数小仓的窝
§ QM-22 附录
2025-05-22
量子力学原理习题解
00

目录

附录
A.1 矩阵求逆
A.2 Gaussian 积分
A.3 复数
A.4 $\mathrm{i}\varepsilon$ 处方

附录

A.1 矩阵求逆

习题 A.1.1 利用上面介绍的方法，证明

(213012−111)−1=(1−212−54−13−2)\begin{pmatrix} 2 & 1 & 3\\ 0 & 1 & 2\\ -1 & 1 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 & -2 & 1\\ 2 & -5 & 4\\ -1 & 3 & -2 \end{pmatrix}

以及

(2134120−12)−1=112(−4518−4−84−22)\begin{pmatrix} 2 & 1 & 3\\ 4 & 1 & 2\\ 0 & -1 & 2 \end{pmatrix}^{-1}=\dfrac{1}{12} \begin{pmatrix} -4 & 5 & 1\\ 8 & -4 & -8\\ 4 & -2 & 2 \end{pmatrix}

解答: (1) 对于给定矩阵

M=(213012−111)M = \begin{pmatrix} 2 & 1 & 3 \\ 0 & 1 & 2 \\ -1 & 1 & 1 \end{pmatrix}

定义行矢量：

A=(2,1,3),B=(0,1,2),C=(−1,1,1)\mathbf{A} = (2, 1, 3),\quad \mathbf{B} = (0, 1, 2),\quad \mathbf{C} = (-1, 1, 1)

计算倒易矢量

AR=B×C=∣ijk012−111∣=i(1⋅1−2⋅1)−j(0⋅1−2⋅(−1))+k(0⋅1−1⋅(−1))=i(−1)−j(2)+k(1)=(−1,−2,1)\begin{align*} \mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 2 \\ -1 & 1 & 1 \end{vmatrix} \\ &= \mathbf{i}(1 \cdot 1 - 2 \cdot 1) - \mathbf{j}(0 \cdot 1 - 2 \cdot (-1)) + \mathbf{k}(0 \cdot 1 - 1 \cdot (-1)) \\ &= \mathbf{i}(-1) - \mathbf{j}(2) + \mathbf{k}(1) = (-1, -2, 1) \end{align*}
BR=C×A=∣ijk−111213∣=i(1⋅3−1⋅1)−j(−1⋅3−1⋅2)+k(−1⋅1−1⋅2)=i(2)−j(−5)+k(−3)=(2,5,−3)\begin{align*} \mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 1 \\ 2 & 1 & 3 \end{vmatrix} \\ &= \mathbf{i}(1 \cdot 3 - 1 \cdot 1) - \mathbf{j}(-1 \cdot 3 - 1 \cdot 2) + \mathbf{k}(-1 \cdot 1 - 1 \cdot 2) \\ &= \mathbf{i}(2) - \mathbf{j}(-5) + \mathbf{k}(-3) = (2, 5, -3) \end{align*}
CR=A×B=∣ijk213012∣=i(1⋅2−3⋅1)−j(2⋅2−3⋅0)+k(2⋅1−1⋅0)=i(−1)−j(4)+k(2)=(−1,−4,2)\begin{align*} \mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ 0 & 1 & 2 \end{vmatrix} \\ &= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 0) + \mathbf{k}(2 \cdot 1 - 1 \cdot 0) \\ &= \mathbf{i}(-1) - \mathbf{j}(4) + \mathbf{k}(2) = (-1, -4, 2) \end{align*}

构造余因子矩阵

M‾=(−12−1−25−41−32)\overline{M} = \begin{pmatrix} -1 & 2 & -1 \\ -2 & 5 & -4 \\ 1 & -3 & 2 \end{pmatrix}

计算行列式

det⁡M=A⋅(B×C)=∣213012−111∣=2(1⋅1−2⋅1)−1(0⋅1−2⋅(−1))+3(0⋅1−1⋅(−1))=2(−1)−1(2)+3(1)=−2−2+3=−1\begin{align*} \det M &= \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \\ &= \begin{vmatrix} 2 & 1 & 3 \\ 0 & 1 & 2 \\ -1 & 1 & 1 \end{vmatrix} \\ &= 2(1 \cdot 1 - 2 \cdot 1) - 1(0 \cdot 1 - 2 \cdot (-1)) + 3(0 \cdot 1 - 1 \cdot (-1)) \\ &= 2(-1) - 1(2) + 3(1) = -2 - 2 + 3 = -1 \end{align*}

逆矩阵为

M−1=M‾det⁡M=1−1(−12−1−25−41−32)=(1−212−54−13−2)M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-1} \begin{pmatrix} -1 & 2 & -1 \\ -2 & 5 & -4 \\ 1 & -3 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -2 & 1 \\ 2 & -5 & 4 \\ -1 & 3 & -2 \end{pmatrix}

(2) 对于给定矩阵

M=(2134120−12)M = \begin{pmatrix} 2 & 1 & 3 \\ 4 & 1 & 2 \\ 0 & -1 & 2 \end{pmatrix}

定义行矢量为

A=(2,1,3),B=(4,1,2),C=(0,−1,2)\mathbf{A} = (2, 1, 3),\quad \mathbf{B} = (4, 1, 2),\quad \mathbf{C} = (0, -1, 2)

计算倒易矢量

AR=B×C=∣ijk4120−12∣=i(1⋅2−2⋅(−1))−j(4⋅2−2⋅0)+k(4⋅(−1)−1⋅0)=i(4)−j(8)+k(−4)=(4,−8,−4)\begin{align*} \mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 1 & 2 \\ 0 & -1 & 2 \end{vmatrix} \\ &= \mathbf{i}(1 \cdot 2 - 2 \cdot (-1)) - \mathbf{j}(4 \cdot 2 - 2 \cdot 0) + \mathbf{k}(4 \cdot (-1) - 1 \cdot 0) \\ &= \mathbf{i}(4) - \mathbf{j}(8) + \mathbf{k}(-4) = (4, -8, -4) \end{align*}
BR=C×A=∣ijk0−12213∣=i((−1)⋅3−2⋅1)−j(0⋅3−2⋅2)+k(0⋅1−(−1)⋅2)=i(−5)−j(−4)+k(2)=(−5,4,2)\begin{align*} \mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 2 \\ 2 & 1 & 3 \end{vmatrix} \\ &= \mathbf{i}((-1) \cdot 3 - 2 \cdot 1) - \mathbf{j}(0 \cdot 3 - 2 \cdot 2) + \mathbf{k}(0 \cdot 1 - (-1) \cdot 2) \\ &= \mathbf{i}(-5) - \mathbf{j}(-4) + \mathbf{k}(2) = (-5, 4, 2) \end{align*}
CR=A×B=∣ijk213412∣=i(1⋅2−3⋅1)−j(2⋅2−3⋅4)+k(2⋅1−1⋅4)=i(−1)−j(−8)+k(−2)=(−1,8,−2)\begin{align*} \mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ 4 & 1 & 2 \end{vmatrix} \\ &= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 4) + \mathbf{k}(2 \cdot 1 - 1 \cdot 4) \\ &= \mathbf{i}(-1) - \mathbf{j}(-8) + \mathbf{k}(-2) = (-1, 8, -2) \end{align*}

构造余因子矩阵：

M‾=(4−5−1−848−42−2)\overline{M} = \begin{pmatrix} 4 & -5 & -1 \\ -8 & 4 & 8 \\ -4 & 2 & -2 \end{pmatrix}

行列式为

det⁡M=A⋅(B×C)=∣2134120−12∣=2(1⋅2−2⋅(−1))−1(4⋅2−2⋅0)+3(4⋅(−1)−1⋅0)=2(4)−1(8)+3(−4)=8−8−12=−12\begin{align*} \det M &= \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \\ &= \begin{vmatrix} 2 & 1 & 3 \\ 4 & 1 & 2 \\ 0 & -1 & 2 \end{vmatrix} \\ &= 2(1 \cdot 2 - 2 \cdot (-1)) - 1(4 \cdot 2 - 2 \cdot 0) + 3(4 \cdot (-1) - 1 \cdot 0) \\ &= 2(4) - 1(8) + 3(-4) = 8 - 8 - 12 = -12 \end{align*}

逆矩阵为

M−1=M‾det⁡M=1−12(4−5−1−848−42−2)=112(−4518−4−84−22)M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-12} \begin{pmatrix} 4 & -5 & -1 \\ -8 & 4 & 8 \\ -4 & 2 & -2 \end{pmatrix} = \frac{1}{12} \begin{pmatrix} -4 & 5 & 1 \\ 8 & -4 & -8 \\ 4 & -2 & 2 \end{pmatrix}
 ■~\tag*{$\blacksquare$}

A.2 Gaussian 积分

A.3 复数

A.4 iε\mathrm{i}\varepsilon 处方