近年来, 不确定原理引起了数学家的广泛注意, 并取得了进步.参考文献[1-4]都有关于不确定原理的研究.在时频分析里不确定原理又被称为时间带宽乘积定理, 是关于Fourier变换的一个基本叙述.最早提出解析信号不确定原理思想的是Gabor, 他给出了不确定原理的一个表达式^[5]

$ \sigma_{t}\sigma_{\omega}\geq\frac{1}{2}, $

其中$\sigma_{t}$和$\sigma_{\omega}$是信号$s(t)\in L^{2}(\mathbb{R} )$的持续时间和带宽, 定义如下

$ \sigma^{2}_{t}=\int^{\infty}_{-\infty}(t-\langle t\rangle)^{2}|s(t)|^{2}{\rm d}t , \;\; \sigma^{2}_{\omega}=\int^{\infty}_{-\infty}(\omega-\langle \omega\rangle)^{2}| \hat{s}(\omega)|^{2}{\rm d}\omega. $

这里$\langle t\rangle$和$\langle \omega\rangle$分别是时间$ t $和Fourier频率$ \omega $的平均值, 定义为

$ \langle t\rangle=\int^{\infty}_{-\infty}t|s(t)|^{2}{\rm d}t, \;\; \langle \omega\rangle=\int^{\infty}_{-\infty}\omega|\hat{s}(\omega)|^{2}{\rm d}\omega, $

这里$\hat{s}(\omega)$是$s(t)$的Fourier变换.

$s\in L^{1}(\mathbb{R} ^{n})$的Fourier变换是由以下公式定义的

$ \hat{s}(\omega)=Fs(\omega)=\int_{\mathbb{R} ^{n}}s(t){\rm e}^{-2\pi {\rm i}\omega \cdot t}{\rm d}t. $

Gabor给出的不确定原理是最简洁的形式但不是最好的. Leon Cohen在文献[6]中得到了一个更强的结果

$ \sigma_{t}\sigma_{\omega}\geq\frac{1}{2}\sqrt{1+4Cov^{2}}, $

其中

$ Cov=\langle t\varphi'(t)\rangle-\langle t\rangle \langle\omega\rangle =\int^{\infty}_{-\infty}t\varphi'(t)|s(t)|^{2}{\rm d}t-\langle t\rangle \langle\omega\rangle $

是信号$s(t)$的协方差^[1], 这里的$s(t)$通常写成指数函数的形式: $s(t)=\rho(t){\rm e}^{{\rm i}\varphi(t)}$, $\rho(t)=|s(t)|$, $\varphi(t)$是实值函数, $\varphi'(t)$是$\varphi(t)$的经典导数.而参考文献[1]中得到了一个更强的结果

$ \sigma^{2}_{t}\sigma^{2}_{\omega}\geq\frac{1}{4}+ \bigg[\int|(t-\langle t\rangle)(\varphi'(t)-\langle\omega\rangle)||s(t)|^{2}{\rm d}t\bigg]^{2}. $

为了读者的方便, 在此我们给出本文所需要的一些记号.

设$f(t)\in L^{2}(\mathbb{R} ^{n})$, $g=(g_{1}, g_{2}, \cdots, g_{n})\in L^{2}(\mathbb{R} ^{n})$, 则有

$ \nabla=\Big(\frac{\partial}{\partial t_{1}}, \frac{\partial}{\partial t_{2}}, \cdots, \frac{\partial}{\partial t_{n}}\Big), \;\; \nabla f(t)=\Big(\frac{\partial f}{\partial t_{1}}, \frac{\partial f}{\partial t_{2}}, \cdots, \frac{\partial f}{\partial t_{n}}\Big), $

$ \nabla\cdot g=\frac{\partial g_{1}}{\partial t_{1}}+\frac{\partial g_{2}}{\partial t_{2}}+\cdots+\frac{\partial g_{n}}{\partial t_{n}}, \;\; t\cdot g=t_{1}g_{1}+t_{2}g_{2}+\cdots+t_{n}g_{n}, $

$ t\circ f=(t_{1}f, t_{2}f, \cdots, t_{n}f), \;\; \tilde{t}=(|t_{1}|, |t_{2}|, \cdots, |t_{n}|), $

以及

$ |\tilde{t}|^{2}=|t_{1}|^{2}+|t_{2}|^{2}+\cdots+|t_{n}|^{2}. $

对于解析信号$s(t)=\rho(t){\rm e}^{{\rm i}\varphi(t)}\in L^{2}(\mathbb{R} )$定义了时间和频率的平均值以及它们的方差, 协方差$\langle t\rangle, \quad \langle \omega\rangle, \quad \sigma^{2}_{t}, \quad \sigma^{2}_{\omega}, \quad Cov.$现在我们将这些定义推广到$L^{2}(\mathbb{R} ^{n})$上.

定义2.1 设$f(t)=\rho(t){\rm e}^{{\rm i}\varphi(t)}\in L^{2}(\mathbb{R} ^{n})$.则有

$ \langle t\rangle=\int_{\mathbb{R} ^{n}}t|f(t)|^{2}{\rm d}t =\Big(\int_{\mathbb{R} ^{n}}t_{1}|f(t)|^{2}{\rm d}t, \int_{R^{n}}t_{2}|f(t)|^{2}{\rm d}t, \cdots, \int_{\mathbb{R} ^{n}}t_{n}|f(t)|^{2}{\rm d}t\Big), \nonumber $

$ \langle \omega\rangle=\int_{\mathbb{R} ^{n}}\omega|\hat{f}(\omega)|{\rm d}\omega\nonumber =\Big(\int_{\mathbb{R} ^{n}}\omega_{1}|\hat{f}(\omega)|^{2}{\rm d}\omega, \int_{\mathbb{R} ^{n}}\omega_{2}|\hat{f}(\omega)|^{2}{\rm d}\omega, \cdots, \int_{\mathbb{R} ^{n}}\omega_{n}|\hat{f}(\omega)|^{2}{\rm d}\omega\Big), \nonumber $

$ \sigma^{2}_{t}=\int_{\mathbb{R} ^{n}}|t-\langle t\rangle|^{2}|f|^{2}{\rm d}t\nonumber =\sum\limits_{k=1}^{n}{{}}\int_{\mathbb{R} ^{n}}|t_{k}-\langle t_{k}\rangle|^{2}|f(t)|^{2}{\rm d}t, \nonumber $

其中

$ \langle t_{k}\rangle=\int_{\mathbb{R} ^{n}}t_{k}|f(t)|^{2}{\rm d}t, \ k=1, 2, \cdots, n. $

$ \sigma^{2}_{\omega}=\int_{\mathbb{R} ^{n}}|\omega-\langle\omega\rangle|^{2}|\hat{f}(\omega)|^{2}{\rm d}\omega\nonumber =\sum\limits_{k=1}^{n}{{}}\int_{\mathbb{R} ^{n}}|\omega_{k}-\langle \omega_{k}\rangle|^{2}|\hat{f}(\omega)|^{2}{\rm d}\omega, \nonumber $

其中$\hat{f}(\omega)$是函数$f(t)$的Fourier变换,

$ \langle \omega_{k}\rangle=\int_{\mathbb{R} ^{n}}\omega_{k}|\hat{f}(\omega)|^{2}{\rm d}\omega, \ k=1, 2, \cdots, n. $

$ \begin{eqnarray} Cov&=&\int_{\mathbb{R} ^{n}}\big( t\cdot\bigtriangledown\varphi(t)\big)|f(t)|^{2}{\rm d}t-2\pi\langle t\rangle\langle\omega\rangle\nonumber\\ &=&\int_{\mathbb{R} _{n}}\sum^{n}_{k=1}t_{k}\frac{\partial\varphi}{\partial t_{k}}|f(t)|^{2}{\rm d}t-2\pi\langle t\rangle\langle\omega\rangle.\nonumber \end{eqnarray} $

接下来, 我们介绍Heisenberg不确定原理, 它的物理背景是粒子的位置和动量不能同时被确定. Heisenberg不确定原理是量子力学最基本的一个结果.以下定理就是Heisenberg不确定原理的数学表达式.

定理2.1^[2] 设$f\in L^{2}(\mathbb{R} ^{n}).$其中$x_{0}, \xi_{0}\in \mathbb{R} ^{n}$.则有

$ \Big(\int_{\mathbb{R} ^{n}}|\xi-\xi_{0}|^{2}|\hat{f}(\xi)|^{2}{\rm d}\xi\Big) \Big(\int_{\mathbb{R} ^{n}}|x-x_{0}|^{2}|f(x)|^{2}{\rm d}x\Big)\geq\frac{n^{2}}{16\pi^{2}}\|f\|^{4}. $

等号成立当且仅当$f(x)=c{\rm e}^{2\pi {\rm i}x\cdot\xi_{0}}{\rm e}^{-\alpha|x-x_{0}|^{2}/2}, $其中$\alpha>0$以及$c\in {\Bbb C}.$

在本文中, 我们得到了比Heisenberg不确定原理更强的结果, 要证明本文的结果我们需要以下引理.

引理2.1 设$f(t)=\rho(t){\rm e}^{{\rm i}\varphi(t)}\in L^{2}(\mathbb{R} ^{n})$, 如果$\nabla \rho(t)$几乎处处存在, 并且$\nabla \rho(t)\in L^{2}(\mathbb{R} ^{n}).$则有

$ \bigg|\int_{\mathbb{R} ^{n}}(t\circ \rho(t))\cdot\nabla\rho(t){\rm d}t\bigg|^{2}=\frac{n^{2}}{4}\|f\|^{4}. $

证 由以上记号计算可得

$ \begin{eqnarray*} &&\nabla\cdot(t\circ\rho(t))-t\cdot\nabla\rho(t)\\ &=&\nabla\cdot(t_{1}\rho(t), t_{2}\rho(t), \cdots, t_{n}\rho(t))\nonumber -t\Big(\frac{\partial\rho(t)}{\partial t_{1}}, \frac{\partial\rho(t)}{\partial t_{2}}, \cdots, \frac{\partial\rho(t)}{\partial t_{n}}\Big)\nonumber\\ &=&\rho(t)+t_{1}\frac{\partial\rho(t)}{\partial t_{1}}+\rho(t)+t_{2}\frac{\partial\rho(t)}{\partial t_{2}}+, \cdots, +\rho(t)+t_{n}\frac{\partial\rho(t)}{\partial t_{n}}\nonumber\\ &&-\Big(t_{1}\frac{\partial\rho(t)}{\partial t_{1}}+t_{2}\frac{\partial\rho(t)}{\partial t_{2}}+, \cdots, +t_{n}\frac{\partial\rho(t)}{\partial t_{n}}\Big)\nonumber\\ &=&n\rho(t).\nonumber \end{eqnarray*} $

由上式并运用分部积分可得

$ \begin{eqnarray} \|f\|^{2}&=&\|\rho\|^{2}\nonumber\\ &=&\frac{1}{n}\int_{\mathbb{R} ^{n}}[\nabla\cdot(t\circ\rho)-t\cdot\nabla\rho]\bar{\rho}{\rm d}t\nonumber\\ &=&\frac{1}{n}\int_{\mathbb{R} ^{n}}[\nabla\cdot(t\circ\rho)-t\cdot\nabla\rho]\rho{\rm d}t\nonumber\\ &=&-\frac{1}{n}\int_{\mathbb{R} ^{n}}[(t\circ\rho)\cdot\nabla\rho+\nabla\rho\cdot(t\circ\rho)]{\rm d}t\nonumber\\ &=&-\frac{2}{n}\int_{\mathbb{R} ^{n}}(t\circ\rho)\cdot\nabla\rho{\rm d}t.\nonumber \end{eqnarray} $

故有

$ \bigg|\int_{\mathbb{R} ^{n}}(t\circ\rho)\cdot\nabla\rho{\rm d}t\bigg|^{2}=\frac{n^{2}}{4}\|f\|^{4}. $

引理2.1证毕.

引理2.2 设$f(t)=\rho(t){\rm e}^{{\rm i}\varphi(t)}\in L^{2}(\mathbb{R} ^{n}), \|f\|^{2}=1.$假设$\nabla\rho(t), \nabla\varphi(t), \nabla f(t)$几乎处处存在, 并且$\frac{\partial f(t)}{\partial t_{k}}\in L^{2}(\mathbb{R} ^{n}), \ k=1, 2, \cdots, m$.则有

$ \sigma^{2}_{\omega}=\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho|^{2}{\rm d}t+\frac{1}{4\pi^{2}} \int_{\mathbb{R} ^{n}}\rho^{2}|\nabla\varphi-2\pi\langle\omega\rangle|^{2}{\rm d}t. $

证 令

$ g(t)={\rm e}^{-2\pi {\rm i}(t+\langle t\rangle)\cdot\langle\omega\rangle}f(t+\langle t\rangle)=\rho_{1}(t){\rm e}^{{\rm i}\varphi_{1}(t)}, $

其中

$ \langle t\rangle=\int_{\mathbb{R} ^{n}}t\circ|f(t)|^{2}{\rm d}t, \;\; \langle\omega\rangle=\int_{\mathbb{R} ^{n}}\omega\circ|\hat{f}(\omega)|^{2}{\rm d}\omega, $

$ \rho_{1}(t)=\rho(t+\langle t\rangle), (\nabla \rho_{1}(t))=(\nabla \rho)(t+\langle t\rangle), $

$ \varphi_{1}(t)=\varphi(t+\langle t\rangle)-2\pi (t+\langle t\rangle)\cdot\langle\omega\rangle, (\nabla\varphi_{1})(t)=(\nabla\varphi)(t+\langle t\rangle)-2\pi\langle\omega\rangle. $

则$\langle t\rangle_{g}=0, \langle\omega\rangle_{g}=0, \|g\|^{2}=\|f\|^{2}$, 并且有

$ \int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t=\int_{\mathbb{R} ^{n}}|t-\langle t\rangle|^{2}|f(t)|^{2}{\rm d}t=\sigma^{2}_{t}, $

$ \int_{\mathbb{R} ^{n}}|\omega|^{2}|\hat{g}(\omega)|^{2}{\rm d}\omega=\int_{\mathbb{R} ^{n}}|\omega-\langle \omega\rangle|^{2}|\hat{f}(\omega)|^{2}{\rm d}\omega=\sigma^{2}_{\omega}. $

于是有

$ \begin{eqnarray} \sigma^{2}_{\omega}&=&\int_{\mathbb{R} ^{n}}|\omega|^{2}|\hat{g}(\omega)|^{2}{\rm d}\omega\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\widehat{\nabla g}|^{2}{\rm d}\omega\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla g|^{2}{\rm d}t\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\sum^{n}_{k=1}\Big|\frac{\partial g} {\partial t_{k}}\Big|^{2}{\rm d}t\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\sum^{n}_{k=1}\Big|\frac{\partial\rho_{1}} {\partial t_{k}}+i\rho_{1}\frac{\partial\varphi_{1}}{\partial t_{k}}\Big|^{2}{\rm d}t\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\sum^{n}_{k=1}\Big[\Big(\frac{\partial\rho_{1}} {\partial t_{k}}\Big)^{2}+\rho_{1}^{2}\Big(\frac{\partial\varphi_{1}}{\partial t_{k}}\Big)^{2}\Big]{\rm d}t\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\sum^{n}_{k=1}\Big(\frac{\partial\rho_{1}} {\partial t_{k}}\Big)^{2}{\rm d}t+\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}} \sum^{n}_{k=1}\rho^{2}\Big(\frac{\partial\varphi_{1}}{\partial t_{k}}\Big)^{2}{\rm d}t\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho_{1}|^{2}{\rm d}t+\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho_{1}^{2}|\nabla\varphi_{1}|^{2}{\rm d}t\nonumber\\ &=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho|^{2}{\rm d}t+\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho^{2}|\nabla\varphi-2\pi\langle\omega\rangle|^{2}{\rm d}t.\nonumber \end{eqnarray} $

因此有

$ \sigma^{2}_{\omega}=\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho|^{2}{\rm d}t+\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho^{2}|\nabla\varphi-2\pi\langle\omega\rangle|^{2}{\rm d}t. $

引理2.2证毕.

有了以上两个引理, 我们就可以得到下面的定理.

定理2.2 假设$f(t)=\rho(t){\rm e}^{{\rm i}\varphi(t)}\in L^{2}(\mathbb{R} ^{n}), \|f\|^{2}=1, $ $\nabla\rho(t), \nabla\varphi(t)$, $\nabla f(t)$几乎处处存在, 并且$\frac{\partial f}{\partial t_{k}}, t_{k} f(t)\in L^{2}(\mathbb{R} ^{n}), \ k=1, 2, \cdots, n$. 则有

$ \sigma^{2}_{t}\sigma^{2}_{\omega}\geq\frac{n^{2}}{16\pi^{2}}+\frac{1}{4\pi^{2}} \bigg[\int_{\mathbb{R} ^{n}}\sum\limits_{k=1}^{n}{{}}\Big|(t_{k}-\langle t_{k}\rangle) \Big(\frac{\partial\varphi(t)}{\partial t_{k}}-2\pi\langle\omega_{k}\rangle\Big)\Big|\rho^{2}(t){\rm d}t\bigg]^{2}. $

如果$\frac{\partial\varphi}{\partial t_{k}}, \ k=1, 2, \cdots, n.$是连续的且$\rho(t)$几乎处处不为$0$, 则等号成立当且仅当$f(t)$为下面$2^{n}$种情况的一种

$ f(t)=d_{1}{\rm e}^{-\lambda_{1}|t-\langle t\rangle|^{2}/2}{\rm e}^{\frac{\rm i}{2} \lambda_{2}\sum\limits^{n}_{k=1}(-1)^{l_{k}}(t_{k}-\langle t_{k}\rangle)^{2}+c+2\pi{\rm i}t\cdot\langle\omega\rangle}, \;\; k=1, 2, \cdots, n, $

其中$\lambda_{1}>0, \ \lambda_{2}>0, \ l_{k}\in{\Bbb N}_{+}$, 并且$d_{1}, \lambda_{1}$满足$d_{1}^{\frac{2}{n}}\sqrt{\frac{\pi}{\lambda_{1}}}=1.$

证 我们同样令

$ g(t)={\rm e}^{-2\pi {\rm i}(t+\langle t\rangle)\cdot\langle\omega\rangle}f(t+\langle t\rangle)=\rho_{1}(t){\rm e}^{{\rm i}\varphi_{1}(t)}. $

由引理2.2可得

$ \sigma^{2}_{t_{g}}\sigma^{2}_{\omega_{g}}=\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho_{1}|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t+ \frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho_{1}^{2}|\nabla\varphi_{1}|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t. $

因此要证明此定理, 我们可以分两部分来证明

$ \frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho_{1}|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t\geq\frac{n^{2}}{16\pi^{2}}\|g\|^{4} $

和

$ \frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho_{1}^{2}|\nabla\varphi|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t \geq\frac{1}{4\pi^{2}}\bigg[\int_{\mathbb{R} ^{n}}\sum\limits_{k=1}^{n}{{}}|t_{k}\frac{\partial\varphi_{1}}{\partial t_{k}}|\rho_{1}^{2}{\rm d}t\bigg]^{2}. $

第一个不等式证明如下:有Cauchy-Schwarz不等式和引理2.1可得

$ I_{1}=\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho_{1}|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t \\ \geq\frac{1}{4\pi^{2}}\bigg|\int_{\mathbb{R} ^{n}}(\tilde{t}\circ\rho_{1}(t))\cdot\widetilde{\nabla\rho_{1}(t)}{\rm d}t\bigg|^{2} $

(2.1)

$ \geq\frac{1}{4\pi^{2}}\bigg|\int_{\mathbb{R} ^{n}}(t\circ\rho_{1}(t))\cdot\nabla\rho_{1}(t){\rm d}t\bigg|^{2}\\ =\frac{n^{2}}{16\pi^{2}}\|g\|^{4}. $

(2.2)

有Hölder不等式可以得到第二个不等式

$ \begin{eqnarray} I_{2}&=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho_{1}^{2}|\nabla\varphi_{1}|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|g|^{2}{\rm d}t\nonumber\\ &\geq&\frac{1}{4\pi^{2}}\bigg[\int_{\mathbb{R} ^{n}}(\tilde{t}\cdot\widetilde{\nabla\varphi_{1}})\rho_{1}^{2}(t){\rm d}t\bigg]^{2}\\ &=&\frac{1}{4\pi^{2}}\bigg[\int_{\mathbb{R} ^{n}}\sum^{n}_{k=1}|t_{k} \frac{\partial\varphi_{1}(t)}{\partial t_{k}}|\rho_{1}^{2}(t){\rm d}t\bigg]^{2}.\nonumber \end{eqnarray} $

(2.3)

结合$I_{1}$和$I_{2}$可得

$ \sigma^{2}_{t_{g}}\sigma^{2}_{\omega_{g}}\geq\frac{n^{2}}{16\pi^{2}}\|g\|^{4}+\frac{1}{4\pi^{2}} \bigg[\int_{\mathbb{R} ^{n}}\sum\limits_{k=1}^{n}{{}}|t_{k} \frac{\partial\varphi_{1}(t)}{\partial t_{k}}|\rho_{1}^{2}(t){\rm d}t\bigg]^{2}. $

由于

$ \int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t=\int_{\mathbb{R} ^{n}}|t-\langle t\rangle|^{2}|f(t)|^{2}{\rm d}t, \\ \int_{\mathbb{R} ^{n}}|\omega|^{2}|\hat{g}(\omega)|^{2}{\rm d}\omega=\int_{\mathbb{R} ^{n}}|\omega-\langle \omega\rangle|^{2}|\hat{f}(\omega)|^{2}{\rm d}\omega $

以及

$ \|g\|^{2}=\|f\|^{2}=1, $

$ \int_{\mathbb{R} ^{n}}\sum\limits_{k=1}^{n}{{}}\bigg|t_{k} \frac{\partial\varphi_{1}(t)}{\partial t_{k}}\bigg|\rho_{1}^{2}(t){\rm d}t =\int_{\mathbb{R} ^{n}}\sum\limits_{k=1}^{n}{{}}\bigg|(t_{k}-\langle t_{k}\rangle) \Big(\frac{\partial\varphi }{\partial t_{k}}-2\pi\langle\omega_{k}\rangle\Big)\bigg|\rho^{2}(t){\rm d}t. $

所以有

$ \sigma^{2}_{t}\sigma^{2}_{\omega}\geq\frac{n^{2}}{16\pi^{2}} +\frac{1}{4\pi^{2}}\bigg|\int_{\mathbb{R} ^{n}}\sum\limits_{k=1}^{n}{{}} \bigg|(t_{k}-\langle t_{k}\rangle)\Big(\frac{\partial\varphi }{\partial t_{k}}-2\pi\langle\omega_{k}\rangle\Big)\bigg|\rho^{2}(t){\rm d}t\bigg|^{2}. $

接下来我们讨论等式成立的条件, 注意到定理2.1等式成立当且仅当$I_{1}$和$I_{2}$均成为等式.而$I_{1}$等式成立当且仅当$(2.1)$和$(2.2)$式同时成为等式.而$(2.1)$等式成立当且仅当存在$\lambda_{1}\in \mathbb{R}, \lambda_{1}>0, $使得对任意的$t\in \mathbb{R} ^{n}$有

$ \widetilde{\nabla\rho_{1}}=\lambda_{1} \tilde{t}\circ\rho_{1}(t), $

也就是

$ \bigg|\frac{\partial\rho_{1}}{\partial t_{k}}\bigg|=\lambda_{1}|t_{k}|\rho_{1}(t), $

其中$ k=1, 2, \cdots, n.$

由于$t\circ\rho_{1}(t)$和$\nabla\rho_{1}(t)$都是向量, (2.2)等式成立当且仅当$t\circ\rho_{1}(t)$和$\nabla\rho_{1}(t)$的所有分量同时有相同或相反的符号.若$t\circ\rho(t)$和$\nabla\rho(t)$的所有分量同时有相同的符号, 则有

$ \nabla\rho_{1}(t)=\lambda_{1} t\circ\rho_{1}(t), $

由此可得

$ \rho_{1}(t)=d{\rm e}^{\lambda_{1}|t|^{2}/2}. $

显然, $\rho_{1}(t)=d{\rm e}^{\lambda_{1}|t|^{2}/2}$不属于$L^{2}(\mathbb{R} ^{n}).$因此, $t\circ\rho_{1}(t)$和$\nabla\rho_{1}(t)$的所有分量不可能同时是相同的符号, 只能是所有分量的符号同时相反, 因此有

$ \nabla\rho_{1}(t)=-\lambda_{1} t\circ\rho_{1}(t), $

由此可得

$ \rho_{1}(t)=d_{1}{\rm e}^{-\lambda_{1}|t|^{2}/2}. $

由于我们假设$\|f\|^{2}=1, $所以$d_{1}, \lambda_{1}$需满足$d_{1}^{\frac{2}{n}}\sqrt{\frac{\pi}{\lambda_{1}}}=1.$

$I_{2}$成为等式当且仅当$(2.3)$式成为等式.而$(2.3)$式成为等式等价于存在$\lambda_{2}>0$使得

$ \widetilde{\nabla\varphi_{1}}\circ\rho_{1}(t)=\lambda_{2}\tilde{t}\circ\rho_{1}(t). $

在$\rho_{1}(t)$几乎处处不为0和$\frac{\partial\varphi_{1}}{\partial t_{k}}$连续的条件下, 我们有

$ \bigg|\frac{\partial\varphi_{1}}{\partial t_{1}}\bigg|=\lambda_{2}|t_{1}|, $

(2.4)

$ \bigg|\frac{\partial\varphi_{1}}{\partial t_{2}}\bigg|=\lambda_{2}|t_{2}|, $

(2.5)

$ \bigg|\frac{\partial\varphi_{1}}{\partial t_{3}}\bigg|=\lambda_{2}|t_{3}|, $

(2.6)

$ \;\;\;\;\;\;\;\;\cdots\nonumber\\ \bigg|\frac{\partial\varphi_{1}}{\partial t_{n}}\bigg|=\lambda_{2}|t_{n}|. $

(2.7)

解(2.4)式, 直接去绝对值可得

$ \frac{\partial\varphi_{1}}{\partial t_{1}}=\lambda_{2}t_{1}, \;\;\mbox{或}\;\; \frac{\partial\varphi_{1}}{\partial t_{1}}=-\lambda_{2}t_{1}, $

解常微分方程可得

$ \varphi_{1}(t)=\frac{1}{2}\lambda_{2}t^{2}_{1}+c_{1}, \;\;\mbox{或}\;\; \varphi_{1}(t)=-\frac{1}{2}\lambda_{2}t^{2}_{1}+c_{2}. $

(2.5)式去绝对值同样有两种情况

$ \frac{\partial\varphi_{1}}{\partial t_{2}}=\lambda_{2}t_{2}, \;\;\mbox{或}\;\; \frac{\partial\varphi_{1}}{\partial t_{2}}=-\lambda_{2}t_{2}, $

将$(2.4)$式得到的$\varphi_{1}(t)$分别代入这两种情况我们可以得到

$ c_{1}=\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{3}, \;\;\mbox{或}\;\; c_{1}=-\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{4}. $

$ c_{2}=\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{5}, \;\;\mbox{或}\;\; c_{2}=-\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{6}. $

由此可得

$ \varphi_{1}(t)=\frac{1}{2}\lambda_{2}t^{2}_{1}+\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{3}, \;\;\mbox{或}\;\; \varphi_{1}(t)=\frac{1}{2}\lambda_{2}t^{2}_{1}-\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{4}, $

$ \varphi_{1}(t)=-\frac{1}{2}\lambda_{2}t^{2}_{1}+\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{5}, \;\;\mbox{或}\;\; \varphi_{1}(t)=-\frac{1}{2}\lambda_{2}t^{2}_{1}-\frac{1}{2}\lambda_{2}t^{2}_{2}+c_{6}. $

同样将得到的$\varphi_{1}(t)$代入(2.5)式可得

$ c_{3}=\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{7}, \;\;\mbox{或}\;\; c_{3}=-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{8}. $

$ c_{4}=\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{9}, \;\;\mbox{或}\;\; c_{4}=-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{10}. $

$ c_{5}=\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{11}, \;\;\mbox{或}\;\; c_{5}=-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{12}, $

$ c_{6}=\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{13}, \;\;\mbox{或}\;\; c_{6}=-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{14}, $

由此可得

$ \varphi_{1}(t)=\frac{1}{2}\lambda_{2}t_{1}^{2}+\frac{1}{2}\lambda_{2}t^{2}_{2}+\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{7}, $

或

$ \varphi_{1}(t)=\frac{1}{2}\lambda_{2}t_{1}^{2}+\frac{1}{2}\lambda_{2}t^{2}_{2}-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{8}, $

或

$ \varphi_{1}(t)=\frac{1}{2}\lambda_{2}t_{1}^{2}-\frac{1}{2}\lambda_{2}t^{2}_{2}+\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{9}, $

或

$ \varphi_{1}(t)=\frac{1}{2}\lambda_{2}t_{1}^{2}-\frac{1}{2}\lambda_{2}t^{2}_{2}-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{10}, $

或

$ \varphi_{1}(t)=-\frac{1}{2}\lambda_{2}t_{1}^{2}+\frac{1}{2}\lambda_{2}t^{2}_{2}+\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{11}, $

或

$ \varphi_{1}(t)=-\frac{1}{2}\lambda_{2}t_{1}^{2}+\frac{1}{2}\lambda_{2}t^{2}_{2}-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{12}, $

或

$ \varphi_{1}(t)=-\frac{1}{2}\lambda_{2}t_{1}^{2}-\frac{1}{2}\lambda_{2}t^{2}_{2}+\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{13}, $

或

$ \varphi_{1}(t)=-\frac{1}{2}\lambda_{2}t_{1}^{2}-\frac{1}{2}\lambda_{2}t^{2}_{2}-\frac{1}{2}\lambda_{2}t^{2}_{3}+c_{14}. $

依次类推当$k=n$时

$ \varphi_{1}(t)=(-1)^{l_{1}}\frac{1}{2}\lambda_{2}t^{2}_{1}+(-1)^{l_{2}}\frac{1}{2}\lambda_{2}t^{2}_{2}+(-1)^{l_{3}}\frac{1}{2}\lambda_{2}t^{2}_{3}+\cdots+ (-1)^{l_{n}}\frac{1}{2}\lambda_{2}t^{2}_{n}+c. $

其中$l_{k}\in{\Bbb N}_{+}$ $k=1, 2, \cdots n$. $\varphi_{1}(t)$有$2^{n}$中情况, 结合$\rho_{1}(t)$可得

$ g(t)=d_{1}{\rm e}^{-\lambda_{1}|t|^{2}/2}{\rm e}^{\frac{\rm i}{2}\lambda_{2} \sum\limits^{n}_{k=1}(-1)^{l_{k}}t^{2}_{k}+c},\ k=1,2,\cdots,n, $

其中$\lambda_{1}>0, \ \lambda_{2}>0, \ l_{k}\in{\Bbb N}_{+}, d^{\frac{2}{n}}_{1}\sqrt{\frac{\pi}{\lambda_{1}}}=1$.

由于$f(t)=g(t-\langle t\rangle){\rm e}^{2\pi {\rm i}(t-\langle t\rangle) \langle\omega\rangle}, $所以定理2.1的等式成立当且仅当$f(t)$下面$2^{n}$种情况的一种.

$ f(t)=d_{1}{\rm e}^{-\lambda_{1}|t-\langle t\rangle|^{2}/2}{\rm e}^{\frac{\rm i}{2} \lambda_{2}\sum\limits^{n}_{k=1}(-1)^{l_{k}}(t_{k}-\langle t_{k}\rangle)^{2} +c+2\pi {\rm i}t\cdot\langle\omega\rangle}, $

其中$\lambda_{1}>0, \ \lambda_{2}>0, \ l_{k}\in{\Bbb N}_{+}, d^{\frac{2}{n}}_{1}\sqrt{\frac{\pi}{\lambda_{1}}}=1$.定理2.2证毕.

有了定理$2.2$可以得到下面的推论.

推论2.1 假设$f(t)=\rho(t){\rm e}^{{\rm i}\varphi(t)}\in L^{2}(\mathbb{R} ^{n}), \|f\|^{2}=1, $ $\nabla\rho(t), \nabla\varphi(t)$ , $\nabla f(t)$几乎处处存在, 并且$\frac{\partial f}{\partial t_{k}}, t_{k} f(t)\in L^{2}(\mathbb{R} ^{n}), \ k=1, 2, \cdots, n$.则有

$ \sigma^{2}_{t}\sigma^{2}_{\omega}\geq\frac{n^{2}}{16\pi^{2}}\|f\|^{4}+\frac{1}{4\pi^{2}}Cov^{2}. $

如果$\frac{\partial\varphi}{\partial t_{k}}$ $k=1, 2, \cdots, n$是连续的且$\rho(t)$几乎处处不为0, 则等式成立当且仅当

$ f(t)=d_{1}{\rm e}^{-\lambda_{1}|t-\langle t\rangle|^{2}/2}{\rm e}^{\frac{\rm i}{2}\lambda_{2}|t-\langle t\rangle|^{2}+c_{2}}{\rm e}^{2\pi {\rm i}t\cdot\langle \omega\rangle}, $

或

$ f(t)=d_{1}{\rm e}^{-\lambda_{1}|t-\langle t\rangle|^{2}/2}{\rm e}^{-\frac{\rm i}{2}\lambda_{2}|t-\langle t\rangle|^{2}+c_{2}}{\rm e}^{2\pi {\rm i}t\cdot\langle \omega\rangle}, $

其中$\lambda_{1}, \lambda_{2}\in \mathbb{R}, $ $\lambda_{1}>0, \lambda_{2}>0, $并且$d_{1}, \lambda_{1}$满足$d_{1}^{\frac{2}{n}}\sqrt{\frac{\pi}{\lambda_{1}}}=1.$

证 不失一般性不妨假设$\langle t \rangle=0$, $\langle\omega\rangle=0$. 由引理2.2可得

$ \sigma^{2}_{t}\sigma^{2}_{\omega}=\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|f(t)|^{2}{\rm d}t+ \frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho^{2}|\nabla\varphi|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|f(t)|^{2}{\rm d}t. $

因此我们也分两部分来证明此推论.

$ \frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}|\nabla\rho|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|f(t)|^{2}{\rm d}t\geq\frac{n^{2}}{16\pi^{2}}\|f\|^{4}, $

$ \frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho^{2}|\nabla\varphi|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|f(t)|^{2}{\rm d}t\geq\frac{1}{4\pi^{2}}Cov^{2}. $

由定理2.2的证明过程可知, 第一个不等式成立.并且等号成立当且仅当

$ \rho(t)=d_{1}{\rm e}^{-\lambda_{1}|t|^{2}/2}, $

其中$\lambda_{1}\in \mathbb{R}, \lambda_{1}>0, $并且$d_{1}, \lambda_{1}$, 需满足$d_{1}^{\frac{2}{n}}\sqrt{\frac{\pi}{\lambda_{1}}}=1.$

由Hölder不等式可得第二个不等式

$ \begin{eqnarray} I_{2}&=&\frac{1}{4\pi^{2}}\int_{\mathbb{R} ^{n}}\rho^{2}|\nabla\varphi|^{2}{\rm d}t\int_{\mathbb{R} ^{n}}|t|^{2}|f(t)|^{2}{\rm d}t\nonumber\\ &\geq&\frac{1}{4\pi^{2}}\bigg[\int_{\mathbb{R} ^{n}}(\tilde{t}\cdot\widetilde{\nabla\varphi})\rho^{2}(t){\rm d}t\bigg]^{2}\nonumber\\ &\geq&\frac{1}{4\pi^{2}}\bigg[\int_{\mathbb{R} ^{n}}(t\cdot\nabla\varphi(t))\rho^{2}(t){\rm d}t\bigg]^{2}\\ &=&\frac{1}{4\pi^{2}}Cov^{2}.\nonumber \end{eqnarray} $

(2.8)

所以有

$ \sigma^{2}_{t}\sigma^{2}_{\omega}\geq\frac{n^{2}}{16\pi^{2}}\|f\|^{4}+\frac{1}{4\pi^{2}}Cov^{2}. $

下面讨论等号成立的条件, 我们只需考虑$(2.8)$式成为等式的情况而$(2.2)$式等号成立当且仅当$t$和$\nabla\varphi$所有分量同时有相同或相反的符号.再结合定理2.2中$I_{2}$的讨论可得:若$t$和$\nabla\varphi$所有分量同时有相同的符号则有

$ \nabla\varphi=\lambda_{2}t. $

由此可得

$ \varphi(t)=\frac{1}{2}\lambda_{2}|t|^{2}+c_{2}. $

如果$t$和$\nabla\varphi$有相反的符号则有

$ \nabla\varphi=-\lambda_{2}t, $

由此可得

$ \varphi(t)=-\frac{1}{2}\lambda_{2}|t|^{2}+c_{3}. $

因此当$\varphi(t)=\frac{1}{2}\lambda_{2}|t|^{2}+c_{2}$时, 有

$ f_{1}(t)=d_{1}{\rm e}^{-\lambda_{1}|t|^{2}/2}{\rm e}^{\frac{i}{2}\lambda_{2}|t|^{2}+c_{2}}, $

当$\varphi(t)=-\frac{1}{2}\lambda_{2}|t|^{2}+c_{3}$时, 有

$ f_{2}(t)=d_{1}{\rm e}^{-\lambda_{1}|t|^{2}/2}{\rm e}^{-\frac{i}{2}\lambda_{2}|t|^{2}+c_{3}}. $

最后, 对$f(t)\in L^{2}(\mathbb{R} ^{n})$, 令$g(t)={\rm e}^{-2\pi {\rm i}(t+\langle t\rangle)\cdot\langle\omega\rangle}f(t+\langle t\rangle), $则有

$ \int_{\mathbb{R} ^{n}}|t|^{2}|g(t)|^{2}{\rm d}t=\int_{\mathbb{R} ^{n}}|t-\langle t\rangle|^{2}|f(t)|^{2}{\rm d}t, \\ \int_{\mathbb{R} ^{n}}|\omega|^{2}|\hat{g}(\omega)|^{2}{\rm d}\omega=\int_{\mathbb{R} ^{n}}|\omega-\langle \omega\rangle|^{2}|\hat{f}(\omega)|^{2}{\rm d}\omega, $

并且有$\|g\|^{2}=\|f\|^{2}, Cov_{g}=Cov_{f}$.

因此推论2.1等式成立当且仅当$f(t)$是下面两种形式中的一种,

$ f_{1}(t)=d_{1}{\rm e}^{-\lambda_{1}|t-\langle t\rangle|^{2}/2}{\rm e}^{\frac{\rm i}{2}\lambda_{2}|t-\langle t\rangle|^{2}+c_{2}}{\rm e}^{2\pi {\rm i}t\cdot\langle\omega\rangle}, $

$ f_{2}(t)=d_{1}{\rm e}^{-\lambda_{1}|t-\langle t\rangle|^{2}/2}{\rm e}^{-\frac{\rm i}{2}\lambda_{2}|t-\langle t\rangle|^{2}+c_{2}}{\rm e}^{2\pi {\rm i}t\cdot\langle\omega\rangle}. $

推论2.1证毕.