1.   引　言

觉察物理相似性是前进的因素. 麦克斯韦善于从类比中悟出共性, 他写道: “为了不通过一种物理理论而获得物理思想, 我们就应当熟悉现存的物理相似性. 所谓物理相似性, 我认为是在一种科学定律和一些能够相互阐明的定律之间存在着的局部相似.” 从量子力学诞生日起, 它的经典对应(或类比)一直是物理学家关心的话题. 量子力学中的很多概念都有经典对应或类比, 如平移、转动和宇称等. 狄拉克认为量子幺正变换是经典正则变换的对应, 但也有不存在经典对应的例子, 如自旋. 如麦克斯韦所说, 物理类比是发展物理学的一个途径, 那么量子纠缠有没有经典对应(或类比)呢?

本文将要指出: 在介观电路量子化的框架中, 带有互感的两个介观电容-电感($\rm {LC}$)回路, 其互感是产生量子纠缠的源头, 用量子力学方法可以求出其特征频率的公式, 该公式与如下描述的经典系统的小振动频率的表达式有相似之处, 可见两者有可比拟之处. 该经典系统如图1所示, 两个墙壁之间各连一个相同的弹簧, 弹簧系数是k, 两个弹簧之间接着一个可以在光滑的桌面上运动的滑动小车$m_{1}$, 小车挂有一根长为l的单摆, 摆球质量是$m_{2}$. 单摆的摆动会造成小车来回振动, 摆、小车和弹簧的互相牵制, 晃动效应反映了小车和摆的“纠缠”.

在固态物理中, 当输运尺度与电荷非弹性相干长度可以比拟时, 电路中的量子效应必须被计入, 这种情形下的电路便称为介观电路. 集成电路向原子尺度的趋小化刺激了电路理论的研究进入量子领域^[1,2]. 历史上, 一个单电容-电感($\rm {LC}$)回路的状态, 作为一个电路的“元胞”, 被Louisell^[3]在1973年量子化, 他认电荷q为正则坐标, 取电流$I = {\rm{d}}q/{\rm{d}}t$乘上电感L为正则动量, $p = L{\rm{d}}q/{\rm{d}}t$, 进一步将$(q, p)$加上量子化条件$[\hat{q}, \hat {p}] = {\rm{i}}\hbar$, 则${\rm{LC}}$电路被视为一个量子谐振子. 从那以后, 很多有关介观电路量子化的研究论文陆续发表^[4-6].

本文的内容安排如下: 先指出量子纠缠存在于带有互感m的两个介观电容-电感(${\rm{ LC }}$)回路中, 如图1所示. 给出这个系统的量子化形式. 然后得到系统哈密顿量的退纠缠算符并得到介观电路的量子纠缠态及其本征频率. 之后我们引入弹簧约束滑动小车-单摆系统的微振动系统并同时给出其振动频率, 最终得到两种体系相似点的类比.

2.   带有互感的两个介观电容-电感(${\rm{LC}}$)回路的量子化

在分析力学中, 带有互感(系数$m)$的两个介观电容-电感(${\rm{LC}}$)回路的经典拉格朗日量为^[7-9]

$$\begin{align} {\cal{L}} = \frac{1}{2}\left( l_{1}I_{1}^{2}+l_{2}I_{2}^{2}\right) +mI_{1}I_{2}-\frac{1}{2}\left( \frac{q_{1}^{2}}{c_{1}}+\frac{q_{2}^{2}} {c_{2}}\right) , \end{align}$$

(1)

这里$mI_{1}I_{2}$代表两个单回路中的电流相互作用; $l_{1}$和$l_{2}$是两个单回路的电感, 在无漏磁情形下, $0 < m <$$\sqrt {l_{1}l_{2}}$. 取$q_{1}$, $q_{2}$为正则坐标, 他们的共轭量为

$$\begin{split} p_{1} = \frac{\partial{\cal{L}}}{\partial\dot{q}_{1}} = l_{1}I_{1}+mI_{2}, \end{split}$$

(2)

$$\begin{split} p_{2} = \frac{\partial{\cal{L}}}{\partial\dot{q}_{2}} = l_{2}I_{2}+mI_{1}. \end{split}$$

(3)

相应的哈密顿量是

$$\begin{split} &{\cal{H}} = p_{1}\dot{q}_{1}+p_{2}\dot{q}_{2}-{\cal{L}} \\ =& \frac{1}{2}\left( l_{1}I_{1}^{2}+l_{2}I_{2}^{2}\right) +mI_{1} I_{2}+\frac{1}{2}\left( \frac{q_{1}^{2}}{c_{1}}+\frac{q_{2}^{2}}{c_{2} }\right) \\ =& \frac{1}{2A}\left( \frac{p_{1}^{2}}{l_{1}} + \frac{p_{2}^{2}}{l_{2} }\right) -\frac{m}{Al_{1}l_{2}}p_{1}p_{2} + \frac{1}{2}\left( \frac{q_{1}^{2} }{c_{1}}{+}\frac{q_{2}^{2}}{c_{2}}\right) , \end{split}$$

(4)

其中定义了

$$\begin{align} A = 1-\frac{m^{2}}{l_{1}l_{2}},\;m^{2} < l_{1}l_{2}. \end{align}$$

(5)

将$q_{i}, p_{i}$作为共轭对进行正则量子化为算符$\hat{q}_{i}, \hat{p}_{i}$, 加上量子化条件$\left[ \hat{q} _{i}, \hat{p}_{j}\right] = {\rm{i}}\hbar\delta_{i, j},$ ${\cal{H\rightarrow \hat{H}}}$是哈密顿算符. $\dfrac{m}{Al_{1}l_{2}} \hat{p}_{1}\hat{p}_{2}$是引起量子纠缠的项. 为什么如此说呢? 因为根据文献可知^[10,11], 此项会导致双模压缩态的产生, 非简并参量放大器输出的双模压缩态的信号模和闲置模是纠缠在一起的.

3.   哈密顿量${\cal{H}}$的退纠缠算符U

为了在理论上化去${\cal{\hat {H}}}$含的耦合项, 本文试图找到一个幺正算符U, 将${\cal{\hat{H}}}$对角化^[12,13]. 采用坐标表象$\vert q_{i}\rangle$, 记U为

$$\begin{split} {\boldsymbol{U}} = { \iint_{-\infty}^{\infty}} {\rm{d}}q_{1}{\rm{d}}q_{2}\left\vert {\boldsymbol{u}}\left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right) \right \rangle\left \langle \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right) \right\vert ,\;\det {\boldsymbol{u}} = 1, \end{split}$$

(6)

这里$\left \langle \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right) \right\vert$ $\equiv\langle q_{1}, q_{2}\vert$是双模坐标本征态,

$$\begin{split} \hat{q}_{i}\left\vert \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right) \right \rangle = q_{i}\left\vert \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right)\right \rangle ,\;{\boldsymbol{u}} = \left( \begin{array} [c]{cc} u_{11} & u_{12}\\ u_{21} & u_{22} \end{array} \right) , \end{split}$$

(7)

其中, u是一个待定的$2 \times 2$矩阵, 由对角化的要求决定. 方程(7)明显地体现了经典矩阵u映射为Hilbert空间中的量子幺正算符U. 用(6)式和(7)式得到$\hat{q} _{i}$的变换性质

$$\begin{split} {\boldsymbol{U}}\left( \begin{array} [c]{c} \hat{q}_{1}\\ \hat{q}_{2} \end{array} \right) {\boldsymbol{U}}^{\dagger} = {\boldsymbol{u}}^{-1}\left( \begin{array} [c]{c} \hat{q}_{1}\\ \hat{q}_{2} \end{array} \right) , \end{split}$$

(8)

转到动量表象, 用其完备性^[14-16]

$$\begin{split} { \iint_{-\infty}^{\infty}} {\rm{d}}p_{1}{\rm{d}}p_{2}\left\vert \left( \begin{array} [c]{c} p_{1}\\ p_{2} \end{array} \right)\right \rangle \left\langle \left( \begin{array} [c]{c} p_{1}\\ p_{2} \end{array} \right) \right\vert = 1, \end{split}$$

(9)

和

$$\begin{split}\bigg\langle \begin{pmatrix} p_{1}\\ p_{2} \end{pmatrix} \bigg \vert {\boldsymbol{u}} \begin{pmatrix} q_{1}\\ q_{2} \end{pmatrix} \bigg\rangle = \frac{1}{2\pi}\exp\left[ -{\rm{i}}\left( {\boldsymbol{u}}^{{\rm{T}}}p\right) _{j}q_{j}\right] \end{split}$$

(10)

(这里重复指标暗示求和)得到U的动量表示:

$$\begin{split} {\boldsymbol{U}} & = \frac{1}{2\pi} { \iint_{-\infty}^{\infty}} {\rm{d}}p_{1}{\rm{d}}p_{2} { \iint_{-\infty}^{\infty}} {\rm{d}}q_{1}{\rm{d}}q_{2}\\&\times\left\vert \left( \begin{array} [c]{c} p_{1}\\ p_{2} \end{array} \right)\right \rangle \left\langle \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right) \right\vert \exp\left[ -{\rm{i}}\left( {\boldsymbol{u}}^{{\rm{T}}}p\right) _{j}q_{j}\right] \\ & = { \iint_{-\infty}^{\infty}} {\rm{d}}p_{1}{\rm{d}}p_{2}\left\vert \left( \begin{array} [c]{c} p_{1}\\ p_{2} \end{array} \right) \right\rangle \left\langle {\boldsymbol{u}}^{{\rm{T}}}\left( \begin{array} [c]{c} p_{1}\\ p_{2} \end{array} \right) \right\vert , \end{split}$$

(11)

于是就得到

$$\begin{split} {\boldsymbol{U}}\left( \begin{array} [c]{c} \hat{p}_{1}\\ \hat{p}_{2} \end{array} \right) {\boldsymbol{U}}^{\dagger} = {\boldsymbol{u}}^{{\rm{T}}}\left( \begin{array} [c]{c} \hat{p}_{1}\\ \hat{p}_{2} \end{array} \right) . \end{split}$$

(12)

假设u的形式是

$$\begin{split} {\boldsymbol{u}} = \left( \begin{array} [c]{cc} 1 & E\\ G & H \end{array} \right) ,~~~\det {\boldsymbol{u}} = H-EG = 1, \end{split}$$

(13)

这里$H, E, G$是待定的, 则

$$\begin{split} \left( {\boldsymbol{u}}^{{\rm{T}}}\right) ^{-1} = \left( \begin{array} [c]{cc} H & -G\\ -E & 1 \end{array} \right) . \end{split}$$

(14)

在${\boldsymbol{U}}^{\dagger}$变换下,

$$\begin{split} \hat{q}_{1} & \rightarrow {\boldsymbol{U}}^{\dagger}\hat{q}_{1}{\boldsymbol{U}} = \hat{q}_{1}+E\hat{q} _{2}, \\ \hat{q}_{2} & \rightarrow {\boldsymbol{U}}^{\dagger}\hat{q}_{2}{\boldsymbol{U}} = G\hat{q}_{1}+H\hat{q} _{2},\\ \hat{p}_{1} & \rightarrow {\boldsymbol{U}}^{\dagger}\hat{p}_{1}{\boldsymbol{U}} = H\hat{p}_{1}-G\hat{p} _{2},\\ \hat{p}_{2} & \rightarrow {\boldsymbol{U}}^{\dagger}\hat{p}_{2}{\boldsymbol{U}} = -E\hat{p}_{1}+\hat{p} _{2}, \end{split}$$

(15)

故而${\cal{\hat{H}}}$在${\boldsymbol{U}}^{\dagger}$变换下变成

$$\begin{split} {\boldsymbol{U}}^{\dagger}{\cal{\hat{H}}}{\boldsymbol{U}} & = \frac{1}{2A}\bigg[ \frac{\left( Hp_{1}-Gp_{2}\right) ^{2}}{l_{1}}+\frac{\left( -Ep_{1}+p_{2}\right) ^{2} }{l_{2}}\bigg] \\ & -\frac{m}{Al_{1}l_{2}}\left( Hp_{1}-Gp_{2}\right) \left( -Ep_{1} +p_{2}\right) \\ & +\frac{1}{2}\bigg[ \frac{\left( q_{1}+Eq_{2}\right) ^{2}}{c_{1}} +\frac{\left( Gq_{1}+Hq_{2}\right) ^{2}}{c_{2}}\bigg] .\\[-15pt] \end{split}$$

(16)

对角化要求(16)式中含有$\hat {p}_{1}\hat{p}_{2}$和$\hat{q}_{1}\hat{q}_{2}$的项消失, 即要求

$$\begin{split} l_{2}HG+l_{1}E+m\left( GE+H\right) = 0, \end{split}$$

(17)

$$\begin{align} c_{2}E+c_{1}GH = 0. \end{align}$$

(18)

这意味着退纠缠, 联立$H-EG = 1$可知

$$\begin{align} H = {c_{2}}/({c_{2}+c_{1}G^{2}}), \end{align}$$

(19)

于是

$$\begin{align} E = {-Gc_{1}}/({c_{2}+c_{1}G^{2}}), \end{align}$$

(20)

接着有

$$\begin{split} HG = \frac{c_{2}G}{c_{2}+c_{1}G^{2}} = -\frac{c_{2}}{c_{1}}E. \end{split}$$

(21)

将(21)式代入(17)式可以导出

$$\begin{split} -l_{2}\dfrac{c_{2}}{c_{1}}E+l_{1}E+m\left( 2EG+1\right) = 0, \end{split}$$

(22)

(21)式和(22)式联立给出

$$\begin{align} mc_{1}G^{2}+G\left( l_{1}c_{1}-l_{2}c_{2}\right) -mc_{2} = 0. \end{align}$$

(23)

此方程的通解是

$$\begin{align} G = \frac{l_{2}c_{2}-l_{1}c_{1}\pm\sqrt{\left( l_{1}c_{1}-l_{2}c_{2}\right) ^{2}+4m^{2}c_{1}c_{2}}}{2mc_{1}}. \end{align}$$

(24)

不失一般性, (24)式中取负号, 并令

$$\begin{align} \left( c_{2}l_{2}-c_{1}l_{1}\right) ^{2}+4m^{2}c_{2}c_{1} = \varDelta, \end{align}$$

(25)

可得

$$\begin{split} G & = \frac{c_{2}l_{2}-c_{1}l_{1}-\sqrt{\varDelta}}{2mc_{1}} = \frac{-2mc_{2} }{\sqrt{\varDelta}+c_{2}l_{2}-c_{1}l_{1}},\\ E & = \frac{c_{1}m}{\sqrt{\varDelta}},\;\;H = \frac{\sqrt{\varDelta}-\left( c_{1}l_{1}-c_{2}l_{2}\right) }{2\sqrt{\varDelta}}, \end{split}$$

(26)

于是相应的退纠缠算符为

$$\begin{split} {\boldsymbol{U}} = { \iint_{-\infty}^{\infty}} {\rm{d}}q_{1}{\rm{d}}q_{2}\left\vert \left( \begin{array} [c]{cc} 1 & E\\ G & H \end{array} \right) \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right)\right \rangle \left\langle \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right) \right\vert . \end{split}$$

(27)

4.   介观电路的量子纠缠态

用有序算符内的积分理论及双模坐标本征态的Fock表象^[17-19]

$$\begin{split}\left \langle q_{1},q_{2}\right\vert & = \sqrt{\frac{1}{\pi}}\left\langle 00\right\vert \exp\left\{ -\frac{1}{2}\left( q_{1}^{2}+q_{2}^{2}\right) \right. \\ & \left. +\sqrt{2}\left( q_{1}a_{1}+q_{2}a_{2}\right) -\frac{1}{2} a_{1}^{2}-\frac{1}{2}a_{2}^{2}\right\}, \end{split}$$

(28)

$[ a_{i}, a_{j}^{\dagger}] = \delta_{i, j},$以及真空投影算符的正规乘积表示^[20-22]

$$\begin{align} \left\vert 00\rangle \langle 00\right\vert = \colon\exp\big( -a_{1}^{\dagger}a_{1}-a_{2}^{\dagger}a_{2}\big) \colon \end{align}$$

(29)

对U的表达式积分得到

$$\begin{split} {\boldsymbol{U}} & = { \iint_{-\infty}^{\infty}} {\rm{d}}q_{1}{\rm{d}}q_{2}\left\vert \left( \begin{array} [c]{cc} {\bf{1}} & E\\ G & H \end{array} \right) \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right)\right \rangle \left\langle \left( \begin{array} [c]{c} q_{1}\\ q_{2} \end{array} \right) \right\vert = \frac{1}{\pi} { \iint_{-\infty}^{\infty}} {\rm{d}}q_{1}{\rm{d}}q_{2}\colon\exp\left\{ -\frac{1}{2}\left[ \left( q_{1} +Eq_{2}\right) ^{2}+\left( Gq_{1}+Hq_{2}\right) ^{2}\right] \right. \\ & \quad -\sqrt{2}\left( q_{1}+Eq_{2}\right) a_{1}^{\dagger}+\sqrt{2}\left( Gq_{1}+Hq_{2}\right) a_{2}^{\dagger}-\frac{1}{2}\left( q_{1}^{2}+q_{2} ^{2}\right) \left. +\sqrt{2}\left( q_{1}a_{1}+q_{2}a_{2}\right) -\frac{1}{2}\left( a_{1}+a_{1}^{\dagger}\right) ^{2}-\frac{1}{2}\left( a_{2}+a_{2}^{\dagger}\right) ^{2}\right\} \colon\\ & = \frac{2}{\sqrt{L}}\exp\left\{ \frac{1}{2L}[\left( 1+E^{2}-G^{2} -H^{2}\right) \left( a_{1}^{\dagger2}-a_{2}^{\dagger2}\right) +4\left( G+EG\right) a_{1}^{\dagger}a_{2}^{\dagger}]\right\} \colon\exp\left\{ \left( a_{1}^{\dagger}\ a_{2}^{\dagger}\right) \left( g-{\bf{1}}\right) \left( \begin{array} [c]{c} a_{1}\\ a_{2} \end{array} \right) \right\} \colon\\ & \quad \times \exp\left\{ \frac{1}{2L}[\left( E^{2}+H^{2}-1-G^{2}\right) \left( a_{1}^{2}-a_{2}^{2}\right) +4\left( G+EH\right) a_{1}a_{2}]\right\} , \\[-18pt]\end{split}$$

(30)

其中

$$\begin{split} L & = E^{2}+G^{2}+H^{2}+3,\\ {\boldsymbol{g}} & = \frac{2}{L}\left( \begin{array} [c]{cc} {\bf{1}}+H & E-G\\ G-E & {\bf{1}}+H \end{array} \right) ,\;\mathbf{1 = }\left( \begin{array} [c]{cc} 1 & 0\\ 0 & 1 \end{array} \right) . \end{split}$$

(31)

可见${\boldsymbol{U}} \vert 00\rangle$是一个双模压缩态

$$\begin{split} & {\boldsymbol{U}}|00\rangle =\frac{2}{\sqrt{L}} \exp \Big\{\frac { 1 } { 2 L } \Big[\left(1+E^{2}-G^{2}-H^{2}\right) \\ & \quad \times \left(a_{1}^{\dagger 2}-a_{2}^{\dagger 2}\right)+4(G+E H) a_{1}^{\dagger} a_{2}^{\dagger}\Big]\Big\}|00\rangle, \end{split}$$

(32)

同时它也是一个纠缠态. 可见互感的存在导致量子纠缠. 该结果表示的是两个介观回路处于纠缠态, 注意到当前是在双模坐标表象下表示出来, 而此处的广义坐标$q_{1}, q_{2}$对应两个介观回路的电容各自携带的电量. 这就意味着测量其中一个回路电容上的电量后, 另外一个回路的电容电量也会塌缩到某个特定值上. 当然如果换成动量表象(对应两个回路中各自的电流), 也依然有纠缠的特性.

5.   带互感的两个介观电容-电感(${\rm{LC}}$)电路的特征频率

去除含有$\hat{p}_{1}\hat{p}_{2}$和$\hat{q}_{1}\hat {q}_{2}$的项后, 方程(18)变成

$$\begin{split} {\boldsymbol{U}}^{\dagger}{\cal{\hat{H}}}{\boldsymbol{U}} =\;& \frac{p_{1}^{2}}{2Al_{1}l_{2}}\left( l_{2}H^{2}+l_{1}E^{2}+2mHE\right) \\ & +\frac{p_{2}^{2}}{2Al_{1}l_{2}}\left( l_{2}G^{2}+l_{1}+2mG\right) \\ & +\frac{q_{1}^{2}}{2}\left( \frac{1}{c_{1}}{+}\frac{G^{2}}{c_{2}}\right) {+}\frac{q_{1}^{2}}{2}\left( \frac{E^{2}}{c_{1}}{+}\frac{H^{2}}{c_{2}}\right) . \end{split}$$

(33)

用(33)式算出

$$\begin{align} l_{2}H^{2}+l_{1}E^{2}+2mHE = \frac{m^{2}c_{1}G-mc_{2}l_{2}}{G\sqrt{\varDelta}}, \end{align}$$

(34)

$$\begin{align} l_{2}G^{2}+l_{1}+2mG = -\frac{\left( l_{2}G+m\right) \sqrt{\varDelta}}{mc_{1}}, \end{align}$$

(35)

$$\begin{align} \frac{1}{c_{1}}+\frac{G^{2}}{c_{2}} = \frac{c_{2}+c_{1}G^{2}}{c_{1}c_{2}} = -\frac{G}{Ec_{2}}, \end{align}$$

(36)

$$\begin{align} \frac{E^{2}}{c_{1}}+\frac{H^{2}}{c_{2}} = -\frac{m}{G\sqrt{\varDelta}}, \end{align}$$

(37)

于是(33)式变成

$$\begin{split} {\boldsymbol{U}}^{\dagger}{\cal{\hat{H}}}{\boldsymbol{U}} =\;& \frac{\left( m^{2}c_{1}G-mc_{2}l_{2}\right) }{2Al_{1}l_{2}G\sqrt{\varDelta}}p_{1}^{2}-\frac{G}{2Ec_{2}}q_{1}^{2} \\&-\frac{\left( l_{2}G+m\right) \sqrt{\varDelta}}{2Al_{1}l_{2}mc_{1}}p_{2} ^{2}-\frac{m}{2G\sqrt{\varDelta}}q_{2}^{2}.\nonumber \end{split}$$

与谐振子哈密顿量的标准形式$\dfrac{p^{2}}{2{\cal{M}}}+\dfrac{{\cal{M}}\omega^{2}q^{2}}{2}$比较, 可得两个特征频率为

$$\begin{align} -\frac{G}{Ec_{2}}\frac{m^{2}c_{1}G-mc_{2}l_{2}}{Al_{1}l_{2}G\sqrt{\varDelta} } = \frac{c_{2}l_{2}+c_{1}l_{1}+\sqrt{\varDelta}}{2Al_{1}l_{2}c_{1}c_{2}} \equiv\omega_+^{2} \end{align}$$

(38)

以及

$$\begin{split} \dfrac{\dfrac{m}{G\sqrt{\varDelta}}}{Al_{1}l_{2}\dfrac{mc_{1}}{\left( l_{2}G+m\right) \sqrt{\varDelta}}} = \dfrac{c_{2}l_{2}+c_{1}l_{1}-\sqrt{\varDelta} }{2Al_{1}l_{2}c_{2}c_{1}}\equiv\omega_-^{2}. \end{split}$$

(39)

再用$A = 1- {m^{2}}/({l_{1}l_{2}}),$可见

$$\begin{align} \omega_{\pm}^{2} = \frac{c_{2}l_{2}+c_{1}l_{1}\pm\sqrt{\left( c_{2}l_{2} -c_{1}l_{1}\right) ^{2}+4m^{2}c_{2}c_{1}}}{2c_{2}c_{1}(l_{1}l_{2}-m^{2})}, \end{align}$$

(40)

或

$$\begin{split} \omega_{\pm}^{2} =\;& \frac{c_{2}l_{2}+c_{1}l_{1}\pm\sqrt{\left( c_{2} l_{2}+c_{1}l_{1}\right) ^{2}+4c_{2}c_{1}\left( m^{2}-l_{2}l_{1}\right) } }{2c_{2}c_{1}(l_{1}l_{2}-m^{2})} \\=\;& \frac{c_{2}l_{2}+c_{1}l_{1}}{2c_{2}c_{1}(l_{1}l_{2}-m^{2})}pm \sqrt{\left[ \frac{\left( c_{2}l_{2}+c_{1}l_{1}\right) }{2c_{2}c_{1} (l_{1}l_{2}-m^{2})}\right] ^{2}-\frac{1}{c_{2}c_{1}(l_{1}l_{2}-m^{2})}}\;, \end{split}$$

(41)

这是可以用实验验证的.

6.   弹簧约束滑动小车-单摆系统的微振动频率

本节讨论自由滑动小车-单摆系统的微振动频率. 从分析力学观点出发, 可以根据摆线偏离竖直线的角度θ以及对滑块的坐标x写出系统的动能与势能. 其中动能为

$$\begin{split} T {=} \frac{1}{2}m_{1}\dot{x}^{2}+\frac{m_{2}}{2}\big[ (\dot{x}+l\dot{\theta }\cos\theta)^{2}+l^{2}\dot{\theta}^{2}\sin^{2}\theta\big], \end{split}$$

(42)

其中第一项是滑动小车动能, 第二项是摆球动能, 反映了摆球同时参与滑动和摆动的速度合成规则, 即三角形余弦定理,

$$\begin{align} (\dot{x}+l\dot{\theta}\cos\theta)^{2}+l^{2}\dot{\theta}^{2}\sin^{2}\theta = \dot{x}^{2}+l^{2}\dot{\theta}^{2}+2\dot{x}l\dot{\theta}\cos\theta. \end{align}$$

(43)

势能是

$$\begin{align} V = -m_{2}gl\cos\theta+2 \times \frac{1}{2}kx^{2}, \end{align}$$

(44)

从$L = T-V$以及

$$\frac{{\rm{d}}}{{\rm{d}}t}\frac{\partial L}{\partial\dot{x}} = \frac{\partial L}{\partial x},~~\frac{{\rm{d}}}{{\rm{d}}t}\frac{\partial L}{\partial\dot{\theta}} = \frac{\partial L}{\partial\theta}$$

(45)

导出动力学方程

$$\begin{align} \left( m_{1}+m_{2}\right) \ddot{x}+m_{2}l\ddot{\theta}\cos\theta-m_{2} l\dot{\theta}^{2}\sin\theta+2kx = 0, \end{align}$$

(46)

$$\begin{split} l\ddot{\theta}+\ddot{x}\cos\theta+g\sin\theta = 0. \end{split}$$

(47)

在小振动时, $\cos\theta\approx 1, ~\sin \theta\approx \theta,$故(46)式和(47)式分别约化为

$$\begin{split} \left( m_{1}+m_{2}\right) \ddot{x}+m_{2}l\ddot{\theta}+2kx = 0, \end{split}$$

(48)

$$\begin{align} l\ddot{\theta}+\ddot{x}+g\theta = 0, \end{align}$$

(49)

即

$$\begin{align} m_{1}\ddot{x} = m_{2}g\theta-2kx, \end{align}$$

(50)

$$\begin{align} m_{1}l\ddot{\theta} = 2kx-g\left( m_{1}+m_{2}\right) \theta. \end{align}$$

(51)

晃动过程中, 小车与摆有相同的频率, 故可令

$$\begin{align} x = Y\sin(\omega t), \end{align}$$

(52)

$$\begin{align} \theta = Z\sin(\omega t), \end{align}$$

(53)

代入(50)式和(51)式得到

$$\begin{align} \left( \omega^{2}m_{1}-2k\right) Y+m_{2}gZ = 0, \end{align}$$

(54)

$$\begin{align} 2kY+\left[ m_{1}l\omega^{2}-g\left( m_{1}+m_{2}\right) \right] Z = 0, \end{align}$$

(55)

其系数行列式为零才有非平庸解, 即

$$\begin{split} \left\vert \begin{array} [c]{cc} \omega^{2}m_{1}-2k & m_{2}g\\ 2k & -g\left( m_{1}+m_{2}\right) +m_{1}l\omega^{2} \end{array} \right\vert = 0, \end{split}$$

(56)

也就是

$$\begin{align} m_{1}^{2}l\omega^{4}-m_{1}\left[ 2kl+g\left( m_{1}+m_{2}\right) \right] \omega^{2}+2kgm_{1} = 0. \end{align}$$

(57)

由此解出

$$\begin{split}\,& \omega^{2} = \dfrac1{2lm_{1}^{2}}\Big\{m_{1}\left[ g\left( m_{1}+m_{2}\right) +2kl\right] \\ & ~ \pm\sqrt{\left[ g\left( m_{1} {+} m_{2}\right) {+} 2kl\right] ^{2}m_{1} ^{2}-4lm_{1}^{2}2km_{1}g}\Big\}. \end{split}$$

(58)

所以弹簧约束滑动小车-单摆系统的微振动频率是

$$\begin{split} \omega^{2} =\;& \frac{\left( m_{1}+m_{2}\right) g+2kl}{2m_{1}l}\\&\pm\sqrt{\left[ \frac{\left( m_{1}+m_{2}\right) g+2kl}{2m_{1}l}\right] ^{2}-\frac {2kg}{m_{1}l}}\;, \end{split}$$

(59)

这里的两个根都是正定的, 都是物理解, $\dfrac {2 kg}{m_{1}l} =$$\dfrac{2 k}{m_{1}} \times \dfrac{g}{l}$代表弹簧振子带动小车运动(以$\dfrac {2 k}{m_{1}}$表征)与单摆运动(以$\dfrac {g}{l}$表征)之间的耦合, 是量子纠缠的经典类比.

7.   量子纠缠的经典类比

将介观电路的本征频率(41)式改写为

$$\begin{split} & \omega_{\pm}^{2}\\ =\;&\frac{c_{2}l_{2}{+} c_{1}l_{1} {\pm} \sqrt{\left( c_{2} l_{2}+c_{1}l_{1}\right) ^{2} {+} 4c_{2}c_{1}\left( m^{2}-l_{2}l_{1}\right) } }{2c_{2}c_{1}(l_{1}l_{2}-m^{2})} \\ =\;& \frac{c_{2}l_{2}+c_{1}l_{1}}{2c_{2}c_{1}(l_{1}l_{2}-m^{2})}\\&\pm \sqrt{\left[ \frac{\left( c_{2}l_{2}+c_{1}l_{1}\right) }{2c_{2}c_{1} (l_{1}l_{2}-m^{2})}\right] ^{2} {-} \frac{1}{c_{2}c_{1}(l_{1}l_{2}-m^{2})}}\;, \end{split}$$

(60)

再和上述力学系统的频率(59)式作比较, 就可得到如下的对应:

$$\begin{align} \frac{\left( m_{1}+m_{2}\right) g+2kl}{2m_{1}l}\rightarrow \frac{c_{1} l_{1}+c_{2}l_{2}}{2c_{1}c_{2}\left( l_{1}l_{2}-m^{2}\right) }, \end{align}$$

(61)

$$\begin{align} \frac{2kg}{m_{1}l} = \frac{2k}{m_{1}} \times \frac{g}{l}\rightarrow \frac{1} {c_{1}c_{2}\left( l_{1}l_{2}-m^{2}\right) }. \end{align}$$

(62)

于是找到了一个鲜明的例子, 即量子纠缠可以有经典力学模拟或对应. 我们期望有更多的例子出现.

8   结　论

本文首次讨论了量子纠缠有没有经典类比(或对应) 的问题, 指出在介观电路量子化的框架中, 带有互感的两个介观电容-电感 (LC) 电路与两个弹簧之间夹一个小滑车在光滑的地面上附带一个单摆的运动可比拟. 先用有序算符内的积分理论证明第一个系统的互感是产生量子纠缠的源头, 再推导出求其特征频率的公式, 就发现它与第二个系统的小振动频率公式类似. 第二个系统中单摆的摆动会造成小车来回振动, 摆、小车和弹簧的互相牵制效应反映了小车和摆的“纠缠”. 从两个系统的振动频率对比中发现有类似, 这是严格数学推导的结果, 而不是哲学观点的逻辑推理. 时至今日, 我们还不能武断有量子纠缠的系统就不存在可以类比的经典力学系统, 真理是在探索讨论中渐渐显露的, 希望本文严密正确的推导能起抛砖引玉的作用.