二次量子化 (Second Quantization) PT. 2 - 产生与湮灭

• 原文: https://zhuanlan.zhihu.com/p/97661115
• 发布日期: 2019-12-16
• 分类: 量子力学基础与量子信息

二次量子化三连:

[1~3] 東雲正樹: 二次量子化 (Second Quantization) PT. 1 - 对称与反对称基矢

[4~5] 東雲正樹: 二次量子化 (Second Quantization) PT. 2 - 产生与湮灭

[6~9] 東雲正樹: 二次量子化 (Second Quantization) PT. 3 - 二次量子化形式

目録:

4. 对称化基矢量的正交归一化关系与完全性关系

4.1. $n$ 粒子系统对称或反对称基矢的正交归一性 4.2. $n$ 粒子系统对称或反对称基矢的完全性关系

5. 产生算符与湮灭算符

5.1. 产生算符的定义式 5.2. 产生算符之间的对易关系 5.3. 湮灭算符的定义式 5.4. 湮灭算符之间的对易关系 5.5. 产生湮灭算符之间的对易关系 5.6. 我们这样定义产生湮灭算符会不会导致一些运算不自洽

4. 对称化基矢量的正交归一化关系与完全性关系:

4.1.$n$粒子系统对称或反对称基矢的正交归一性:

我们先用对称或反对称的基矢做内积, 按照定义有:

\begin{align} & \ \ \ \ \ \left\langle n;{\lambda }_{a}'}{\lambda }_{b}'}\cdot \cdot \cdot {\lambda }_{z}'} | n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & =\frac{1}{n!\sqrt{n}_{a}!{n}_{a}'}!\cdot \cdot \cdot {n}_{z}!{n}_{z}'}!}\sum\limits_{s}'}{P_{s}'}^{'}{\varepsilon }^{p_{s}'}^{'}{\left\langle {\lambda }_{a}'} \right|}_{1}{\left\langle {\lambda }_{b}'} \right|}_{2}\cdot \cdot \cdot {\left\langle {\lambda }_{z}'} \right|}_{n}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\left| {\lambda }_{a} \right\rangle }_{1}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot {\left| {\lambda }_{z} \right\rangle }_{n} \\ & =\frac{1}{n!\sqrt{n}_{a}!{n}_{a}'}!\cdot \cdot \cdot {n}_{z}!{n}_{z}'}!}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}\sum\limits_{s}'}{P_{s}'}^{'}{\varepsilon }^{p_{s}'}^{'}{\left\langle {\lambda }_{a}'} \right|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}{\left\langle {\lambda }_{b}'} \right|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot {\left\langle {\lambda }_{z}'} \right|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ \end{align}

到这一步就需要稍微思考一下结构了, 我们先不考虑系数:

\begin{align} & \ \ \ \ \sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}\sum\limits_{s}'}{P_{s}'}^{'}{\varepsilon }^{p_{s}'}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & =\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}P_{1}^{'}{\varepsilon }^{p_{1}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & \ \ \ \ +\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}P_{2}^{'}{\varepsilon }^{p_{2}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & \ \ \ \ +\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}P_{3}^{'}{\varepsilon }^{p_{3}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & \ \ \ \ \cdot \cdot\ \cdot \\ & \ \ \ \ +\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}P_{n!}^{'}{\varepsilon }^{p_{n!}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ \end{align}

其中 P_{s}'}^{'}\langle {\lambda }_{a}'}{|}_{1}\langle {\lambda }_{b}'}{|}_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n} 是对 {\left\langle {\lambda }_{a}'} \right|}_{1}{\left\langle {\lambda }_{b}'} \right|}_{2}\cdot \cdot \cdot {\left\langle {\lambda }_{z}'} \right|}_{n} 取的一种排列. 包括原顺序, 全排列共有 $n!$ 种, 所以排列算符 P_{s}'}^{'}共 $n!$ 个, $s^{\prime }\in \left\{1,2,3,\cdot \cdot \cdot ,n!\right\}$. 我们不妨令$P_1^{{}^{\prime }}$表示不做交换, 即$P_1^{{}^{\prime }}=I$.

像上面这样拆开写, 实际上你会发现每一项都是完全相等的. 其实原理很简单: 你对左矢先取一个固定排列, 而右矢取全排列的话, 无论如何都是要把全部可能的组合都过一遍的. 所以无论左矢取哪个排列最后结果都是一样的.

系数 {\varepsilon }^{p_{s}'}^{'}+{p}_{s}为何也全相等? 因为从最初 \langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} 这一项变换到确定的一项, 比如说 \langle {\lambda }_{z}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{z} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{a}'}{|}_{n}{\left| {\lambda }_{b} \right\rangle }_{n} , 无论如何去交换, 无论是交换左矢还是交换右矢所需要的交换次数总数只会相差一个偶数.

是不是听不懂? 笑了, 这个我也没办法帮你思考, 但可以先看一个 $n=3$ 实例直观来感受一下:

这里我们将 $3!$ 个排列算符这样设置: \left\{ \begin{align} & P_{1}^{'}\to 123\ \ \ \ \ P_{2}^{'}\to 132 \\ & P_{3}^{'}\to 213\ \ \ \ \ P_{4}^{'}\to 231 \\ & P_{5}^{'}\to 312\ \ \ \ \ P_{6}^{'}\to 321 \\ \end{align} \right. 即 P_{3}^{'}{\left\langle {\lambda }_{a}'} \right|}_{1}{\left\langle {\lambda }_{b}'} \right|}_{2}{\left\langle {\lambda }_{c}'} \right|}_{3}={\left\langle {\lambda }_{a}'} \right|}_{2}{\left\langle {\lambda }_{b}'} \right|}_{1}{\left\langle {\lambda }_{c}'} \right|}_{3}={\left\langle {\lambda }_{b}'} \right|}_{1}{\left\langle {\lambda }_{a}'} \right|}_{2}{\left\langle {\lambda }_{c}'} \right|}_{3}. 显然 {\varepsilon }^{p_{3}^{'}={\varepsilon }^{1}.

下面分别取 $s^{\prime }=1,s^{\prime }=3$ 两项来感受一下:

\begin{align} & \ \ \ \ \sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}P_{1}^{'}{\varepsilon }^{p_{1}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3} \\ & =\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\varepsilon }^{0}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3} \\ & ={\varepsilon }^{0}{\varepsilon }^{0}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3}+{\varepsilon }^{1}{\varepsilon }^{0}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{c} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{b} \right\rangle }_{3} \\ & \ +{\varepsilon }^{1}{\varepsilon }^{0}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{b} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{a} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3}+{\varepsilon }^{2}{\varepsilon }^{0}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{b} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{c} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{a} \right\rangle }_{3} \\ & \ +{\varepsilon }^{2}{\varepsilon }^{0}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{c} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{a} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{b} \right\rangle }_{3}+{\varepsilon }^{1}{\varepsilon }^{0}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{c} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{a} \right\rangle }_{3} \\ \end{align}

\begin{align} & \ \ \ \ \sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}P_{3}^{'}{\varepsilon }^{p_{3}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3} \\ & =\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\varepsilon }^{1}\langle {\lambda }_{b}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{a}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3} \\ & \ +{\varepsilon }^{0}{\varepsilon }^{1}\langle {\lambda }_{b}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{a}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3}+{\varepsilon }^{1}{\varepsilon }^{1}\langle {\lambda }_{b}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{a}'}{|}_{2}{\left| {\lambda }_{c} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{b} \right\rangle }_{3} \\ & \ +{\varepsilon }^{1}{\varepsilon }^{1}\langle {\lambda }_{b}'}{|}_{1}{\left| {\lambda }_{b} \right\rangle }_{1}\langle {\lambda }_{a}'}{|}_{2}{\left| {\lambda }_{a} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{c} \right\rangle }_{3}+{\varepsilon }^{2}{\varepsilon }^{1}\langle {\lambda }_{b}'}{|}_{1}{\left| {\lambda }_{b} \right\rangle }_{1}\langle {\lambda }_{a}'}{|}_{2}{\left| {\lambda }_{c} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{a} \right\rangle }_{3} \\ & \ +{\varepsilon }^{2}{\varepsilon }^{1}\langle {\lambda }_{b}'}{|}_{1}{\left| {\lambda }_{c} \right\rangle }_{1}\langle {\lambda }_{a}'}{|}_{2}{\left| {\lambda }_{a} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{b} \right\rangle }_{3}+{\varepsilon }^{1}{\varepsilon }^{1}\langle {\lambda }_{b}'}{|}_{1}{\left| {\lambda }_{c} \right\rangle }_{1}\langle {\lambda }_{a}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\langle {\lambda }_{c}'}{|}_{3}{\left| {\lambda }_{a} \right\rangle }_{3} \\ \end{align}

你可以发现 上$1$下$3$, 上$2$下$5$, 上$3$下$1$, 上$4$下$6$, 上$5$下$2$, 上$6$下$4$ 是完全相等的. 其原理就是我前面讲的那些. 看书很多时候都不怎么需要动脑, 看着就行, 但学一门科目总有那么几个点要停下来细想, 听别人讲是听不懂的, 这里就是这种情况, 自己仔细揣摩一下吧.

如果你想通了就想通了, 想不通你只能相信我. 最后我们可以得到 $n!$ 个相等的项, 故有:

\begin{align} & \ \ \ \ \ \sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}\sum\limits_{s}'}{P_{s}'}^{'}{\varepsilon }^{p_{s}'}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & =n!\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}P_{1}^{'}{\varepsilon }^{p_{1}^{'}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & =n!\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}\langle {\lambda }_{a}'}{|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}\langle {\lambda }_{b}'}{|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot \langle {\lambda }_{z}'}{|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ \end{align}

综上所述:

\begin{align} & \ \ \ \ \ \left\langle n;{\lambda }_{a}'}{\lambda }_{b}'}\cdot \cdot \cdot {\lambda }_{z}'} | n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & =\frac{1}{n!\sqrt{n}_{a}!{n}_{a}'}!\cdot \cdot \cdot {n}_{z}!{n}_{z}'}!}\sum\limits_{s}'}{P_{s}'}^{'}{\varepsilon }^{p_{s}'}^{'}{\left\langle {\lambda }_{a}'} \right|}_{1}{\left\langle {\lambda }_{b}'} \right|}_{2}\cdot \cdot \cdot {\left\langle {\lambda }_{z}'} \right|}_{n}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\left| {\lambda }_{a} \right\rangle }_{1}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot {\left| {\lambda }_{z} \right\rangle }_{n} \\ & =\frac{1}{n!\sqrt{n}_{a}!{n}_{a}'}!\cdot \cdot \cdot {n}_{z}!{n}_{z}'}!}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}\sum\limits_{s}'}{P_{s}'}^{'}{\varepsilon }^{p_{s}'}^{'}{\left\langle {\lambda }_{a}'} \right|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}{\left\langle {\lambda }_{b}'} \right|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot {\left\langle {\lambda }_{z}'} \right|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & =\frac{1}{\sqrt{n}_{a}!{n}_{a}'}!\cdot \cdot \cdot {n}_{z}!{n}_{z}'}!}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\left\langle {\lambda }_{a}'} \right|}_{1}{\left| {\lambda }_{a} \right\rangle }_{1}{\left\langle {\lambda }_{b}'} \right|}_{2}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot {\left\langle {\lambda }_{z}'} \right|}_{n}{\left| {\lambda }_{z} \right\rangle }_{n} \\ & =\frac{1}{\sqrt{n}_{a}!{n}_{a}'}!\cdot \cdot \cdot {n}_{z}!{n}_{z}'}!}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\delta }_{a}'a}{\delta }_{b}'b}\cdot \cdot \cdot {\delta }_{z}'z} \\ \end{align}

这里注意 $P_s$ 只交换下标的右边那个数字.

若 $a^{\prime }=a,b^{\prime }=b,\cdot \cdot \cdot ,z^{\prime }=z$ 则上式自然是归一的.

从最初我们系数的设置原因来看就该如此, 但我们仍然可以验证一下, 此时式子变为: \left\langle n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} | n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle =\frac{1}{n}_{a}!{n}_{b}!\cdot \cdot \cdot {n}_{z}!}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\delta }_{a}'a}{\delta }_{b}'b}\cdot \cdot \cdot {\delta }_{z}'z}=1 后面那个求和中的 {\delta }_{a}'a}{\delta }_{b}'b}\cdot \cdot \cdot {\delta }_{z}'z}=1 , 要想交换一项仍不为零就只有交换相同的项, 那么全排列中一共有几项不为零呢? 显然有 $n_a!n_b!\cdot \cdot \cdot n_z!$ 项.

若两个矢量内的包含的态[1]不相同的话则 \left\langle n;{\lambda }_{a}'}{\lambda }_{b}'}\cdot \cdot \cdot {\lambda }_{z}'}|n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle =0

我们可以从反证法的角度来看, 假如两个矢量内的各个态的数目不相同, 无论你怎么交换下标, 你可能找到一项不为零的 {\delta }_{a}'a}{\delta }_{b}'b}\cdot \cdot \cdot {\delta }_{z}'z} 吗? 当然不能. 如果包含的态相同的话我们可以通过关系 \left\{ \begin{align} & {\left| n;{\lambda }_{a}{\lambda }_{b}{\lambda }_{c}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S}={\left| n;{\lambda }_{a}{\lambda }_{c}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & {\left| n;{\lambda }_{a}{\lambda }_{b}{\lambda }_{c}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=-{\left| n;{\lambda }_{a}{\lambda }_{c}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ \end{align} \right. 化为第一种情况.

上面, 就是对称或反对称基矢的正交归一化关系的证明.

4.2.$n$粒子系统对称或反对称基矢的完全性关系:

我们前面构造了 $n$ 个全同粒子系统所有的对称或反对称的基矢. 理论上系统的任何一个态 (对称或反对称的) 都可以被这些基矢展开. 尽管这个结论十分显然, 我们仍然要就此给出一个证明, 事实上这个证明的思路也十分有趣:

任何一个$n$ 个全同粒子系统的对称或反对称的态 $|\psi ⟩$ 一定满足这个关系: $P_s|\psi ⟩={\epsilon }_s^p|\psi ⟩$

从复合量子系统的概念来看, 我们至少一定可以把任何一个总系统的态 (哪怕非对称的) 用各个分系统的基矢张成的总空间的普通基矢展开:

$|\psi ⟩=\sum _{a,b,\cdot \cdot \cdot ,z}{c}_{a,b,\cdot \cdot \cdot ,z}{|{\lambda }_a⟩}_1{|{\lambda }_b⟩}_2\cdot \cdot \cdot {|{\lambda }_z⟩}_n$ 其中 $c_{a,b,\cdot \cdot \cdot ,z}$ 是系数, 当然, 这有超妈多项.

我们对上式两边进行 $\sum _s{P}_s{\epsilon }_s^p$ 操作即得到:

\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}\left| \psi \right\rangle }=\sum\limits_{a,b,\cdot \cdot \cdot ,z}{c}_{a,b,\cdot \cdot \cdot ,z}\sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}{\left| {\lambda }_{a} \right\rangle }_{1}{\left| {\lambda }_{b} \right\rangle }_{2}\cdot \cdot \cdot {\left| {\lambda }_{z} \right\rangle }_{n}

等式左边: \sum\limits_{s}{P}_{s}{\varepsilon }^{p}_{s}\left| \psi \right\rangle }=\sum\limits_{s}{\varepsilon }^{2{p}_{s}\left| \psi \right\rangle }=n!\left| \psi \right\rangle

等式右边: \sum\limits_{a,b,\cdot \cdot \cdot ,z}{c}_{a,b,\cdot \cdot \cdot ,z}\sqrt{n!{n}_{a}!\cdot \cdot \cdot {n}_{z}!}\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }

所以就得到了结论: \left| \psi \right\rangle =\sum\limits_{a,b,\cdot \cdot \cdot ,z}{c}_{a,b,\cdot \cdot \cdot ,z}\sqrt{\frac{n}_{a}!\cdot \cdot \cdot {n}_{z}!}{n!}\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }

也就是说任何一个具有对称或反对称性的矢量都可以用这些对称或反对称化的基矢展开.

只要我想的话, 其实我还可以把这个完全性关系抽离出来:

\begin{align} & \ \ \ \ \ \left| \psi \right\rangle =\sum\limits_{a,b,\cdot \cdot \cdot ,z}{c}_{a,b,\cdot \cdot \cdot ,z}\sqrt{\frac{n}_{a}!\cdot \cdot \cdot {n}_{z}!}{n!}\left| n;{\lambda }_{a}{\lambda }_{b},\cdot \cdot \cdot {\lambda }_{z} \right\rangle } \\ & \Rightarrow {c}_{a,b,\cdot \cdot \cdot ,z}=\sqrt{\frac{n!}{n}_{a}!\cdot \cdot \cdot {n}_{z}!}\left\langle n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z}|\psi \right\rangle \\ \end{align} \begin{align} & \Rightarrow \left| \psi \right\rangle =\sum\limits_{a,b,\cdot \cdot \cdot ,z}{c}_{a,b,\cdot \cdot \cdot ,z}\sqrt{\frac{n}_{a}!\cdot \cdot \cdot {n}_{z}!}{n!}\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle } \\ & \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{a,b,\cdot \cdot \cdot ,z}{\sqrt{\frac{n!}{n}_{a}!\cdot \cdot \cdot {n}_{z}!}\left\langle n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z}|\psi \right\rangle \sqrt{\frac{n}_{a}!\cdot \cdot \cdot {n}_{z}!}{n!}\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle } \\ & \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{a,b,\cdot \cdot \cdot ,z}{\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \left\langle n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z}|\psi \right\rangle } \\ & \Rightarrow \sum\limits_{a,b,\cdot \cdot \cdot ,z}{\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \langle n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z}|}=I \\ \end{align}

总而言之对称或反对称的矢量具有完全性关系, 且表达如下:

$\sum _{a,b,\cdot \cdot \cdot ,z}|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩⟨n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z|=I$

上式也被称作封闭性关系式.

到此为止, 就是以单粒子算符 $M$ 建立了一个 $n$ 粒子系统的对称或反对称化希尔伯特空间. 这个空间的对称或反对称的基矢为 $\left\{|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩|a,b,\cdot \cdot \cdot ,z\in N\right\}$ , 我们也将以此为基矢的表象称作对称或反对称化的 $M$ 表象.

5. 产生算符与湮灭算符:

5.1. 产生算符的定义式:

我们希望沿用初等量子力学中处理谐振子的那一套产生湮灭算符与粒子数算符的逻辑, 至少希望尽可能地向其靠近, 所以我们希望产生算符具有这样的性质:

\left\{ \begin{align} & {a}^{\dagger }\left( {\lambda }_{i} \right)\left| 0 \right\rangle =\left| 1;{\lambda }_{i} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle =\sqrt{n}_{i}+1}\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ \end{align} \right.

注意这里的$n_i$指的是 $a^{†}\left({\lambda }_i\right)$ 作用前的矢量$|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$中${\lambda }_i$的数量.

我们定义产生算符 $a^{†}\left({\lambda }_i\right)$ 使对称或反对称矢量产生一个确定态 $|{\lambda }_i⟩$ 的粒子. 上面的 $|0⟩$ 表示真空态, 是一个没有粒子的空间的唯一状态. 由式子 \left\{ \begin{align} & {\left| n;{\lambda }_{a}{\lambda }_{b}{\lambda }_{c}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S}={\left| n;{\lambda }_{a}{\lambda }_{c}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & {\left| n;{\lambda }_{a}{\lambda }_{b}{\lambda }_{c}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=-{\left| n;{\lambda }_{a}{\lambda }_{c}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ \end{align} \right. 可知反对称矢量本征值的排布顺序是重要的, 所以我们人为统一规定产生的态必须放在最左边.[[2]](#ref_2) 有人问道: 这个产生算符好像是凭空出现的, 我们不用证明它是否真的存在吗? 但其实我们在数学证明过程中为了方便而构造的函数不也是凭空构造的吗？产生算符是非厄米的, 它不是一个真实存在的物理量, 我们为了更方便的处理体系而构造了这样一个线性变换, 也就是说满足前面所述性质的就是产生算符. 而这个算符的存在性是显然的, 这个变换实际上就只是在巨希尔伯特空间把一个矢量旋转之后拉伸了一下罢了, 是一个很平凡的变换.

根据定义, 我们也可以进一步推得下述关系:

\left\{ \begin{align} & \left| \text{0} \right\rangle \to \left| \text{6};{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{3} \right)\left| 0 \right\rangle =\left| 1;{\lambda }_{3} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{3} \right)\left| 1;{\lambda }_{3} \right\rangle =\sqrt{2}\left| 2;{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{2} \right)\left| 2;{\lambda }_{3}{\lambda }_{3} \right\rangle =\left| 3;{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{2} \right)\left| 3;{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle =\sqrt{2}\left| 4;{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{2} \right)\left| 4;{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle =\sqrt{3}\left| 5;{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{1} \right)\left| 5;{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle =\left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ \end{align} \right.

\begin{align} & \Rightarrow \left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle ={a}^{\dagger }\left( {\lambda }_{1} \right)\frac{1}{\sqrt{3}\sqrt{3}\left| 5;{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={a}^{\dagger }\left( {\lambda }_{1} \right)\frac{1}{\sqrt{3}{a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}\sqrt{2}\left| 4;{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={a}^{\dagger }\left( {\lambda }_{1} \right)\frac{1}{\sqrt{3}{a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}{a}^{\dagger }\left( {\lambda }_{2} \right)\left| 3;{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={a}^{\dagger }\left( {\lambda }_{1} \right)\frac{1}{\sqrt{3}{a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}{a}^{\dagger }\left( {\lambda }_{2} \right){a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}\sqrt{2}\left| 2;{\lambda }_{3}{\lambda }_{3} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={a}^{\dagger }\left( {\lambda }_{1} \right)\frac{1}{\sqrt{3}{a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}{a}^{\dagger }\left( {\lambda }_{2} \right){a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}{a}^{\dagger }\left( {\lambda }_{3} \right)\left| 1;{\lambda }_{3} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={a}^{\dagger }\left( {\lambda }_{1} \right)\frac{1}{\sqrt{3}{a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}{a}^{\dagger }\left( {\lambda }_{2} \right){a}^{\dagger }\left( {\lambda }_{2} \right)\frac{1}{\sqrt{2}{a}^{\dagger }\left( {\lambda }_{3} \right){a}^{\dagger }\left( {\lambda }_{3} \right)\left| 0 \right\rangle \\ & \Rightarrow \left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}{\lambda }_{3} \right\rangle =\frac{1}{\sqrt{3!}\frac{1}{\sqrt{2}{a}^{\dagger }\left( {\lambda }_{1} \right){a}^{\dagger }\left( {\lambda }_{2} \right){a}^{\dagger }\left( {\lambda }_{2} \right){a}^{\dagger }\left( {\lambda }_{2} \right){a}^{\dagger }\left( {\lambda }_{3} \right){a}^{\dagger }\left( {\lambda }_{3} \right)\left| 0 \right\rangle \\ \end{align}

这样我们就建立了所有对称或反对称矢量与真空态 $|0⟩$ 的联系:

$|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩=\frac{1}{{\sqrt{n}}_a!n_b!\cdot \cdot \cdot n_z!}{a}^{†}\left({\lambda }_a\right)a^{†}\left({\lambda }_b\right)\cdot \cdot \cdot a^{†}\left({\lambda }_z\right)|0⟩$

很显然上面 $n_a{n}_b\cdot \cdot \cdot n_z$ 指的是左边 $|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$ 中各个态的数目.

其实产生算符的定义式还有一种稍微复杂一些, 但形式上更完善的写法:

$i\in \left\{ a,b,\cdot \cdot \cdot ,z \right\}$ 时: $\ \ \ \ {a}^{\dagger }\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle =\left[ \begin{align} & \ \ \ \ \ {\delta }_{ia}\sqrt{n}_{a}+1}\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & +{\delta }_{ib}\sqrt{n}_{b}+1}\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & +\cdot \cdot \cdot \\ & +{\delta }_{iz}\sqrt{n}_{z}+1}\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ \end{align} \right]$

$i\notin \left\{a,b,\cdot \cdot \cdot ,z\right\}$ 时: $a^{†}\left({\lambda }_i\right)|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩=|n+1;{\lambda }_i{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$

这应该是很好理解的, $i\in \left\{a,b,\cdot \cdot \cdot ,z\right\}$ 时右边只保留一项, 就和前面定义相同, 如果 $i\notin \left\{a,b,\cdot \cdot \cdot ,z\right\}$ 的话, 自然就是添加一个处于 $|{\lambda }_i⟩$ 的粒子罢了.

5.2. 产生算符之间的对易关系:

\left. \begin{align} & {a}^{\dagger }\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right)\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle ={a}^{\dagger }\left( {\lambda }_{i} \right)\sqrt{n}_{j}+1}\left| n+1;{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{n}_{i}+1}\sqrt{n}_{j}+1}\left| n+2;{\lambda }_{i}{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & {a}^{\dagger }\left( {\lambda }_{j} \right){a}^{\dagger }\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle ={a}^{\dagger }\left( {\lambda }_{j} \right)\sqrt{n}_{i}+1}\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{n}_{j}+1}\sqrt{n}_{i}+1}\left| n+2;{\lambda }_{j}{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\varepsilon \sqrt{n}_{i}+1}\sqrt{n}_{j}+1}\left| n+2;{\lambda }_{i}{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ \end{align} \right\}

$\Rightarrow a^{†}\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩=\epsilon a^{†}\left({\lambda }_j\right)a^{†}\left({\lambda }_i\right)|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$

$\Rightarrow a^{†}\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)-\epsilon a^{†}\left({\lambda }_j\right)a^{†}\left({\lambda }_i\right)=0$

也就是说在 $i\ne j$ 时: 对于对称态而言 $a^{†}\left({\lambda }_i\right),a^{†}\left({\lambda }_j\right)$ 是可对易的; 而对于反对称态而言有 $a^{†}\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)+a^{†}\left({\lambda }_j\right)a^{†}\left({\lambda }_i\right)=0$ , 即二者是反对易的.

而在 $i=j$ 时, 对称态的可对易性是显然的, 而反对称态是不允许有两个及以上的粒子处于相同的态的, 所以反对称态的产生算符具有关系 $a^{†}\left({\lambda }_i\right)a^{†}\left({\lambda }_i\right)\equiv 0$ .

5.3. 湮灭算符的定义式:

我们知道湮灭算符 $a\left({\lambda }_i\right)$ 是产生算符 $a^{†}\left({\lambda }_i\right)$ 的厄米共轭算符, 所以我们取前面产生算符满足的式子的伴随式就可以得到湮灭算符满足的式子:

\begin{align} & {a}^{\dagger }\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle =\sqrt{n}_{i}+1}\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & \Rightarrow \langle n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z}|a\left( {\lambda }_{i} \right)=\sqrt{n}_{i}+1}\langle n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z}| \\ \end{align}

两边右乘 $|n+1;{\lambda }_i{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$ 得到:

\langle n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z}|a\left( {\lambda }_{i} \right)\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle =\sqrt{n}_{i}+1}

所以只可能是 a\left( {\lambda }_{i} \right)\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle =\sqrt{n}_{i}+1}\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle

注意这里$n_i$指的是$a\left({\lambda }_i\right)$作用后的矢量$|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$中${\lambda }_i$出现的次数.

但为了让形式更加统一, 请允许我任性地再修改一个地方:

规定 $n_i$ 表示 $a\left({\lambda }_i\right)$ 作用前的矢量 $|n+1;{\lambda }_i{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$ 中 ${\lambda }_i$ 出现的次数.

则上式将修改为: $a\left({\lambda }_i\right)|n+1;{\lambda }_i{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩={\sqrt{n}}_i|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$

其中$n_i$ 表示 $a\left({\lambda }_i\right)$ 作用前的矢量 $|n+1;{\lambda }_i{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$ 中 ${\lambda }_i$ 出现的次数.

假如 ${\lambda }_i$ 不在最左边, 对于对称的态来说是与上面等价的情况, 但对于反对称的态我们就要先把 ${\lambda }_i$ 移到最左边, 每移一次就要填上一个系数 $-1$ . 所以对称和反对称情况统一可以写成:

**注意下面**${n}_{i}$**指的是**$\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{i}\cdot \cdot \cdot {\lambda }_{z} \right\rangle$]**中**${\lambda }_{i}$**出现的次数.**[[3]](#ref_3)

$a\left({\lambda }_i\right)|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_i\cdot \cdot \cdot {\lambda }_z⟩={\epsilon }^m{\sqrt{n}}_i|n-1;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩$

其中 $m$ 是将 ${\lambda }_i$ 移到最左边一共移动的次数, 如果作用的态没有包含 ${\lambda }_i$ 就直接等于 $0$ .

其实产生算符的定义式还有一种稍微复杂一些, 但形式上更完善的写法:

$a\left( {\lambda }_{i} \right)\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{y}{\lambda }_{z} \right\rangle =\left[ \begin{align} & \ \ \ \ \ \sqrt{n}_{a}{\varepsilon }^{0}{\delta }_{ia}\left| n-1;{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{y}{\lambda }_{z} \right\rangle \\ & +\sqrt{n}_{b}{\varepsilon }^{1}{\delta }_{ib}\left| n-1;{\lambda }_{a}\cdot \cdot \cdot {\lambda }_{y}{\lambda }_{z} \right\rangle \\ & +\cdot \cdot \cdot \\ & +\sqrt{n}_{y}{\varepsilon }^{n-2}{\delta }_{iy}\left| n-1;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle \\ & +\sqrt{n}_{z}{\varepsilon }^{n-1}{\delta }_{iz}\left| n-1;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{y} \right\rangle \\ \end{align} \right]$ 其中 $n_a,\cdot \cdot \cdot ,n_z$ 分别是 $|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_y{\lambda }_z⟩$ 中 ${\lambda }_a,\cdot \cdot \cdot ,{\lambda }_z$ 出现的次数.

复杂一点儿, 但经过前面内容的洗礼应该很容易理解: 无论 $i$ 取值如何, 右边最多保存一项. 如果所有的克罗内克delta全部为 $0$ 就表明 $|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_y{\lambda }_z⟩$ 不存在 ${\lambda }_i$ , 结果归 $0$

5.4. 湮灭算符之间的对易关系:

湮灭算符之间的对易关系看起来不好整, 实际上很显然, 因为:

$a^{†}\left({\lambda }_a\right)a^{†}\left({\lambda }_b\right)-\epsilon a^{†}\left({\lambda }_b\right)a^{†}\left({\lambda }_a\right)=0\Rightarrow a\left({\lambda }_b\right)a\left({\lambda }_a\right)-\epsilon a\left({\lambda }_a\right)a\left({\lambda }_b\right)=0$

$\Rightarrow a\left({\lambda }_a\right)a\left({\lambda }_b\right)-\epsilon a\left({\lambda }_b\right)a\left({\lambda }_a\right)=0$

5.5. 产生湮灭算符之间的对易关系:

$a\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)-\epsilon a^{†}\left({\lambda }_j\right)a\left({\lambda }_i\right)={\delta }_{ij}$

分析:

研究的是 $a^{†}\left({\lambda }_j\right)a\left({\lambda }_i\right)$ 与 $a\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)$ 间的关系. 当 $i=j$ 且为对称态情况时: \left. \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S}=a\left( {\lambda }_{i} \right)\sqrt{n}_{i}+1}{\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{n}'}_{i}\sqrt{n}_{i}+1}{\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {n}_{i}+1 \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & {a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S}={a}^{\dagger }\left( {\lambda }_{i} \right)\sqrt{n}_{i}{\left| n-1;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{n}''}_{i}+1}\sqrt{n}_{i}{\left| n;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={n}_{i}{\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ \end{align} \right\} $\Rightarrow a\left({\lambda }_i\right)a^{†}\left({\lambda }_i\right)-a^{†}\left({\lambda }_i\right)a\left({\lambda }_i\right)=1$ 你也许会问, 万一 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_S$ 没有 ${\lambda }_i$ 上式是否还能成立? 其实令前面的 $n_i=0$ 就是这个情况了.

当 $i\ne j$ 且为对称态情况时: \left. \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S}=a\left( {\lambda }_{i} \right)\sqrt{n}_{j}+1}{\left| n+1;{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{n}_{i}\sqrt{n}_{j}+1}{\left| n;{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & {a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S}={a}^{\dagger }\left( {\lambda }_{j} \right)\sqrt{n}_{i}{\left| n-1;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{n}_{j}+1}\sqrt{n}_{i}{\left| n;{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{S} \\ \end{align} \right\} $\Rightarrow a\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)-a^{†}\left({\lambda }_j\right)a\left({\lambda }_i\right)=0$ 你也许会问, 万一 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_S$ 没有 ${\lambda }_i$ 上式是否还能成立? 其实令前面的 $n_i=0$ 就是这个情况了.

当 $i=j$ 且 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_A$ 中不包含 ${\lambda }_i$ 时: \left. \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=a\left( {\lambda }_{i} \right)\sqrt{0+1}{\left| n+1;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & {a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=0 \\ \end{align} \right\} \Rightarrow \left. \left\{ \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{i} \right)=1 \\ & {a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right)=0 \\ \end{align} \right. \right\}\Rightarrow a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{i} \right)+{a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right)=1

当 $i=j$ 且 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_A$ 中包含 ${\lambda }_i$ 时: \left. \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=0 \\ & {a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}={a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right){\left( -1 \right)}^{m}{\left| n;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={a}^{\dagger }\left( {\lambda }_{i} \right)\sqrt{1}{\left( -1 \right)}^{m}{\left| n-1;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sqrt{0+1}\sqrt{1}{\left( -1 \right)}^{m}{\left| n;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={\left( -1 \right)}^{m}{\left( -1 \right)}^{m}{\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ \end{align} \right\} \Rightarrow \left. \left\{ \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{i} \right) \\ & {a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right)=1 \\ \end{align} \right. \right\}\Rightarrow a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{i} \right)+{a}^{\dagger }\left( {\lambda }_{i} \right)a\left( {\lambda }_{i} \right)=1

当 $i\ne j$ 且 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_A$ 中既不包含 ${\lambda }_i$ 又不包含 ${\lambda }_j$ 时: \left. \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=\text{0} \\ & {a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=\text{0} \\ \end{align} \right\}\Rightarrow a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right)\text{+}{a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right)=0

当 $i\ne j$ 且 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_A$ 中包含 ${\lambda }_i$ 而不包含 ${\lambda }_j$ 时: \left. \begin{align} & \ \ \ \ \ a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & =a\left( {\lambda }_{i} \right)\sqrt{0+1}{\left| n+1;{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & =a\left( {\lambda }_{i} \right)\sqrt{0+1}{\left( -1 \right)}^{m+1}{\left| n+1;{\lambda }_{i}{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & =\sqrt{1}\sqrt{0+1}{\left( -1 \right)}^{m+1}{\left| n;{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & \ \ \ \ \ {a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & ={a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & ={a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right){\left( -1 \right)}^{m}{\left| n;{\lambda }_{i}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & ={a}^{\dagger }\left( {\lambda }_{j} \right)\sqrt{1}{\left( -1 \right)}^{m}{\left| n-1;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ & =\sqrt{0+1}\sqrt{1}{\left( -1 \right)}^{m}{\left| n;{\lambda }_{j}{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A} \\ \end{align} \right\} $\Rightarrow a\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)-\left(-1\right)a^{†}\left({\lambda }_j\right)a\left({\lambda }_i\right)=0\Rightarrow a\left({\lambda }_i\right)a^{†}\left({\lambda }_j\right)+a^{†}\left({\lambda }_j\right)a\left({\lambda }_i\right)=0$

当 $i\ne j$ 且 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_A$ 中不包含 ${\lambda }_i$ 而包含 ${\lambda }_j$ 时: \left. \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=0 \\ & {a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=0 \\ \end{align} \right\}\Rightarrow a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right)\text{+}{a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right)=0

当 $i\ne j$ 且 ${|n;{\lambda }_a{\lambda }_b\cdot \cdot \cdot {\lambda }_z⟩}_A$ 中既包含 ${\lambda }_i$ 又包含 ${\lambda }_j$ 时: \left. \begin{align} & a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=0 \\ & {a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right){\left| n;{\lambda }_{a}{\lambda }_{b}\cdot \cdot \cdot {\lambda }_{z} \right\rangle }_{A}=0 \\ \end{align} \right\}\Rightarrow a\left( {\lambda }_{i} \right){a}^{\dagger }\left( {\lambda }_{j} \right)\text{+}{a}^{\dagger }\left( {\lambda }_{j} \right)a\left( {\lambda }_{i} \right)=0

综上所述, 全部情况可以总结为: $a\left({\lambda }_i\right)a^{†}\left({\lambda }_i\right)-\epsilon a^{†}\left({\lambda }_i\right)a\left({\lambda }_i\right)={\delta }_{ij}$ 其中$\epsilon =\pm 1$ , 且 $\epsilon =1$ 表征对称态; $\epsilon =-1$ 表征反对称态.

5.6. 我们这样定义产生湮灭算符会不会导致一些运算不自洽:

比如说 $⟨5;{\lambda }_1{\lambda }_2{\lambda }_2{\lambda }_2{\lambda }_3|a\left({\lambda }_2\right)|6;{\lambda }_1{\lambda }_2{\lambda }_2{\lambda }_2{\lambda }_2{\lambda }_3⟩$ 中 $a\left({\lambda }_2\right)$ 是否无论向左还是向右作用结果都相同呢? 答案是肯定的, 结果都是 $2$ , 运算过程如下所示:

向右作用:

\begin{align} & \ \ \ \ \ \langle 5;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}|\left[ a\left( {\lambda }_{2} \right)\left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle \right] \\ & =\langle 5;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}|\sqrt{4}\left| 5;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle =2 \\ \end{align}

向左作用:

\begin{align} & \ \ \ \ \ \left[ \langle 5;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}|a\left( {\lambda }_{2} \right) \right]\left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle \\ & ={\left[ {a}^{\dagger }\left( {\lambda }_{2} \right)\left| 5;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle \right]}^{\dagger }\left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle \\ & ={\left[ \sqrt{3+1}\left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle \right]}^{\dagger }\left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle \\ & =\langle 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3}|\sqrt{4}\left| 6;{\lambda }_{1}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{2}{\lambda }_{3} \right\rangle =2 \\ \end{align}

参考

• ^不考虑顺序
• ^可以统一规定为放在任何一个位置, 这种统一只是为了大家交流的方便.
• ^抱歉反反复复地强调这个地方, 因为我这个修改实在是太冒险了.