泡利矩阵

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泡利矩阵

泡利矩阵的一些性质

泡利矩阵

定义

\[\sigma_{x}=\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right), \quad \sigma_{y}=\left(\begin{array}{cc} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{array}\right), \quad \sigma_{z}=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\]

这些都是厄米矩阵, 它们的特征方程都是:

\[\lambda^{2}-1=0\]

因而 $\sigma_{x} 、 \sigma_{y}$ 及 $\sigma_{z}$ 的本征值都是:

\[\lambda=\pm 1\]

本征矢

\[\begin{array}{l} \sigma_{x}|\pm\rangle_{x}=\pm|\pm\rangle_{x} \\ \sigma_{y}|\pm\rangle_{y}=\pm|\pm\rangle_{y} \\ \sigma_{z}|\pm\rangle=\pm|\pm\rangle \end{array}\] \[\begin{aligned} |\pm\rangle_{x} &=\frac{1}{\sqrt{2}}[|+\rangle \pm|-\rangle] \\ |\pm\rangle_{y} &=\frac{1}{\sqrt{2}}[|+\rangle \pm \mathrm{i}|-\rangle] \end{aligned}\]

性质

\[\operatorname{Det}\left(\sigma_{j}\right)=-1\] \[\operatorname{Tr}\left(\sigma_{j}\right)=0\] \[\sigma_{j} \sigma_{k}=\mathrm{i} \sum_{l} \varepsilon_{j k l} \sigma_{l}+\delta_{j k} I\] \[[\sigma_j,\sigma_k]=2i\varepsilon_{jkl}\sigma_l\] \[\sigma_{x} \sigma_{y}+\sigma_{y} \sigma_{x}=0\] \[\sigma_{x} \sigma_{y} \sigma_{z}=\mathrm{i} I\] \[(\boldsymbol{\sigma} \cdot \boldsymbol{A})(\boldsymbol{\sigma} \cdot \boldsymbol{B})=\boldsymbol{A} \cdot \boldsymbol{B} I+\mathrm{i} \boldsymbol{\sigma} \cdot(\boldsymbol{A} \times \boldsymbol{B})\]

2x2矩阵空间的基

我们来考虑一个任意的 $2 \times 2$ 矩阵

\[M=\left(\begin{array}{ll} m_{11} & m_{12} \\ m_{21} & m_{22} \end{array}\right)\]

我们总可以 将它写作下列四个矩阵:

\[I, \sigma_{x}, \sigma_{y}, \sigma_{z}\] \[M=\frac{m_{11}+m_{22}}{2} I+\frac{m_{11}-m_{22}}{2} \sigma_{z}+\frac{m_{12}+m_{21}}{2} \sigma_{x}+\mathrm{i} \frac{m_{12}-m_{21}}{2} \sigma_{y}\] \[M=a_{0} I+\boldsymbol{a} \cdot \boldsymbol{\sigma}\]

可以看出, 当而且仅当系数$a_{0}$ 及 $\boldsymbol{a}$ 都是实数时, $M$ 才是厄米矩阵. 利用矩阵 $M$, 我们可将这些系数在形式上写作:

\[a_{0}=\frac{1}{2} \operatorname{Tr}(M) \qquad \boldsymbol{a}=\frac{1}{2} \operatorname{Tr}(M \sigma)\]

对角化

\[(H)=\left(\begin{array}{ll} H_{11} & H_{12} \\ H_{21} & H_{22} \end{array}\right)\] \[H_{12}=H_{21}^{*}\]

为厄密矩阵。表示一个厄密算符,可以写作

\[(H)=\left(\begin{array}{cc} \frac{1}{2}\left(H_{11}+H_{22}\right) & 0 \\ 0 & \frac{1}{2}\left(H_{11}+H_{22}\right) \end{array}\right)+\left(\begin{array}{cc} \frac{1}{2}\left(H_{11}-H_{22}\right) & H_{12} \\ H_{21} & -\frac{1}{2}\left(H_{11}-H_{22}\right) \end{array}\right)\]

算符为

\[H=\frac{1}{2}\left(H_{11}+H_{22}\right) \mathbb{1}+\frac{1}{2}\left(H_{11}-H_{22}\right) K\] \[(K)=\left(\begin{array}{cc} 1 & \frac{2 H_{12}}{H_{11}-H_{22}} \\ \frac{2 H_{21}}{H_{11}-H_{22}} & -1 \end{array}\right)\]

H和K有相同的本征矢。记为$\ket{\psi_\pm}$,相应本征值为$E_{\pm}$和$\kappa_{\pm}$

\[E_{\pm}=\frac{1}{2}\left(H_{11}+H_{22}\right)+\frac{1}{2}\left(H_{11}-H_{22}\right) \kappa_{\pm}\]

定义:

\[\tan \theta=\frac{2\left|H_{21}\right|}{H_{11}-H_{22}} \quad(0 \leqslant \theta<\pi)\] \[H_{21}=\left|H_{21}\right| \mathrm{e}^{\mathrm{i} \varphi} \quad(0 \leqslant \varphi<2 \pi)\] \[(K)=\left(\begin{array}{cc} 1 & \tan \theta \mathrm{e}^{-\mathrm{i} \varphi} \\ \tan \theta \mathrm{e}^{\mathrm{i} \varphi} & -1 \end{array}\right)\]

特征方程:

\[\operatorname{Det}[(K)-\kappa I]=\kappa^{2}-1-\tan ^{2} \theta=0\] \[\kappa_{+}=+\frac{1}{\cos \theta}\] \[\kappa_{-}=-\frac{1}{\cos \theta}\] \[\frac{1}{\cos \theta}=\frac{\sqrt{\left(H_{11}-H_{22}\right)^{2}+4\left|H_{12}\right|^{2}}}{H_{11}-H_{22}}\] \[E_{+}=\frac{1}{2}\left(H_{11}+H_{22}\right)+\frac{1}{2} \sqrt{\left(H_{11}-H_{22}\right)^{2}+4\left|H_{12}\right|^{2}}\] \[E_{-}=\frac{1}{2}\left(H_{11}+H_{22}\right)-\frac{1}{2} \sqrt{\left(H_{11}-H_{22}\right)^{2}+4\left|H_{12}\right|^{2}}\]

如果要使 $E_{+}=E_{-}$, 就必须使 $\left(H_{11}-H_{22}\right)^{2}+4\lvert H_{12}\rvert^{2}=0$, 也就是 应该使 $H_{11}=H_{22}$ 及 $H_{12}=H_{21}=0$. 由此可见, 具有简并本征值的 $2 \times 2$ 厄米矩阵一定和单位矩阵成正比.

本征矢

\[\left(\begin{array}{cc} 1 & \tan \theta \mathrm{e}^{-\mathrm{i} \varphi} \\ \tan \theta \mathrm{e}^{\mathrm{i} \varphi} & -1 \end{array}\right)\left(\begin{array}{l} a \\ b \end{array}\right)=\frac{1}{\cos \theta}\left(\begin{array}{l} a \\ b \end{array}\right)\] \[\left(1-\frac{1}{\cos \theta}\right) a+\tan \theta \mathrm{e}^{-\mathrm{i} \varphi} b=0\] \[-\left(\sin \frac{\theta}{2} \mathrm{e}^{\mathrm{i} \varphi / 2}\right) a+\left(\cos \frac{\theta}{2} \mathrm{e}^{-\mathrm{i} \varphi / 2}\right) b=0\] \[\left|\psi_{+}\right\rangle=\cos \frac{\theta}{2} \mathrm{e}^{-\mathrm{i} \varphi / 2}\left|\varphi_{1}\right\rangle+\sin \frac{\theta}{2} \mathrm{e}^{\mathrm{i} \varphi / 2}\left|\varphi_{2}\right\rangle\] \[\left|\psi_{-}\right\rangle=-\sin \frac{\theta}{2} \mathrm{e}^{-\mathrm{i} \varphi / 2}\left|\varphi_{1}\right\rangle+\cos \frac{\theta}{2} \mathrm{e}^{\mathrm{i} \varphi / 2}\left|\varphi_{2}\right\rangle\]