和達三樹《物理用數學》習題2.3.3，Pauli矩陣

==問題==

令$\sigma_1 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right], \sigma_2 = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right], \sigma_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$，求證：
(i) $\sigma_1, \sigma_2, \sigma_3$為Hermite矩陣。
(ii) $\sigma_1 \sigma_1 = I, \sigma_2 \sigma_2 = I, \sigma_3 \sigma_3 = I$。
(iii) $\sigma_1 \sigma_2 = i \sigma_3, \sigma_2 \sigma_3 = i \sigma_1, \sigma_3 \sigma_1 = i \sigma_2$。
$\sigma_1, \sigma_2, \sigma_3$稱為Pauli矩陣。

==解答==

(i) 首先確認2階Hermite矩陣的形式。

設${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$，其中$z_{11}, z_{12}, z_{21}, z_{22} \in \mathbb{C}$。

若${\rm A}$為Hermite矩陣，則意味著${\rm A}$滿足
$${\rm A}^{\dagger} = {\rm A}.$$
其中${\rm A}^{\dagger} = \left( \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]^* \right)^{\rm T} = \left( \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{12}} \\ \bar{z_{21}} & \bar{z_{22}} \end{array} \right] \right)^{\rm T} = \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right]$，於是有
$$\left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right] = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right].$$
從而得到
$$\bar{z_{11}} = z_{11}, \bar{z_{21}} = z_{12}, \bar{z_{12}} = z_{21}, \bar{z_{22}} = z_{22}.$$
所以$z_{11}, z_{22}$皆為實數，而$z_{12}$與$z_{21}$互為共軛複數(因此實部相同)。

總結上述，我們可以得到：${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$為Hermite矩陣，若且唯若${\rm A}$形如
$${\rm A} = \left[ \begin{array}{cc} a & p+qi \\ p-qi & b \end{array} \right], a, b, p, q \in \mathbb{R}.$$
(主對角線元素為實數，副對角線元素為共軛複數對)

現在回頭來判定Pauli矩陣是否為Hermite矩陣。

顯然$\sigma_1, \sigma_2, \sigma_3$都是Hermite矩陣。

(ii) 直接計算。

$\sigma_1 \sigma_1 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I.$

注意對矩陣$\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$左乘$\sigma_1 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$的效果相當於「交換1st row與2nd row」，因此可以利用此性質立刻寫出$\sigma_1 \sigma_1$的乘積。

$\sigma_2 \sigma_2 = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = i \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = i \left[ \begin{array}{cc} -i & 0 \\ 0 & -i \end{array} \right] = \left[ \begin{array}{cc} -i^2 & 0 \\ 0 & -i^2 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I.$

這裡一樣使用了基本矩陣(Elementary matrix)的性質。對矩陣$\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$左乘$\left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right]$的效果相當於「先將2nd row的元素添上負號後，再與1st row交換」。

$\sigma_3 \sigma_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ -0 & -(-1) \end{array} \right] = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] = I.$

這裡一樣使用了基本矩陣(Elementary matrix)的性質。對矩陣$\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]$左乘$\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$的效果相當於「將2nd row的元素添上負號」。

(iii) 與(ii)相同，直接計算，但一樣使用基本矩陣的性質來化簡計算量。

$\sigma_1 \sigma_2 = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = \left[ \begin{array}{cc} i & 0 \\ 0 & -i \end{array} \right] = i \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = i \sigma_3$。

$\sigma_2 \sigma_3 = \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = i \left[ \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] = i \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = i \sigma_1$。

$\sigma_3 \sigma_1 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & -i^2 \\ i^2 & 0 \end{array} \right] = i \left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right] = i \sigma_2$。

(解答結束)

==附註==

維基百科關於Pauli矩陣的介紹：https://zh.wikipedia.org/wiki/%E6%B3%A1%E5%88%A9%E7%9F%A9%E9%99%A3