全文HTML

在量子物理中算符是一个重要的基本概念. 一般来说, 由于算符的不可对易性, 譬如玻色湮灭算符与产生算符的对易关系为$ \left[a, {a}^\dagger \right]=a{a}^\dagger -{a}^\dagger a=1 $, 使得涉及算符的计算颇为棘手. 如果能够把算符转化为有序排列形式, 则相应的计算就方便得多. 例如一个正规乘积排序的算符$ :f(a, {a}^\dagger ): $, 其在相干态表象$\left|z\right \rangle =\mathrm{e}\mathrm{x}\mathrm{p}\left(-\dfrac{1}{2}z{z}^{*}+z{a}^\dagger \right)\left|0\right \rangle$中的矩阵元^[1,2]为

$\langle z|:f(a,{a}^\dagger ):|{z}' \rangle =f({z}',{z}^{\mathrm{*}}){\mathrm{e}}^{-\frac{1}{2}z{z}^{*}-\tfrac{1}{2}{z}'{z}^{\prime*}+{z}^{*}{z}'}.$

又如, 密度算符$ \rho $的Glauber-Sudarshan$ P $表示可由其反正规乘积排序形式直接得到^[3,4], 即

$\rho =\int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}P(z,{z}^{*})\left|z\right \rangle \left \langle z\right|=\vdots P(a,{a}^\dagger )\vdots.$

(1)式中记号:: 表示正规乘积排序(即所有的产生算符位于所有的湮灭算符的左侧), (2)式中记号$ \vdots \mathrm{ }\mathrm{ }\vdots $表示反正规乘积排序(即所有的湮灭算符位于所有的产生算符的左侧). 除了算符的这两种有序化排列形式外, 还有一些其他的有序化排列方式, 如坐标-动量排序和动量-坐标排序^[5-12]. 坐标-动量排序算符和动量-坐标排序算符在$ x\text-p $相空间的矩阵元可方便地得到, 即

$\begin{split}&\left \langle x \right|{\cal{Q}}g\left( {X,P} \right){\cal{Q}}\left| p \right \rangle = g\left( {x,p} \right)\left \langle {x{\rm{|}}p} \right \rangle ,\;\\&\left \langle p \right|{\cal{P}}h\left( {X,P} \right){\cal{P}}\left| x \right \rangle = h\left( {x,p} \right)\left \langle {p{\rm{|}}x} \right \rangle ,\end{split}$

这里记号${\cal{Q}} \cdots {\cal{Q}}$表示坐标-动量排序(即所有的坐标算符$ X $位于所有的动量算符$ P $的左侧), 记号$ {\cal{P}}\cdots {\cal{P}} $表示动量-坐标排序(即所有的动量算符$ P $位于所有的坐标算符$ X $的左侧). 因此, 尽可能直接地得到算符的有序化排列形式就是一项重要的数理任务.
在量子力学和量子光学中, 人们经常会遇到形如$ {\left(XP\right)}^{n} $和$ {\left(XP\right)}^{-n} $的正负指数幂算符($ n=0,\; \mathrm{ }1,\; \mathrm{ }2,\; \mathrm{ }3, \cdots $). 对于低次正指数幂算符, 可以利用对易关系将其转换为有序化排列形式. 而对于负指数幂算符和高次正指数幂算符, 其处理是非常棘手的. 本文将利用特殊函数和算符的正规与反正规乘积排序间的互换法则^[4,13]导出幂算符$ {\left(a{a}^\dagger \right)}^{\pm n} $和$ {\left({a}^\dagger a\right)}^{\pm n} $的正规与反正规乘排序式. 进一步, 利用类比法得到算符$ {\left(XP\right)}^{\pm n} $和$ {\left(PX\right)}^{\pm n} $的坐标-动量排序与动量-坐标排序式. 最后, 将对新得到的这些结果的应用进行一些讨论.

这一节来讨论指数算符$ {\mathrm{e}}^{\lambda {a}^\dagger a} $和$ {\mathrm{e}}^{\lambda a{a}^\dagger } $的正规乘积排序与反正规乘积排序. 将充分利用有序算符内的积分技术和正规乘积排序与反正规乘积排序间的互换法则^[4,13].
首先来导出真空投影子$ \left|0\right. \rangle \langle \left.0\right| $的正规乘积排序形式. 令$ \left|0\right. \rangle \langle \left.0\right|=:f(a, {a}^\dagger ): $, 并用相干态左矢$ \langle \left.z\right| $和右矢$ \left|z\right. \rangle $夹乘之, 得$ f\left(z, {z}^{*}\right)= \langle z|0 \rangle \langle 0 |z \rangle ={\mathrm{e}}^{-z{z}^{*}} $. 于是得到

$\left|0\right. \rangle \langle \left.0\right|= :{\mathrm{e}}^{-a{a}^\dagger }:, $

这就是真空投影子$ \left|0\right. \rangle \langle \left.0\right| $的正规乘积排序形式. 利用福克空间的完备性关系式$\displaystyle \sum _{n=0}^{\infty }\left|n\right. \rangle \langle \left.n\right|=1 $, $ {a}^\dagger a\left|n\right. \rangle =n\left|n\right. \rangle $以及真空投影子的正规乘积排序形式, 便得到

$\begin{split} {\mathrm{e}}^{\lambda {a}^\dagger a}=\;&\sum\limits _{n=0}^{\infty }{\mathrm{e}}^{\lambda {a}^\dagger a}\left|n\right. \rangle \langle \left.n\right|=\sum\limits _{n=0}^{\infty }{\mathrm{e}}^{\lambda n}\left|n\right. \rangle \langle \left.n\right| \\ =\;&\sum\limits _{n=0}^{\infty }{\mathrm{e}}^{\lambda n}\frac{{a}^{\dagger n}}{\sqrt{n!}}\left|0\right. \rangle \langle \left.0\right|\frac{{a}^{n}}{\sqrt{n!}}\\ =\;&\sum\limits _{n=0}^{\infty }{\mathrm{e}}^{\lambda n}\frac{{a}^{\dagger n}}{\sqrt{n!}}:{\mathrm{e}}^{-a{a}^\dagger }:\frac{{a}^{n}}{\sqrt{n!}} \\ =\;& :\sum\limits _{n=0}^{\infty }{\mathrm{e}}^{\lambda n}\frac{{\left({aa}^\dagger \right)}^{n}}{n!}{\mathrm{e}}^{-a{a}^\dagger }\!: \\ =\;& :\mathrm{e}\mathrm{x}\mathrm{p}\left[{aa}^\dagger ({\mathrm{e}}^{\lambda }\!-\!1)\right]:, \end{split}$

这就是指数算符$ {\mathrm{e}}^{\lambda {a}^\dagger a} $的正规乘积排序式. 作参数变换$ {\mathrm{e}}^{\lambda }-1=\mu $, 则有

$\begin{split}:\mathrm{e}\mathrm{x}\mathrm{p}\left(\mu {aa}^\dagger \right):=\;&\mathrm{e}\mathrm{x}\mathrm{p}\left[{a}^\dagger a\mathrm{l}\mathrm{n}(1+\mu )\right]\\=\;&\frac{1}{1+\mu }\mathrm{e}\mathrm{x}\mathrm{p}\left[a{a}^\dagger \mathrm{l}\mathrm{n}(1+\mu )\right], \end{split}$

这就是脱掉正规乘积排序指数算符$ :\mathrm{e}\mathrm{x}\mathrm{p}\left(\mu {aa}^\dagger \right): $的正规乘积排序记号$::$的公式.
进一步利用对易关系式$ \left[a, {a}^\dagger \right]=a{a}^\dagger -{a}^\dagger a=1 $和(5)式可得

${\mathrm{e}}^{\lambda a{a}^\dagger }={\mathrm{e}}^{\lambda }{\mathrm{e}}^{\lambda {a}^\dagger a}={\mathrm{e}}^{\lambda }:\mathrm{e}\mathrm{x}\mathrm{p}\left[{aa}^\dagger ({\mathrm{e}}^{\lambda }-1)\right]:, $

这就是指数算符$ {\mathrm{e}}^{\lambda a{a}^\dagger } $的正规乘积排序式.
利用(5)式和算符的正规乘积排序与反正规乘积排序间的互换法则

$:f(a,{a}^\dagger )\mathrm{ }:=\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }\right)f(a,{a}^\dagger )\vdots, $

可得到指数算符$ {\mathrm{e}}^{\lambda {a}^\dagger a} $的反正规乘积排序, 即

$\begin{split}{\mathrm{e}}^{\lambda {a}^\dagger a}=\;& :\mathrm{e}\mathrm{x}\mathrm{p}\left[{aa}^\dagger ({\mathrm{e}}^{\lambda }-1)\right]:\\=\;& \vdots \; \mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }\right)\mathrm{e}\mathrm{x}\mathrm{p}\left[{aa}^\dagger ({\mathrm{e}}^{\lambda }-1)\right]\vdots \\ =\;&\vdots \; \mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }\right)\int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}\bigg(-z{z}^{*}\\ &+za\sqrt{{\mathrm{e}}^{\lambda }-1}+{z}^{*}{a}^\dagger \sqrt{{\mathrm{e}}^{\lambda }-1}\bigg)\vdots \\ =\;&\vdots \int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}\bigg(-{\mathrm{e}}^{\lambda }z{z}^{*}+za\sqrt{{\mathrm{e}}^{\lambda }-1}\\ &+{z}^{*}{a}^\dagger \sqrt{{\mathrm{e}}^{\lambda }-1}\bigg)\vdots . \end{split}$

在(9)式的计算中为了方便约化掉微分算子$\mathrm{e}\mathrm{x}\mathrm{p}\left(-\dfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }\right)$, 对反正规乘积排序算符

$ \vdots \; \mathrm{e}\mathrm{x}\mathrm{p}\left[{aa}^\dagger ({\mathrm{e}}^{\lambda }-1)\right]\vdots $

中指数上的函数采用了积分降次法, 也就是将湮灭算符与产生算符的“乘积”形式通过积分法降成了“和”的形式, 并且用到了积分公式

$\int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}\left(-\zeta z{z}^{*}+\eta z+\xi {z}^{*}\right)=\frac{1}{\zeta }\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\eta \xi }{\zeta }\right),$

其收敛条件为$ {\rm Re}\left(\zeta \right)>0 $, 否则该积分是发散的.
如果$ \mathrm{R}\mathrm{e}\left({\mathrm{e}}^{\lambda }\right)>0 $, 利用积分公式(10)式完成(9)式中的积分, 得

${\mathrm{e}}^{\lambda {a}^\dagger a}={\mathrm{e}}^{-\lambda } \; \vdots \; \mathrm{e}\mathrm{x}\mathrm{p}\left[{aa}^\dagger (1-{\mathrm{e}}^{-\lambda })\right]\vdots,$

这就是在$ \mathrm{R}\mathrm{e}\left({\mathrm{e}}^{\lambda }\right)>0 $的条件下指数算符$ {\mathrm{e}}^{\lambda {a}^\dagger a} $的反正规乘积排序式. 若作参数替换$ 1-{\mathrm{e}}^{-\lambda }=\mu $, 则可得到

$\begin{split}\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left(\mu {aa}^\dagger \right)\vdots =\;&\frac{1}{1-\mu }\mathrm{e}\mathrm{x}\mathrm{p}\left[-{a}^\dagger a\mathrm{l}\mathrm{n}\left(1-\mu \right)\right]\\=\;&\mathrm{e}\mathrm{x}\mathrm{p}\left[-a{a}^\dagger \mathrm{l}\mathrm{n}\left(1-\mu \right)\right],\end{split}$

这就是脱掉反正规乘积排序指数算符$ \vdots \; \mathrm{e}\mathrm{x}\mathrm{p}\left(\mu {aa}^\dagger \right)\vdots $的反正规乘积排序记号$\vdots~ \vdots$的公式, 它要求

$\mathrm{R}\mathrm{e}\left(\mu \right) < 1. $

如果$ {\mathrm{e}}^{\lambda }=-1 $, 也就是$ \lambda =\mathrm{i}(2 n+1)\mathrm{\pi } $, $n=0,\; \pm 1,\; \pm 2,\; \pm 3, \cdots$, 则(9)式化为

$\begin{split} \;& {\mathrm{e}}^{\lambda {a}^\dagger a}=\vdots \int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}(z{z}^{*}+za\sqrt{-2}+{z}^{*}\mathrm{i}z{a}^\dagger \sqrt{-2})\vdots = \vdots \int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}(z{z}^{*}+\mathrm{i}za\sqrt{2}+\mathrm{i}{z}^{*}\mathrm{i}z{a}^\dagger \sqrt{2})\vdots \\ =\;&\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{1}{2}\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }\right)\int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}(\mathrm{i}za\sqrt{2} +\mathrm{i}{z}^{*}\mathrm{i}z{a}^\dagger \sqrt{2})\vdots = \frac{\mathrm{\pi }}{2}\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{1}{2}\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }\right)\mathrm{\delta }\left(a\right)\mathrm{\delta }\left({a}^\dagger \right)\vdots . \\[-10pt] \end{split}$

如果$ \mathrm{R}\mathrm{e}\left({\mathrm{e}}^{\lambda }\right)<0 $且$ {\mathrm{e}}^{\lambda }\ne -1 $, 则(9)式化为

$\begin{split} {\mathrm{e}}^{\lambda {a}^\dagger a}=\;&\vdots \int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}\bigg(-{\mathrm{e}}^{\lambda }z{z}^{*}+za\sqrt{{\mathrm{e}}^{\lambda }-1}+{z}^{*}{a}^\dagger \sqrt{{\mathrm{e}}^{\lambda }-1}\bigg)\vdots \\ =\;&\vdots \int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}\left[(1-{\mathrm{e}}^{\lambda })z{z}^{*}\right]\mathrm{e}\mathrm{x}\mathrm{p}\bigg(-z{z}^{*}+za\sqrt{{\mathrm{e}}^{\lambda }-1}+{z}^{*}{a}^\dagger \sqrt{{\mathrm{e}}^{\lambda }-1}\bigg)\vdots \\ =\;&\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }\right)\mathrm{e}\mathrm{x}\mathrm{p}\left[a{a}^\dagger ({\mathrm{e}}^{\lambda }-1)\right]\vdots = \vdots \sum _{n=0}^{\infty }\frac{{(-)}^{n}}{n!}\frac{{\partial }^{n}}{\partial {a}^{n}}\frac{{\partial }^{n}}{\partial {a}^{\dagger n}}\mathrm{e}\mathrm{x}\mathrm{p}\left[-a{a}^\dagger (1-{\mathrm{e}}^{\lambda })\right]\vdots \\ =\;&\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left[-a{a}^\dagger (1-{\mathrm{e}}^{\lambda })\right]\sum _{n=0}^{\infty }\frac{{(-)}^{n}}{n!} \mathrm{e}\mathrm{x}\mathrm{p}\left[a{a}^\dagger (1-{e}^{\lambda })\right]\frac{{\partial }^{n}}{\partial {a}^{n}}\frac{{\partial }^{n}}{\partial {a}^{\dagger n}}\mathrm{e}\mathrm{x}\mathrm{p}\left[-a{a}^\dagger (1-{\mathrm{e}}^{\lambda })\right]\vdots \\ =\;&\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left[-a{a}^\dagger (1-{\mathrm{e}}^{\lambda })\right]\sum _{n=0}^{\infty }\frac{{({\mathrm{e}}^{\lambda }-1)}^{n}}{n!}{H}_{n,n}\left(a\sqrt{1-{\mathrm{e}}^{\lambda }},{a}^\dagger \sqrt{1-{\mathrm{e}}^{\lambda }}\right)\vdots \\ =\;&\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left[a{a}^\dagger ({\mathrm{e}}^{\lambda }-1)\right]\sum _{n=0}^{\infty }\frac{{(1-{\mathrm{e}}^{\lambda })}^{n}}{n!}{L}_{n}\left[a{a}^\dagger (1-{\mathrm{e}}^{\lambda })\right]\vdots .\\[-10pt] \end{split}$

(15)式的最后一步计算中用到了同下标双变量厄米多项式与拉盖尔多项式的关系, 即

${H}_{n,n}(x,y)={(-)}^{n}{L}_{n}\left(xy\right). $

在$\mathrm{R}\mathrm{e}\left({\mathrm{e}}^{\lambda }\right)\!> \!0$时, 利用对易关系式$\left[a, {a}^\dagger \right]=a{a}^\dagger - {a}^\dagger a=1$和(11)式可得

${\mathrm{e}}^{\lambda a{a}^\dagger }={\mathrm{e}}^{\lambda }{\mathrm{e}}^{\lambda {a}^\dagger a}=\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left[{aa}^\dagger (1-{\mathrm{e}}^{-\lambda })\right]\vdots, $

这一节来导出幂算符$ {\left({a}^\dagger a\right)}^{n} $和$ {\left(a{a}^\dagger \right)}^{n} $的正规乘积排序和反正规乘积排序, 其中$ n $是正整数.
2

3.1.幂算符$ {\left(\mathit{a}{\mathit{a}}^\dagger \right)}^{\mathit{n}} $的正规乘积排序和反正规乘积排序

$ t $

$ {\left(a{a}^\dagger \right)}^{n} $

$ t=0 $

$ {T}_{n}\left(\xi \right) $

$ {T}_{n}\left(\xi \right) $

$ \mathrm{e}\mathrm{x}\mathrm{p}\left[\xi \left({\mathrm{e}}^{t}-1\right)\right] $

$ t=0 $

$ {T}_{n}\left(\xi \right) $

$ {x}_{n}\left(\xi \right) $

$ {T}_{n}\left(\xi \right) $

$ {x}_{n}\left(\xi \right) $

$ {\left(a{a}^\dagger \right)}^{m} $

$ t=0 $

$ \mathrm{R}\mathrm{e}\left({\mathrm{e}}^{t}\right)\approx 1>0 $

$ {\left(a{a}^\dagger \right)}^{n} $

$\displaystyle\int \dfrac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}{\mathrm{e}}^{-\zeta z{z}^{*}+\eta z+\xi {z}^{*}}=\dfrac{1}{\zeta }{\mathrm{e}}^{\eta \xi /\zeta }$

$ \mathrm{R}\mathrm{e}\left(\zeta \right)>0 $

3.2.幂算符$ {\left({\mathit{a}}^\dagger \mathit{a}\right)}^{\mathit{n}} $的正规乘积排序和反正规乘积排序

$ {\left({a}^\dagger a\right)}^{n} $

$ {\left({a}^\dagger a\right)}^{n} $

$ {\left(a{a}^\dagger \right)}^{n} $

$ {\left({a}^\dagger a\right)}^{n} $

$ [{a}^\dagger , -a]=1=[a, {a}^\dagger ] $

$ {\left({a}^\dagger a\right)}^{n} $

$ {\left({a}^\dagger a\right)}^{n} $

$ \left[a, {a}^\dagger \right]=a{a}^\dagger -{a}^\dagger a=1 $

$ {x}_{n}\left(\xi \right) $

$ {T}_{n}\left(\xi \right) $

基于上一节的结果, 利用类比法可导出幂算符$ {\left(XP\right)}^{n} $和$ {\left(PX\right)}^{n} $的坐标-动量排序和动量-坐标排序式. 鉴于$\left[X, -\dfrac{\mathrm{i}}{\hslash }P\right]=1=[a, {a}^\dagger ]$, 通过与(28)式类比可得

$\begin{split}{\left(XP\right)}^{n}=\;&{\left(\mathrm{i}\hslash \right)}^{n}{\left[X\left(-\frac{\mathrm{i}}{\hslash }P\right)\right]}^{n}\\=\;&{(-\mathrm{i}\hslash )}^{n}{\cal{Q}} {T}_{n}\left(\frac{\mathrm{i}}{\hslash }XP\right) {\cal{Q}}, \end{split}$

此即幂算符$ {\left(XP\right)}^{n} $的坐标-动量排序展开式. 通过与(30)式类比, 能得到幂算符$ {\left(XP\right)}^{n} $的动量-坐标排序展开式, 即

$\begin{split}{\left(XP\right)}^{n}=\;&{(-\mathrm{i}\hslash )}^{n}{\left[X\left(\frac{\mathrm{i}}{\hslash }P\right)\right]}^{n} \\= \;&{\left(\mathrm{i}\hslash \right)}^{n} {\cal{P}} {x}_{n}\left(-\frac{\mathrm{i}}{\hslash }XP\right){\cal{P}}. \end{split}$

同样道理, 基于$ \left[\dfrac{1}{\mathrm{i}\hslash }P, X\right]=-1=[{a}^\dagger , a] $, 与(30)式作类比可得

$\begin{split}{\left(PX\right)}^{n}=\;&{\left(\mathrm{i}\hslash \right)}^{n}{\left[\left(\frac{1}{\mathrm{i}\hslash }P\right)X\right]}^{n}\\=\;&{(-\mathrm{i}\hslash )}^{n} {\cal{Q}} {x}_{n}\left(\frac{\mathrm{i}}{\hslash }XP\right){\cal{Q}},\end{split}$

以及与(29)式作类比得

${\left(PX\right)}^{n}\!=\!{\left(\mathrm{i}\hslash \right)}^{n}\!{\left[\left(\frac{1}{\mathrm{i}\hslash }P\right)\!X\right]}^{n}\!=\!{\left(\mathrm{i}\hslash \right)}^{n} {\cal{P}} {T}_{n}\!\left(\frac{1}{\mathrm{i}\hslash }QP\!\right) {\cal{P}}.$

进一步利用指数函数的Taylor展开式和(32)式—(35)式可得到

$\begin{split} &{\mathrm{e}}^{\lambda XP}\!=\!\sum\limits _{n=0}^{\infty }\!\frac{{\lambda }^{n}}{n!}{\left(XP\right)}^{n}\!=\!\sum\limits _{n=0}^{\infty }\!\frac{{(-\mathrm{i}\hslash \lambda )}^{n}}{n!} {\cal{Q}} {T}_{n}\!\left(\frac{\mathrm{i}}{\hslash }XP\!\right){\cal{Q}}, \\ &{\mathrm{e}}^{\lambda XP}\!=\!\sum\limits _{n=0}^{\infty }\!\frac{{\lambda }^{n}}{n!}{\left(XP\right)}^{n}\!=\!\sum\limits _{n=0}^{\infty }\frac{\!{\left(\mathrm{i}\hslash \lambda \!\right)}^{n}}{n!} {\cal{P}} {x}_{n}\left(-\frac{\mathrm{i}}{\hslash }XP\right){\cal{P}},\\ &{\mathrm{e}}^{\lambda PX}\!=\!\sum\limits _{n=0}^{\infty }\!\frac{{\lambda }^{n}}{n!}{\!\left(PX\right)}^{n}\!=\!\sum\limits _{n=0}^{\infty }\!\frac{{(-\mathrm{i}\hslash \lambda )}^{n}}{n!} {\cal{Q}} {x}_{n}\!\left(\frac{\mathrm{i}}{\hslash }XP\!\right) {\cal{Q}},\\ & {\mathrm{e}}^{\lambda PX}\!=\!\sum\limits _{n=0}^{\infty }\!\frac{{\lambda }^{n}}{n!}{\left(PX\right)}^{n}\!=\!\sum\limits _{n=0}^{\infty }\!\frac{{\left(\mathrm{i}\hslash \lambda \right)}^{n}}{n!} {\cal{P}} {T}_{n}\!\left(-\frac{\mathrm{i}}{\hslash }XP\!\right) {\cal{P}}, \end{split}$

这就是指数算符$ {\mathrm{e}}^{\lambda XP} $和$ {\mathrm{e}}^{\lambda PX} $的坐标-动量排序及动量-坐标排序的基本形式. 由于在$ \lambda =0 $的邻域上以上各级数是收敛的, 所以根据(20)式和(25)式便可得到

$\begin{split}{\mathrm{e}}^{\lambda XP}=\;& {\cal{Q}} \mathrm{e}\mathrm{x}\mathrm{p}\left[\frac{\mathrm{i}}{\hslash }XP({\mathrm{e}}^{-\mathrm{i}\hslash \lambda }-1)\right] {\cal{Q}} \\=\;& {\cal{P}} \mathrm{e}\mathrm{x}\mathrm{p}\left[\mathrm{i}\hslash \lambda +\frac{\mathrm{i}}{\hslash }XP(1-{\mathrm{e}}^{\mathrm{i}\hslash \lambda })\right]{\cal{P}},\end{split}$

$\begin{split}{\mathrm{e}}^{\lambda PX}=\;& {\cal{P}} \mathrm{e}\mathrm{x}\mathrm{p}\left[-\frac{\mathrm{i}}{\hslash }XP({\mathrm{e}}^{\mathrm{i}\hslash \lambda }-1)\right] {\cal{P}}\\=\;& {\cal{Q}} \mathrm{e}\mathrm{x}\mathrm{p}\left[-\mathrm{i}\hslash \lambda +\frac{\mathrm{i}}{\hslash }XP({\mathrm{e}}^{-\mathrm{i}\hslash \lambda }-1)\right]{\cal{Q}}.\end{split}$

令$\dfrac{\mathrm{i}}{\hslash }({\mathrm{e}}^{-\mathrm{i}\hslash \lambda }-1)=\mu$, 也就是作参数替换$\lambda =\dfrac{\mathrm{i}}{\hslash }\mathrm{l}\mathrm{n}(1- \mathrm{i}\hslash \mu )$, 从(37)式可得

$\begin{split}\;&{\cal{Q}} \mathrm{e}\mathrm{x}\mathrm{p}\left(\mu XP\right) {\cal{Q}} \\=\;&\mathrm{e}\mathrm{x}\mathrm{p}\left[\frac{\mathrm{i}}{\hslash }XP\mathrm{l}\mathrm{n}(1-\mathrm{i}\hslash \mu )\right]\\=\;&\frac{1}{1-\mathrm{i}\hslash \mu }\mathrm{e}\mathrm{x}\mathrm{p}\left[\frac{\mathrm{i}}{\hslash }PX\mathrm{l}\mathrm{n}(1-\mathrm{i}\hslash \mu )\right],\end{split}$

这是脱掉坐标-动量排序算符$ {\cal{Q}} \mathrm{e}\mathrm{x}\mathrm{p}\left(\mu XP\right) {\cal{Q}}$的坐标-动量排序记号的公式. 若令$-\dfrac{\mathrm{i}}{\hslash }({\mathrm{e}}^{\mathrm{i}\hslash \lambda }-1)=\sigma$, 亦即$\lambda =-\dfrac{\mathrm{i}}{\hslash }\mathrm{l}\mathrm{n}(1+\mathrm{i}\hslash \sigma )$, 则从(38)式可得到

$\begin{split}\;& {\cal{P}} \mathrm{e}\mathrm{x}\mathrm{p}\left(\sigma XP\right) {\cal{P}} \\ =\;&\mathrm{e}\mathrm{x}\mathrm{p}\left[-\frac{\mathrm{i}}{\hslash }PX\mathrm{l}\mathrm{n}(1+\mathrm{i}\hslash \sigma )\right]\\=\;&\frac{1}{1+\mathrm{i}\hslash \sigma }\mathrm{e}\mathrm{x}\mathrm{p}\left[-\frac{\mathrm{i}}{\hslash }XP\mathrm{l}\mathrm{n}(1+\mathrm{i}\hslash \sigma )\right],\end{split}$

这是脱掉动量-坐标排序算符${\cal{P}} \mathrm{e}\mathrm{x}\mathrm{p}\left(\sigma XP\right) {\cal{P}}$的动量-坐标排序记号的公式.

这一节来导出形如$ {\left(AB\right)}^{-1} $的负指数幂算符的有序排列式. 由于

$\left({a}^\dagger a\right)\left|n\right. \rangle =n\left|n\right. \rangle ,\;n=\mathrm{0,1},\mathrm{2,3},\cdots$

其中$\left|n\right. \rangle =\dfrac{{a}^{\dagger n}}{\sqrt{n!}}\left|0\right. \rangle$是粒子数算符$ \left({a}^\dagger a\right) $的本征矢. 也就是说, 粒子数算符$ \left({a}^\dagger a\right) $的本征值是离散的且包括零, 可知数算符$ \left({a}^\dagger a\right) $不可逆, 或者说$ {\left({a}^\dagger a\right)}^{-1} $不存在.
2

5.1.算符$ {\left(\mathit{a}{\mathit{a}}^\dagger \right)}^{-1} $的正规乘积排序和反正规乘积排序

$ \left({a}^\dagger a\right)\left|n\right. \rangle =n\left|n\right. \rangle $

$\left(a{a}^\dagger \right)\left|n\right. \rangle =({a}^\dagger a+ 1) \left|n\right. \rangle = (n+1)\left|n\right. \rangle$

$ n=0, \mathrm{ }1, \mathrm{ }2, \mathrm{ }3, \mathrm{ }\cdots $

$ \left(a{a}^\dagger \right) $

${\left(a{a}^\dagger \right)}^{-1}\equiv {1}/{a{a}^\dagger }$

$ \left|0\right. \rangle \langle \left.0\right|=:{\mathrm{e}}^{-a{a}^\dagger }: $

$ {\left(a{a}^\dagger \right)}^{-1} $

${\mathrm{e}}^{-x}=\displaystyle\sum _{l=0}^{\infty }\dfrac{{(-)}^{l}}{l!}{x}^{l}$

$ {\left(a{a}^\dagger \right)}^{-1} $

$ {H}_{m, n}\left(x, y\right) $

$ {L}_{m}\left(\xi \right)={\mathrm{e}}^{\xi }\dfrac{{\partial }^{m}}{\partial {\xi }^{m}}\left({\xi }^{m}{\mathrm{e}}^{-\xi }\right) $

$ {H}_{m, m}\left(x, y\right) $

5.2.负指数幂算符$ {\left(\mathit{X}\mathit{P}\right)}^{-1} $和$ {\left(\mathit{P}\mathit{X}\right)}^{-1} $的坐标-动量排序及动量-坐标排序

$ (XP+PX) $

$ \left(XP\right)=(XP+PX)/2+\mathrm{i}\hslash /2 $

$ \left(XP\right) $

$ \left(XP\right) $

${\left(XP\right)}^{-1}\equiv {1}/{XP}$

$\left[X, \dfrac{1}{\mathrm{i}\hslash }P]=1=[a, {a}^\dagger \right]$

$ {\left(XP\right)}^{-1} $

$ {\left(XP\right)}^{-1} $

$\left[\dfrac{\mathrm{i}}{\hslash }P, X\right]=1=[a, {a}^\dagger ]$

$ {\left(PX\right)}^{-1} $

由(2)式可知, 一旦得到了密度算符$ \rho $的反正规乘积排序就是相当于得到了其Glauber-Sudarshan$ P $表示^[18], 而密度算符满足的海森堡方程(算符方程)就可转化为相应的$ c $数方程, 这给某些问题的求解带来一定的方便. 同时, 有序算符方法可应用于量子统计.
作为第一个例子, 下面讨论处于热平衡的辐射场(混沌光场). 按照统计物理理论, 表示该辐射光场的密度算符为

$ \rho =\sum _{n=0}^{\infty }\left|n\right. \rangle {p}_{n} \langle \left.n\right|, $

式中${p}_{n}=\dfrac{1}{\mathcal{Z}}\mathrm{e}\mathrm{x}\mathrm{p}\left(-\dfrac{\hslash \omega n}{kT}\right)$是处于$ \left|n\right. \rangle \langle \left.n\right| $态的概率, $\mathcal{Z}=\displaystyle\sum _{n=0}^{\infty }\mathrm{e}\mathrm{x}\mathrm{p}\left(-\dfrac{\hslash \omega n}{kT}\right)=\dfrac{1}{1-\mathrm{e}\mathrm{x}\mathrm{p}\left(-\hslash \omega /kT\right)}$是配分函数, $ k $是玻尔兹曼常数, $ T $是热力学温度. 利用$ \left({a}^\dagger a\right)\left|n\right. \rangle =n\left|n\right. \rangle $可得

$\begin{split}\rho =\;&\left[1-\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{\hslash \omega }{kT}\right)\right]\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{\hslash \omega }{kT}{a}^\dagger a\right) \\ =\;&\left[1-\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{\hslash \omega }{kT}\right)\right]\sum _{k=0}^{\infty }\frac{{\left(-\hslash \omega /kT\right)}^{k}}{k!}{\left({a}^\dagger a\right)}^{k}. \end{split}$

式中幂算符$ {\left({a}^\dagger a\right)}^{k} $的反正规乘积排序已被求出, 也就是(30)式. 把(30)式代入并注意到(20)式, 可得

$\begin{split}\rho =\;&\left[1-\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{\hslash \omega }{kT}\right)\right]\sum _{k=0}^{\infty }\frac{{\left(\hslash \omega /kT\right)}^{k}}{k!}\vdots {x}_{k}(-a{a}^\dagger )\vdots \\ =\;&\left[\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\hslash \omega }{kT}\right)\!-\!1\right]\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left\{\!-\!a{a}^\dagger \left[\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\hslash \omega }{kT}\right)\!-\!1\right]\right\}\vdots \\ =\;&\frac{1}{\bar{n}}\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left(-a{a}^\dagger \frac{1}{\bar{n}}\right)\vdots, \\[-15pt] \end{split}$

这就是混沌光场密度算符的反正规乘积排序形式, 式中

$ \bar{n}\equiv \displaystyle\sum _{n=0}^{\infty }n{p}_{n}=\dfrac{1}{\mathrm{e}\mathrm{x}\mathrm{p}(\hslash \omega /kT)-1}. $

基于(2)式和(49)式, 便得到

$P(z,{z}^{*})=\frac{1}{\bar{n}}\mathrm{exp}\Big(-z{z}^{*}\frac{1}{\bar{n}}\Big),$

此即混沌光场的Glauber-Sudarshan$ P $表示. 混沌光场存在$ P $表示, 意味着存在相应的“经典”光场. 而对于相干态$ \left|\zeta \right. \rangle $, 密度算符$ \left|\zeta \right. \rangle \langle \left.\zeta \right| $的Glauber-Sudarshan$ P $表示则为

$ P(z,{z}^{*})=\mathrm{\pi }\mathrm{\delta }(z-\zeta )\mathrm{\delta }({z}^{*}-{\zeta }^{*}). $

现在来计算混沌光的能量涨落, 能量算符为$H= \mathrm{\hslash }\omega {a}^\dagger a$, 能量的均方涨落定义为

$\langle {{(\Delta H)}^{2}} \rangle \!\equiv\! \langle {{(H-\langle {H} \rangle )}^{2}} \rangle \!=\! \big[\langle {{H}^{2}} \rangle \!-\! {\langle {H} \rangle }^{2}\big]g(\omega ),$

其中$ g\left(\omega \right) $为能级密度因子. 因为

$\begin{split} &\left\langle {{H}^{2}} \right\rangle = \mathrm{t}\mathrm{r}\left(\rho {H}^{2}\right) \\=\;&\int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\frac{1}{\bar{n}}\mathrm{e}\mathrm{x}\mathrm{p}\left(-z{z}^{*}\frac{1}{\bar{n}}\right) \langle z|{H}^{2}|z \rangle \\ =\;&\frac{{\hslash }^{2}{\omega }^{2}}{\bar{n}}\int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}(z{z}^{*}+{z}^{2}{z}^{*2})\mathrm{e}\mathrm{x}\mathrm{p}\left(-z{z}^{*}\frac{1}{\bar{n}}\right) \\ =\;&{\left(\hslash \omega \right)}^{2}\left[\bar{n}+2{\bar{n}}^{2}\right]. \end{split} $

以及$ \left\langle {H} \right\rangle =\mathrm{t}\mathrm{r}\left(\rho H\right)=\hslash \omega \bar{n} $, 故有

$\big\langle {{(\Delta H)}^{2}} \big\rangle ={\left(\hslash \omega \right)}^{2}\left[\bar{n}+{\bar{n}}^{2}\right]g\left(\omega \right). $

另一方面, 由(51)式可得到相干光场的能量涨落, 即$ {\big\langle {{(\Delta H)}^{2}} \big\rangle }_{\text{相干态}}={\left(\hslash \omega \right)}^{2}\zeta {\zeta }^{*}g\left(\omega \right) $. 比较可知, 若把$ \bar{n} $视同为$ \sqrt{\zeta {\zeta }^{*}} $时, 混沌光的能量涨落大于相干光.
作为第二个例子, 下面讨论互换法则(8)式的应用. 考虑正规排序算符$ {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}} $, 利用该互换法则可得其反正规排序式为

$\begin{split} {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}=\;&\vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}{\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}\vdots \\ =\;&\vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}\sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{a}^{\dagger 2m}{a}^{2n}\vdots \\ =\;&\vdots \sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{\mathrm{e}}^{-\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}{a}^{\dagger 2m}{a}^{2n}\vdots \\ =\;&\vdots \sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{H}_{2m,2n}({a}^\dagger ,a)\vdots, \end{split}$

这里再一次用到了双变量Hermite多项式. 另一方面, 也可以采用积分降次法进行计算, 即

$\begin{split} {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}=\;&\vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}{\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}\vdots = \vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}\int \frac{{\mathrm{d}}^{2}{z}_{1}{\mathrm{d}}^{2}{z}_{2}}{{\mathrm{\pi }}^{2}}{\mathrm{e}}^{-{z}_{1}{z}_{1}^{*}+{z}_{1}\sqrt{\lambda }{a}^\dagger +{z}_{1}^{*}\sqrt{\lambda }{a}^\dagger -{z}_{2}{z}_{2}^{*}+{z}_{2}\sqrt{v}a+{z}_{2}^{*}\sqrt{v}a}\vdots \\ =\;&\vdots \int \frac{{\mathrm{d}}^{2}{z}_{1}{\mathrm{d}}^{2}{z}_{2}}{{\mathrm{\pi }}^{2}}{\mathrm{e}}^{-{z}_{1}{z}_{1}^{*}+{z}_{1}\sqrt{\lambda }{a}^\dagger +{z}_{1}^{*}\sqrt{\lambda }{a}^\dagger -{z}_{2}{z}_{2}^{*}+{z}_{2}\sqrt{v}a+{z}_{2}^{*}\sqrt{v}a-\sqrt{\lambda }\sqrt{v}({z}_{1}+{z}_{1}^{*})({z}_{2}+{z}_{2}^{*})}\vdots \\ =\;&\vdots {\mathrm{e}}^{\lambda {a}^{\dagger 2}}\int \frac{{\mathrm{d}}^{2}{z}_{2}}{\mathrm{\pi }}{\mathrm{e}}^{-(1-2\lambda v){z}_{2}{z}_{2}^{*}+{z}_{2}\sqrt{v}(a-2\lambda {a}^\dagger )+{z}_{2}^{*}\sqrt{v}(a-2\lambda {a}^\dagger )+\lambda v{z}_{2}^{2}+\lambda v{z}_{2}^{*2}}\vdots \\ =\;&\frac{1}{\sqrt{1-4\lambda v}}\vdots \mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\lambda {a}^{\dagger 2}-4\lambda va{a}^\dagger +v{a}^{2}}{1-4\lambda v}\right)\vdots, \end{split}$

这是算符$ {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}} $反正规乘积排序的另一种形式, 是指数形式的, 它要求$ \mathrm{R}\mathrm{e}\left(1-4\lambda v\right)>0 $. 在上面的计算中用到了如下积分公式:

$ \begin{split}&\int \frac{{\mathrm{d}}^{2}z}{\mathrm{\pi }}\mathrm{e}\mathrm{x}\mathrm{p}(-\zeta z{z}^{*}+\eta z+\xi {z}^{*}+t{z}^{2}+\tau {z}^{*2}) = \frac{1}{\sqrt{{\zeta }^{2}-4t\tau }}\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\zeta \eta \xi +t{\xi }^{2}+\tau {\eta }^{2}}{{\zeta }^{2}-4t\tau }\right), \end{split}$

其收敛条件为 $ \mathrm{R}\mathrm{e}\left(\zeta +t+\tau \right)>0 $, $\mathrm{R}\mathrm{e}\left(\dfrac{{\zeta }^{2}-4 t\tau }{\zeta +t+\tau }\right) > 0$ 或 $ \mathrm{R}\mathrm{e}\left(\zeta -t-\tau \right)>0 $, $\mathrm{R}\mathrm{e}\left(\dfrac{{\zeta }^{2}-4 t\tau }{\zeta -t-\tau }\right) > 0$. 比较(49)式和(50)式, 并做替换 $ {a}^\dagger \to x $, $ a\to y $, 便得到偶次双变量Hermite多项式$ {H}_{2 m, 2 n}(x, y) $的生成函数如下

$\sum _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{H}_{2m,2n}(x,y)=\;\frac{1}{\sqrt{1-4\lambda v}}\mathrm{e}\mathrm{x}\mathrm{p}\left[\frac{1}{1-4\lambda v}(\lambda {x}^{2}-4\lambda vxy+v{y}^{2})\right].$

同样方法, 正规乘积排序算符$ \left({a}^\dagger {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}a\right) $的反正规乘积排序为

$\begin{split} {a}^\dagger {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}a=\;&\vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}{a}^\dagger {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}a\vdots = \vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}\sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{a}^{\dagger 2m+1}{a}^{2n+1}\vdots \\ =\;& \vdots \sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{\mathrm{e}}^{-\frac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}{a}^{\dagger 2m+1}{a}^{2n+1}\vdots = \vdots \sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{H}_{2m+\mathrm{1,2}n+1}({a}^\dagger ,a)\vdots . \end{split}$

另一方面, 利用互换法则(8)式可得

$\begin{split}\;& {a}^\dagger {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}a=\vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}{a}^\dagger {\mathrm{e}}^{\lambda {a}^{\dagger 2}}{\mathrm{e}}^{v{a}^{2}}a\vdots \\ =\;&{\left.\vdots {\mathrm{e}}^{-\tfrac{{\partial }^{2}}{\partial a\partial {a}^\dagger }}\frac{{\partial }^{2}}{\partial t\partial \tau }\int \frac{{\mathrm{d}}^{2}{z}_{1}{d}^{2}{z}_{2}}{{\mathrm{\pi }}^{2}}{\mathrm{e}}^{-{z}_{1}{z}_{1}^{*}+{z}_{1}\sqrt{\lambda }{a}^\dagger +{z}_{1}^{*}\sqrt{\lambda }{a}^\dagger +t{a}^\dagger -{z}_{2}{z}_{2}^{*}+{z}_{2}\sqrt{v}a+{z}_{2}^{*}\sqrt{v}a+\tau a}\vdots \right|}_{t=\tau =0} \\ =\;&\vdots \frac{{\partial }^{2}}{\partial t\partial \tau }\int \frac{{\mathrm{d}}^{2}{z}_{1}{\mathrm{d}}^{2}{z}_{2}}{{\mathrm{\pi }}^{2}}{\mathrm{e}}^{-({z}_{1}\sqrt{\lambda }+{z}_{1}^{*}\sqrt{\lambda }+t)({z}_{2}\sqrt{v}+{z}_{2}^{*}\sqrt{v}+\tau )}\\ & {\left.\times {\mathrm{e}}^{-{z}_{1}{z}_{1}^{*}+{z}_{1}\sqrt{\lambda }{a}^\dagger +{z}_{1}^{*}\sqrt{\lambda }{a}^\dagger +t{a}^\dagger -{z}_{2}{z}_{2}^{*}+{z}_{2}\sqrt{v}a+{z}_{2}^{*}\sqrt{v}a+\tau a}\vdots \right|}_{t=\tau =0} \\ =\;&\vdots \frac{{\partial }^{2}}{\partial t\partial \tau }{\mathrm{e}}^{t{a}^\dagger +\tau a-t\tau }\int \frac{{\mathrm{d}}^{2}{z}_{1}{\mathrm{d}}^{2}{z}_{2}}{{\mathrm{\pi }}^{2}}{\mathrm{e}}^{-{z}_{1}{z}_{1}^{*}+{z}_{1}\sqrt{\lambda }({a}^\dagger -{z}_{2}\sqrt{v}-{z}_{2}^{*}\sqrt{v}-\tau )+{z}_{1}^{*}\sqrt{\lambda }({a}^\dagger -{z}_{2}\sqrt{v}-{z}_{2}^{*}\sqrt{v}-\tau )} \\ & {\left.\times {\mathrm{e}}^{-{z}_{2}{z}_{2}^{*}+{z}_{2}\sqrt{v}(a-t)+{z}_{2}^{*}\sqrt{v}(a-t)}\vdots \right|}_{t=\tau =0} \\ =\;&\vdots \left[\frac{(a-2\lambda {a}^\dagger )({a}^\dagger -2va)}{{(1-4\lambda v)}^{5/2}}-\frac{1}{{(1-4\lambda v)}^{3/2}}\right]\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\lambda {a}^{\dagger 2}-4\lambda va{a}^\dagger +v{a}^{2}}{1-4\lambda v}\right)\vdots, \end{split}$

这要求$ \mathrm{R}\mathrm{e}\left(1-4\lambda v\right)>0 $. 比较(57)式和(58)式, 并做替换 $ {a}^\dagger \to x $, $ a\to y $, 便得到

$\begin{split}&\sum _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{H}_{2m+\mathrm{1,2}n+1}(x,y) =\left[\frac{\left(x-2vy\right)\left(y-2\lambda x\right)}{{(1-4\lambda v)}^{5/2}}-\frac{1}{{(1-4\lambda v)}^{3/2}}\right] \mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\lambda {x}^{2}-4\lambda vxy+v{y}^{2}}{1-4\lambda v}\right), \end{split}$

这就是奇次双变量Hermite多项式$ {H}_{2 m+\mathrm{1, 2}n+1}(x, y) $的生成函数. 同样可得到奇偶次和偶奇次双变量Hermite多项式的生成函数, 即

$\begin{split} &\sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{H}_{2m+\mathrm{1,2}n}(x,y)\\ =\;&\frac{x-2vy}{{(1-4\lambda v)}^{3/2}}\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\lambda {x}^{2}-4\lambda vxy+v{y}^{2}}{1-4\lambda v}\right), \\ &\sum\limits _{m,n=0}^{\infty }\frac{{\lambda }^{m}{v}^{n}}{m!n!}{H}_{2m,2n+1}(x,y)\\ =\;&\frac{y-2\lambda x}{{(1-4\lambda v)}^{3/2}}\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{\lambda {x}^{2}-4\lambda vxy+v{y}^{2}}{1-4\lambda v}\right). \end{split}$

这一些关于双变量Hermite多项式的生成函数都将在数学物理中有着各自的应用.

利用特殊函数和正规乘积排序与反正规乘积排序间的一般互换法则将幂算符$ {\mathrm{ }\left(a{a}^\dagger \right)}^{n} $, $ {\left({a}^\dagger a\right)}^{n} $及$ {\left(a{a}^\dagger \right)}^{-1} $化为了正规乘积排序和反正规乘积排序形式. 进一步利用类比法得到了幂算符$ {\left(XP\right)}^{n} $, $ {\left(PX\right)}^{n} $及$ {\left(XP\right)}^{-1} $的坐标-动量排序和动量-坐标排序式. 最后, 利用幂算符$ {\left({a}^\dagger a\right)}^{n} $的有序排列结果讨论了混沌光场, 并通过正规乘积与反正规乘积排序间的一般互换法则得到了在数学物理中有着应用价值的偶次及奇次双变量Hermite多项式的生成函数.

从多项式$ {x}_{n}\left(\xi \right) $的定义(19)式可得

$ {x}_{n}\left(\xi \right)={\left.\frac{{\partial }^{n}}{\partial {t}^{n}}{\mathrm{e}}^{t+\xi ({\mathrm{e}}^{t}-1)}\right|}_{t=0} = {\left.{\mathrm{e}}^{-\xi }\frac{{\partial }^{n}}{\partial {t}^{n}}{\mathrm{e}}^{t}{\mathrm{e}}^{\xi {\mathrm{e}}^{t}}\right|}_{t=0}={\left.{\mathrm{e}}^{-\xi }\frac{\partial }{\partial \xi }\frac{{\partial }^{n}}{\partial {t}^{n}}{\mathrm{e}}^{\xi {\mathrm{e}}^{t}}\right|}_{t=0}. \tag{A1} $

做参数替换 $ {\mathrm{e}}^{t}=\tau $, 则$\dfrac{\partial }{\partial t}=\dfrac{\partial \tau }{\partial t}\dfrac{\partial }{\partial \tau }=\tau \dfrac{\partial }{\partial \tau }$, 于是有

$\begin{split} \;& {x}_{n}\left(\xi \right)={{\mathrm{e}}^{-\xi }\frac{\partial }{\partial \xi }\left.{\left(\tau \frac{\partial }{\partial \tau }\right)}^{n}{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1} ={\left.{\mathrm{e}}^{-\xi }{\frac{\partial }{\partial \xi }}\underbrace{\left(\tau \frac{\partial }{\partial \tau }\right)\left(\tau \frac{\partial }{\partial \tau }\right)\cdots \left(\tau \frac{\partial }{\partial \tau }\right)\left(\tau \frac{\partial }{\partial \tau }\right)}_{n{\text{次}}}{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1} \\ =\;&{\left.{\mathrm{e}}^{-\xi }{\frac{\partial }{\partial \xi }}\underbrace{\left(\tau \frac{\partial }{\partial \tau }\right)\left(\tau \frac{\partial }{\partial \tau }\right)\cdots \left(\tau \frac{\partial }{\partial \tau }\right)}_{n-1{\text{次}}}\left(\xi \frac{\partial }{\partial \xi }\right){\mathrm{e}}^{\xi \tau }\right|}_{\tau =1} = {\left.{\mathrm{e}}^{-\xi }{\frac{\partial }{\partial \xi }\left(\xi \frac{\partial }{\partial \xi }\right)}\underbrace{\left(\tau \frac{\partial }{\partial \tau }\right)\left(\tau \frac{\partial }{\partial \tau }\right)\cdots \left(\tau \frac{\partial }{\partial \tau }\right)}_{n-1{\text{次}}}{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1} \\ =\;&{\left.{\mathrm{e}}^{-\xi }{\frac{\partial }{\partial \xi }}\underbrace{\left(\xi \frac{\partial }{\partial \xi }\right)\left(\xi \frac{\partial }{\partial \xi }\right)\cdots \left(\xi \frac{\partial }{\partial \xi }\right)\left(\xi \frac{\partial }{\partial \xi }\right)}_{n{\text{次}}}{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1} = {\left.{\mathrm{e}}^{-\xi }\underbrace{{\left(\frac{\partial }{\partial \xi }\xi \right)\left(\frac{\partial }{\partial \xi }\xi \right)\cdots \left(\frac{\partial }{\partial \xi }\xi \right)}\left(\frac{\partial }{\partial \xi }\xi \right)}_{n{\text{次}}}\frac{\partial }{\partial \xi }{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1} ={\mathrm{e}}^{-\xi }{\left(\frac{\partial }{\partial \xi }\xi \right)}^{n}{\mathrm{e}}^{\xi }. \end{split}\tag{A2}$

易见, $ {x}_{n}\left(\xi \right) $不同于Touchard多项式${T}_{n}\left(\xi \right)={\mathrm{e}}^{-\xi }{\left(\xi \dfrac{\partial }{\partial \xi }\right)}^{n}{\mathrm{e}}^{\xi }$, 亦不同于Laguerre多项式${L}_{n}\left(\xi \right)={\mathrm{e}}^{\xi }{\left(\dfrac{\partial }{\partial \xi }\right)}^{n}{\xi }^{n}{\mathrm{e}}^{-\xi }$. 从(A2)式可得

$\begin{split}&n=0, {x}_{0}\left(\xi \right)=1,\\ &n=1, {x}_{1}\left(\xi \right)=\xi +1,\\ & n=2, {x}_{2}\left(\xi \right)={\xi }^{2}+3\xi +1,\\ & n=3, {x}_{3}\left(\xi \right)={\xi }^{3}+6{\xi }^{2}+7\xi +1,\\ &n=4, {x}_{4}\left(\xi \right)={\xi }^{4}+10{\xi }^{3}+25{\xi }^{2}+15\xi +1,\\ & n=5, {x}_{5}\left(\xi \right)={\xi }^{5}+15{\xi }^{4}+65{\xi }^{3}+90{\xi }^{2}+31\xi +1,\cdots .\end{split}$

进一步可得到多项式$ {x}_{n}\left(\xi \right) $的递推公式和幂级数展开式分别为

$ {x}_{n+1}\left(\xi \right)=\left(1+\xi \right){x}_{n}\left(\xi \right)+\xi \frac{\partial {x}_{n}\left(\xi \right)}{\partial \xi }. \tag{A3}$

$ {x}_{n}\left(\xi \right)=\sum _{m=0}^{n}{\xi }^{m}\sum _{l=0}^{m}{(-)}^{m+l}\frac{{(l+1)}^{n}}{l!\left(m-l\right)!}. \tag{A4}$

多项式$ {x}_{n}\left(\xi \right) $跟Touchard多项式的关系为

$ {x}_{n}\left(\xi \right)=\frac{\partial {T}_{n}\left(\xi \right)}{\partial \xi }+{T}_{n}\left(\xi \right).\tag{A5}$

另外, 如果作一特殊规定, 即 ${x}_{-1}\left(\xi \right)=\dfrac{1}{\xi }$, 则有

$ {T}_{n+1}\left(\xi \right)=\xi {x}_{n}\left(\xi \right). \tag{A6}$

在(B5)式中若取$ {y}_{1}=2\xi $, $ {y}_{2}=-2 $, ${y}_{m\geqslant 3}=0$, 则有

${\mathrm{e}}^{-{t}^{2}+2t\xi }=\sum _{n\geqslant 0}{B}_{n}\left(\xi \right)\frac{{t}^{n}}{n!}.\tag{A7}$

将(A7)跟Hermite多项式的定义 ${\mathrm{e}}^{\!-\!{t}^{2}\!+\!2 t\xi }=\sum\limits _{n\geqslant 0}{H}_{n}\left(\xi \right)\dfrac{{t}^{n}}{n!}$ 比较可知, Hermite多项式是Bell多项式的一个特例. 同样可以证明, Laguerre多项式以及本文引入的$ {x}_{n}\left(\xi \right) $ 多项式也可以由Bell多项式派生. 也就是说, Bell多项式是一个多项式家族, 在一定的条件下可以派生出有着特定物理用途的多项式.

从Touchard多项式$ {T}_{n}\left(\xi \right) $的定义可得

$ {T}_{n}\left(\xi \right)={\left.\frac{{\partial }^{n}}{\partial {t}^{n}}{\mathrm{e}}^{\xi ({\mathrm{e}}^{t}-1)}\right|}_{t=0}={\mathrm{e}}^{-\xi }{\left.\frac{{\partial }^{n}}{\partial {t}^{n}}{\mathrm{e}}^{\xi {\mathrm{e}}^{t}}\right|}_{t=0}. \tag{B1}$

做参数替换$ {\mathrm{e}}^{t}=\tau $, 则有

$\begin{split} {T}_{n}\left(\xi \right)=\;&{\left.{\mathrm{e}}^{-\xi }{\left(\frac{\partial \tau }{\partial t}\frac{\partial }{\partial \tau }\right)}^{n}{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1}={\left.{\mathrm{e}}^{-\xi }{\left(\tau \frac{\partial }{\partial \tau }\right)}^{n}{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1}\\ =\;&{\left.{\mathrm{e}}^{-\xi }{\left(\xi \frac{\partial }{\partial \xi }\right)}^{n}{\mathrm{e}}^{\xi \tau }\right|}_{\tau =1}={\mathrm{e}}^{-\xi }{\left(\xi \frac{\partial }{\partial \xi }\right)}^{n}{\mathrm{e}}^{\xi }.\\[-10pt] \end{split}\tag{B2}$

由(B2)式容易得到

$\begin{split}&n=0,{T}_{0}\left(\xi \right)=1, \\ & n=1,{T}_{1}\left(\xi \right)=\xi, \\ &n=2,{T}_{2}\left(\xi \right)={\xi }^{2}+\xi, \\ & n=3,{T}_{3}\left(\xi \right)={\xi }^{3}+3{\xi }^{2}+\xi,\\ &n=4,{T}_{4}\left(\xi \right)={\xi }^{4}+6{\xi }^{3}+7{\xi }^{2}+\xi, \\ &n=5,{T}_{5}\left(\xi \right)={\xi }^{5}+10{\xi }^{4}+25{\xi }^{3}+15{\xi }^{2}+\xi ,\cdots .\end{split}$

由(B2)式还可得出其递推公式, 即

$\begin{split}{T}_{n+1}\left(\xi \right)=\;&{\mathrm{e}}^{-\xi }{\left(\xi \frac{\partial }{\partial \xi }\right)}^{n+1}{\mathrm{e}}^{\xi }={\mathrm{e}}^{-\xi }\left(\xi \frac{\partial }{\partial \xi }\right){\mathrm{e}}^{\xi }{\mathrm{e}}^{-\xi }{\left(\xi \frac{\partial }{\partial \xi }\right)}^{n}{\mathrm{e}}^{\xi } \\ =\;&{\mathrm{e}}^{-\xi }\left(\xi \frac{\partial }{\partial \xi }\right){\mathrm{e}}^{\xi }{T}_{n}\left(\xi \right)\\ =\;&{\mathrm{e}}^{-\xi }\xi \left({\mathrm{e}}^{\xi }{T}_{n}\left(\xi \right)+{\mathrm{e}}^{\xi }\frac{\partial {T}_{n}\left(\xi \right)}{\partial \xi }\right) \\ =\;&\xi {T}_{n}\left(\xi \right)+\xi \frac{\partial {T}_{n}\left(\xi \right)}{\partial \xi }. \\[-15pt]\end{split}\tag{B3}$

另外, 还可得到$ {T}_{n}\left(\xi \right) $的幂级数展开式, 即

$ {T}_{n}\left(\xi \right)=\sum _{k=0}^{n}{\xi }^{k}\sum _{l=0}^{k}{(-)}^{l}\frac{{(k-l)}^{n}}{l!\left(k-l\right)!}=\sum _{k=0}^{n}S(n,k){\xi }^{k}, \tag{B4}$

式中$S\left(n, k\right)=\sum\limits _{l=0}^{k}{(-)}^{l}\dfrac{{(k-l)}^{n}}{l!\left(k-l\right)!}$是第二类Stirling数^[12]. 事实上, 完全型Bell多项式的定义为^[12]

$\mathrm{e}\mathrm{x}\mathrm{p}\left(\sum _{m\geqslant 1}{y}_{m}\frac{{t}^{m}}{m!}\right)=\sum _{n\geqslant 0}{B}_{n}({y}_{1},{y}_{2},\cdots,{y}_{n})\frac{{t}^{n}}{n!},\tag{B5}$

其中已规定$ {B}_{0}=1 $. 如果取$ {y}_{m}=\xi $, $ m=\mathrm{1, 2}, 3, \cdots $, 则(B5)式约化为

$ {\mathrm{e}}^{\xi ({\mathrm{e}}^{t}-1)}=\sum _{n\geqslant 0}{B}_{n}(\xi,\xi,\cdots,\xi )\frac{{t}^{n}}{n!},\tag{B6}$

比较(B6)式和(25)式可知Touchard多项式是Bell多项式的一个特例, 即$ {T}_{n}\left(\xi \right)={B}_{n}(\xi, \xi, \cdots, \xi ) $.