle i}$  个玻色子的湮灭算符

${\displaystyle \left[{\hat {b}}_{i}^{\dagger },{\hat {b}}_{j}^{\dagger }\right]_{-}=0}$
${\displaystyle \left[{\hat {b}}_{i},{\hat {b}}_{j}\right]_{-}=0}$
${\displaystyle \left[{\hat {b}}_{i},{\hat {b}}_{j}^{\dagger }\right]_{-}=\delta _{ij}}$

#### 例子

1.对于两个玻色子 (${\displaystyle N=2}$ ) ，有：

${\displaystyle :{\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{2}:\,={\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{2}}$
${\displaystyle :{\hat {b}}_{2}\,{\hat {b}}_{1}^{\dagger }:\,={\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{2}}$

2. 对三个玻色子 (${\displaystyle N=3}$ ) ，有：

${\displaystyle :{\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{2}\,{\hat {b}}_{3}:\,={\hat {b}}_{1}^{\dagger }\,{\hat {b}}_{2}\,{\hat {b}}_{3}}$

## 费米子

### 单个费米子

• ${\displaystyle {\hat {f}}^{\dagger }}$ ：费米子的产生算符
• ${\displaystyle {\hat {f}}}$ ：费米子的湮灭算符

${\displaystyle \left[{\hat {f}}^{\dagger },{\hat {f}}^{\dagger }\right]_{+}=0}$
${\displaystyle \left[{\hat {f}},{\hat {f}}\right]_{+}=0}$
${\displaystyle \left[{\hat {f}},{\hat {f}}^{\dagger }\right]_{+}=1}$

#### 例子

1. 最简单的例子是：

${\displaystyle :{\hat {f}}^{\dagger }\,{\hat {f}}:\,={\hat {f}}^{\dagger }\,{\hat {f}}}$

${\displaystyle :{\hat {f}}\,{\hat {f}}^{\dagger }:\,=-{\hat {f}}^{\dagger }\,{\hat {f}}}$

${\displaystyle {\hat {f}}\,{\hat {f}}^{\dagger }-:{\hat {f}}\,{\hat {f}}^{\dagger }:=1.}$

2. 其它情形下的正规序都是零，因为此时同一个湮灭算符或产生算符至少连续出现了两次。根据费米子的性质，此时结果为零，例如：

${\displaystyle :{\hat {f}}\,{\hat {f}}^{\dagger }\,{\hat {f}}{\hat {f}}^{\dagger }:\,={\hat {f}}^{\dagger }\,{\hat {f}}^{\dagger }\,{\hat {f}}\,{\hat {f}}=0}$

### 多个费米子

${\displaystyle N}$  个费米子有 ${\displaystyle 2N}$  个产生湮灭算符，设：

• ${\displaystyle {\hat {f}}_{i}^{\dagger }}$ 为第 ${\displaystyle i}$  个费米子的产生算符
• ${\displaystyle {\hat {f}}_{i}}$ 为第 ${\displaystyle i}$ 个费米子的湮灭算符

${\displaystyle \left[{\hat {f}}_{i}^{\dagger },{\hat {f}}_{j}^{\dagger }\right]_{+}=0}$
${\displaystyle \left[{\hat {f}}_{i},{\hat {f}}_{j}\right]_{+}=0}$
${\displaystyle \left[{\hat {f}}_{i},{\hat {f}}_{j}^{\dagger }\right]_{+}=\delta _{ij}}$

#### 例子

1. 对两个费米子 (${\displaystyle N=2}$ ) ，有：

${\displaystyle :{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}:\,={\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}}$

${\displaystyle :{\hat {f}}_{2}\,{\hat {f}}_{1}^{\dagger }:\,=-{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}}$

${\displaystyle :{\hat {f}}_{2}\,{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}^{\dagger }:\,={\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}^{\dagger }\,{\hat {f}}_{2}=-{\hat {f}}_{2}^{\dagger }\,{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}}$

2. 对三个费米子 (${\displaystyle N=3}$ ) ，有：

${\displaystyle :{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}\,{\hat {f}}_{3}:\,={\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}\,{\hat {f}}_{3}=-{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{3}\,{\hat {f}}_{2}}$

${\displaystyle :{\hat {f}}_{2}\,{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{3}:\,=-{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}\,{\hat {f}}_{3}={\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{3}\,{\hat {f}}_{2}}$
${\displaystyle :{\hat {f}}_{3}{\hat {f}}_{2}\,{\hat {f}}_{1}^{\dagger }:\,={\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{3}\,{\hat {f}}_{2}=-{\hat {f}}_{1}^{\dagger }\,{\hat {f}}_{2}\,{\hat {f}}_{3}}$

## 参考文献

1. ^ 尹道乐，尹澜. 2. 凝聚态量子理论. ISBN 9787301161609.