量子游荡作为经典随机游荡概念的量子化版本, 最早由Aharonov等人于1993年提出^[1], 它是构成量子算法的有力工具^[2-3]. 量子游荡包括连续时间量子游荡和离散时间量子游荡. 连续时间量子游荡最早于1998年由Farhi和Gutmann提出^[4]; 离散时间量子游荡最早于2001年由Watrous提出^[5]. 其中, 线性、格上、环上和高维空间上的离散时间量子游荡受到了广泛关注^[6]. 作为数学模型, 学者们主要研究其基础性质, 包括概率分布、平稳测度、对称性、命中时间、周期性、渐进性等^[7-12].

近年来, 学者们也关注了非齐次量子游荡, 主要有交替式量子游荡、时间非齐次量子游荡和空间非齐次量子游荡. 这些模型主要特点是系统演化酉算子会随时间或空间变化而变化. 2009年, N.Linden等人最早引入非齐次量子游荡模型, 将硬币演化算子设为与状态和空间有关的量, 由此推导出有界性等相关性质^[13]. 2010年, Shikano等人引入一维格上自对偶非齐次量子游荡, 研究其局域性和分形(霍夫斯塔特蝴蝶)^[14]. 对于交替式量子游荡, 2017年Rousseva等人研究以相同角度顺时针和逆时针进行交替旋转的非齐次量子游荡, 并使用相位法近似计算游荡概率^[15]. 对于空间非齐次量子游荡, 2013年Konno等人研究空间非齐次量子游荡并提出平均时间极限、弱极限和平稳等三类测度^[16]; 2019年韩琦提出空间非齐次三态量子游荡的平稳测度^[17]. 对于时间非齐次量子游荡, 2013年Machida讨论时间非齐次二态量子游荡的几种极限概率分布^[18].

该文针对时间非齐次二态量子游荡分析其演化过程, 建立相应的数学模型, 通过归纳硬币演化序列给出位置概率分布的计算表达式, 运用谱理论分析系统量子态的演化过程, 从随机微分方程角度引出时间非齐次二态量子游荡的伊藤引理和伊藤公式, 并对公式进行矩阵分解和解释.

定义2.1  设一个量子系统的状态空间为$ {\cal H} = {\cal H}_C\otimes{\cal H}_P $, 其中$ {\cal H}_C = span\{|0\rangle, |1\rangle\} $表示为硬币状态空间, $ {\cal H}_P = span\{|x\rangle: x\in {\Bbb Z}\} $表示为位置状态空间, 系统的酉算子为$ \boldsymbol{ U }_t = \boldsymbol{ S }\cdot(\boldsymbol{ C }_{t}\otimes \boldsymbol{ I }), t = 1, 2, \cdots, N $, 其中$ \boldsymbol{ C }_{t} = \left( \begin{array}{ccc} a_{t} & b_{t} \\ c_{t} & d_{t} \\ \end{array} \right) $为硬币演化算子; $ \boldsymbol{ S } = \sum\limits_{x\in{\Bbb Z}} [|0\rangle\langle 0|\otimes|x-1\rangle\langle x|+|1\rangle\langle 1|\otimes|x+1\rangle\langle x|] $为移位算子, 称该系统为时间非齐次二态量子游荡.

该文简称时间非齐次二态量子游荡为非齐次量子游荡. 如果硬币演化算子不会随着时间或者空间改变而变化, 则称为齐次量子游荡.

将硬币演化算子$ \boldsymbol{ C }_t $分解为$ \boldsymbol{ C }_{t} = \boldsymbol{ P }_{t}+\boldsymbol{ Q }_t $, 其中$ \boldsymbol{ P }_{t} = \left( \begin{array}{ccc} a_{t} & b_{t} \\ 0 & 0 \\ \end{array} \right) $表示为向左演化算子; $ \boldsymbol{ Q }_{t} = \left( \begin{array}{ccc} 0 & 0 \\ c_{t} & d_{t} \\ \end{array} \right) $表示向右演化算子. 如果硬币演化算子定义为$ \boldsymbol{ C }_{t} = \left( \begin{array}{ccc} \cos\theta_{t} & \sin\theta_{t} \\ \sin\theta_{t} & -\cos\theta_{t} \\ \end{array} \right) $, 则称为硬币翻转算子, 其中$ \theta_{t}\in[0, 2\pi] $为翻转角度. 该文采用硬币翻转算子作为硬币演化算子. 设系统的量子初态为$ |\psi_{0}\rangle $, 将酉算子$ \boldsymbol{ U }_1, \boldsymbol{ U }_2, \cdots, \boldsymbol{ U }_t $依次作用在$ |\psi_0\rangle $上, 得到$ t $次演化后的量子态

(2.1) $ \begin{equation} |\psi_{t}\rangle = \boldsymbol{ {U} }_t\boldsymbol{ {U} }_{t-1}\cdots \boldsymbol{ {U} }_1|\psi_{0}\rangle = \prod\limits^t_{k = 1} \boldsymbol{ {U} }_k|\psi_{0}\rangle = \sum\limits_{x\in {\Bbb Z}}|\psi_{t}(x)\rangle. \end{equation} $

该式给出了系统经过$ t $次演化后的量子态表达式, 它与次数$ t $有关. 其中, $ |\psi_{t}(x)\rangle = \psi^L_{t}(x)|0\rangle+\psi^R_{t}(x)|1\rangle, x\in{\Bbb Z} $. 由于$ |\psi_{t}(x)\rangle $与位置$ x $有关, 故称它为关于位置$ x $的状态波函数, 其中概率振幅$ \psi^L_{t}(x) $和$ \psi^R_{t}(x) $分别计算如下

(2.2) $ \begin{equation} \begin{array}{ll} \psi^L_{t}(x) = \cos\theta_{t-1}\psi^L_{t-1}(x+1)+\sin\theta_{t-1}\psi^R_{t-1}(x+1), \\ \psi^R_{t}(x) = \sin\theta_{t-1}\psi^L_{t-1}(x-1)-\cos\theta_{t-1}\psi^R_{t-1}(x-1). \end{array} \end{equation} $

给定函数$ f_j:{\Bbb Z}\rightarrow {\Bbb C} $, 定义离散傅里叶变换为$ \hat{f}_t:[-\pi, \pi]\rightarrow {\Bbb C} $使得

$ \hat{f}_t(s) = \sum\limits_jf_j(t){\rm e}^{-{\rm i}js}, $

其中$ {\rm i} $是虚根, $ s\in[-\pi, \pi] $, $ j\in{\Bbb Z} $; 同时定义离散傅里叶逆变换为

$ f_j(t) = \frac{1}{2\pi}\int_{-\pi}^{\pi}\hat{f}_t(s){\rm e}^{{\rm i}js}{\rm d}s. $

对系统经过$ t $次演化后处于位置$ x $的状态波函数$ |\psi_{t}(x)\rangle $分别做离散傅里叶变换和逆变换, 记为

(2.3) $ \begin{equation} \widehat{\psi}_t(\xi) = \sum\limits_{x\in {\Bbb Z}}\widehat{\psi}_t(\xi)(x) = \sum\limits_{x\in {\Bbb Z}}{\rm e}^{{\rm i}\xi x}|\psi_t(x)\rangle, |\psi_t(x)\rangle = \int^{\pi}_{-\pi} \frac{d\xi}{2\pi}{\rm e}^{-{\rm i}\xi x}\widehat{\psi}_t(\xi), x\in{\Bbb Z}. \end{equation} $

函数$ \widehat{\psi}_{t}(\xi) $与$ \widehat{\psi}_{t-1}(\xi) $的关系为

(2.4) $ \begin{equation} \widehat{\psi}_{t}(\xi) = \left( \begin{array}{c} \widehat{\psi}^L_t(\xi) \\ \widehat{\psi}^R_t(\xi) \\ \end{array} \right) = \left( \begin{array}{cc} \cos\theta_{t}{\rm e}^{-{\rm i}\xi} & \sin\theta_{t}{\rm e}^{-{\rm i}\xi} \\ \sin\theta_{t}{\rm e}^{{\rm i}\xi} & -\cos\theta_{t}{\rm e}^{{\rm i}\xi} \\ \end{array} \right)\left( \begin{array}{c} \widehat{\psi}^L_{t-1}(\xi) \\ \widehat{\psi}^R_{t-1}(\xi)\\ \end{array} \right) = \boldsymbol{ {C} }_{t}(\xi)\widehat{\psi}_{t-1}(\xi), \end{equation} $

其中$ \boldsymbol{ C }_t(\xi) = \left( \begin{array}{ccc} {\rm e}^{-{\rm i}\xi} & 0 \\ 0 & {\rm e}^{{\rm i}\xi} \\ \end{array} \right)\boldsymbol{ C }_{t} $. 于是, 系统经过$ t $次演化后得到

(2.5) $ \begin{equation} \widehat{\psi}_{t}(\xi) = \boldsymbol{ {C} }_t(\xi)\boldsymbol{ {C} }_{t-1}(\xi) \cdots\boldsymbol{ {C} }_1(\xi)\widehat{\psi}_{0}(\xi) = \prod\limits^t_{k = 1}\boldsymbol{ {C} }_k(\xi)\widehat{\psi}_{0}(\xi). \end{equation} $

非齐次量子游荡经过$ N $次演化后的位置集合记为$ B_{N} = \{-N, -N+1, \cdots, N-1, N\} $. 基于位置集合构造笛卡尔积$ \Omega_N = B^{N+1}_N = \{-N, -N+1, \cdots, N-1, N\}^{N+1} $.

定义3.1   任取一个向量$ \omega = (\omega(0), \omega(1), \cdots, \omega(N))\in \Omega_N $, 其中分量$ \omega(0), \omega(1), \cdots, $$ \omega(N) $分别表示系统在$ t = 0, 1, \cdots, N $时刻所处的位置, 其中令$ \omega(0) = 0 $, 则称该向量为一条位置路径.

定义3.2   称$ v = \nabla\omega = (v(1), v(2), \cdots, v(N)) = (\omega(1)-\omega(0), \omega(2)-\omega(1), \cdots, \omega(N) -\omega(N-1)) $为一条移动路径.

在该定义中, $ \omega(t)-\omega(t-1)\in\{-1, 1\}(t = 1, 2, \cdots, N) $表示系统所经历的前后位置差, 反映移动的方向. 当$ v(t) = -1 $时, 表示系统在第$ t $时刻向左移动一个位置; 当$ v(t) = 1 $时, 表示系统在第$ t $时刻向右移动一个位置. 因而, 可以用$ v $表示移动的方向序列.

设$ u = (u(1), u(2), \cdots, u(N)) = (I_{\{1\}}(v(1)), I_{\{1\}}(v(2)), \cdots, I_{\{1\}}(v(N))) $, 其中分量$ I_{\{1\}} $$ (x) $表示集合$ \{1\} $的示性函数. 当$ v(t) = 1 $时$ u(t) = 1 $; 当$ v(t) = -1 $时$ u(t) = 0 $. 记$ v_i $为系统在时刻$ t = N $上所选择的第$ i $条移动路径, 其中

(3.1) $ \begin{equation} i = u(N)2^{N-1}+u(N-1)2^{N-2}+\cdots+u(2)2^{1}+u(1)2^{0}\in\{0, 1, 2, \cdots, 2^N-1\}. \end{equation} $

定义3.3  设$ \boldsymbol{ P }_{N}^{v_{i}(t)}\in\{\boldsymbol{ P }_t, \boldsymbol{ Q }_t\}, t = 1, 2, \cdots, N $, 称

$ \boldsymbol{ {P} }_{N}^{v_{i}} = \boldsymbol{ {P} }_{N}^{v_{i}(N)}\cdots\boldsymbol{ {P} }_{N}^{v_{i}(2)} \boldsymbol{ {P} }_{N}^{v_{i}(1)} = \prod\limits_{t = 1}^{N}\boldsymbol{ {P} }_{N}^{v_{i}(t)} $

为一条硬币演化序列.

定义3.3中, $ \boldsymbol{ P }_{N}^{-v_{i}(t)}+\boldsymbol{ P }_{N}^{v_{i}(t)} = \boldsymbol{ C }_t, t = 1, \cdots, N $. 用硬币演化序列$ \boldsymbol{ P }_{N}^{v_{i}} $表示系统经过$ N $次演化后达到的位置为$ \sum\limits_{t = 1}^Nv_{i}(t) $. 硬币演化序列与移动路径是一一对应的, 记所有硬币演化序列为$ {\cal P}_N $, 称为$ N $级路径集, 每个元素简称为路径.

在上述定义基础上, 非齐次量子游荡的酉算子也可以表示为

(3.2) $ \begin{equation} \boldsymbol{ {C} }_0 = \boldsymbol{ {E} }, \boldsymbol{ {C} }_1 = \boldsymbol{ {P} }_1+\boldsymbol{ {Q} }_1, \cdots, \prod\limits_{t = 1}^N\boldsymbol{ {C} }_t = \sum\limits_{i = 0}^{2^N-1}\boldsymbol{ {P} }_{N}^{v_i} = \sum\limits_{i = 0}^{2^N-1}\prod\limits_{t = 1}^N\boldsymbol{ {P} }_{N}^{v_i(t)}. \end{equation} $

经过$ N $次演化后非齐次量子游荡的位置概率分布有以下两个计算表达式

(3.3) $ \begin{equation} P(X = x) = \||\psi_{N}(x)\rangle\|^2 = |\psi^L_{N}(x)|^2+|\psi^R_{N}(x)|^2, x\in{\Bbb Z}, \end{equation} $

(3.4) $ \begin{equation} P(X = x) = \int^{\pi}_{-\pi} \frac{{\rm d}\zeta}{2\pi}{\rm e}^{{\rm i}\zeta x}(\prod\limits^N_{t = 1} \boldsymbol{ {C} }_t(\zeta)\widehat{\psi}_{0}(\zeta))^{*}\int^{\pi}_{-\pi} \frac{{\rm d}\xi}{2\pi} {\rm e}^{-{\rm i}\xi x}\prod\limits^N_{t = 1}\boldsymbol{ {C} }_t(\xi)\widehat{\psi}_{0}(\xi). \end{equation} $

(3.3)式是将$ N $个酉算子$ \boldsymbol{ U }_t $依次作用在量子初态上, 得到量子终态后进行量子测量计算出系统处于每个位置的概率. (3.4)式是利用傅里叶变换将概率分布的计算转化为积分计算问题. 在第5节中, 将提出伊藤公式对$ \prod\limits^N_{t = 1}\boldsymbol{ C }_t({\xi}) $进行分解计算. 系统到达某个位置是由一系列具有共同特征的硬币演化序列叠加作用在量子初态上得到的, 它包括具有相同的向左、向右演化算子个数. 基于此, 下面对硬币演化序列进行归纳计算, 由此给出位置概率分布的计算公式.

定义3.4  任意给定非齐次量子游荡的一条路径$ \boldsymbol{ P }_{N}^{v}\in{\cal P}_N $, 将同类型向左和向右演化算子进行归纳划分, 称之为对$ \boldsymbol{ P }_{N}^{v} $的一种分块, 用$ |\boldsymbol{ P }_{N}^{v}| $表示路径$ \boldsymbol{ P }_{N}^{v} $的分块数.

令$ \nabla^2\omega = (\nabla\omega(1)-\nabla\omega(0), \nabla\omega(2) -\nabla\omega(1), \cdots , \nabla\omega(n)-\nabla\omega(n-1))^T, \nabla\omega(0) = 0. $

引理3.1  任意给定非齐次量子游荡的一条路径$ \boldsymbol{ P }_{N}^{v} = \boldsymbol{ P }_{N}^{v(N)} \cdots \boldsymbol{ P }_{N}^{v(2)} \boldsymbol{ P }_{N}^{v(1)}\in{\cal P}_N $, 则有

(3.5) $ \begin{equation} \boldsymbol{ {P} }_{N}^{v} = (-1)^{\lfloor\frac{s}{2}\rfloor}(\cot\theta_1)^{Mod(s, 2)}\prod\limits_{t = 1}^{N}\cos\theta_t (\tan\theta_t)^{\delta(\nabla^2\omega(t))}\frac{\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)}}{tr(\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)})}, \end{equation} $

其中$ s = \sum\limits_{t = 1}^N\delta(\nabla^2\omega(t)) $, $ \delta(\cdot) $表示示性函数, $ tr(\cdot) $表示矩阵的迹.

证  路径$ \boldsymbol{ P }_{N}^{v} $可分为4类. 下面将对每一类路径进行归纳计算, 其中记$ l $为每条路径中所包含的向右演化算子个数.

(1) $ \boldsymbol{ P }_{N}^{v} = {\cal P}_m{\cal Q}_{m}{\cal P}_{m-1}{\cal Q}_{n-1}\cdots {\cal P}_1{\cal Q}_{1}\in{\cal P}_N $.

该类型将路径划分为$ 2m $块, 每一块均含有若干个向左演化算子或向右演化算子, 记$ {\cal P}_i{\cal Q}_{i} = \prod\limits_{t = 1}^{s_i}\boldsymbol{ P }_t\prod\limits_{t = 1}^{r_i}\boldsymbol{ Q }_t, s_i, r_i\in{\Bbb Z^+}, i = 1, 2, \cdots, m $. 按先后顺序, $ \boldsymbol{ P }_{N}^{v(N)} = \boldsymbol{ P }_N, \boldsymbol{ P }_{N}^{v(1)} = \boldsymbol{ Q }_1 $. 经过归纳计算得到如下式子

(3.6) $ \begin{eqnarray} \boldsymbol{ {P} }_{N}^{v}& = &(-1)^{l-m}\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\left( \begin{array}{cc} 1 & -\cot\theta_1\\ 0 & 0 \\ \end{array} \right){}\\ & = &(-1)^{l-m}\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\frac{\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)}}{tr(\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)})}. \end{eqnarray} $

(2) $ \boldsymbol{ P }_{N}^{v} = {\cal Q}_{m+1}{\cal P}_m{\cal Q}_{m}{\cal P}_{m-1}{\cal Q}_{m-1}\cdots {\cal P}_1{\cal Q}_{1}\in{\cal P}_N $.

该类型将路径划分为$ 2m+1 $块, 每一块均含有若干个向左演化算子或向右演化算子, 记$ {\cal Q}_{m+1} = \prod\limits_{t = 1}^{r_{m+1}}\boldsymbol{ Q }_t, {\cal P}_i{\cal Q}_{i} = \prod\limits_{t = 1}^{s_i}\boldsymbol{ P }_t\prod\limits_{t = 1}^{r_i}\boldsymbol{ Q }_t, s_i, r_i\in{\Bbb Z^+}, i = 1, 2, \cdots, m $. 按先后顺序, $ \boldsymbol{ P }_{N}^{v(N)} = \boldsymbol{ Q }_N, $$ \boldsymbol{ P }_{N}^{v(1)} = \boldsymbol{ Q }_1 $. 经过归纳计算得到如下式子

(3.7) $ \begin{eqnarray} \boldsymbol{ {P} }_{N}^{v}& = &(-1)^{l-m}\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\left( \begin{array}{cc} 0 & 0\\ -1 & \cot\theta_1\\ \end{array} \right){}\\ & = &(-1)^{l-m}\cot\theta_1\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\frac{\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)}}{tr(\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)})}. \end{eqnarray} $

(3) $ \boldsymbol{ P }_{N}^{v} = {\cal Q}_{m}{\cal P}_{m}{\cal Q}_{m-1}{\cal P}_{m-1}\cdots {\cal Q}_{1}{\cal P}_{1}\in{\cal P}_N $.

该类型将路径划分为$ 2m $块, 每一块均含有若干个向左演化算子或向右演化算子$ {\cal Q}_{i}{\cal P}_i = \prod\limits_{t = 1}^{r_i}\boldsymbol{ Q }_t\prod\limits_{t = 1}^{s_i}\boldsymbol{ P }_t, s_i, r_i\in{\Bbb Z^+}, i = 1, 2, \cdots, m $. 按先后顺序, $ \boldsymbol{ P }_{N}^{v(N)} = \boldsymbol{ Q }_N, $$ \boldsymbol{ P }_{N}^{v(1)} = \boldsymbol{ P }_1 $. 经过归纳计算得到如下式子

(3.8) $ \begin{eqnarray} \boldsymbol{ {P} }_{N}^{v}& = &(-1)^{l-m}\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\left( \begin{array}{cc} 0 & 0\\ \cot\theta_1 &1\\ \end{array} \right){}\\& = &(-1)^{l-m}\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\frac{\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)}}{tr(\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)})}. \end{eqnarray} $

(4) $ \boldsymbol{ P }_{N}^{v} = {\cal P}_{m+1}{\cal Q}_{m}{\cal P}_{m} {\cal Q}_{m-1}{\cal P}_{m-1} \cdots{\cal Q}_{1}{\cal P}_{1}\in{\cal P}_N $.

该类型将路径划分为$ 2m+1 $块, 每一块均含有若干个向左演化算子或向右演化算子$ {\cal P}_{m+1} = \prod\limits_{t = 1}^{s_{m+1}}\boldsymbol{ P }_t, {\cal Q}_{i}{\cal P}_i = \prod\limits_{t = 1}^{r_i}\boldsymbol{ Q }_t\prod\limits_{t = 1}^{s_i}\boldsymbol{ P }_t, s_i, r_i\in{\Bbb Z^+}, i = 1, 2, \cdots, m $. 按先后顺序, $ \boldsymbol{ P }_{N}^{v(N)} = \boldsymbol{ P }_N, $$ \boldsymbol{ P }_{N}^{v(1)} = \boldsymbol{ P }_1 $. 经过归纳计算得到如下式子

(3.9) $ \begin{eqnarray} \boldsymbol{ {P} }_{N}^{v}& = &(-1)^{l-m}\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\left( \begin{array}{cc} \cot\theta_1 &1\\ 0 & 0\\ \end{array} \right){}\\ & = &(-1)^{l-m}\cot\theta_1\prod\limits_{t = 1}^{N}\cos\theta_t \tan^{\delta(\nabla^2\omega(t))}\theta_t\frac{\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)}}{tr(\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)})}. \end{eqnarray} $

由于到达同个位置的路径会具有相同向左或向右演化算子个数, 因此对式子(3.6)–(3.9)均去除符号$ (-1)^{l} $使每个式子符号变为$ (-1)^{m} $, 这对位置概率分布计算不会产生影响. 进一步, 对(3.6)–(3.9)式归纳化简如下

$ \boldsymbol{ {P} }_{N}^{v} = (-1)^{\lfloor\frac{s}{2}\rfloor}(\cot\theta_1)^{Mod(s, 2)}\prod\limits_{t = 1}^{N} \cos\theta_t (\tan\theta_t)^{\delta(\nabla^2\omega(t))}\frac{\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)}}{tr(\boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)})}, $

其中$ s = \sum\limits_{t = 1}^N\delta(\nabla^2\omega(t)) $.证毕.

推论3.1  任意给定齐次量子游荡的一条路径$ \boldsymbol{ P }_{N}^{v} = \boldsymbol{ P }_{N}^{v(N)}\cdots\boldsymbol{ P }_{N}^{v(2)} \boldsymbol{ P }_{N}^{v(1)}\in{\cal P}_N $, 则有

(3.10) $ \begin{equation} \boldsymbol{ {P} }_{N}^{v} = \frac{\cos^n\theta({\rm i} \tan\theta)^{2s}}{tr(\boldsymbol{ {P} }_{N}^{v(n)}\boldsymbol{ {P} }_{1}^{v(1)})} \boldsymbol{ {P} }_{N}^{v(N)}\boldsymbol{ {P} }_{1}^{v(1)}, \end{equation} $

其中$ s = \lfloor\frac{\sum\limits_{t = 1}^N\delta(\nabla^2\omega(t))}{2}\rfloor $.

定理3.1  给定非齐次量子游荡的一个量子初态$ |\psi_0\rangle = |\psi_0(0)\rangle\otimes|0\rangle $, $ {\cal P}_{N, k} $是含有$ k $个向左演化算子的$ N $级路径子集, 该集合元素个数为$ C^k_N $, $ X $是位置随机变量, 则系统处于位置$ x = N-2k $的概率为

(3.11) $ \begin{equation} P(X = N-2k) = \bigg\|\sum\limits_{i = 1}^{C^k_N}\boldsymbol{ {P} }_{N, k}^{v_i}|\psi_0\rangle \bigg\|^2, \boldsymbol{ {P} }_{N, k}^{v_i}\in {\cal P}_{N, k}, k = 0, 1, 2, \cdots, N. \end{equation} $

证  给定非齐次量子游荡的一个量子初态$ |\psi_0\rangle $, 系统经过$ N $次演化后所处的位置取值范围为$ \{-N, -N+2, \cdots, N-2, N\} $, 量子态演化为$ |\psi_N\rangle $, 展开为

$ \begin{eqnarray*} |\psi_N\rangle& = &\prod\limits^N_{t = 1}\boldsymbol{ {U} }_t|\psi_0\rangle = \prod\limits^N_{t = 1}\boldsymbol{ {U} }_t|\psi_0(0)\rangle\otimes|0\rangle = \sum\limits_{x = -N}^{N}|\psi_N(x)\rangle\otimes|x\rangle\\ & = &\sum\limits_{x = -N}^{N}\prod\limits_{t = 1}^N\boldsymbol{ {C} }_t|\psi_0(0)\rangle\otimes|x\rangle = \sum\limits_{x = -N}^{N}\sum\limits_{i = 0}^{2^N-1}\boldsymbol{ {P} }_{N}^{v_i}|\psi_0(0)\rangle\otimes|x\rangle\\ & = &\sum\limits_{k = 0}^{N}\sum\limits_{i = 1}^{C^k_N}\boldsymbol{ {P} }_{N, k}^{v_i}|\psi_0(0)\rangle\otimes|N-2k\rangle, \ {其中\ }x = N-2k. \end{eqnarray*} $

因而, 量子系统位于位置$ N-2k $的概率计算为

$ P(X = N-2k) = \langle\psi_N|(\boldsymbol{ {I} }\otimes|N-2k\rangle\langle N-2k|) |\psi_N\rangle = \bigg\|\sum\limits_{i = 1}^{C^k_N}\boldsymbol{ {P} }_{N, k}^{v_i}|\psi_0(0)\rangle\bigg\|^2. $

其中$ \boldsymbol{ P }_{N, k}^{v_i}\in {\cal P}_{N, k} $. 证毕.

利用该定理可推出齐次量子游荡的位置概率分布. 令

$ \boldsymbol{ {P} }_{00} = \left( \begin{array}{ccc} 1 & \tan\theta \\ 0 & 0 \\ \end{array}\right), \ \boldsymbol{ {P} }_{11} = \left( \begin{array}{ccc} 0 & 0 \\ -\tan\theta & 1 \\ \end{array} \right), \ \boldsymbol{ {P} }_{01} = \left( \begin{array}{ccc} 1 & -\cot\theta \\ 0 & 0 \\ \end{array} \right), \ \boldsymbol{ {P} }_{10} = \left( \begin{array}{ccc} 0 & 0 \\ \cot\theta & 1 \\ \end{array} \right). $

考虑一类超几何级数$ {_2F}_1(a, b;c;z) = \sum\limits_{k = 0}^{\infty}\frac{a^{(k)}b^{(k)}}{c^{(k)}}\frac{z^k}{k!} $, 并令$ H(\alpha, \beta, \gamma) = {_2F}_1(\alpha+1-k, \beta+1-l; $$ \gamma+1;-\tan^2\theta)(-\tan^2\theta) $, $ k_1 = (k-1)H(1, 0, 1), k_2 = (l-1)H(0, 1, 1), k_3 = H(0, 0, 0) $.

推论3.2  给定齐次量子游荡的一个量子初态$ |\psi_0\rangle $, 位置$ x $的对应的路径具有$ k $个向左演化算子和$ l $个向右演化算子, 则系统处于位置$ x $的概率为

(3.12) $ \begin{equation} P(X = x) = \left\{ \begin{array}{ll} \cos^{2N}\theta\|\boldsymbol{ {P} }_{00}|\psi_0\rangle\|^2, & x = -N, \\ \cos^{2N}\theta(\|(k_{1}\boldsymbol{ {P} }_{00} + k_3\boldsymbol{ {P} }_{01})|\psi_0 \rangle\|^2+\|(k_{2}\boldsymbol{ {P} }_{11}+k_3\boldsymbol{ {P} }_{10})|\psi_0\rangle\|^2), & -N<x<N, \\ \cos^{2N}\theta\|\boldsymbol{ {P} }_{11}|\psi_0\rangle\|^2, & x = N. \end{array} \right. \end{equation} $

证  根据定理3.1, 系统处于位置$ x $的概率为

$ P(X = x) = \bigg\|\sum\limits_{i = 1}^{C^k_N}\boldsymbol{ {P} }_{N, k}^{v_i}|\psi_0\rangle\bigg\|^2, $

其中到达位置$ x $的路径数共有$ {C^k_N} $条, 式子中$ k $表示路径$ \boldsymbol{ P }_{N, k}^{v_i} $包含了$ k $个向右演化算子$ \boldsymbol{ Q }_{t} $.

同时, 对于齐次量子游荡, 根据推论3.1可知其路径只与分块数$ s $和首尾向左或向右演化算子$ \boldsymbol{ P }_{N}^{v(N)}, \boldsymbol{ P }_{1}^{v(1)} $有关. 这意味着相同分块数、相同的首尾向左或向右演化算子的两条路径$ \boldsymbol{ P }_{N}^{v_1}, \boldsymbol{ P }_{N}^{v_2} $均有$ \|\boldsymbol{ P }_{N}^{v_1}|\psi_0\rangle\|^2 = \|\boldsymbol{ P }_{N}^{v_2}|\psi_0\rangle\|^2. $不妨称这样两条路径是等效的. 故按等效划分, 则有

$ P(X = x) = \bigg\|\sum\limits_{s = 1}^{2k+1}M_s\boldsymbol{ {P} }_{N, s}^{v}|\psi_0\rangle\bigg\|^2, $

其中$ M_s $表示分块数为$ s $的路径个数. 进一步, 利用超几何级数计算可得

$ \sum\limits_{s = 1}^{2k+1}M_s\boldsymbol{ {P} }_{N, s}^{v} = (k_1+1)\cos^N\theta\boldsymbol{ {P} }_{00} +(k_2+1)\cos^N\theta\boldsymbol{ {P} }_{11}+k_3\cos^N\theta(\boldsymbol{ {P} }_{10}+\boldsymbol{ {P} }_{01}). $

因而

$ \begin{eqnarray*} P(X = x) = \left\{ \begin{array}{ll} \cos^{2N}\theta\|\boldsymbol{ {P} }_{00}|\psi_0\rangle\|^2, &x = -N, \\ \cos^{2N}\theta(\|(k_{1}\boldsymbol{ {P} }_{00} + k_3\boldsymbol{ {P} }_{01})|\psi_0\rangle\|^2+\|(k_{2}\boldsymbol{ {P} }_{11}+k_3\boldsymbol{ {P} }_{10})|\psi_0\rangle\|^2), &-N<x<N, \\ \cos^{2N}\theta\|\boldsymbol{ {P} }_{11}|\psi_0\rangle\|^2, &x = N. \end{array} \right. \end{eqnarray*} $

证毕.

定义4.1  称$ \tilde{\psi}_{t}(\xi) = (0\cdots 0 \widehat{\psi}_{t}(\xi)^T(-t) \cdots \widehat{\psi}_{t}(\xi)^T(0) \cdots \widehat{\psi}_{t}(\xi)^T(t) 0\cdots 0)^T $为一个完全状态波函数, $ T $为转置运算, $ t = 0, 1, \cdots, N $.

当$ t = 0 $时, 称$ \tilde{\psi}_{0}(\xi) = (0\cdots0 \widehat{\psi}_{0}(\xi)^T(0) 0\cdots0)^T $为一个初始完全状态波函数. 所有完全状态波函数组成的集合对于向量的普通加法和数乘运算构成数域$ {\Bbb C}^2 $上一个有限维线性空间$ V_{4N+2} $. 取该线性空间的一组基为$ \tilde{\varepsilon}_0(-N), \tilde{\varepsilon}_1(-N), \tilde{\varepsilon}_0(-N+1), \tilde{\varepsilon}_1(-N+1), \cdots, \tilde{\varepsilon}_0(N), \tilde{\varepsilon}_1(N) $, 其中$ \tilde{\varepsilon}_0(x) = (0, \cdots, 0, \widehat{\varepsilon}_0(\xi)^T(x), 0, \cdots, 0)^T, \tilde{\varepsilon}_1(x) = (0, \cdots, 0, \widehat{\varepsilon}_1(\xi)^T(x), 0, \cdots, 0)^T, \widehat{\varepsilon}_0(\xi)(x) = (1, 0)^T, \widehat{\varepsilon}_1(\xi)(x) = (0, 1)^T, x\in\{-N, -N+1, \cdots, N\} $. 任取一个向量$ \tilde{\psi}_t(\xi)\in V_{4N+2} $, 则该向量可以表示为$ \tilde{\psi}_t(\xi) = (\widehat{\psi}_t(\xi)(x)) = \sum\limits_{x = -N}^{N}(\widehat{\psi}_t^L(\xi) \tilde{\varepsilon}_0(x) +\widehat{\psi}_t^R(\xi)\tilde{\varepsilon}_1(x)), $其中$ 0\leq\|\widehat{\psi}_t(\xi)(x)\|\leq 1, t = 0, 1, \cdots, N. $

$ \tilde{\psi}_{t}(\xi) $与$ \tilde{\psi}_{t-1}(\xi) $关系为

(4.1) $ \begin{equation} \left\{ \begin{array}{lll} \widehat{\psi}_{t}(\xi)(-x) & = \widehat{\psi}_{t}(\xi)(x) = 0, & |x|>t, \\ \widehat{\psi}_{t}(\xi)(-x) & = {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{t}\widehat{\psi}_{t-1}(\xi)(-x+1), & x = -t, \\ \widehat{\psi}_{t}(\xi)(x) & = {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}\widehat{\psi}_{t-1}(\xi)(x-1), & x = t, \\ \widehat{\psi}_{t}(\xi)(x) & = {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{t}\widehat{\psi}_{t-1}(\xi)(x+1)+{\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}\widehat{\psi}_{t-1}(\xi)(x-1), & |x|<t.\\ \end{array} \right. \end{equation} $

(4.1)式用矩阵表示为$ \tilde{\psi}_{t}(\xi) = \tilde{\boldsymbol{ C }}(\xi)\tilde{\psi}_{t-1}(\xi), t = 0, 1, \cdots, N $, 其中

(4.2) $ \begin{equation} \widetilde{\boldsymbol{ {C} }}(\xi) = \left( \begin{array}{cccccccccc} 0 & {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{N-1} & \cdots & 0 & 0 & 0 & \cdots & 0 & {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_N \\ {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{N} & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{N-1} & \cdots & 0 & 0 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \cdots & {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{2}& 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 0 & {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{2}& 0 & {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{2} & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 0 & {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 0 & 0 & {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{2} & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \cdots & 0 & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{N-1} & 0 \\ 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{N} \\ {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{N} & 0 & \cdots & 0 & 0 & 0 & \cdots & {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{N-1} & 0 \\ \end{array} \right). \end{equation} $

定义4.2   称$ \widetilde{\boldsymbol{ C }}(\xi) $是一个完全硬币翻转算子.

运用算子$ \widetilde{\boldsymbol{ C }}(\xi) $, 有$ \tilde{\psi}_{t}(\xi) = \tilde{\boldsymbol{ C }}^t(\xi)\tilde{\psi}_{0}(\xi) $, 当$ \xi = 0 $时, 有$ \tilde{\psi}_{t}(0) = \tilde{\boldsymbol{ C }}^t(0)\tilde{\psi}_{0}(0) $.

定义4.3   定义完全硬币翻转算子$ \tilde{\boldsymbol{ C }}(\xi) $的谱值$ \lambda $为$ \tilde{\boldsymbol{ C }}(\xi)\hat{\psi}_t(\xi) = \lambda\hat{\psi}_t(\xi), \hat{\psi}_t(\xi)\in V_{4N+2} $; 称所有$ \tilde{\boldsymbol{ C }}(\xi) $的谱值组成的集合$ \sigma(\tilde{\boldsymbol{ C }}(\xi)) $为$ \tilde{\boldsymbol{ C }}(\xi) $的谱; 定义$ \tilde{\boldsymbol{ C }}(\xi) $的谱半径为

$ r(\tilde{\boldsymbol{ {C} }}(\xi)) = \max\{\lambda|\lambda\in\sigma(\tilde{\boldsymbol{ {C} }}(\xi))\}. $

性质4.1  完全硬币翻转算子的谱是闭的单位圆盘, 即$ r(\tilde{\boldsymbol{ C }}(\xi))\leq 1 $.

证  设$ \tilde{\boldsymbol{ C }}(\xi) $是作用在$ V_{4N+2} $上的一个有界线性算子, $ t $为系统演化次数, $ t = 0, 1, \cdots, N $. $ \tilde{\boldsymbol{ C }}(\xi) $的谱值表示为$ \tilde{\boldsymbol{ C }}(\xi)\tilde{\psi}_t(\xi) = \lambda\tilde{\psi}_t(\xi), \tilde{\psi}_t(\xi)\in V_{4N+2} $. 记$ \tilde{\boldsymbol{ C }}(\xi) $的第$ x $行为$ \tilde{\boldsymbol{ C }}(\xi)(x) $, 则有

$ \tilde{\boldsymbol{ {C} }}(\xi)(x)\tilde{\psi}_t(\xi) = \lambda\hat{\psi}_t(\xi)(x), x\in\{-N, -N+1, \cdots, N\}. $

将该式子展开

$ \begin{eqnarray*} \left\{\begin{array}{lll} \lambda\hat{\psi}_t(\xi)(x) = 0, & |x|>t, \\ \lambda\hat{\psi}_t(\xi)(x) = {\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_t\hat{\psi}_t(\xi)(x+1), &x = -t, \\ \lambda\hat{\psi}_t(\xi)(x) = {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}\hat{\psi}_t(\xi)(x-1), &x = t, \\ \lambda\hat{\psi}_t(\xi)(x) = {\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}\hat{\psi}_t(\xi)(x-1)+{\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{t}\hat{\psi}_t(\xi)(x+1), & |x|<t.\\ \end{array} \right. \end{eqnarray*} $

对位置$ x $累加得到

$ \begin{eqnarray*} \sum\limits_{x = -N}^{N}\lambda\hat{\psi}_t(\xi)(x) & = &{\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{t}\hat{\psi}_t(\xi)(1-t)+{\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}\hat{\psi}_t(\xi)(t-1) \\ &&+\sum\limits_{x = 1-t}^{t-1}{\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_{t}\hat{\psi}_t(\xi)(x+1)+{\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}\hat{\psi}_t(\xi)(x-1), \end{eqnarray*} $

经化简有

$ ({\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_t-\lambda\boldsymbol{ {I} })\hat{\psi}_t(\xi)(t)+({\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}-\lambda\boldsymbol{ {I} })\hat{\psi}_t(\xi)(-t) +\sum\limits_{x = 1-t}^{t-1}({\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_t+{\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_t-\lambda\boldsymbol{ {I} })\hat{\psi}_t(\xi)(x) = 0, $

所以有

$ \left\{ \begin{array}{ll} ({\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_t-\lambda\boldsymbol{ {I} })\hat{\psi}_t(\xi)(t) = 0, \\ ({\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_{t}-\lambda\boldsymbol{ {I} })\hat{\psi}_t(\xi)(-t) = 0, \\ ({\rm e}^{-{\rm i}\xi}\boldsymbol{ {P} }_t+{\rm e}^{{\rm i}\xi}\boldsymbol{ {Q} }_t-\lambda\boldsymbol{ {I} })\hat{\psi}_t(\xi)(x) = 0, \end{array} \right. $

其中$ x = 1-t, 2-t, \cdots, t-1 $.

对于任意的$ x\in[-t, t] $, 有$ 0\leq \|\hat{\psi}_t(\xi)(x)\|\leq 1 $, 所以存在$ [-t, t] $的子集$ X $使得$ \forall x\in X $, 有$ \|\hat{\psi}_t(\xi)(x)\|\neq 0 $. 因而有$ |{\rm e}^{-{\rm i}\xi}\boldsymbol{ P }_t+{\rm e}^{{\rm i}\xi}Q_t-\lambda\boldsymbol{ I }| = 0, |{\rm e}^{-{\rm i}\xi}\boldsymbol{ P }_t-\lambda\boldsymbol{ I }| = 0, |{\rm e}^{{\rm i}\xi}\boldsymbol{ Q }_{t}-\lambda\boldsymbol{ I }| = 0 $. 计算解得$ \lambda^2+{\rm e}^{{\rm i}\xi}\cos\theta_t\lambda = 0, \lambda^2-{\rm e}^{-{\rm i}\xi}\cos\theta_{t}\lambda = 0, \lambda^2+({\rm e}^{{\rm i}\xi}-{\rm e}^{-{\rm i}\xi})\cos\theta_t\lambda-1 = 0 $. 得到谱值为$ 0, -{\rm e}^{{\rm i}\xi}\cos\theta_t; 0, {\rm e}^{-{\rm i}\xi}\cos\theta_{t}; -{\rm i}(\sin(\xi)\cos\theta_t\pm\sqrt{1 - \sin^2(\xi)\cos^2\theta_t}i)\triangleq (-{\rm i}){\rm e}^{\pm {\rm i}\eta_t} $, 其中$ \eta_t = \arccos(\sin(\xi)\cos(\theta_t)) $. 因而有$ \max|\lambda_{1, 2}| = 1 $. 所以$ r(\tilde{\boldsymbol{ C }}(\xi)) = 1 $. 证毕.

性质4.1给出完全硬币翻转算子$ \tilde{\boldsymbol{ C }}(\xi) $的谱值和谱半径. 其中, 谱值为

(4.3) $ 0,-\mathrm{e}^{\mathrm{i} \xi} \cos \theta_{t} ; 0, \mathrm{e}^{-\mathrm{i} \xi} \cos \theta_{t} ;(-\mathrm{i}) \mathrm{e}^{\pm \mathrm{i} \eta_{t}}, t=0,1, \cdots, N $

同时, 对谱向量讨论如下.

(1) 当谱值$ \lambda = (-{\rm i}){\rm e}^{\pm {\rm i}\eta_t} $时, 谱向量为

(4.4) $\begin{equation} \hat{\psi}_t(\xi)(x) = \left\{ \begin{array}{ll} 0, & |x|>t,\\ { } \frac{{\rm e}^{-{\rm i}\xi}}{\lambda}\rm{\bf{ {P} }}_t{\psi}(\xi)(-x+1), & x = -t,\\ { } \frac{{\rm e}^{{\rm i}\xi}}{\lambda}\rm{\bf{ {Q} }}_t{\psi}(\xi)(x-1), & x = t,\\ \rm{\bf{ {T} }}^x{\psi}(\xi)(0),& |x|<t.\\ \end{array} \right. \end{equation} $

其中

(4.5) $ \begin{equation} \boldsymbol{ {T} } = \frac{{\rm e}^{{\rm i}\xi}}{\lambda \cos\theta_t}\left( \begin{array}{cc} \lambda^2-\sin\theta^2_t {\quad} & \sin\theta_t \cos\theta_{t} \\ \sin\theta_{t}\cos\theta_{t}{\quad} & -\cos\theta^2_t \\ \end{array} \right). \end{equation} $

(2) 当谱值$ \lambda = -{\rm e}^{{\rm i}\xi}\cos\theta_t, {\rm e}^{-{\rm i}\xi}\cos\theta_t $时, 谱向量为

(4.6) $ \begin{equation} \hat{\psi}_t(\xi)(x) = \left\{ \begin{array}{ll} 0, & |x|>t, \\ { } \frac{{\rm e}^{-{\rm i}\xi}}{\lambda}\boldsymbol{ {P} }_t\hat{\psi}_t(\xi)(-x+1), & x = -t, \\ { } \frac{{\rm e}^{{\rm i}\xi}}{\lambda}\boldsymbol{ {Q} }_t\hat{\psi}_t(\xi)(x-1), & x = t, \\ \boldsymbol{ {T} }^x{\psi}(\xi)(0), & |x|<t.\\ \end{array} \right. \end{equation} $

(3) 当谱值$ \lambda = 0 $时, 谱向量为

(4.7) $ \begin{equation} \hat{\psi}_t(\xi)(x) = \left\{ \begin{array}{ll} 0, & |x|\neq t-1, \\ (a, b)^T, & x = -t+1, \\ (c, d)^T, & x = t-1, \\ \end{array} \right. \end{equation} $

其中$ \cos\theta a +\sin\theta b = 0, \sin\theta c-\cos\theta d = 0. $

定义5.1^[19]  设随机过程$ \{X(t), t\geq 0\} $满足

(5.1) $ \begin{equation} \begin{array}{l} {\rm d}(X(t)) = \mu(t, X(t)){\rm d}t+\sigma(t, X(t)){\rm d}W_t, \\ { } X(t) = X(t_0)+\int_{t_0}^t\mu(t, X(t)){\rm d}t+\int_{t_0}^t\sigma(t, X(t)){\rm d}W_t, \end{array} \end{equation} $

则称它为一个伊藤过程, 称(5.1)式为伊藤随机微分方程.

伊藤过程也称为扩散过程. 在该定义中, $ X(t) $表示位移随机变量, $ {\rm d}(X(t)) = X(t)-X(t-1) $表示位移增量, $ \mu(t, X(t)) $是漂移率, $ \mu(t, X(t)){\rm d}t $表示每单位时间内的漂移量, $ \sigma^2(t, X(t)) $表示方差率, $ {\rm d}W_t $表示噪声服从标准布朗过程, $ \sigma(t, X(t)){\rm d}W_t $表示噪声引起的随机波动量.

引理5.1  设$ X(t) $是一个伊藤过程, 给定一个具有二阶连续可微的二元函数$ Y(t) = f(t, X(t))\in C^2([0, \infty]) $, 则随机过程$ Y(t) = f(t, X(t)) $满足

(5.2) $ \begin{eqnarray} {\rm d}Y(t)& = & \bigg(\frac{\partial f(t, X(t))}{\partial t}+\mu(t, X(t))\frac{\partial f(t, X(t))}{\partial x}+\frac{\sigma^2(t, X(t))}{2}\frac{\partial^2 f(t, X(t))}{\partial x^2} \bigg){\rm d}t {}\\ &&+\sigma(t, X(t))\frac{\partial f(t, X(t))}{\partial x}{\rm d}W_t. \end{eqnarray} $

该引理称为伊藤引理. 其中, $ \frac{\partial f(t, X(t))}{\partial t}+\mu(t, X(t))\frac{\partial f(t, X(t))}{\partial x} $$ +\frac{\sigma^2(t, X(t))}{2}\frac{\partial^2 f(t, X(t))}{\partial x^2} $表示漂移率, $ (\sigma(t, X(t))\frac{\partial f(t, X(t))}{\partial x})^2 $表示方差率. 文献[19-21]研究了齐次量子游荡的伊藤引理和伊藤公式, 下面给出非齐次量子游荡的伊藤公式.

定理5.1

(5.3) $ \begin{eqnarray} \prod\limits^N_{t = 1}\boldsymbol{ {C} }_t({\xi})& = &\sum\limits^{2^N-1}_{k = 0}{\rm e}^{{\rm i}\xi\omega_k(t)}\boldsymbol{ {P} }_{N}^{v_{k}} {}\\ & = &\prod\limits^N_{t = 1}\boldsymbol{ {C} }_t+ {\rm i}\sin{\xi}\sum\limits^{2^N-1}_{k = 0}\sum\limits^{N-1}_{t = 0}{\rm e}^{{\rm i}\xi\omega_k(t)}(\omega_k(t+1)-\omega_k(t))\boldsymbol{ {P} }_{N}^{v_{k}}{}\\ &&+(\cos{\xi}-1)\sum\limits^{2^N-1}_{k = 0}\sum\limits^{N-1}_{t = 0}{\rm e}^{{\rm i}\xi\omega_k(t)}\boldsymbol{ {P} }_{N}^{v_{k}}, \end{eqnarray} $

其中$ \prod\limits^N_{t = 1}\boldsymbol{ C }_t = \sum\limits^{2^N-1}_{k = 0}\boldsymbol{ P }_{N}^{v_{k}} $, $ \boldsymbol{ P }_{N}^{v_{k}} = \prod\limits_{t = 1}^{N}\boldsymbol{ P }_{t}^{v_{k}(t)} $.

证  首先, 取$ X(t) = \omega_k(t) $表示第$ k $条位置路径在$ t $时刻的位置. 令

$ Y(t) = f(t, X(t)) = {\rm e}^{{\rm i}\xi\omega_k(t)}, k = 0, 1, \cdots, 2^N-1, t = 1, 2, \cdots, N. $

根据(5.2)式, 位移增量为

$ {\rm d}Y(t) = f(t, X(t))-f(t-1, X(t-1)) = {\rm e}^{{\rm i}\xi\omega_k(t)}-{\rm e}^{{\rm i}\xi\omega_k(t-1)}; $

漂移率为

$ \frac{\partial f(t, X(t))}{\partial t}+\mu(t, X(t))\frac{\partial f(t, X(t))}{\partial x}+\frac{\sigma^2(t, X(t))}{2}\frac{\partial^2 f(t, X(t))}{\partial x^2}; $

方差率为

$ (\sigma(t, X(t))\frac{\partial f(t, X(t))}{\partial x})^2. $

其中

$ \frac{\partial f(t, X(t))}{\partial t} = \frac{\partial {\rm e}^{{\rm i}\xi\omega_k(t)}}{\partial t} = 0, $

$ \mu(t, X(t)) = \frac{\{\omega_k(t)+1-\omega_k(t)\}+\{\omega_k(t)-1-\omega_k(t)\}}{2} = 0, $

$ \frac{\partial f(t, X(t))}{\partial x} = \frac{{\rm e}^{{\rm i}\xi(\omega_k(t)+1)}-{\rm e}^{{\rm i}\xi(\omega_k(t)-1)}}{(\omega_k(t)+1)-(\omega_k(t)-1)} = \frac{{\rm e}^{{\rm i}\xi(\omega_k(t)+1)}-{\rm e}^{{\rm i}\xi(\omega_k(t)-1)}}{2}, $

$ \sigma^2(t, X(t)) = \frac{1}{2}\{(\omega_k(t)+1)-\omega_k(t)\}^2+\frac{1}{2}\{(\omega_k(t)-1)-\omega_k(t)\}^2 = 1, $

$ \begin{eqnarray*} \frac{\partial^2 f(t, X(t))}{\partial x^2}& = &\frac{\{{\rm e}^{{\rm i}\xi(\omega_k(t)+1)} -{\rm e}^{{\rm i}\xi\omega_k(t)}\}-\{{\rm e}^{{\rm i}\xi\omega_k(t)} -{\rm e}^{{\rm i}\xi(\omega_k(t)-1)}\}}{(\omega_k(t)+1)-\omega_k(t)}\\ & = &{\rm e}^{{\rm i}\xi(\omega_k(t)+1)}-2{\rm e}^{{\rm i}\xi\omega_k(t)}+{\rm e}^{{\rm i}\xi(\omega_k(t)-1)}, \end{eqnarray*} $

$ {\rm d}t = 1, {\qquad} {\rm d}W_t = \omega_k(t+1)-\omega_k(t). $

将上式代入(5.2)式, 得到

$ \begin{eqnarray*} {\rm e}^{{\rm i}\xi\omega_k(t)}-{\rm e}^{{\rm i}\xi\omega_k(t-1)}& = &\frac{1}{2}\{{\rm e}^{{\rm i}\xi(\omega_k(t)+1)}-{\rm e}^{{\rm i}\xi(\omega_k(t)-1)}\} (\omega_k(t)-\omega_k(t-1))\\ &&+\frac{1}{2}\{{\rm e}^{{\rm i}\xi(\omega_k(t)+1)}-2{\rm e}^{{\rm i}\xi\omega_k(t)}+{\rm e}^{{\rm i}\xi(\omega_k(t)-1)}\}\\ & = &{\rm i}\sin\xi {\rm e}^{{\rm i}\xi\omega_k(t)}(\omega_k(t+1)-\omega_k(t))+(\cos\xi-1){\rm e}^{{\rm i}\xi\omega_k(t)}. \end{eqnarray*} $

对$ t = 0, 1, \cdots, N-1 $进行累加, 则有

$ {\rm e}^{{\rm i}\xi\omega_k(N)}-{\rm e}^{{\rm i}\xi\omega_k(0)} = {\rm i}\sin\xi\sum\limits^{N-1}_{t = 0} {\rm e}^{{\rm i}\xi\omega_k(t)}(\omega_k(t)-\omega_k(t-1))+(\cos\xi-1)\sum\limits^{N-1}_{t = 0}{\rm e}^{{\rm i}\xi\omega_k(t)}. $

对该式子两边同时乘以$ \boldsymbol{ P }_{N}^{v_{k}} $并对$ k = 0, 1, \cdots, 2^N-1 $进行累加, 则有

$ \begin{eqnarray*} &&\sum\limits^{2^N-1}_{k = 0}{\rm e}^{{\rm i}\xi\omega_k(N)}\boldsymbol{ {P} }_{N}^{v_{k}}- \sum\limits^{2^N-1}_{k = 0}{\rm e}^{{\rm i}\xi\omega_k(0)}\boldsymbol{ {P} }_{N}^{v_{k}}\\ & = &{\rm i}\sin\xi\sum\limits^{2^N-1}_{k = 0}\sum\limits^{N-1}_{t = 0} {\rm e}^{{\rm i}\xi\omega_k(t)}(\omega_k(t+1)-\omega_k(t))\boldsymbol{ {P} }_{N}^{v_{k}}+(\cos\xi-1)\sum\limits^{2^N-1}_{k = 0}\sum\limits^{N-1}_{t = 0}{\rm e}^{{\rm i}\xi\omega_k(t)}\boldsymbol{ {P} }_{N}^{v_{k}}. \end{eqnarray*} $

化简得到

$ \begin{eqnarray*} \prod\limits^N_{t = 1}\boldsymbol{ {C} }_t({\xi}) & = &\prod\limits^N_{t = 1}\boldsymbol{ {C} }_t+ i\sin{\xi}\sum\limits^{2^N-1}_{k = 0}\sum\limits^{N-1}_{t = 0}{\rm e}^{{\rm i}\xi\omega_k(t)}(\omega_k(t+1)-\omega_k(t))\boldsymbol{ {P} }_{N}^{v_{k}}\\ &&+(\cos{\xi}-1)\sum\limits^{2^N-1}_{k = 0}\sum\limits^{N-1}_{t = 0}{\rm e}^{{\rm i}\xi\omega_k(t)}\boldsymbol{ {P} }_{N}^{v_{k}}. \end{eqnarray*} $

所以结论成立. 证毕.

定理5.1从随机微分方程角度出发推导出非齐次量子游荡的伊藤公式, 给出$ \prod\limits^N_{t = 1}\boldsymbol{ C }_t({\xi}) $的分解式. 当$ \boldsymbol{ C }_i = \boldsymbol{ C }_j, i, j = 1, 2, \cdots, N $时, 即系统是齐次量子游荡, 定理5.1的结论与文献[19]的结论是一致的. 不过与文献[19-21]一样, 伊藤公式(5.3)不但表达式复杂以致难以计算和应用, 而且缺乏数学或物理上的含义. 为此, 下面对伊藤公式进行化简和解释.

首先, 对$ \prod\limits^N_{t = 1}\boldsymbol{ C }_t({\xi}) $化简如下:

(5.4) $ \begin{equation} \prod\limits^N_{t = 1}\boldsymbol{ {C} }_t({\xi}) = \sum\limits^{2^N-1}_{k = 0}\boldsymbol{ {C} }^T_0{\rm e}^{{\rm i}\xi\omega_k}\boldsymbol{ {P} }_{N}^{v_{k}} +\sum\limits^{2^N-1}_{k = 0}\boldsymbol{ {C} }_{1}^T{\rm e}^{{\rm i}\xi\omega_k}\boldsymbol{ {P} }_{N}^{v_{k}} -\sum\limits^{2^N-1}_{k = 0}\boldsymbol{ {C} }_{-1}^T{\rm e}^{{\rm i}\xi\omega_k}\boldsymbol{ {P} }_{N}^{v_{k}}. \end{equation} $

其中$ \boldsymbol{ C }_0 = ({\rm e}^{{\rm i}\xi(0)}, 0, \cdots, 0)^T, \boldsymbol{ C }_{1} = ({\rm e}^{{\rm i}\xi\nabla\omega_k(0)}, {\rm e}^{{\rm i}\xi\nabla\omega_k(1)}, \cdots, {\rm e}^{{\rm i}\xi\nabla\omega_k(N-1)})^T, \boldsymbol{ C }_{-1} = ({\rm e}^{{\rm i}\xi(0)}, {\rm e}^{{\rm i}\xi(0)}, $$ \cdots, {\rm e}^{{\rm i}\xi(0)})^T, {\rm e}^{{\rm i}\xi\omega_k} = ({\rm e}^{{\rm i}\xi\omega_k(0)}, {\rm e}^{{\rm i}\xi\omega_k(1)}, \cdots, {\rm e}^{{\rm i}\xi\omega_k(N-1)})^T $.

其次, 使用完全硬币翻转算子$ \widetilde{\boldsymbol{ C }}(\xi) $对(5.4)式进行矩阵改写. 完全硬币翻转算子$ \widetilde{\boldsymbol{ C }}^N(\xi) $可分解为

(5.5) $ \begin{equation} \widetilde{\boldsymbol{ {C} }}^N(\xi) = \widetilde{\boldsymbol{ {C} }}^N(0)+\widetilde{\boldsymbol{ {C} }}^N_{\leftarrow}(\xi) -\widetilde{\boldsymbol{ {C} }}^N_{\rightarrow }(\xi), \end{equation} $

其中$ \widetilde{\boldsymbol{ C }}^N_{\leftarrow}(\xi) = \sum\limits_{t = 1}^N\widetilde{\boldsymbol{ C }}^{N-t}(0)\widetilde{\boldsymbol{ C }}^t(\xi), \widetilde{\boldsymbol{ C }}^N_{\rightarrow }(\xi) = \sum\limits_{t = 1}^N\widetilde{\boldsymbol{ C }}^{N-t}(\xi)\widetilde{\boldsymbol{ C }}^t(0) $.

引理5.2  在伊藤公式中, 式子$ \prod\limits^N_{t = 1}\boldsymbol{ C }_t({\xi}) $、$ \sum\limits^{2^N-1}_{k = 0}\boldsymbol{ C }^T_0{\rm e}^{{\rm i}\xi\omega_k}\boldsymbol{ P }_{N}^{v_{k}} $、$ \sum\limits^{2^N-1}_{k = 0}\boldsymbol{ C }^T_1{\rm e}^{{\rm i}\xi\omega_k}\boldsymbol{ P }_{N}^{v_{k}} $和$ \sum\limits^{2^N-1}_{k = 0}\boldsymbol{ C }_{-1}^T{\rm e}^{{\rm i}\xi\omega_k} $$ \boldsymbol{ P }_{N}^{v_{k}} $分别对应于完全硬币翻转算子的$ \widetilde{\boldsymbol{ C }}^N(\xi) $、$ \widetilde{\boldsymbol{ C }}^N(0) $、$ \widetilde{\boldsymbol{ C }}^N_{\leftarrow}(\xi) $和$ \widetilde{\boldsymbol{ C }}^N_{\rightarrow }(\xi) $.

经过化简后, 伊藤公式(5.5)作用在初始完成状态波函数$ \widetilde{\psi}_0(\xi) $上就相应改写为

(5.6) $ \begin{equation} \widetilde{\boldsymbol{ {C} }}^N(0)\widetilde{\psi}_0(\xi) +\widetilde{\boldsymbol{ {C} }}^N_{\leftarrow}(\xi)\widetilde{\psi}_0(\xi) -\widetilde{\boldsymbol{ {C} }}^N_{\rightarrow }(\xi)\widetilde{\psi}_0(\xi). \end{equation} $

该式不但表达式简洁, 而且分解式具有一定数学含义. 解释如下

(1) $ \widetilde{\boldsymbol{ C }}^N(0)\widetilde{\psi}_0(\xi)(x) $, 表示未演化前系统处于位置$ x $的初始信息;

(2) $ \widetilde{\boldsymbol{ C }}^N_{\leftarrow}(\xi)\widetilde{\psi}_0(\xi)(x) $, 表示系统在演化过程中所有移入到位置$ x $的路径信息. 这体现在式子$ \sum\limits_{t' = 0}^{N-1}\sum\limits_{k = 0}^{2^N-1}{\rm e}^{{\rm i}\xi(\omega_k(t') +\nabla\omega_k(t'))}\boldsymbol{ P }_N^{v_{k}}\hat{\psi}_0(\xi) $中的$ \omega_k(t')+\nabla\omega_k(t') $, 它表示从当前位置开始向前移入一个单位. 在图论角度来看, 可以用下式计算它的入度$ \delta^2_+(x) $

(5.7) $ \begin{equation} \delta^2_+(x) = \int^{\pi}_{-\pi} \frac{{\rm d}\zeta}{2\pi}{\rm e}^{{\rm i}\zeta x}(\widetilde{\boldsymbol{ {E} }}^N_{\leftarrow}(\zeta)\widehat{\psi}_{0}(\zeta))^{*}\int^{\pi}_{-\pi} \frac{{\rm d}\xi}{2\pi}{\rm e}^{-{\rm i}\xi x}\widetilde{\boldsymbol{ {E} }}^N_{\leftarrow}(\xi)\widehat{\psi}_{0}(\xi), \end{equation} $

其中$ \widetilde{\boldsymbol{ E }}^N_{\leftarrow}(\xi) $是将$ \widetilde{\boldsymbol{ C }}^N_{\leftarrow}(\xi) $中$ \boldsymbol{ P }_t, \boldsymbol{ Q }_t $替换为单位矩阵$ \boldsymbol{ E } $;

(3) $ \widetilde{\boldsymbol{ C }}^N_{\rightarrow }(\xi)\widetilde{\psi}_0(\xi)(x) $, 表示系统在演化过程中所有从位置$ x $移出的路径信息. 这体现在式子$ \sum\limits_{t' = 0}^{N-1}\sum\limits_{k = 0}^{2^N-1}{\rm e}^{{\rm i}\xi(\omega_k(t') -\nabla\omega_k(t'))}\boldsymbol{ P }_N^{v_{k}}\hat{\psi}_0(\xi) $中的$ \omega_k(t')-\nabla\omega_k(t') $, 它表示从当前位置开始向后移出一个单位. 在图论角度来看, 可以用下式计算它的出度$ \delta^2_-(x) $

(5.8) $ \begin{equation} \delta^2_-(x) = \int^{\pi}_{-\pi} \frac{{\rm d}\zeta}{2\pi}{\rm e}^{{\rm i}\zeta x}(\widetilde{\boldsymbol{ {E} }}^N_{\rightarrow }(\zeta)\widehat{\psi}_{0}(\zeta))^{*}\int^{\pi}_{-\pi} \frac{{\rm d}\xi}{2\pi}{\rm e}^{-{\rm i}\xi x}\widetilde{\boldsymbol{ {E} }}^N_{\rightarrow }(\xi)\widehat{\psi}_{0}(\xi), \end{equation} $

其中$ \widetilde{\boldsymbol{ E }}^N_{\rightarrow }(\xi) $是将$ \widetilde{\boldsymbol{ C }}^N_{\leftarrow}(\xi) $中$ \boldsymbol{ P }_t, \boldsymbol{ Q }_t $替换为单位矩阵$ \boldsymbol{ E } $.

例5.1  考虑一个非齐次量子游荡, 量子初态为$ \hat{\psi}_0(\xi)(0) = |\psi_0\rangle $, 初始完全状态波函数为

$ \widetilde{\psi}_0(\xi) = (0, 0, 0, \hat{\psi}_0(\xi)(0), 0, 0, 0), $

取$ N = 3 $, 令$ \boldsymbol{ C }_t = \boldsymbol{ P }_t+\boldsymbol{ Q }_t, t = 1, 2, 3 $, 分别计算$ \widetilde{\boldsymbol{ C }}^N(0)\widetilde{\psi}_0(\xi)(x) $、$ \widetilde{\boldsymbol{ C }}^N_{\leftarrow}(\xi)\widetilde{\psi}_0(\xi)(x) $和$ \widetilde{\boldsymbol{ C }}^N_{\rightarrow }(\xi)\widetilde{\psi}_0(\xi)(x) $如下.

(1)

$ \begin{eqnarray*} &&\widetilde{\boldsymbol{ {C} }}^3(0)\widetilde{\psi}_0(\xi) (-3 ) = \boldsymbol{ {P} }_3\boldsymbol{ {P} }_2\boldsymbol{ {P} }_1|\psi_0\rangle; \\ && \widetilde{\boldsymbol{ {C} }}^3(0)\widetilde{\psi}_0(\xi) (-1 ) = (\boldsymbol{ {P} }_3\boldsymbol{ {P} }_2\boldsymbol{ {Q} }_1 + \boldsymbol{ {P} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {P} }_1 + \boldsymbol{ {Q} }_3\boldsymbol{ {P} }_2\boldsymbol{ {P} }_1)|\psi_0\rangle; \\ && \widetilde{\boldsymbol{ {C} }}^3(0)\widetilde{\psi}_0(\xi)(1) = (\boldsymbol{ {Q} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {P} }_1+\boldsymbol{ {Q} }_3\boldsymbol{ {P} }_2\boldsymbol{ {Q} }_1 +\boldsymbol{ {P} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {Q} }_1)|\psi_0\rangle;\\ && \widetilde{\boldsymbol{ {C} }}^3(0)\widetilde{\psi}_0(\xi)(3) = \boldsymbol{ {Q} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {Q} }_1|\psi_0\rangle. \end{eqnarray*} $

(2)

$ \begin{eqnarray*} &&\widetilde{\boldsymbol{ {C} }}^3_{\leftarrow}(\xi)\widetilde{\psi}_0(\xi)(-3) = ({\rm e}^{-{\rm i}\xi}+{\rm e}^{-2{\rm i}\xi}+{\rm e}^{-3{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {P} }_2\boldsymbol{ {P} }_1|\psi_0\rangle; \\ && \widetilde{\boldsymbol{ {C} }}^3_{\leftarrow}(\xi)\widetilde{\psi}_0(\xi)(-1) = [({\rm e}^{{\rm i}\xi}+{\rm e}^{0{\rm i}\xi} +{\rm e}^{-{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {P} }_2\boldsymbol{ {Q} }_1+({\rm e}^{-{\rm i}\xi}+{\rm e}^{0{\rm i}\xi}+{\rm e}^{-{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {P} }_1 \\ && \hskip 3.1cm +({\rm e}^{-{\rm i}\xi}+{\rm e}^{-2{\rm i}\xi}+{\rm e}^{-{\rm i}\xi})\boldsymbol{ {Q} }_3\boldsymbol{ {P} }_2\boldsymbol{ {P} }_1]|\psi_0\rangle; \\ && \widetilde{\boldsymbol{ {C} }}^3_{\leftarrow}(\xi)\widetilde{\psi}_0(\xi)(1) = [({\rm e}^{-{\rm i}\xi} +{\rm e}^{0{\rm i}\xi}+{\rm e}^{{\rm i}\xi})\boldsymbol{ {Q} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {P} }_1+({\rm e}^{{\rm i}\xi}+{\rm e}^{0{\rm i}\xi}+{\rm e}^{{\rm i}\xi})\boldsymbol{ {Q} }_3 \boldsymbol{ {P} }_2\boldsymbol{ {Q} }_1\\ &&\hskip 2.8cm +({\rm e}^{{\rm i}\xi}+{\rm e}^{2{\rm i}\xi}+{\rm e}^{{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {Q} }_1] |\psi_0\rangle; \\ && \widetilde{\boldsymbol{ {C} }}^3_{\leftarrow}(\xi)\widetilde{\psi}_0(\xi)(3) = ({\rm e}^{{\rm i}\xi}+{\rm e}^{2{\rm i}\xi} +{\rm e}^{3{\rm i}\xi})\boldsymbol{ {Q} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {Q} }_1|\psi_0\rangle. \end{eqnarray*} $

其中, $ \delta^2_+(-3) = 1, \delta^2_+(-1) = 5, \delta^2_+(1) = 5, \delta^2_+(3) = 1 $. 结果表明在系统演化过程中分别有$ 1 $条、$ 5 $条、$ 5 $条、$ 1 $条路径移入位置$ -3, -1, 1, 3 $.

(3)

$ \begin{eqnarray*} &&\widetilde{\boldsymbol{ {C} }}^3_{\rightarrow }(\xi)\widetilde{\psi}_0(\xi)(-3) = ({\rm e}^{-0{\rm i}\xi} +{\rm e}^{-{\rm i}\xi}+{\rm e}^{-2{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {P} }_2\boldsymbol{ {P} }_1|\psi_0\rangle; \\ && \widetilde{\boldsymbol{ {C} }}^3_{\rightarrow }(\xi)\widetilde{\psi}_0(\xi)(-1) = [({\rm e}^{0{\rm i}\xi} +{\rm e}^{{\rm i}\xi}+{\rm e}^{0{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {P} }_2\boldsymbol{ {Q} }_1 + ({\rm e}^{0{\rm i}\xi} + {\rm e}^{-{\rm i}\xi} + {\rm e}^{0{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {P} }_1 \\ &&\hskip 3.1cm + ({\rm e}^{0{\rm i}\xi} + {\rm e}^{-{\rm i}\xi} + {\rm e}^{-2{\rm i}\xi})\boldsymbol{ {Q} }_3\boldsymbol{ {P} }_2\boldsymbol{ {P} }_1] |\psi_0\rangle; \\ && \widetilde{\boldsymbol{ {C} }}^3_{\rightarrow }(\xi)\widetilde{\psi}_0(\xi)(1) = [({\rm e}^{0{\rm i}\xi} +{\rm e}^{-{\rm i}\xi}+{\rm e}^{0{\rm i}\xi})\boldsymbol{ {Q} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {P} }_1 +({\rm e}^{0{\rm i}\xi}+{\rm e}^{{\rm i}\xi}+{\rm e}^{0{\rm i}\xi})\boldsymbol{ {Q} }_3\boldsymbol{ {P} }_2\boldsymbol{ {Q} }_1 \\ &&\hskip 2.8cm+({\rm e}^{0{\rm i}\xi}+{\rm e}^{{\rm i}\xi}+{\rm e}^{2{\rm i}\xi})\boldsymbol{ {P} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {Q} }_1]|\psi_0\rangle; \\ \;\;\;\;\;\; && \widetilde{\boldsymbol{ {C} }}^3_{\rightarrow }(\xi)\widetilde{\psi}_0(\xi)(3) = ({\rm e}^{0{\rm i}\xi} +{\rm e}^{{\rm i}\xi}+{\rm e}^{2{\rm i}\xi})\boldsymbol{ {Q} }_3\boldsymbol{ {Q} }_2\boldsymbol{ {Q} }_1|\psi_0\rangle. \end{eqnarray*} $

其中, $ \delta^2_-(-3) = 0, \delta^2_-(-1) = 2, \delta^2_-(1) = 2, \delta^2_-(3) = 0 $. 结果表明在系统演化过程中分别有$ 0 $条、$ 2 $条、$ 2 $条、$ 0 $条路径移出位置$ -3, -1, 1, 3 $.