作为量子力学的“语言”, 狄拉克符号俨然诠释了量子世界的抽象思想和普遍概念, 并促进了人们对量子力学的深刻理解.同时, 量子力学中的很多概念全然不同于经典力学, 为了阐释特定的物理意义, 量子力学必须发展其自身的数学语言.例如, 在数学上第一类Fredholm方程^[1]为$g\left( t\right) =\int_{-a}^{b}k\left( t, s\right) f\left( s\right){\rm d}s, $连续函数$ k\left( t, s\right) $为积分核.在文献[2-3]中算符Fredholm方程定义为$G\left( a, {{a}^{\dagger }}\right) =\int_{-a}^{b}K\left( a, {{a}^{\dagger }}, q\right) F\left( q\right){\rm d}q, $其中积分核$K\left( a,{{a}^{\dagger }} ,q \right)$是一个量子力学算符, $q$为实变量, $a$和${{a}^{\dagger }}$表示量子辐射场的湮灭与产生算符.众所周知, 牛顿-莱布尼兹(Newton-Leibniz)积分规则并不能直接应用于形如$\left \vert {}\right \rangle \left \langle {}\right \vert $的算符积分.考虑到如上算符积分的问题, 文献[4]介绍了范洪义教授提出的有序算符内积分技术(the technique of integration within an ordered product of operators, IWOP), 在该积分规则下能直接将牛顿-莱布尼兹积分规则应用于狄拉克(Dirac)的算符积分.有序算符内积分技术表明如果算符$K\left( a, {{a}^{\dagger }}, q\right) $是编序算符, 则算符Fredholm方程就能被直接积分.例如, $K\left( a, {{a}^{\dagger }}, q\right) =\colon \exp \left( q-Q\right) \colon, $则

$ \int_{-\infty }^{\infty }{\frac{\text{d}q}{\sqrt{\pi }}}:\exp \left[ -{{\left( q-Q \right)}^{2}} \right]:f\left( q \right)=:G\left( Q \right):,\text{ }\ \ Q=\frac{a+{{a}^{\dagger }} }{\sqrt{2}}, $

(1.1)

这里$\colon \exp \left[-\left( q-Q\right) ^{2}\right] \colon $是积分核, 在正规编序算符“$\colon \colon $”内, 算符$a$和${{a}^{\dagger }}$是对易的.注意到$\frac{1}{\sqrt{\pi }}\colon \exp \left[-\left( q-Q\right) ^{2}\right] \colon =\left \vert q\right \rangle \left \langle q\right \vert $和坐标表象的完备性关系$\int_{-\infty }^{\infty }{\rm d}q\left \vert q\right \rangle \left \langle q\right \vert =1, $得到

$ \begin{eqnarray} \int_{-\infty }^{\infty }\frac{{\rm d}q}{\sqrt{\pi }}\colon \exp \left[-\left( q-Q\right) ^{2}\right] \colon f\left( q\right) &=&\int_{-\infty }^{\infty }{\rm d}q\left \vert q\right \rangle \left \langle q\right \vert f\left( q\right) \nonumber\\ &=&\int_{-\infty }^{\infty }{\rm d}q\left \vert q\right \rangle \left \langle q\right \vert f\left( Q\right) =f\left( Q\right), \label{eq:a2} \end{eqnarray} $

(1.2)

这里$\left| q \right\rangle ={{\pi }^{-1/4}}\exp \left( -{{q}^{2}}/2+\sqrt{2}q{{a}^{\dagger }}-{{a}^{\dagger }}^{2}/2 \right)\left| 0 \right\rangle $为坐标表象, 坐标算符$Q$的本征方程为$Q\left \vert q\right \rangle =q\left \vert q\right \rangle, $ $\colon G\left( Q\right) \colon $是算符$f\left( Q\right) $的正规乘积编序形式.当$f\left( q\right) =H_{n}\left( q\right), $则有

$ \begin{equation} \int_{-\infty }^{\infty }\frac{{\rm d}q}{\sqrt{\pi }}\colon \exp \left[-\left( q-Q\right) ^{2}\right] \colon H_{n}\left( q\right) =2^{n}\colon Q^{n}\colon, \label{eq:a3} \end{equation} $

(1.3)

这里$2^{n}\colon Q^{n}\colon $是算符厄米特多项式$H_{n}\left( Q\right) $的正规乘积形式. $H_{n}\left( q\right) $为单变量的厄米特多项式, 其产生函数为

$ \begin{equation} \sum_{n=0}^{\infty }\frac{q^{n}}{n!}H_{n}\left( t\right) =\exp \left( 2tq-q^{2}\right) . \label{eq:a4} \end{equation} $

(1.4)

$H_{n}\left( q\right) $张开了一个正交和完备的函数空间, 即$\int_{-\infty }^{\infty }\frac{{\rm d}q}{\sqrt{\pi }} {\rm e}^{-q^{2}}H_{n}\left( q\right) H_{m}\left( q\right) =2^{n}n!\delta _{nm}.$注意到湮灭算符$a= \left( Q+{\rm i}P\right) /\sqrt{2}$和$ \left \langle q\right \vert {{a}^{\dagger }}=2^{-1/2}\left( q-\frac{\rm d}{{\rm d}q}\right) \left \langle q\right \vert, $则$H_{n}\left( Q\right) $在坐标表象与真空态中的矩阵元表示为

$ \begin{eqnarray*} \left \langle q\right \vert H_{n}\left( Q\right) \left \vert 0\right \rangle =2^{n}\left \langle q\right \vert \colon Q^{n}\colon \left \vert 0\right \rangle, \end{eqnarray*} $

进而有

$ \begin{matrix} {{H}_{n}}\left( q \right)\text{ }={{2}^{n/2}}{{\text{e}}^{{{q}^{2}}/2}}\langle q|:{{\left( {{a}^{\dagger }}+a \right)}^{n}}:\left| 0 \right\rangle ={{2}^{n/2}}{{\text{e}}^{{{q}^{2}}/2}}\langle q|{{a}^{\dagger 2}}\left| 0 \right\rangle \\ ={{\text{e}}^{{{q}^{2}}/2}}{{\left( q-\frac{\text{d}}{\text{d}q} \right)}^{n}}{{\text{e}}^{-{{q}^{2}}/2}}={{\left( -1 \right)}^{n}}{{\text{e}}^{{{q}^{2}}}}\frac{{{\text{d}}^{n}}}{\text{d}{{q}^{n}}}{{\text{e}}^{-{{q}^{2}}}}, \\ \end{matrix} $

(1.5)

这正是单模厄米特多项式的微分表示形式.上述推掉表明了狄拉克符号与它的自身运算法则能促进基本量子理论的不断发展.

文献[5]作者介绍了一个定义于复空间的双变量厄米特多项式

$ {{H}_{m,n}}\left( \alpha ,{{\alpha }^{*}} \right)=\sum\limits_{l=0}^{\min \left( m,n \right)}{\frac{m!n!}{l!\left( m-l \right)!\left( n-l \right)!}}{{\left( -1 \right)}^{l}}{{\alpha }^{m-l}}{{\alpha }^{*n-l}},\text{ }\alpha =q+\text{i}p, $

(1.6)

它的生成函数为

$ \begin{equation} \sum_{m, n=0}^{\infty }\frac{t^{m}\tau ^{n}}{m!n!}H_{m, n}\left( \alpha, \alpha ^{\ast }\right) =\exp \left( -t\tau +t\alpha +\tau \alpha ^{\ast }\right) . \label{eq:a7} \end{equation} $

(1.7)

双变量厄米特多项式可在物理学的多个领域得到应用, 如已得到证明$H_{m, n}\left( \alpha, \alpha ^{\ast }\right) $复分数傅里叶变换的本征模^[6-8], 该本征模能在梯度折射率介质中的光传播中观察到, 且梯度折射率介质中的两维太保(Talbot)效应中也能观察到这个本征模^[9].

由于在单变量厄米特多项式与双变量复厄米特多项式间存在着很多相似性, 那么单、双变量厄米特多项式的微分形式是否也具有相似性呢?本文借助Weyl编序规则和量子算符厄米特多项式, 得到了双变量厄米特多项式的微分表示和一个交叉性的广义微分形式, 在形式上, 该表示类似于单变量厄米特多项式.本文内容如下:第2节简要介绍基于Weyl编序和Weyl编序乘积算符积分技术(the technique of integration within Weyl ordered product of operators, IWWP)的Weyl对应规则, 该对应规则揭示了经典函数$f\left( p, q\right) $的量子对应算符可通过$Q$和$P$直接代替$q$和$p$而得到, 且保持函数的形式不变.第3节主要介绍双变量厄米特多项式, 通过Weyl对应规则的公式化, 可以推导厄米特多项式$H_{m, n}\left( \alpha, \alpha ^{\ast }\right) $的微分表示.由于Weyl对应具有相似变换不变的特征, 一类广义的厄米特多项式微分表示将会在第4节中得到.第5节将利用上述微分等式研究激发压缩真空场的非经典特征, 该量子光场一般可在受限量子体系中产生.

众所周知, Weyl对应规则^[10-11], 即

$ \begin{equation} F\left( P, Q\right) =\int\!\!\! \int {\rm d}q{\rm d}pf\left( p, q\right) \Delta \left( q, p\right) \label{eq:b1} \end{equation} $

(2.1)

为经典相空间函数$f\left( p, q\right) $与其量子对应算符$F\left( P, Q\right) $间建立了一座桥梁, $\Delta \left( q, p\right) $表示Wigner算符^[12], 定义为

$ \begin{equation} \Delta \left( p, q\right) =\int_{-\infty }^{\infty }\frac{{\rm d}u}{2\pi } {\rm e}^{{\rm i}pu}\left \vert q+\frac{u}{2}\right \rangle \left \langle q-\frac{u}{2} \right \vert . \label{eq:b2} \end{equation} $

(2.2)

当$f\left( p, q\right) =q^{m}p^{r}$, 等式(2.1)得到其对应的量子算符为

$ \begin{eqnarray} q^{m}p^{r} &\rightarrow &\left( \frac{1}{2}\right) ^{m}\sum_{l=0}^{m}\frac{m!}{ l!\left( m-l\right) !}Q^{m-l}P^{r}Q^{l}\nonumber\\ &=&\left( \frac{1}{2}\right) ^{m}\sum_{l=0}^{m}\frac{m!}{l!\left( m-l\right) !} \begin{array}{c} \colon \\ \colon \\ \end{array} Q^{m-l}P^{r}Q^{l} \begin{array}{c} \colon \\ \colon \\ \end{array} = \begin{array}{c} \colon \\ \colon \\ \end{array} Q^{m}P^{r} \begin{array}{c} \colon \\ \colon \\ \end{array} , \label{eq:b3} \end{eqnarray} $

(2.3)

这里$ \begin{array}{c} \colon \\\colon \\ \end{array} $ $\begin{array}{c} \colon \\\colon \\ \end{array} $表示Weyl编序, 从等式(2.3)可以看出:经典函数$f\left( p, q\right) $的量子对应算符可保持函数形式不变的情况下, 由$Q$和$P$直接代替变量$q$和$p$而得到.在量子力学中, 算符编序有多种(正规编序、反正规编序和Weyl编序), 而Weyl编序非常特殊且有重要的应用, 特别是利用有序算符内的积分技术(IWOP), 文献[13]提出了Weyl编序算符的积分技术(IWWP).公式(2.2) Wigner算符的Weyl编序为

$ \begin{equation} \Delta \left( p, q\right) = \begin{array}{c} \colon \\ \colon \\ \end{array} \delta \left( p-P\right) \delta \left( q-Q\right) \begin{array}{c} \colon \\ \colon \\ \end{array}% . \label{eq:b4} \end{equation} $

(2.4)

因此, 直接用$q\rightarrow Q, $ $p\rightarrow P $替换$h\left( p, q\right) $, 即可得经典函数$f\left( p, q\right) $的量子对应算符

$ \begin{equation} F\left( P, Q\right) =% \begin{array}{c} \colon \\ \colon \\ \end{array}% h\left( P, Q\right) \begin{array}{c} \colon \\ \colon \\ \end{array}% =\int\!\!\! \int_{-\infty }^{\infty }{\rm d}p{\rm d}qh\left( p, q\right) \Delta \left( p, q\right) . \label{eq:b5} \end{equation}% $

(2.5)

对比式(1.1), 不难看出Weyl对应规则是一个算符Fredholm方程(OFE).注意到

$ \begin{eqnarray} \mbox{Tr}\left[\Delta \left( p_{1}, q_{1}\right) \Delta \left( p_{2}, q_{2}\right) \right] &=&\int \frac{{\rm d}^{2}z}{\pi }\left \langle z\right \vert \left[\Delta \left( p_{1}, q_{1}\right) \Delta \left( p_{2}, q_{2}\right) \right] \left \vert z\right \rangle \nonumber \\ &=&\frac{1}{2\pi }\delta \left( q_{1}-q_{2}\right) \delta \left( p_{1}-p_{2}\right), \label{eq:b6} \end{eqnarray}% $

(2.6)

同时, Weyl对应规则的逆变换为

$ 2\pi \text{Tr}\left[ F\left( P,Q \right)\Delta \left( q,p \right) \right]=2\pi \text{Tr}\left[ \int{}\int{\text{d}}{{q}_{1}}\text{d}{{p}_{1}}h\left( {{p}_{1}},{{q}_{1}} \right)\Delta \left( {{p}_{1}},{{q}_{1}} \right)\Delta \left( p,q \right) \right]=h\left( p,q \right). $

(2.7)

在多数情况下, 取$\alpha =\left( q+{\rm i}p\right) /\sqrt{2}$, 式(2.2)的Wigner算符可表示为

$ \begin{eqnarray} \Delta \left( p, q\right) &\rightarrow &\Delta \left( \alpha, \alpha ^{\ast }\right) =\int \frac{{\rm d}^{2}z}{\pi ^{2}}\left \vert \alpha +z\right \rangle \left \langle \alpha -z\right \vert {\rm e}^{\alpha z^{\ast }-\alpha ^{\ast }z} \nonumber \\ &=&\frac{1}{\pi }\colon \exp \left[-2\left( a^{\dagger }-\alpha ^{\ast }\right) \left( a-\alpha \right) \right] \colon =\frac{1}{2}% \begin{array}{c} \colon \\ \colon \\ \end{array}% \delta \left( a^{\dagger }-\alpha ^{\ast }\right) \delta \left( a-\alpha \right) \begin{array}{c} \colon \\ \colon \\ \end{array}% . \label{eq:b8} \end{eqnarray}% $

(2.8)

以致于(2.5)式的Weyl对应规则可表示为

$ \begin{equation} G\left( a, a^{\dagger }\right) =% \begin{array}{c} \colon \\ \colon \\ \end{array}% f\left( a, a^{\dagger }\right) \begin{array}{c} \colon \\ \colon \\ \end{array}% =2\int\!\!\! \int {\rm d}q{\rm d}pf\left( \alpha, \alpha ^{\ast }\right) \Delta \left( \alpha , \alpha ^{\ast }\right), \label{eq:b9} \end{equation}% $

(2.9)

于是, 其逆变换的公式为

$ \begin{equation} 2\pi {\rm Tr}\left[G\left( a, a^{\dagger }\right) \Delta \left( \alpha, \alpha ^{\ast }\right) \right] =2\pi {\rm Tr}\left[ \begin{array}{c} \colon \\ \colon \\ \end{array}% f\left( a, a^{\dagger }\right) \begin{array}{c} \colon \\ \colon \\ \end{array}% \Delta \left( \alpha, \alpha ^{\ast }\right) \right] =f\left( \alpha, \alpha ^{\ast }\right) . \label{eq:b10} \end{equation}% $

(2.10)

当$G\left( a, a^{\dagger }\right) =\rho \left( a, a^{\dagger }\right) $, 即描述了量子系统的密度算符, 则有上式的逆变换关系可得

$ \begin{equation} 2\pi {\rm Tr}\left[\rho \left( a, a^{\dagger }\right) \Delta \left( \alpha, \alpha ^{\ast }\right) \right] =W\left( \alpha, \alpha ^{\ast }\right), \label{eq:b11} \end{equation} $

(2.11)

$W\left( \alpha, \alpha ^{\ast }\right) $表示准概率分布Wigner函数, 式(2.11)也称为Wigner-Weyl对应规则.

在双体连续纠缠表象的量子理论中^[14], 双变量厄米特多项式$H_{m, n}\left( \alpha, \alpha ^{\ast }\right) $是双模粒子数态的广义Bargmann表示, 即

$ \begin{eqnarray*} \left \vert m, n\right \rangle =\frac{a^{\dagger m}b^{\dagger m}}{\sqrt{m!n!}} \left \vert 0, 0\right \rangle \rightarrow \frac{1}{\sqrt{m!n!}}H_{m, n}\left( \xi, \xi ^{\ast }\right) {\rm e}^{-\frac{\left \vert \xi \right \vert ^{2}}{2}}, \end{eqnarray*} $

且该表示构成了一个正交完备的函数空间

$ \begin{equation} 2\int\!\!\! \int \frac{{\rm d}^{2}\xi }{\pi }{\rm e}^{-2\left \vert \xi \right \vert ^{2}}H_{m, n}\left( \sqrt{2}\xi, \sqrt{2}\xi ^{\ast }\right) H_{m^{\prime }, n^{\prime }}^{\ast }\left( \sqrt{2}\xi, \sqrt{2}\xi ^{\ast }\right) =\sqrt{% m!n!m^{\prime }!n^{\prime }!}\delta _{m, m^{\prime }}\delta _{n, n^{\prime }}. \label{eq:c1} \end{equation} $

(3.1)

式(3.1)表明任意函数$f\left( \alpha, \alpha ^{\ast }\right) $都可在该正交基下进行展开

$ \begin{equation} f\left( \alpha, \alpha ^{\ast }\right) =\sum_{m, n=0}^{\infty }C_{m, n}H_{m, n}^{\ast }\left( \sqrt{2}\alpha, \sqrt{2}\alpha ^{\ast }\right), \label{eq:c2} \end{equation} $

(3.2)

这里系数$C_{m, n}$可由下面的推导得到.由双变量厄米特多项式的生成函数(式(1.7)), 并考虑Wigner算符的正规乘积编序, 可以得到

$ \begin{matrix} {} \\ . \\ \end{matrix}\begin{matrix} \Delta \left( \alpha ,{{\alpha }^{*}} \right)\text{ }=\frac{1}{\pi }:\exp \left[ -2\left( {{a}^{\dagger }}-{{\alpha }^{*}} \right)\left( a-\alpha \right) \right]: \\ =\frac{1}{\pi }{{\text{e}}^{-2{{\left| \alpha \right|}^{2}}}}:\sum\limits_{m,n=0}^{\infty }{\frac{\sqrt{{{2}^{m+n}}}{{a}^{\dagger m}}{{a}^{n}}}{m!n!}}{{H}_{m,n}}\left( \sqrt{2}\alpha ,\sqrt{2}{{\alpha }^{*}} \right):. \\ \end{matrix} $

(3.3)

将式(3.2)和(3.3)代入式(2.9)得

$ \begin{eqnarray} G\left( {{a}^{\dagger }}, a\right) &=&2\int \frac{{\rm d}^{2}\alpha }{\pi }{\rm e}^{-2\left \vert \alpha \right \vert ^{2}}\colon \sum_{m, n=0}^{\infty }\frac{\sqrt{ 2^{m+n}}{{a}^{\dagger m}}a^{n}}{m!n!}H_{m, n}\left( \sqrt{2}\alpha, \sqrt{2}\alpha ^{\ast }\right) \colon \nonumber \\ &&\sum_{m^{\prime }, n^{\prime }=0}^{\infty }C_{m^{\prime }, n^{\prime }}H_{m^{\prime }, n^{\prime }}^{\ast }\left( \sqrt{2 }\alpha, \sqrt{2}\alpha ^{\ast }\right) \nonumber \\ &=&2\ \colon \sum_{m, n=0}^{\infty }\sum_{m^{\prime }, n^{\prime }=0}^{\infty }C_{m^{\prime }, n^{\prime }}\frac{\sqrt{2^{m+n}}{{a}^{\dagger m}}a^{n}}{m!n!}\ \colon \nonumber \\ && \int \frac{{\rm d}^{2}\alpha }{\pi }{\rm e}^{-2\left \vert \alpha \right \vert ^{2}}H_{m, n}\left( \sqrt{2}\alpha, \sqrt{2}\alpha ^{\ast }\right) H_{m^{\prime }, n^{\prime }}^{\ast }\left( \sqrt{2}\alpha, \sqrt{2}\alpha ^{\ast }\right) \nonumber\\ &=&\colon \sum_{m, n=0}^{\infty }C_{m, n}\sqrt{2^{m+n}}{{a}^{\dagger m}}a^{n}\colon \equiv \colon F\left( {{a}^{\dagger }}, a\right) \colon . \label{eq:c4} \end{eqnarray}% $

(3.4)

求解式(3.4)算符在相干表象中的期望值, 得

$ \begin{eqnarray*} \left \langle \alpha \right \vert \colon F\left( {{a}^{\dagger }}, a\right) \colon \left \vert \alpha \right \rangle &=&F\left( \alpha ^{\ast }, \alpha \right) =\left \langle \alpha \right \vert \colon \sum_{m, n=0}^{\infty }C_{m, n}\sqrt{% 2^{m+n}}{{a}^{\dagger m}}a^{n}\colon \left \vert \alpha \right \rangle \\ &=&\sum_{m, n=0}^{\infty }C_{m, n}\sqrt{2^{m+n}}\left( \alpha ^{\ast }\right) ^{m}\alpha ^{n}, \end{eqnarray*}% $

并且

$ \begin{eqnarray*} C_{m, n}=\frac{1}{m!n!\sqrt{2^{m+n}}}\left. \frac{\partial {}^{m}}{\partial \alpha ^{\ast }{}^{m}}\frac{\partial {}^{n}}{\partial \alpha {}^{n}}F\left( \alpha ^{\ast }, \alpha \right) \right \vert _{\alpha =0}. \end{eqnarray*}% $

于是, 由式(3.2)得

$ F\left( \alpha, \alpha ^{\ast }\right) =\sum_{m, n=0}^{\infty }H_{m, n}\left( \sqrt{2}\alpha, \sqrt{2}\alpha ^{\ast }\right) \frac{1}{m!n!\sqrt{2^{m+n}}}% \left. \frac{\partial {}^{m}}{\partial \alpha ^{\ast }{}^{m}}\frac{\partial {}^{n}}{\partial \alpha {}^{n}}F\left( \alpha ^{\ast }, \alpha \right) \right \vert _{\alpha =0}. $

(3.5)

上式是推导正规乘积算符的Weyl经典对应函数的新公式.若以$G_{1}\left( {{a}^{\dagger }}, a\right) =\colon F_{1}\left( {{a}^{\dagger }}, a\right) \colon =\colon {{a}^{\dagger m}}a^{n}\colon $为例, 从式(3.5)得

$ g_{1}\left( \alpha, \alpha ^{\ast }\right) =\frac{1}{\sqrt{2^{m+n}}}% H_{m, n}\left( \sqrt{2}\alpha, \sqrt{2}\alpha ^{\ast }\right) . $

(3.6)

同时, 考虑到(2.10)式中的逆变换关系, 则有

$ \begin{eqnarray*} g_{1}\left( \alpha, \alpha ^{\ast }\right)&=&2\pi \mbox{Tr}\left[G_{1}\left( {{a}^{\dagger }}, a\right) \Delta \left( \alpha, \alpha ^{\ast }\right) \right] =2\pi \mbox{Tr}\left[{{a}^{\dagger m}}a^{n}\Delta \left( \alpha, \alpha ^{\ast }\right) \right] \\ &=&\frac{1}{\sqrt{2^{m+n}}}H_{m, n}\left( \sqrt{2}\alpha, \sqrt{2}\alpha ^{\ast }\right). \end{eqnarray*} $

另一方面, 若给定任意算符$F\left( {{a}^{\dagger }}, a\right) $, 其经典对应函数可通过下面公式得到(具体推导过程见附录)

$ f\left( \alpha ^{\ast }, \alpha \right) ={\rm e}^{2\left \vert \alpha \right \vert ^{2}}\int \frac{{\rm d}^{2}\beta }{\pi }\left \langle -\beta \right \vert F\left( a^{\dagger }, a\right) \left \vert \beta \right \rangle {\rm e}^{2\left( \alpha \beta ^{\ast }-\alpha ^{\ast }\beta \right) }. $

(3.7)

将$G_{1}\left( {{a}^{\dagger }}, a\right) =\colon {{a}^{\dagger m}}a^{n}\colon $代入到式(3.7)中, 得

$ \begin{matrix} {{g}_{1}}\left( \alpha ,{{\alpha }^{*}} \right)\text{ }={{\text{e}}^{2{{\left| \alpha \right|}^{2}}}}\int{\frac{{{\text{d}}^{2}}\beta }{\pi }}\langle -\beta |{{G}_{1}}\left( {{a}^{\dagger }},a \right)\left| \beta \right\rangle {{\text{e}}^{2\left( \alpha {{\beta }^{*}}-{{\alpha }^{*}}\beta \right)}} \\ ={{\text{e}}^{2{{\left| \alpha \right|}^{2}}}}\int{\frac{{{\text{d}}^{2}}\beta }{\pi }}\langle -\beta |:{{a}^{\dagger m}}{{a}^{n}}:\left| \beta \right\rangle {{\text{e}}^{2\left( \alpha {{\beta }^{*}}-{{\alpha }^{*}}\beta \right)}} \\ ={{\text{e}}^{2{{\left| \alpha \right|}^{2}}}}\int{\frac{{{\text{d}}^{2}}\beta }{\pi }}{{\left( -{{\beta }^{*}} \right)}^{m}}{{\beta }^{n}}{{\text{e}}^{-2{{\left| \beta \right|}^{2}}+2\left( \alpha {{\beta }^{*}}-{{\alpha }^{*}}\beta \right)}}. \\ \end{matrix} $

考虑$\sqrt{2}\beta \rightarrow \beta, \sqrt{2}\beta ^{\ast }\rightarrow \beta ^{\ast }, $则上式变为

$ \begin{matrix} g_{1}\left( \alpha, \alpha ^{\ast }\right) &=&\frac{\left( -1\right) ^{m}{\rm e}^{2\left \vert \alpha \right \vert ^{2}}}{\left( \sqrt{2}\right) ^{m+n}}% \int \frac{{\rm d}^{2}\beta }{\pi }\beta ^{\ast m}\beta ^{n}\exp \left[-\left \vert \beta \right \vert ^{2}+\sqrt{2}\left( \alpha \beta ^{\ast }-\alpha ^{\ast }\beta \right) \right] \\ &=&\frac{\left( -1\right) ^{m-n}{\rm e}^{2\left \vert \alpha \right \vert ^{2}}}{% 2^{m+n}}\frac{\partial {}^{m}}{\partial \alpha {}^{m}}\frac{\partial {}^{n}}{% \partial \alpha ^{\ast }{}^{n}}\int \frac{{\rm d}^{2}\beta }{\pi }\exp \left[ -\left \vert \beta \right \vert ^{2}+\sqrt{2}\left( \alpha \beta ^{\ast }-\alpha ^{\ast }\beta \right) \right] . \end{matrix} $

利用如下积分公式

$ \int{\frac{{{\text{d}}^{2}}\alpha }{\pi }}\exp \left[ h{{\left| \alpha \right|}^{2}}+s\alpha +\eta {{\alpha }^{*}} \right]=\frac{1}{h}\exp \left[ -\frac{s\eta }{h} \right],\text{ }Re\left[ h \right] < 0, $

则有

$ g_{1}\left( \alpha, \alpha ^{\ast }\right) =\frac{\left( -1\right) ^{m-n}{\rm e}^{2\left \vert \alpha \right \vert ^{2}}}{2^{m+n}}\frac{\partial {}^{m}}{\partial \alpha {}^{m}}\frac{\partial {}^{n}}{\partial \alpha ^{\ast }{}^{n}}\exp \left( -2\left \vert \alpha \right \vert ^{2}\right) . $

(3.8)

比较式(3.6)和(3.8), 并令$\sqrt{2} \alpha \rightarrow \alpha, \sqrt{2}\alpha ^{\ast }\rightarrow \alpha ^{\ast } $, 则

$ H_{m, n}\left( \alpha, \alpha ^{\ast }\right) =\left( -1\right) ^{m-n}{\rm e}^{\left \vert \alpha \right \vert ^{2}}\frac{\partial {}^{m}}{ \partial \alpha ^{\ast }{}^{m}}\frac{\partial {}^{n}}{\partial \alpha {}^{n}} \exp \left( -\left \vert \alpha \right \vert ^{2}\right), $

(3.9)

这正是双变量厄米特多项式生成函数的微分形式, 若$m=n$

$ H_{m, m}\left( \alpha, \alpha ^{\ast }\right) =\exp \left( \left \vert \alpha \right \vert ^{2}\right) \frac{\partial {}^{2m}}{\partial \alpha ^{\ast }{}{}^{m}\alpha ^{m}}\exp \left( -\left \vert \alpha \right \vert ^{2}\right) . $

为了产生非对称量子力学表象, 文献[18]引入非幺正算符$\hat{U}$, 其定义为

$ \begin{matrix} \hat{U}=\frac{1}{\sqrt{\mu }}\int{\frac{{{\text{d}}^{2}}z}{\pi }}\left| \left( \begin{matrix} \mu \nu \\ \sigma \tau \\ \end{matrix} \right)\left( \begin{matrix} z \\ {{z}^{*}} \\ \end{matrix} \right) \right\rangle \langle \left( \begin{matrix} z \\ {{z}^{*}} \\ \end{matrix} \right)|\text{ } \\ =\frac{1}{\sqrt{\mu }}:\exp \left[ -\frac{\nu }{2\mu }{{a}^{\dagger 2}}+\left( \frac{1}{\mu }-1 \right){{a}^{\dagger }}a+\frac{\sigma }{2\mu }{{a}^{2}} \right]:, \\ \end{matrix} $

(4.1)

这是经典相空间辛变换$\left( z, z^{\ast }\right) \rightarrow \left( \mu z+\nu z^{\ast }, \sigma z+\tau z^{\ast }\right) $的量子对应算符, 上式中的$\left| \left( \begin{matrix} z \\ {{z}^{*}} \\ \end{matrix} \right) \right\rangle =\exp \left( z{{a}^{\dagger }}-{{z}^{*}}a \right)\left| 0 \right\rangle $表示相干态^[19-20], 且

$ \left| \left( \begin{matrix} \mu & \nu \\ \sigma & \tau \\ \end{matrix} \right)\left( \begin{matrix} z \\ {{z}^{*}} \\ \end{matrix} \right) \right\rangle =\left| \left( \begin{matrix} \mu z+\nu {{z}^{*}} \\ \sigma z+\tau {{z}^{*}} \\ \end{matrix} \right) \right\rangle =\exp \left[ \left( \mu z+\nu {{z}^{*}} \right){{a}^{\dagger }}-\left( \sigma z+\tau {{z}^{*}} \right)a \right]\left| 0 \right\rangle . $

$\hat{U}$的逆算符定义为

$ \begin{matrix} {{{\hat{U}}}^{-1}}\text{ }=\sqrt{\mu }\int{\frac{{{\text{d}}^{2}}z}{\pi }}\left| \left( \begin{matrix} z \\ {{z}^{*}} \\ \end{matrix} \right) \right\rangle \langle \left( \begin{matrix} \tau & -\nu \\ -\sigma & \mu \\ \end{matrix} \right)\left( \begin{matrix} z \\ {{z}^{*}} \\ \end{matrix} \right)|\text{ } \\ \ \ \ =\frac{1}{\sqrt{\tau }}:\exp \left[ \frac{\nu }{2\tau }{{a}^{\dagger 2}}+\left( \frac{1}{\tau }-1 \right){{a}^{\dagger }}a-\frac{\sigma }{2\tau }{{a}^{2}} \right]:. \\ \end{matrix}$

(4.2)

从式(4.1)和(4.2)不难发现：${{\hat{U}}^{\dagger }}\ne {{\hat{U}}^{-1}}$, 且$\hat{U}$产生的相似变换为

$ \hat{U}a{{\hat{U}}^{-1}}=\mu a+\nu {{a}^{\dagger }},\ \hat{U}{{a}^{\dagger }}{{\hat{U}}^{-1}}=\sigma a+\tau {{a}^{\dagger }} $

(4.3)

和它的逆变换

$ {{\hat{U}}^{-1}}a\hat{U}=\tau a-\nu {{a}^{\dagger }},\ {{\hat{U}}^{-1}}{{a}^{\dagger }}\hat{U}=\mu {{a}^{\dagger }}-\sigma a, $

(4.4)

这里四个复参量满足$\mu \tau -\nu \sigma =1$, 从而保证等式的幺正性, 即$\left[\mu a+\nu a^{\dagger }, \ \sigma a+\tau a^{\dagger }\right] =1.$需要着重指出的是:文献[4, 15-16]中已经证明了Weyl编序具有相似变换不变的性质, 即

$ \hat{U}G\left( {{a}^{\dagger }},a \right){{\hat{U}}^{-1}}=\hat{U}\begin{matrix} : \\ : \\ \end{matrix}g\left( {{a}^{\dagger }},a \right)\begin{matrix} : \\ : \\ \end{matrix}{{\hat{U}}^{-1}}=\begin{matrix} : \\ : \\ \end{matrix}g\left( \sigma a+\tau {{a}^{\dagger }},\mu a+\nu {{a}^{\dagger }} \right)\begin{matrix} : \\ : \\ \end{matrix}. $

(4.5)

假定$G_{2}\left( {{a}^{\dagger }}, a\right) =a^{\dagger m}\hat{U} \left \vert 0\right \rangle \left \langle 0\right \vert \hat{U}^{-1}a^{n}, $由式(4.4)和(2.10)可得$G_{2}\left( a^{\dagger }, a\right) $的Weyl经典对应是

$ {{g}_{2}}\left( {{\alpha }^{*}},\alpha \right)\text{ }=2\pi \text{Tr}\left[ {{G}_{2}}\left( {{a}^{\dagger }},a \right)\Delta \left( \alpha ,{{\alpha }^{*}} \right) \right]=2\pi \text{Tr}\left[ {{a}^{\dagger m}}\hat{U}\left| 0 \right\rangle \langle 0|{{{\hat{U}}}^{-1}}{{a}^{n}}\Delta \left( \alpha ,{{\alpha }^{*}} \right) \right]\text{ } \\ =2\pi \text{Tr}\left[ \hat{U}{{\left( \mu {{a}^{\dagger }}-\sigma a \right)}^{m}}\left| 0 \right\rangle \langle 0|{{\left( \tau a-\nu {{a}^{\dagger }} \right)}^{n}}{{{\hat{U}}}^{-1}}\Delta \left( \alpha ,{{\alpha }^{*}} \right) \right]. $

(4.6)

参考公式

$ \left( fa+g{{a}^{\dagger }}\right) ^{n}=\left( -{\rm i}\sqrt{\frac{fg}{2}}\right) ^{n}\colon H_{n}\left( {\rm i}\sqrt{\frac{f}{2g}}a+{\rm i}\sqrt{\frac{g}{2f}}{{a}^{\dagger }}\right) \colon, $

则得到

$ {{\left( \mu {{a}^{\dagger }}-\sigma a \right)}^{m}}\text{ }={{\left( \sqrt{\frac{\sigma \mu }{2}} \right)}^{m}}:{{H}_{m}}\left( -\sqrt{\frac{\sigma }{2\mu }}a-\sqrt{%\frac{\mu }{2\sigma }}{{a}^{\dagger }} \right):, $

(4.7)

$ {{\left( \tau a-\nu {{a}^{\dagger }} \right)}^{n}}\text{ }={{\left( \sqrt{\frac{\nu \tau }{2%}} \right)}^{n}}:{{H}_{n}}\left( -\sqrt{\frac{\tau }{2\nu }}a-\sqrt{\frac{%\nu }{2\tau }}{{a}^{\dagger }} \right):. $

(4.8)

于是

$ {{\left( -\sigma a+\mu {{a}^{\dagger }} \right)}^{m}}\left| 0 \right\rangle \langle 0|{{\left( \tau a-\nu {{a}^{\dagger }} \right)}^{n}} \\ ={{\left( \sqrt{\frac{\sigma \mu }{2}} \right)}^{m}}{{\left( \sqrt{\frac{\nu \tau }{2}} \right)}^{n}}:{{H}_{m}}\left( -\sqrt{\frac{\mu }{2\sigma }}{{a}^{\dagger }} \right)\exp \left[ -{{a}^{\dagger }}a \right]{{H}_{n}}\left( -\sqrt{\frac{%\tau }{2\nu }}a \right):, $

(4.9)

这里真空投影算符$\left| 0 \right\rangle \langle 0|=:\exp \left[ -{{a}^{\dagger }}a \right]:.$利用公式(该公式推导详见附录)得

$ \begin{equation} F\left( a^{\dagger }, a\right) = \begin{array}{c} \colon \\ \colon \\ \end{array}% f\left( a^{\dagger }, a\right) \begin{array}{c} \colon \\ \colon \\ \end{array}% =2\int \frac{{\rm d}^{2}z}{\pi }% \begin{array}{c} \colon \\ \colon \\ \end{array}% \left \langle -z\right \vert F\left( a^{\dagger }, a\right) \left \vert z\right \rangle \exp \left[2\left( az^{\ast }-{{a}^{\dagger }}z+{{a}^{\dagger }}a\right) % \right] \begin{array}{c} \colon \\ \colon \\ \end{array}% , \label{eq:d10} \end{equation}% $

(4.10)

则$\left( -\sigma a+\mu a^{\dagger }\right) ^{m}\left \vert 0\right \rangle \left \langle 0\right \vert \left( \tau a-\nu a^{\dagger }\right) ^{n}$的Weyl编序形式为

$ \begin{eqnarray} &&\left( -\sigma a+\mu a^{\dagger }\right) ^{m}\left \vert 0\right \rangle \left \langle 0\right \vert \left( \tau a-\nu a^{\dagger }\right) ^{n}=2\left( \sqrt{\frac{\sigma \mu }{2}}\right) ^{m}\left( -\sqrt{\frac{\nu \tau }{2}}\right) ^{n} \nonumber \\ &&\times \begin{array}{c} \colon \\ \colon \\ \end{array}% \int \frac{{\rm d}^{2}z}{\pi }H_{m}\left( \sqrt{\frac{\mu }{2\sigma }}z^{\ast }\right) H_{n}\left( \sqrt{\frac{\tau }{2\nu }}z\right) \exp \left[-\left \vert z\right \vert ^{2}-2{{a}^{\dagger }}z+2az^{\ast }+2{{a}^{\dagger }}a\right] \begin{array}{c} \colon \\ \colon \\ \end{array}% . \label{eq:d11} \end{eqnarray}% $

(4.11)

文献[11]中曾证明

$ \begin{eqnarray} &&\int \frac{{\rm d}^{2}z}{\pi }H_{m}\left( z^{\ast }\right) H_{n}\left( z\right) \exp \left[-\left( z-\lambda \right) \left( z^{\ast }-\lambda ^{\ast }\right) \right] \nonumber \\ &=&\sum_{l=0}^{\min \left[m, n\right] }\frac{2^{2l}m!n!}{% l!\left( m-l\right) !\left( n-l\right) !}H_{m-l}\left( \lambda ^{\ast }\right) H_{n-l}\left( \lambda \right), \label{eq:d12} \end{eqnarray} $

(4.12)

运用上述公式可进一步获得如下积分公式

$ \begin{eqnarray} &&\int \frac{{\rm d}^{2}z}{\pi }H_{m}\left( \alpha z^{\ast }\right) H_{n}\left( \beta z\right) \exp \left[-\left( z-\lambda \right) \left( z^{\ast }-\lambda ^{\ast }\right) \right] \nonumber \\ &=&\sum_{l=0}^{\min \left[m, n\right] }\frac{% \left( 4\alpha \beta \right) ^{l}m!n!}{l!\left( m-l\right) !\left( n-l\right) !}H_{m-l}\left( \alpha \lambda ^{\ast }\right) H_{n-l}\left( \beta \lambda \right), \label{eq:d13} \end{eqnarray} $

(4.13)

于是

$ \begin{eqnarray} &&\left( -\sigma a+\mu a^{\dagger }\right) ^{m}\left \vert 0\right \rangle \left \langle 0\right \vert \left( \tau a-\nu a^{\dagger }\right) ^{n}=2m!n!\left( -\sqrt{\frac{\sigma \mu }{2}}\right) ^{m}\left( -\sqrt{% \frac{\nu \tau }{2}}\right) ^{n} \nonumber \\ &&\times \begin{array}{c} \colon \\ \colon \\ \end{array}% \sum_{l=0}^{\min \left[m, n\right] }\frac{\left( -2\sqrt{\frac{\mu \tau }{% \sigma \nu }}\right) ^{l}}{l!\left( m-l\right) !\left( n-l\right) !}% H_{m-l}\left( \sqrt{\frac{2\mu }{\sigma }}{{a}^{\dagger }}\right) H_{n-l}\left( \sqrt{\frac{2\tau }{\nu }}a\right) \exp \left[-2{{a}^{\dagger }}a\right] \begin{array}{c} \colon \\ \colon \\ \end{array}% . \label{eq:d14} \end{eqnarray}% $

(4.14)

注意到Weyl编序的相似变换不变性特征(见(4.5)式), 并利用式(2.10)中的Weyl对应规则, 可推导算符$G_{2}\left( {{a}^{\dagger }}, a\right) $的经典对应

$ \begin{eqnarray} && g_{2}\left( \alpha ^{\ast }, \alpha \right)\nonumber \\ &=&2\pi \mbox{Tr}\left[ G_{2}\left( a^{\dagger }, a\right) \Delta \left( \alpha, \alpha ^{\ast }\right) \right] =2\pi \mbox{Tr}\left[a^{\dagger m}\hat{U}\left \vert 0\right \rangle \left \langle 0\right \vert \hat{U}^{-1}a^{n}\Delta \left( \alpha, \alpha ^{\ast }\right) \right] \nonumber \\ &=&2m!n!\left( -\sqrt{\frac{\sigma \mu }{2}}\right) ^{m}\left( -\sqrt{\frac{% \nu \tau }{2}}\right) ^{n}\sum_{l=0}^{\min \left[m, n\right] }\frac{\left( -2% \sqrt{\frac{\mu \tau }{\sigma \nu }}\right) ^{l}}{l!\left( m-l\right) !\left( n-l\right) !}H_{m-l}\left[\sqrt{\frac{2\mu }{\sigma }}\left( \sigma \alpha +\tau \alpha ^{\ast }\right) \right] \nonumber \\ &&\times H_{n-l}\left[\sqrt{\frac{2\tau }{\nu }}\left( \mu \alpha +\nu \alpha ^{\ast }\right) \right] \exp \left[-2\left( \sigma \alpha +\tau \alpha ^{\ast }\right) \left( \mu \alpha +\nu \alpha ^{\ast }\right) \right] .~~ \label{eq:d15} \end{eqnarray} $

(4.15)

再由式(3.7)得

$ \begin{eqnarray} && g_{2}\left( \alpha ^{\ast }, \alpha \right)\nonumber \\ &=&{\rm e}^{2\left \vert \alpha \right \vert ^{2}}\int \frac{{\rm d}^{2}\beta }{\pi }\left \langle -\beta \right \vert a^{\dagger m}\hat{U}\left \vert 0\right \rangle \left \langle 0\right \vert \hat{U}^{-1}a^{n}\left \vert \beta \right \rangle {\rm e}^{2\left( \alpha \beta ^{\ast }-\alpha ^{\ast }\beta \right) } \nonumber \\ &=&\frac{1}{\sqrt{\mu \tau }}{\rm e}^{2\left \vert \alpha \right \vert ^{2}}\int \frac{{\rm d}^{2}\beta }{\pi }\left( -\beta ^{\ast }\right) ^{m}\beta ^{n}\exp % \left[-\left \vert \beta \right \vert ^{2}-2\alpha ^{\ast }\beta +2\alpha \beta ^{\ast }-\frac{\sigma }{2\tau }\beta ^{2}-\frac{\nu }{2\mu }\beta ^{\ast }{}^{2}\right] \nonumber \\ &=&\frac{{\rm e}^{2\left \vert \alpha \right \vert ^{2}}}{\sqrt{\mu \tau }}\frac{ \left( -1\right) ^{m+n}}{2^{m+n}}\frac{\partial ^{m}}{\partial \alpha ^{m}} \frac{\partial ^{n}}{\partial \alpha ^{\ast n}}\int \frac{{\rm d}^{2}\beta }{\pi } \exp \left[-\left \vert \beta \right \vert ^{2}-2\alpha ^{\ast }\beta +2\alpha \beta ^{\ast }-\frac{\sigma }{2\tau }\beta ^{2}-\frac{\nu }{2\mu } \beta ^{\ast }{}^{2}\right] .~~ \label{eq:d16} \end{eqnarray} $

(4.16)

利用满足收敛条件Re$\left( h\pm f\pm g\right) <0$和Re$ \left( \frac{h^{2}-4fg}{h\pm f\pm g}\right) <0$的积分公式

$ \begin{eqnarray*} \int \frac{{\rm d}^{2}\alpha }{\pi }\exp \left[h\left \vert \alpha \right \vert ^{2}+s\alpha +\eta \alpha ^{\ast }+f\alpha ^{2}+g\alpha ^{\ast 2}\right] = \frac{1}{\sqrt{h^{2}-4fg}}\exp \left[\frac{-hs\eta +s^{2}g+\eta ^{2}f}{ h^{2}-4fg}\right] \end{eqnarray*}% $

得到

$ \begin{equation} g_{2}\left( \alpha ^{\ast }, \alpha \right) ={\rm e}^{2\left \vert \alpha \right \vert ^{2}}\frac{\left( -1\right) ^{m+n}}{2^{m+n}}\frac{\partial ^{m}}{ \partial \alpha ^{m}}\frac{\partial ^{n}}{\partial \alpha ^{\ast n}}\exp \left[-4\mu \tau \left \vert \alpha \right \vert ^{2}-2\nu \tau \alpha ^{\ast 2}-2\sigma \mu \alpha ^{2}\right] . \label{eq:d17} \end{equation}% $

(4.17)

对比式(4.15)与(4.17)得

$ \begin{eqnarray} &&\exp \left[4\mu \tau \left \vert \alpha \right \vert ^{2}+2\nu \tau \alpha ^{\ast 2}+2\sigma \mu \alpha ^{2}\right] \frac{\partial ^{m}}{% \partial \alpha ^{m}}\frac{\partial ^{n}}{\partial \alpha ^{\ast n}}\exp % \left[-4\mu \tau \left \vert \alpha \right \vert ^{2}-2\nu \tau \alpha ^{\ast 2}-2\sigma \mu \alpha ^{2}\right] \nonumber \\ &=&2\left( 2\mu \sigma \right) ^{\frac{m}{2}}\left( 2\tau \nu \right) ^{% \frac{n}{2}}\sum_{l=0}^{\min \left[m, n\right] }\left(\begin{array}{c}m\\ l\end{array}\right)\left(\begin{array}{c}n\\ l\end{array}\right)% l!\left( -\sqrt{\frac{4\mu \tau }{\sigma \nu }}\right) ^{l} \nonumber \\ &&\times H_{m-l}\left[\sqrt{\frac{2\mu }{\sigma }}\left( \sigma \alpha +\tau \alpha ^{\ast }\right) \right] H_{n-l}\left[\sqrt{\frac{2\tau }{\nu }% }\left( \mu \alpha +\nu \alpha ^{\ast }\right) \right] . \label{eq:d18} \end{eqnarray}% $

(4.18)

这是两个单变量厄米特多项式乘积的广义微分生成函数.如果取$\tau =\mu ^{\ast }, $ $\sigma =\nu ^{\ast }$满足$\left \vert \mu \right \vert ^{2}-\left \vert \nu \right \vert ^{2}=1, $保证算符$ \hat{U}$的幺正性, 则上式可简写为

$ \begin{eqnarray} &&\exp \left[4\left \vert \mu \right \vert ^{2}\left \vert \alpha \right \vert ^{2}+2\mu ^{\ast }\nu \alpha ^{\ast 2}+2\mu \nu ^{\ast }\alpha ^{2}% \right] \frac{\partial ^{m}}{\partial \alpha ^{m}}\frac{\partial ^{n}}{% \partial \alpha ^{\ast n}}\exp \left[-4\left \vert \mu \right \vert ^{2}\left \vert \alpha \right \vert ^{2}-2\mu ^{\ast }\nu \alpha ^{\ast 2}-2\mu \nu ^{\ast }\alpha ^{2}\right] \nonumber \\ &=&2\left( 2\mu \nu ^{\ast }\right) ^{\frac{m}{2}}\left( 2\mu ^{\ast }\nu \right) ^{\frac{n}{2}}\sum_{l=0}^{\min \left[m, n\right] }\left(\begin{array}{c}m\\ l\end{array}\right)% \left(\begin{array}{c}n\\ l\end{array}\right)l!\left( -\frac{2\left \vert \mu \right \vert }{\left \vert \nu \right \vert }\right) ^{l} \nonumber \\ &&\times H_{m-l}\left[\sqrt{\frac{2\mu }{\nu ^{\ast }}}\left( \nu ^{\ast }\alpha +\mu ^{\ast }\alpha ^{\ast }\right) \right] H_{n-l}\left[\sqrt{ \frac{2\mu ^{\ast }}{\nu }}\left( \mu \alpha +\nu \alpha ^{\ast }\right) \right] . \label{eq:d19} \end{eqnarray}% $

(4.19)

当$m=n$时, 上式则进一步简化为

$ \exp \left[ 4{{\left| \mu \right|}^{2}}{{\left| \alpha \right|}^{2}}+2{{\mu }^{*}}\nu {{\alpha }^{*2}}+2\mu {{\nu }^{*}}{{\alpha }^{2}} \right]\frac{{{\partial }^{m}}}{\partial {{\alpha }^{m}}}\frac{{{\partial }^{m}}}{\partial {{\alpha }^{*m}}}\exp \left[ -4{{\left| \mu \right|}^{2}}{{\left| \alpha \right|}^{2}}-2{{\mu }^{*}}\nu {{\alpha }^{*2}}-2\mu {{\nu }^{*}}{{\alpha }^{2}} \right]\text{ } \\ ={{2}^{m+1}}{{\left| \mu \right|}^{m}}{{\left| \nu \right|}^{m}}\sum\limits_{l=0}^{m}{\left( \begin{matrix} m \\ l \\ \end{matrix} \right)}\left( \begin{matrix} m \\ l \\ \end{matrix} \right)l!{{\left( -\frac{2\left| \mu \right|}{\left| \nu \right|} \right)}^{l}}{{\left| {{H}_{m-l}}\left[ \sqrt{\frac{2\mu }{{{\nu }^{*}}}}\left( {{\nu }^{*}}\alpha +{{\mu }^{*}}{{\alpha }^{*}} \right) \right] \right|}^{2}}. $

(4.20)

本节将推导一类双变量厄米特多项式的广义微分生成函数.由式(4.7)与(4.8), 算符$G_{2}\left( {{a}^{\dagger }}, a\right) =a^{\dagger m}\hat{U}\left \vert 0\right \rangle \left \langle 0\right \vert \hat{U}^{-1}a^{n}\ $的经典对应为

$ \begin{eqnarray*} && g_{2}\left( \alpha ^{\ast }, \alpha \right)\\ &=&2\pi {\rm Tr}\left[\hat{U}\left( \mu a^{\dagger }-\sigma a\right) ^{m}\left \vert 0\right \rangle \left \langle 0\right \vert \left( \tau a-\nu a^{\dagger }\right) ^{n}\hat{U}% ^{-1}\Delta \left( \alpha, \alpha ^{\ast }\right) \right] \\ &=&2\pi \left( -\sqrt{\frac{\sigma \mu }{2}}\right) ^{m}\left( -\sqrt{\frac{% \nu \tau }{2}}\right) ^{n}\mbox{Tr}\left[\hat{U}H_{m}\left( \sqrt{\frac{\mu }{2\sigma }}{{a}^{\dagger }}\right) \left \vert 0\right \rangle \left \langle 0\right \vert H_{n}\left( \sqrt{\frac{\tau }{2\nu }}a\right) \hat{U}% ^{-1}\Delta \left( \alpha, \alpha ^{\ast }\right) \right] . \end{eqnarray*}% $

由于

$ \begin{eqnarray*} H_{m}\left( x\right) =\sum_{k=0}^{\left[m/2\right] }\frac{\left( -1\right) ^{k}m!}{k!\left( m-2k\right) !}\left( 2x\right) ^{m-2k}, \end{eqnarray*}% $

则

$ {{g}_{2}}\left( {{\alpha }^{*}},\alpha \right)\text{ }=2\pi {{\left( -\frac{\mu }{2} \right)}^{m}}{{\left( -\frac{\tau }{2} \right)}^{n}}\sum\limits_{k=0}^{\left[ m/2 \right]}{\sum\limits_{l=0}^{\left[ n/2 \right]}{\frac{{{\left( -\frac{\sigma }{2\mu } \right)}^{k}}m!}{k!\sqrt{\left( m-2k \right)!}}}}\frac{{{\left( -\frac{\nu }{2\tau } \right)}^{l}}n!}{l!\sqrt{\left( n-2l \right)!}} \\ \times \text{Tr}\left[ \hat{U}\left| m-2k \right\rangle \langle n-2l|{{{\hat{U}}}^{-1}}\Delta \left( \alpha ,{{\alpha }^{*}} \right) \right]. $

(4.21)

对于粒子数投影算符$\left \vert m\right \rangle \left \langle n\right \vert \ $来说, 利用式(4.10)可得其Weyl编序乘积形式

$ \begin{align} & \left| m \right\rangle \langle n|\text{ }=2\int{\frac{{{\text{d}}^{2}}z}{\pi }}\begin{matrix} : \\ : \\ \end{matrix}\langle -z|\left. m \right\rangle \left\langle n \right.\left| z \right\rangle \exp \left[ 2\left( a{{z}^{*}}-{{a}^{\dagger }}z+{{a}^{\dagger }}a \right) \right]\begin{matrix} : \\ : \\ \end{matrix}\ \ \\ & \ \ \ \ \ \ \ \ \ \ \ \ =2\int{\frac{{{\text{d}}^{2}}z}{\pi }}\begin{matrix} : \\ : \\ \end{matrix}\frac{{{\left( -{{z}^{*}} \right)}^{m}}{{z}^{n}}}{\sqrt{n!m!}}\exp \left[ -2{{\left| z \right|}^{2}}+2\left( a{{z}^{*}}-{{a}^{\dagger }}z+{{a}^{\dagger }}a \right) \right]\begin{matrix} : \\ : \\ \end{matrix}. \\ \end{align} $

(4.22)

利用积分公式

$ \begin{eqnarray*} \int \frac{{\rm d}^{2}\beta }{\pi }\beta ^{\ast k}\beta ^{l}\exp \left[-h\left \vert \beta \right \vert ^{2}+s\beta +f\beta ^{\ast }\right] =\left( -{\rm i}\right) ^{k+l}h^{-\frac{k+l}{2}-1}{\rm e}^{\frac{sf}{h}}H_{k, l}\left( \frac{{\rm i}s}{ \sqrt{h}}, \frac{{\rm i}f}{\sqrt{h}}\right), \end{eqnarray*} $

则

$ \left| m \right\rangle \langle n|=\frac{2}{\sqrt{n!m!}}\begin{matrix} : \\ : \\ \end{matrix}{{H}_{m,n}}\left( 2{{a}^{\dagger }},2a \right)\exp \left( -2{{a}^{\dagger }}a \right)\begin{matrix} : \\ : \\ \end{matrix}. $

(4.23)

将式(4.5)与(4.23)比较, 得

$ \begin{eqnarray*} g_{2}\left( \alpha ^{\ast }, \alpha \right) &=&4\pi \left( -\frac{\mu }{2}% \right) ^{m}\left( -\frac{\tau }{2}\right) ^{n}\sum_{k=0}^{\left[m/2\right] }\sum_{l=0}^{\left[n/2\right] }\frac{\left( -\frac{\sigma }{2\mu }\right) ^{k}m!}{k!\left( m-2k\right) !}\frac{\left( -\frac{\nu }{2\tau }\right) ^{l}n!}{l!\left( n-2l\right) !} \\ &&\times \mbox{Tr}\Big[ \begin{array}{c} \colon \\ \colon \\ \end{array}% H_{m-2k, n-2l}\left[2\left( \sigma a+\tau a^{\dagger }\right), 2\left( \mu a+\nu a^{\dagger }\right) \right] \\ &&\times \exp \left[-2\left( \sigma a+\tau a^{\dagger }\right) \left( \mu a+\nu a^{\dagger }\right) \right] \begin{array}{c} \colon \\ \colon \\ \end{array}% \Delta \left( \alpha, \alpha ^{\ast }\right) \Big] . \end{eqnarray*}% $

利用Weyl对应规则(式(2.10))可得出$G_{2}\left( {{a}^{\dagger }}, a\right) $的经典对应为

$ {{g}_{2}}\left( {{\alpha }^{*}},\alpha \right)\text{ }=2{{\left( -\frac{\mu }{2} \right)}^{m}}{{\left( -\frac{\tau }{2} \right)}^{n}}\exp \left[ -2\left( \sigma \alpha +\tau {{\alpha }^{*}} \right)\left( \mu \alpha +\nu {{\alpha }^{*}} \right) \right]\text{} \\ \times \sum\limits_{k=0}^{\left[ m/2 \right]}{\sum\limits_{l=0}^{\left[ n/2 \right]}{\frac{{{\left( -\frac{\sigma }{2\mu } \right)}^{k}}m!}{k!\left( m-2k \right)!}}}\frac{{{\left( -\frac{\nu }{2\tau } \right)}^{l}}n!}{l!\left( n-2l \right)!}{{H}_{m-2k,n-2l}}\left[ 2\left( \sigma \alpha +\tau {{\alpha }^{*}} \right),2\left( \mu \alpha +\nu {{\alpha }^{*}} \right) \right].\text{ } $

(4.24)

比较式(4.24)与(4.17), 可得一个简约的表达式, 即

$ \exp \left[ 4\mu \tau {{\left| \alpha \right|}^{2}}+2\nu \tau {{\alpha }^{*2}}+2\sigma \mu {{\alpha }^{2}} \right]\frac{{{\partial }^{m}}}{\partial {{\alpha }^{m}}}\frac{{{\partial }^{n}}}{\partial {{\alpha }^{*n}}}\exp \left[ -4\mu \tau {{\left| \alpha \right|}^{2}}-2\nu \tau {{\alpha }^{*2}}-2\sigma \mu {{\alpha }^{2}} \right] \\ =2{{\mu }^{m}}{{\tau }^{n}}\sum\limits_{k=0}^{\left[ m/2 \right]}{\sum\limits_{l=0}^{\left[ n/2 \right]}{\frac{{{\left( -\frac{\sigma }{2\mu } \right)}^{k}}m!}{k!\left( m-2k \right)!}}}\frac{{{\left( -\frac{\nu }{2\tau } \right)}^{l}}n!}{l!\left( n-2l \right)!}{{H}_{m-2k,n-2l}}\left[ 2\left( \sigma \alpha +\tau {{\alpha }^{*}} \right),2\left( \mu \alpha +\nu {{\alpha }^{*}} \right) \right]. $

(4.25)

式(4.24)的右边为双变量厄米特多项式的求和形式, 而式(4.17)则为两个单变量厄米特多项式的乘积形式, 式(4.17)和(4.24)都为一类新的厄米特多项式的广义微分生成函数.对于$m=n$情况下, 式(4.24)可简化为

$ \begin{eqnarray} &&\exp \left[4\mu \tau \left \vert \alpha \right \vert ^{2}+2\nu \tau \alpha ^{\ast 2}+2\sigma \mu \alpha ^{2}\right] \frac{\partial ^{m}}{% \partial \alpha ^{m}}\frac{\partial ^{m}}{\partial \alpha ^{\ast m}}\exp % \left[-4\mu \tau \left \vert \alpha \right \vert ^{2}-2\nu \tau \alpha ^{\ast 2}-2\sigma \mu \alpha ^{2}\right] \nonumber \\ &=&2\left( \mu \tau \right) ^{m}\sum_{k, l=0}^{\left[m/2\right] }\frac{% \left( -\frac{\sigma }{2\mu }\right) ^{k}m!}{k!\left( m-2k\right) !}\frac{% \left( -\frac{\nu }{2\tau }\right) ^{l}m!}{l!\left( m-2l\right) !}% H_{m-2k, m-2l}\left[2\left( \sigma \alpha +\tau \alpha ^{\ast }\right) , 2\left( \mu \alpha +\nu \alpha ^{\ast }\right) \right] . \nonumber \end{eqnarray}% $

如果幺正算符$\hat{U}=\exp \left[ \frac{r}{2}\left( {{a}^{\dagger 2}}-{{a}^{2}} \right) \right] $, 并令$\ \tau =\mu ^{\ast }=\cosh r, $ $\sigma =\nu ^{\ast }=\sinh r$, 则

$ \begin{eqnarray*} &&\exp \left[4\cosh ^{2}r\left \vert \alpha \right \vert ^{2}+2\sinh r\cosh r\left( \alpha ^{2}+\alpha ^{\ast 2}\right) \right] \frac{\partial ^{m}}{% \partial \alpha ^{m}}\frac{\partial ^{n}}{\partial \alpha ^{\ast n}}\\ &&\times \exp % \left[-4\cosh ^{2}r\left \vert \alpha \right \vert ^{2}-2\sinh r\cosh r\left( \alpha ^{2}+\alpha ^{\ast 2}\right) \right] \nonumber \\ &=&2\cosh ^{m+n}r\sum_{k=0}^{\left[m/2\right] }\sum_{l=0}^{\left[n/2\right] }\frac{\left( -\frac{\tanh r}{2}\right) ^{k+l}}{k!\left( m-2k\right) !}\frac{% m!n!}{l!\left( n-2l\right) !}H_{m-2k, n-2l} \\ &&\times \left[2\left( \alpha ^{\ast }\cosh r+\alpha \sinh r\right), 2\left( \alpha \cosh r+\alpha ^{\ast }\sinh r\right) \right] . \nonumber \end{eqnarray*} $

对于弱相干场驱动的多能级原子或者量子点腔系统, 原子与腔的相互作用吸收与辐射光子过程中, 也伴随着原子激发态与基态的变化^[21-24].文献[25]中, 作者利用弱相干输入场, 输出光场的量子态可以表示为一系列激发相干叠加态形式$\sum_{m}C_{m}{{a}^{\dagger m}}\left \vert \alpha \right \rangle $, 或者也可表示为粒子态的叠加态形式$ \left \vert \psi \right \rangle =\sum_{n}C_{n}\left \vert n\right \rangle $(由于$\left \vert \alpha \right \rangle =\sum_{n}\frac{\alpha ^{n}}{\sqrt{ n!}}\left \vert n\right \rangle $).在强耦合相互作用情况下, 输入态单模压缩真空态, 基于对输出光场的有效测量(如光子数统计方法), 由非线性相互作用引起的多光子过程输出状态可表示为

$ \rho \left( r,n \right)=C_{n}^{-1}{{a}^{\dagger }}^{n}S\left( r \right)\left| 0 \right\rangle \langle 0|{{S}^{-1}}\left( r \right){{a}^{n}}. $

(5.1)

该输出态体现了与激发量子态(如${{a}^{\dagger m}}\left \vert \varphi \right \rangle $), 归一化常数

$ C_{n}={\rm Tr}\left[\rho \left( r, n\right) \right] =n!\cosh ^{n}rP_{n}\left( \cosh r\right), $

$P_{n}\left( \cosh r\right) $为勒让德(Legendre)多项式, $S\left( r \right)=\exp \left[ \frac{r}{2}\left( {{a}^{\dagger 2}}-{{a}^{2}} \right) \right] $为幺正算符, 其幺正变换关系为

$ {{S}^{-1}}\left( r \right)aS\left( r \right)=a\cosh r+{{a}^{\dagger }}\sinh r, \\ \text{ }{{S}^{-1}}\left( r \right){{a}^{\dagger }}S\left( r \right)={{a}^{\dagger }}\cosh r+a\sinh r. $

(5.2)

为了研究$\rho \left( r, n\right) $的非经典性质, 利用式(2.11)得其准概论分布Wigner函数为

$ W\left( \alpha ,{{\alpha }^{*}} \right)\text{ }=2\pi \text{Tr}\left[ \rho \left( r,n \right)\Delta \left( \alpha ,{{\alpha }^{*}} \right) \right]\text{ } \\ =2\pi C_{n}^{-1}\text{Tr}[S\left( r \right){{\left( {{a}^{\dagger }}\cosh r+a\sinh r \right)}^{n}}\left| 0 \right\rangle \langle 0| \\ \times {{\left( a\cosh r+{{a}^{\dagger }}\sinh r \right)}^{n}}{{S}^{-1}}\left( r \right)\Delta \left( \alpha ,{{\alpha }^{*}} \right)]. $

(5.3)

通过对比式(5.2)和(4.4), 则变量关系为$\mu =\tau \rightarrow \cosh r, $ $\sigma =\nu \rightarrow -\sinh r$.再由式(5.3)和(4.6)得算符$\rho \left( r, n\right) $的Wigner函数为

$ \begin{matrix} W\left( \alpha ,{{\alpha }^{*}} \right)\text{ }=\frac{1}{{{P}_{n}}\left( \cosh r \right)}{{\left( \frac{-\sinh r}{2} \right)}^{n}}\exp \left[ -2{{\left| \alpha \sinh r-{{\alpha }^{*}}\cosh r \right|}^{2}} \right] \\ \times \sum\limits_{l=0}^{n}{\left( \begin{matrix} n \\ l \\ \end{matrix} \right)}\frac{{{2}^{l}}{{\left( -\coth r \right)}^{l}}}{\left( n-l \right)!}{{\left| {{H}_{n-l}}\left[ \text{i}\sqrt{\frac{2}{\tanh r}}\left( {{\alpha }^{*}}\cosh r-\alpha \sinh r \right) \right] \right|}^{2}}. \\ \end{matrix} $

从图中可以看出, 相空间某些特定区域的负值说明了光子增加弱压缩辐射场体现出了明显的非经典光场特征, 在压缩度固定时, 压缩真空态随着增加光子数的变化体现了不同的非经典特征, 这种变化在相互正交的X和Y方向均有展现.上行三个图表示有偶数光子在压缩真空场中产生, 下行则表示奇数个光子在压缩真空场中产生.

作为最为重要的量子相空间理论之一-Glauber-Sudarshan P-表示, 是一种准概论分布函数, 其物理可观测量能被表示成正规乘积形式, 借助相干态表象$\left \vert z\right \rangle =\exp \left( z{{a}^{\dagger }}-z^{\ast }a\right) \left \vert 0\right \rangle$, 量子密度矩阵$\rho $的$P$表示定义为

$ \begin{equation} \rho =\int \frac{{\rm d}^{2}z}{\pi }P\left( z, z^{\ast }\right) \left \vert z\right \rangle \left \langle z\right \vert. \label{eq:e1} \end{equation}% $

(6.1)

式(6.1)的逆变换为

$ P\left( z,{{z}^{*}} \right)={{\text{e}}^{{{\left| z \right|}^{2}}}}\int{\frac{{{\text{d}}^{2}}\beta }{\pi }}\langle -\beta |\rho \left| \beta \right\rangle \exp \text{ }\left( {{\left| \beta \right|}^{2}}+z{{\beta }^{*}}-{{z}^{*}}\beta \right), $

(6.2)

上式首次由Mehta^[26]得到.将式(6.2)代入到式(6.1), 并利用相干态投影算符的Weyl编序乘积, 即$\left\vert z\right\rangle \left\langle z\right\vert =2% \begin{array}{c} \colon \\\colon \\ \end{array}% {\rm e}^{-2\left( z^{\ast }-{{a}^{\dagger }}\right) \left( z-a\right) }% \begin{array}{c} \colon \\\colon \\ \end{array}% $, 于是

$ \rho =2\int{\frac{{{\text{d}}^{2}}z}{\pi }}{{\text{e}}^{{{\left| z \right|}^{2}}}}\int{\frac{%{{\text{d}}^{2}}\beta }{\pi }}\langle -\beta |\rho \left| \beta \right\rangle \exp \left( {{\left| \beta \right|}^{2}}+z{{\beta }^{*}}-{{z}^{*}}\beta \right)\begin{matrix} : \\ : \\ \end{matrix}{{\text{e}}^{-2\left( {{z}^{*}}-{{a}^{\dagger }} \right)\left( z-a \right)}}\begin{matrix} : \\ : \\ \end{matrix}=2\int{\frac{{{\text{d}}^{2}}\beta }{\pi }}\begin{matrix} : \\ : \\ \end{matrix}\langle -\beta |\rho \left| \beta \right\rangle {{\text{e}}^{2\left( a{{\beta }^{*}}-{{a}^{\dagger }}\beta +{{a}^{\dagger }}a \right)}}\begin{matrix} : \\ : \\ \end{matrix}. $

(6.3)