E：期望和方差-探究$\sigma_x$与$\sigma_p$的关系-定域性的例子 in 算符作用在左矢和右矢上 Action des opérateurs sur les kets et les bras
期望和方差-探究\(\sigma_x\)与\(\sigma_p\)的关系-定域性的例子
借助均值，同时应用随时间演变的薛定谔方程
设\(\partial_t \widehat{A}=0\)，证明：
\[ \begin{align} i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\langle A\rangle=\langle[\widehat{A}, \widehat{H}]\rangle\end{align} \]
- 均值的定义：\(\langle A\rangle=\langle\psi(t)|\widehat{A}| \psi(t)\rangle\)
- 计算方程左侧的导数：
\[ i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\langle A\rangle=i \hbar\left(\frac{\mathrm{d}}{\mathrm{d} t}\langle\psi(t)|\right) \widehat{A}|\psi(t)\rangle+i \hbar\langle\psi(t)| \widehat{A}\left(\frac{\mathrm{d}}{\mathrm{d} t}|\psi(t)\rangle\right)+i \hbar\left\langle\psi(t)\left|\left(\frac{\mathrm{d}}{\mathrm{d} t} \widehat{A}\right)\right| \psi(t)\right\rangle=i \hbar\left(\frac{\mathrm{d}}{\mathrm{d} t}\langle\psi(t)|\right) \widehat{A}|\psi(t)\rangle+i \hbar\langle\psi(t)| \widehat{A}\left(\frac{\mathrm{d}}{\mathrm{d} t}|\psi(t)\rangle\right) \]
- 借助体系随时间演变的薛定谔方程（第六定理）：
\[ i \hbar \frac{\mathrm{d}}{\mathrm{d} t}|\psi(t)\rangle=\widehat{H}(t)|\psi(t)\rangle \]
- 由于\(\hat H\)是厄米算子，有：
\[ \langle\psi(t)| \widehat{H}=-i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\langle\psi(t)| \]
- 等式的右侧：
\[ \begin{aligned}\langle[\widehat A, \widehat H]\rangle& = \langle\psi(t)|\widehat A\widehat H|\psi(t)\rangle - \langle\psi(t)|\widehat H\widehat A|\psi(t)\rangle\\& = i \hbar\left(\frac{\mathrm{d}}{\mathrm{d} t}\langle\psi(t)|\right) \widehat{A}|\psi(t)\rangle+i \hbar\langle\psi(t)| \widehat{A}\left(\frac{\mathrm{d}}{\mathrm{d} t}|\psi(t)\rangle\right) \end{aligned} \]
- 由此可见等式左侧与右侧相等。
继续应用\((1)\)进行一些计算：
- \(i \hbar \frac{\mathrm{d}}{\mathrm{d}
t}\langle p\rangle=\langle[\widehat{p},
\widehat{H}]\rangle\)，在自由例子情况下，\(\widehat H = \frac{\hat p^2}{2m}\)，故有：
- \(i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\langle p\rangle=\left\langle\left[\widehat{p}, \frac{\widehat{p}^2}{2 m}\right]\right\rangle=0\)
- \(i \hbar \frac{\mathrm{d}}{\mathrm{d}
t}\langle p\rangle=\langle[\widehat{p},
\widehat{H}]\rangle\)，在自由例子情况下，\(\widehat H = \frac{\hat p^2}{2m}\)，故有：
与标准差相关，探究\(x\)和\(p\)方差
定义\(x\)和\(p\)的方差为标准差的平方：
\[ \begin{aligned}& \sigma_x^2(t)=\left\langle x^2\right\rangle-\langle x\rangle^2=\left\langle(\widehat{x}-\langle x\rangle)^2\right\rangle, \\& \sigma_p^2(t)=\left\langle p^2\right\rangle-\langle p\rangle^2=\left\langle(\widehat{p}-\langle p\rangle)^2\right\rangle .\end{aligned} \]
研究以下表示：
\(i \hbar \frac{\mathrm{d}}{\mathrm{d} t} \sigma_p^2\)
\[ \begin{aligned} i \hbar \frac{\mathrm{d}}{\mathrm{d} t} \sigma_p^2 & =i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left(\left\langle p^2\right\rangle-\langle p\rangle^2\right)=i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left\langle p^2\right\rangle-i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left(\langle p\rangle^2\right)=i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left\langle p^2\right\rangle-2 i \hbar\langle p\rangle \frac{\mathrm{d}}{\mathrm{d} t}\langle p\rangle \\ & \left.=\left\langle\left[\widehat{p}^2, \widehat{H}\right]\right\rangle-2\langle p\rangle\langle\widehat{p}, \widehat{H}]\right\rangle=\left\langle\left[\widehat{p}^2, \frac{\widehat{p}^2}{2 m}\right]\right\rangle-2\langle p\rangle\left\langle\left[\widehat{p}, \frac{\widehat{p}^2}{2 m}\right]\right\rangle = 0 \end{aligned} \]
\(i \hbar \frac{\mathrm{d}}{\mathrm{d} t} \sigma_x^2\)
\[ \begin{aligned} i \hbar \frac{\mathrm{d}}{\mathrm{d} t} \sigma_x^2 & =i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left(\left\langle x^2\right\rangle-\langle x\rangle^2\right)=i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left\langle x^2\right\rangle-i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left(\langle x\rangle^2\right)=i \hbar \frac{\mathrm{d}}{\mathrm{d} t}\left\langle x^2\right\rangle-2 i \hbar\langle x\rangle \frac{\mathrm{d}}{\mathrm{d} t}\langle x\rangle \\ & =\left\langle\left[\widehat{x}^2, \widehat{H}\right]\right\rangle-2\langle x\rangle\langle[\widehat{x}, \widehat{H}]\rangle=\left\langle\left[\widehat{x}^2, \frac{\widehat{p}^2}{2 m}\right]\right\rangle-2\langle x\rangle\left\langle\left[\widehat{x}, \frac{\widehat{p}^2}{2 m}\right]\right\rangle \end{aligned} \]
应用\([\widehat{A}, \widehat{B} \widehat{C}]=\widehat{B}[\widehat{A}, \widehat{C}]+[\widehat{A}, \widehat{B}] \widehat{C},[\widehat{A} \widehat{B}, \widehat{C}]=\widehat{A}[\widehat{B}, \widehat{C}]+[\widehat{A}, \widehat{C}] \widehat{B}\)有：
- \(\left[\widehat{x}^2, \widehat{p}^2\right]=\widehat{x}\left[\widehat{x}, \widehat{p}^2\right]+\left[\widehat{x}, \widehat{p}^2\right] \widehat{x}=2 i \hbar \widehat{x} \widehat{p}+2 i \hbar \widehat{p} \widehat{x}=2 i \hbar(\widehat{x} \widehat{p}+\widehat{p} \widehat{x})\)
- \(\left[\widehat{x}, \widehat{p}^2\right]=\widehat{p}[\widehat{x}, \widehat{p}]+[\widehat{x}, \widehat{p}] \widehat{p}=i \hbar \widehat{p}+i \hbar \widehat{p}=2 i \hbar \widehat{p}\)
代入得：
\[ i \hbar \frac{\mathrm{d}}{\mathrm{d} t} \sigma_x^2=\frac{i \hbar}{m}(\langle x p+p x\rangle-2\langle x\rangle\langle p\rangle) \]
最终没有\(\hat \quad\)可能是因为它们代表的是位置和动量的期望值，而不是算符本身。当我们谈论期望值时，通常会省略算符上的帽子。
\(i \hbar \frac{\mathrm{d}^2}{\mathrm{~d} t^2} \sigma_x^2\)
\[ \begin{aligned}i \hbar \frac{\mathrm{d}^2}{\mathrm{~d} t^2} \sigma_x^2 & =\frac{i \hbar}{m} \frac{\mathrm{d}}{\mathrm{d} t}(\langle x p+p x\rangle-2\langle x\rangle\langle p\rangle)=\frac{i \hbar}{m} \frac{\mathrm{d}}{\mathrm{d} t}\langle x p+p x\rangle-2\langle x\rangle \frac{i \hbar}{m} \frac{\mathrm{d}}{\mathrm{d} t}\langle p\rangle-2\langle p\rangle \frac{i \hbar}{m} \frac{\mathrm{d}}{\mathrm{d} t}\langle x\rangle \\& \left.=\frac{1}{m}\langle[\widehat{x} \widehat{p}+\widehat{p} \widehat{x}, \widehat{H}]\rangle-\frac{2\langle x\rangle}{m}\langle\widehat{p}, \widehat{H}]\right\rangle-\frac{2\langle p\rangle}{m}\langle[\widehat{x}, \widehat{H}]\rangle \\& =\frac{1}{m}\left\langle\left[\widehat{x} \widehat{p}+\widehat{p} \widehat{x}, \frac{\widehat{p}^2}{2 m}\right]\right\rangle-\frac{2\langle x\rangle}{m}\left\langle\left[\widehat{p}, \frac{\widehat{p}^2}{2 m}\right]\right\rangle-\frac{2\langle p\rangle}{m}\left\langle\left[\widehat{x}, \frac{\widehat{p}^2}{2 m}\right]\right\rangle\end{aligned} \]
应用\([\widehat{A}, \widehat{B} \widehat{C}]=\widehat{B}[\widehat{A}, \widehat{C}]+[\widehat{A}, \widehat{B}] \widehat{C},[\widehat{A} \widehat{B}, \widehat{C}]=\widehat{A}[\widehat{B}, \widehat{C}]+[\widehat{A}, \widehat{C}] \widehat{B}\)有：
- \([\widehat x \widehat p +\widehat p \widehat x,\widehat p^2] = 2[\widehat x \widehat p ,\widehat p^2] = \widehat x [\widehat p , \widehat p^2]+[\widehat x, \widehat p^2]\widehat p = 2i\hbar\widehat p^2\)
\[ \begin{align}i \hbar \frac{\mathrm{d}^2}{\mathrm{~d} t^2} \sigma_x^2=\frac{2 i \hbar}{m^2}\left\langle p^2\right\rangle-\frac{2 i \hbar}{m^2}\langle p\rangle^2=\frac{2 i \hbar}{m^2}\left(\left\langle p^2\right\rangle-\langle p\rangle^2\right)=\frac{2 i \hbar}{m^2} \sigma_p^2\end{align} \]
然后我们尝试对\((2)\)进行两次积分，积分下限取0，上限取t:
\[ \begin{align}\sigma_x^2(t)=\frac{\sigma_p^2}{m^2} t^2+\frac{\mathrm{d} \sigma_{x 0}^2}{\mathrm{~d} t} t+\sigma_{x 0}^2\end{align} \]
进一步探究\(\sigma_x(t)\)
我们使用之前的方法来计算\(\langle x p+p x\rangle_0\)：
\[ \langle x p+p x\rangle_0=\int_{-\infty}^{+\infty} \psi^{*}(x)(\widehat{x} \widehat{p}+\widehat{p} \widehat{x}) \psi(x) \mathrm{d} x \]
注意这里与之前计算的不同在于将原本的左右矢的计算转化为了积分计算。这个等式与\(\langle A\rangle_\psi=\langle\psi|\widehat{A}| \psi\rangle\)等价。且由于\(\psi\)为实数，可以省略共轭。
展开\(\widehat p = -i\hbar\frac {\partial}{\partial x}\)，得：
\[ \begin{aligned}\langle x p+p x\rangle_0 & =\int_{-\infty}^{+\infty} \psi(x)\left(x\left(-i \hbar \psi^{\prime}(x)\right)-i \hbar(x \psi(x))^{\prime}\right) \mathrm{d} x \\ & =-i \hbar \int_{-\infty}^{+\infty}\left(\psi(x) x \psi^{\prime}(x)+\psi^2(x)+\psi(x) x \psi^{\prime}(x)\right) \mathrm{d} x \\ & =-i \hbar \int_{-\infty}^{+\infty}\left(\psi^2(x)+2 \psi(x) x \psi^{\prime}(x)\right) \mathrm{d} x=-i \hbar \int_{-\infty}^{+\infty}\left(x \psi^2(x)\right)^{\prime} \mathrm{d} x \end{aligned} \]
由于\(\psi(x, t=0)\)为实数，\(\langle x p+p x\rangle_0\)为纯虚数。然而显而易见的是\((\widehat{x} \widehat{p}+\widehat{p} \widehat{x})^{\dagger}=\widehat{p}^{\dagger} \widehat{x}^{\dagger}+\widehat{x}^{\dagger} \widehat{p}^{\dagger}=\widehat{p} \widehat{x}+\widehat{x} \widehat{p}\)，这意味着\(\widehat{p} \widehat{x}+\widehat{x} \widehat{p}\)是厄米的，即\(\langle x p+p x\rangle_0\)为实数。因此：
\[ \langle x p+p x\rangle_0=0 \]
相同的方法我们处理\(\langle p\rangle_0\)
\[ \begin{aligned}\langle p\rangle_0 & =\int_{-\infty}^{+\infty} \psi(x) \widehat{p} \psi(x) \mathrm{d} x=\int_{-\infty}^{+\infty} \psi(x)\left(-i \hbar \psi^{\prime}(x)\right) \mathrm{d} x=-i \hbar \int_{-\infty}^{+\infty} \psi(x) \psi^{\prime}(x) \mathrm{d} x \\& =-\frac{i \hbar}{2} \int_{-\infty}^{+\infty}\left(\psi^2(x)\right)^{\prime} \mathrm{d} x\end{aligned} \]
因为算符\(\widehat p\)是厄米的，我们可以得到\(\langle p\rangle_0=0\)。
在前一部分中我们计算得到了：
\[ i \hbar \frac{\mathrm{d}}{\mathrm{d} t} \sigma_x^2=\frac{i \hbar}{m}(\langle x p+p x\rangle-2\langle x\rangle\langle p\rangle) \]
带入即可获得：
\[ \left.\frac{\mathrm{d}}{\mathrm{d} t} \sigma_x^2\right|_{t=0}=\frac{1}{m}\left(\langle x p+p x\rangle_0-2\langle x\rangle_0\langle p\rangle_0\right)=0 \]
由此，我们可以得知\((3)\)的第二项为0：
\[ \begin{align}\sigma_x(t)=\sqrt{\frac{\sigma_p^2}{m^2} t^2+\sigma_{x 0}^2}\end{align} \]
\(\sigma_p\)和\(\sigma_{x0}\)都是常数。
然后，我们可以注意到，\(\sigma_x(t)\)是一个增函数，所以在\(t = 0\)时取最小值，又根据海森堡不确定性原理，有：
\[ \sigma_x(t) \sigma_p(t) \geqslant \frac{\hbar}{2} \]
因此我们可以假设\(\sigma_{x 0} \sigma_p \approx \frac{\hbar}{2}\)，\((4)\)在这种假设可以写作:
\[ \begin{align}\sigma_x(t)=\sqrt{\frac{\hbar^2}{\sigma_{x0}^24m^2} t^2+\sigma_{x 0}^2} \end{align} \]
定域性
根据\((5)\)，我们可以研究在这个假设下，影响粒子定域性的因素。定域性可以由以下式子判断:
\[ \sqrt{\sigma_x^2(t)-\sigma_{x 0}^2}=\frac{\sigma_p}{m} t=\frac{\hbar}{2 m \sigma_{x 0}} t \]
可见，粒子定域性与质量和初始位置不确定度分别呈反比。这个结论与德布罗意波长相似。
