C2驾照：科目一

C2 Driver’s License: Subject 1

# 附录

## A.1 矩阵求逆

**习题 A.1.1** 利用上面介绍的方法，证明

$$
\begin{pmatrix}
2 & 1 & 3\\
0 & 1 & 2\\
-1 & 1 & 1
\end{pmatrix}^{-1}=\begin{pmatrix}
1 & -2 & 1\\
2 & -5 & 4\\
-1 & 3 & -2
\end{pmatrix}
$$

以及

$$
\begin{pmatrix}
2 & 1 & 3\\
4 & 1 & 2\\
0 & -1 & 2
\end{pmatrix}^{-1}=\dfrac{1}{12}
\begin{pmatrix}
-4 & 5 & 1\\
8 & -4 & -8\\
4 & -2 & 2
\end{pmatrix}
$$



**解答:** (1) 对于给定矩阵

$$
M = \begin{pmatrix}
2 & 1 & 3 \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{pmatrix}
$$

定义行矢量：

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

计算倒易矢量

$$
\begin{align*}
\mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 1 - 2 \cdot 1) - \mathbf{j}(0 \cdot 1 - 2 \cdot (-1)) + \mathbf{k}(0 \cdot 1 - 1 \cdot (-1)) \\
&= \mathbf{i}(-1) - \mathbf{j}(2) + \mathbf{k}(1) = (-1, -2, 1)
\end{align*}
$$

$$
\begin{align*}
\mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 1 & 1 \\
2 & 1 & 3
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 3 - 1 \cdot 1) - \mathbf{j}(-1 \cdot 3 - 1 \cdot 2) + \mathbf{k}(-1 \cdot 1 - 1 \cdot 2) \\
&= \mathbf{i}(2) - \mathbf{j}(-5) + \mathbf{k}(-3) = (2, 5, -3)
\end{align*}
$$

$$
\begin{align*}
\mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 3 \\
0 & 1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 0) + \mathbf{k}(2 \cdot 1 - 1 \cdot 0) \\
&= \mathbf{i}(-1) - \mathbf{j}(4) + \mathbf{k}(2) = (-1, -4, 2)
\end{align*}
$$

构造余因子矩阵

$$
\overline{M} = \begin{pmatrix}
-1 & 2 & -1 \\
-2 & 5 & -4 \\
1 & -3 & 2
\end{pmatrix}
$$

计算行列式

$$
\begin{align*}
\det M &= \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \\
&= \begin{vmatrix}
2 & 1 & 3 \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{vmatrix} \\
&= 2(1 \cdot 1 - 2 \cdot 1) - 1(0 \cdot 1 - 2 \cdot (-1)) + 3(0 \cdot 1 - 1 \cdot (-1)) \\
&= 2(-1) - 1(2) + 3(1) = -2 - 2 + 3 = -1
\end{align*}
$$

逆矩阵为

$$
M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-1} \begin{pmatrix}
-1 & 2 & -1 \\
-2 & 5 & -4 \\
1 & -3 & 2
\end{pmatrix} = \begin{pmatrix}
1 & -2 & 1 \\
2 & -5 & 4 \\
-1 & 3 & -2
\end{pmatrix}
$$

(2) 对于给定矩阵

$$
M = \begin{pmatrix}
2 & 1 & 3 \\
4 & 1 & 2 \\
0 & -1 & 2
\end{pmatrix}
$$

定义行矢量为

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

计算倒易矢量

$$
\begin{align*}
\mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
4 & 1 & 2 \\
0 & -1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 2 \cdot (-1)) - \mathbf{j}(4 \cdot 2 - 2 \cdot 0) + \mathbf{k}(4 \cdot (-1) - 1 \cdot 0) \\
&= \mathbf{i}(4) - \mathbf{j}(8) + \mathbf{k}(-4) = (4, -8, -4)
\end{align*}
$$

$$
\begin{align*}
\mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & -1 & 2 \\
2 & 1 & 3
\end{vmatrix} \\
&= \mathbf{i}((-1) \cdot 3 - 2 \cdot 1) - \mathbf{j}(0 \cdot 3 - 2 \cdot 2) + \mathbf{k}(0 \cdot 1 - (-1) \cdot 2) \\
&= \mathbf{i}(-5) - \mathbf{j}(-4) + \mathbf{k}(2) = (-5, 4, 2)
\end{align*}
$$

$$
\begin{align*}
\mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 3 \\
4 & 1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 4) + \mathbf{k}(2 \cdot 1 - 1 \cdot 4) \\
&= \mathbf{i}(-1) - \mathbf{j}(-8) + \mathbf{k}(-2) = (-1, 8, -2)
\end{align*}
$$

构造余因子矩阵：

$$
\overline{M} = \begin{pmatrix}
4 & -5 & -1 \\
-8 & 4 & 8 \\
-4 & 2 & -2
\end{pmatrix}
$$

行列式为

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

逆矩阵为

$$
M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-12} \begin{pmatrix}
4 & -5 & -1 \\
-8 & 4 & 8 \\
-4 & 2 & -2
\end{pmatrix} = \frac{1}{12} \begin{pmatrix}
-4 & 5 & 1 \\
8 & -4 & -8 \\
4 & -2 & 2
\end{pmatrix}
$$

$$
~\tag*{$\blacksquare$}
$$

## A.2 Gaussian 积分

## A.3 复数

## A.4 $\mathrm{i}\varepsilon$ 处方

# Appendix
## A.1 Matrix Inversion
**Exercise A.1.1** Using the Method described above, show that
$$
\begin{pmatrix}
2 & 1 & 3\\
0 & 1 & 2\\
-1 & 1 & 1
\end{pmatrix}^{-1}=\begin{pmatrix}
1 & -2 & 1\\
2 & -5 & 4\\
-1 & 3 & -2
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
2 & 1 & 3\\
4 & 1 & 2\\
0 & -1 & 2
\end{pmatrix}^{-1}=\dfrac{1}{12}
\begin{pmatrix}
-4 & 5 & 1\\
8 & -4 & -8\\
4 & -2 & 2
\end{pmatrix}
$$

**Solution.** (1) For given matrix
$$
M = \begin{pmatrix}
2 & 1 & 3 \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{pmatrix}
$$
Define the row vectors:
$$
\mathbf{A} = (2, 1, 3),\quad \mathbf{B} = (0, 1, 2),\quad \mathbf{C} = (-1, 1, 1)
$$
Compute the reciprocal vectors
$$
\begin{align*}
\mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 1 - 2 \cdot 1) - \mathbf{j}(0 \cdot 1 - 2 \cdot (-1)) + \mathbf{k}(0 \cdot 1 - 1 \cdot (-1)) \\
&= \mathbf{i}(-1) - \mathbf{j}(2) + \mathbf{k}(1) = (-1, -2, 1)
\end{align*}
$$
$$
\begin{align*}
\mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 1 & 1 \\
2 & 1 & 3
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 3 - 1 \cdot 1) - \mathbf{j}(-1 \cdot 3 - 1 \cdot 2) + \mathbf{k}(-1 \cdot 1 - 1 \cdot 2) \\
&= \mathbf{i}(2) - \mathbf{j}(-5) + \mathbf{k}(-3) = (2, 5, -3)
\end{align*}
$$
$$
\begin{align*}
\mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 3 \\
0 & 1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 0) + \mathbf{k}(2 \cdot 1 - 1 \cdot 0) \\
&= \mathbf{i}(-1) - \mathbf{j}(4) + \mathbf{k}(2) = (-1, -4, 2)
\end{align*}
$$
Construct the cofactor matrix
$$
\overline{M} = \begin{pmatrix}
-1 & 2 & -1 \\
-2 & 5 & -4 \\
1 & -3 & 2
\end{pmatrix}
$$
Compute the determinant
$$
\begin{align*}
\det M &= \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \\
&= \begin{vmatrix}
2 & 1 & 3 \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{vmatrix} \\
&= 2(1 \cdot 1 - 2 \cdot 1) - 1(0 \cdot 1 - 2 \cdot (-1)) + 3(0 \cdot 1 - 1 \cdot (-1)) \\
&= 2(-1) - 1(2) + 3(1) = -2 - 2 + 3 = -1
\end{align*}
$$
The inverse matrix is
$$
M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-1} \begin{pmatrix}
-1 & 2 & -1 \\
-2 & 5 & -4 \\
1 & -3 & 2
\end{pmatrix} = \begin{pmatrix}
1 & -2 & 1 \\
2 & -5 & 4 \\
-1 & 3 & -2
\end{pmatrix}
$$
(2) For given matrix
$$
M = \begin{pmatrix}
2 & 1 & 3 \\
4 & 1 & 2 \\
0 & -1 & 2
\end{pmatrix}
$$
Define the row vectors as
$$
\mathbf{A} = (2, 1, 3),\quad \mathbf{B} = (4, 1, 2),\quad \mathbf{C} = (0, -1, 2)
$$
Compute the reciprocal vectors
$$
\begin{align*}
\mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
4 & 1 & 2 \\
0 & -1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 2 \cdot (-1)) - \mathbf{j}(4 \cdot 2 - 2 \cdot 0) + \mathbf{k}(4 \cdot (-1) - 1 \cdot 0) \\
&= \mathbf{i}(4) - \mathbf{j}(8) + \mathbf{k}(-4) = (4, -8, -4)
\end{align*}
$$
$$
\begin{align*}
\mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & -1 & 2 \\
2 & 1 & 3
\end{vmatrix} \\
&= \mathbf{i}((-1) \cdot 3 - 2 \cdot 1) - \mathbf{j}(0 \cdot 3 - 2 \cdot 2) + \mathbf{k}(0 \cdot 1 - (-1) \cdot 2) \\
&= \mathbf{i}(-5) - \mathbf{j}(-4) + \mathbf{k}(2) = (-5, 4, 2)
\end{align*}
$$
$$
\begin{align*}
\mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 3 \\
4 & 1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 4) + \mathbf{k}(2 \cdot 1 - 1 \cdot 4) \\
&= \mathbf{i}(-1) - \mathbf{j}(-8) + \mathbf{k}(-2) = (-1, 8, -2)
\end{align*}
$$
Construct the cofactor matrix:
$$
\overline{M} = \begin{pmatrix}
4 & -5 & -1 \\
-8 & 4 & 8 \\
-4 & 2 & -2
\end{pmatrix}
$$
The determinant is
$$
\begin{align*}
\det M &= \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \\
&= \begin{vmatrix}
2 & 1 & 3 \\
4 & 1 & 2 \\
0 & -1 & 2
\end{vmatrix} \\
&= 2(1 \cdot 2 - 2 \cdot (-1)) - 1(4 \cdot 2 - 2 \cdot 0) + 3(4 \cdot (-1) - 1 \cdot 0) \\
&= 2(4) - 1(8) + 3(-4) = 8 - 8 - 12 = -12
\end{align*}
$$
Inverse matrix is 
$$
M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-12} \begin{pmatrix}
4 & -5 & -1 \\
-8 & 4 & 8 \\
-4 & 2 & -2
\end{pmatrix} = \frac{1}{12} \begin{pmatrix}
-4 & 5 & 1 \\
8 & -4 & -8 \\
4 & -2 & 2
\end{pmatrix}
$$
$$
~\tag*{$\blacksquare$}
$$
## A.2 Gaussian Integrals
## A.3 Complex Numbers
## A.4 The $\mathrm{i}\varepsilon$ Prescription

§ QM-22 附录

§ QM-22 Appendix

# 第21章 路径积分 II

## 21.1 路径积分的推导

## 21.2 虚时间形式

## 21.3 自旋与费米子路径积分

## 21.4 总结

# Chapter 21 Path Integrals——II
## 21.1 Derivation of the Path Integral
## 21.2 Imaginary Time Formalism
## 21.3 Spin and Fermion Path Integrals
## 21.4 Summary

§ QM-21 路径积分 II

§ QM-21 Path Integrals——II

# 第20章 Dirac 方程
## 20.1 自由粒子的 Dirac 方程
**习题 20.1.1** 推导连续性方程
$$
\dfrac{\partial P}{\partial t}+\nabla\cdot\mathbf{j}=0
$$
其中
$P=\psi^{\dagger}\psi$
以及
$\mathbf{j}=c\psi^{\dagger}\mathbf{\alpha}\psi$。



**解答。**
从 Dirac 方程出发：
$$
\begin{aligned}
\mathrm{i}\hbar\dfrac{\partial\psi}{\partial t}&=\left(c\mathbf{\alpha}\cdot\mathbf{P}+\beta mc^{2}\right)\psi\\
\dfrac{\partial\psi}{\partial t}&=-c\mathbf{\alpha}\cdot\nabla\psi-\dfrac{\mathrm{i}}{\hbar}\beta mc^{2}\psi
\end{aligned}
$$
其中
$\mathbf{P}=-\mathrm{i}\hbar\nabla$。
对等式两边取 $\dagger$，得到
$$
\dfrac{\partial\psi^{\dagger}}{\partial t}=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}^{\dagger}+\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta^{\dagger} mc^{2}
$$
由于
$$
\begin{aligned}
\mathbf{\alpha}^{\dagger}&=\mathbf{\alpha}\\
\beta^{\dagger}&=\beta
\end{aligned}
$$
因此得到
$$
\dfrac{\partial\psi^{\dagger}}{\partial t}=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}+\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}
$$
于是
$$
\begin{aligned}
\dfrac{\partial P}{\partial t}=\dfrac{\partial}{\partial t}(\psi^{\dagger}\psi)&=\dfrac{\partial\psi^{\dagger}}{\partial t}\psi+\psi^{\dagger}\dfrac{\partial \psi}{\partial t}\\
&=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}\psi+\cancel{\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}\psi}-\psi^{\dagger}c\mathbf{\alpha}\cdot\nabla\psi-\cancel{\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}\psi}\\
&=-c\left(\nabla\psi^{\dagger}\cdot\mathbf{\alpha}\psi+\psi^{\dagger}\mathbf{\alpha}\cdot\nabla\psi\right)\\
&=-\nabla\cdot(c\psi^{\dagger}\mathbf{\alpha}\psi)=-\nabla\cdot\mathbf{j}
\end{aligned}
$$
最终得到连续性方程
$$
\dfrac{\partial P}{\partial t}+\nabla\cdot\mathbf{j}=0
$$
$$
~\tag*{$\blacksquare$}
$$

## 20.2 Dirac 粒子的电磁相互作用
**习题 20.2.1** 推导方程 (20.2.16)。

**解答。**
为了计算
$\boldsymbol{\pi}\times\boldsymbol{\pi}$，
可以先计算它的第 $i$ 个分量：
$$
\begin{aligned}
(\boldsymbol{\pi}\times\boldsymbol{\pi})_{i}&=\varepsilon_{ijk}\pi_{j}\pi_{k}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}+\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}+\dfrac{1}{2}\varepsilon_{ikj}\pi_{k}\pi_{j}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}-\dfrac{1}{2}\varepsilon_{ijk}\pi_{k}\pi_{j}\\
&=\dfrac{1}{2}\varepsilon_{ijk}[\pi_{j},\pi_{k}]
\end{aligned}
$$
因此，需要计算力学动量算符各分量之间的对易关系。根据定义，
$$
\begin{aligned}
[\pi_{j},\pi_{k}]&=\left[P_{j}-\dfrac{q}{c}A_{j},P_{k}-\dfrac{q}{c}A_{k}\right]\\
&=[P_{j},P_{k}]-\dfrac{q}{c}[P_{j},A_{k}]-\dfrac{q}{c}[A_{j},P_{k}]+\left(\dfrac{q}{c}\right)^{2}[A_{j},A_{k}]
\end{aligned}
$$
下面计算这些对易子：
$$
\begin{aligned}
[P_{j},P_{k}]\psi&=(P_{j}P_{k}-P_{k}P_{j})\psi=P_{j}\left(-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{k}}\right)-P_{k}\left(-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{j}}\right)\\
&=(-\mathrm{i}\hbar)^{2}\dfrac{\partial}{\partial x_{j}}\left(\dfrac{\partial\psi}{\partial x_{k}}\right)-(-\mathrm{i}\hbar)^{2}\dfrac{\partial}{\partial x_{k}}\left(\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\hbar^{2}\dfrac{\partial^{2}\psi}{\partial x_{j}\partial x_{k}}+\hbar^{2}\dfrac{\partial^{2}\psi}{\partial x_{k}\partial x_{j}}=0
\end{aligned}
$$
$$
\begin{aligned}
[P_{j},A_{k}]\psi&=(P_{j}A_{k}-A_{k}P_{j})\psi=P_{j}(A_{k}\psi)-A_{k}(P_{j}\psi)\\
&=-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{j}}(A_{k}\psi)-A_{k}\left(-\mathrm{i}\hbar\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\mathrm{i}\hbar\left(\dfrac{\partial A_{k}}{\partial x_{j}}\psi+A_{k}\dfrac{\partial\psi}{\partial x_{j}}\right)-A_{k}\left(-\mathrm{i}\hbar\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\mathrm{i}\hbar\dfrac{\partial A_{k}}{\partial x_{j}}\psi
\end{aligned}
$$
$$
\begin{aligned}
[A_{j},P_{k}]\psi&=(A_{j}P_{k}-P_{k}A_{j})\psi=A_{j}(P_{k}\psi)-P_{k}(A_{j}\psi)\\
&=A_{j}\left(-\mathrm{i}\hbar\dfrac{\partial \psi}{\partial x_{k}}\right)-(-\mathrm{i}\hbar)\dfrac{\partial}{\partial x_{k}}(A_{j}\psi)\\
&=A_{j}\left(-\mathrm{i}\hbar\dfrac{\partial \psi}{\partial x_{k}}\right)+\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\psi+\mathrm{i}\hbar A_{j}\dfrac{\partial\psi}{\partial x_{k}}\\
&=\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\psi
\end{aligned}
$$
$$
[A_{j},A_{k}]\psi=(A_{j}A_{k}-A_{k}A_{j})\psi=(A_{j}A_{k}-A_{j}A_{k})\psi=0
$$
因此，
$$
\begin{aligned}
(\boldsymbol{\pi}\times\boldsymbol{\pi})_{i}&=\dfrac{1}{2}\varepsilon_{ijk}[\pi_{j},\pi_{k}]\\
&=-\dfrac{q}{c}\left(-\mathrm{i}\hbar\dfrac{\partial A_{k}}{\partial x_{j}}+\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\right)\\
&=\dfrac{\mathrm{i}q\hbar}{c}\left(\dfrac{\partial A_{k}}{\partial x_{j}}-\dfrac{\partial A_{j}}{\partial x_{k}}\right)\\
&=\dfrac{\mathrm{i}q\hbar}{c}\varepsilon_{ijk}\dfrac{\partial A_{k}}{\partial x_{j}}\\
&=\dfrac{\mathrm{i}q\hbar}{c}B_{i}
\end{aligned}
$$
于是得到方程 (20.2.16)：
$$
\boldsymbol{\pi}\times\boldsymbol{\pi}=\dfrac{\mathrm{i}q\hbar}{c}\mathbf{B}
$$
$$
~\tag*{$\blacksquare$}
$$

**习题 20.2.2** 求 Dirac 粒子在匀强磁场
$\mathbf{B}=B_{0}\mathbf{k}$
中的精确能级。假设
$\mathbf{A}=(B_{0}/2)(-y\mathbf{i}+x\mathbf{j})$。
参考习题 12.3.8。（写出 $\chi$ 所满足的方程。）

## 20.3 相对论量子力学的进一步讨论

# Chapter 20 The Dirac Equation
## 20.1 The Free-Particle Dirac Equation
**Exercise 20.1.1** Derive the continuity equation
$$
\dfrac{\partial P}{\partial t}+\nabla\cdot\mathbf{j}=0
$$
where $P=\psi^{\dagger}\psi$ and $\mathbf{j}=c\psi^{\dagger}\mathbf{\alpha}\psi$.



**Solution.**
Start from the Dirac equation
$$
\begin{aligned}
\mathrm{i}\hbar\dfrac{\partial\psi}{\partial t}&=\left(c\mathbf{\alpha}\cdot\mathbf{P}+\beta mc^{2}\right)\psi\\
\dfrac{\partial\psi}{\partial t}&=-c\mathbf{\alpha}\cdot\nabla\psi-\dfrac{\mathrm{i}}{\hbar}\beta mc^{2}\psi
\end{aligned}
$$
where $\mathbf{P}=-\mathrm{i}\hbar\nabla$. Take $\dagger$ on the both side, we get 
$$
\dfrac{\partial\psi^{\dagger}}{\partial t}=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}^{\dagger}+\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta^{\dagger} mc^{2}
$$
Since
$$
\begin{aligned}
\mathbf{\alpha}^{\dagger}&=\mathbf{\alpha}\\
\beta^{\dagger}&=\beta
\end{aligned}
$$
We get
$$
\dfrac{\partial\psi^{\dagger}}{\partial t}=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}+\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}
$$
Thus
$$
\begin{aligned}
\dfrac{\partial P}{\partial t}=\dfrac{\partial}{\partial t}(\psi^{\dagger}\psi)&=\dfrac{\partial\psi^{\dagger}}{\partial t}\psi+\psi^{\dagger}\dfrac{\partial \psi}{\partial t}\\
&=-c\nabla\psi^{\dagger}\cdot\mathbf{\alpha}\psi+\cancel{\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}\psi}-\psi^{\dagger}c\mathbf{\alpha}\cdot\nabla\psi-\cancel{\dfrac{\mathrm{i}}{\hbar}\psi^{\dagger}\beta mc^{2}\psi}\\
&=-c\left(\nabla\psi^{\dagger}\cdot\mathbf{\alpha}\psi+\psi^{\dagger}\mathbf{\alpha}\cdot\nabla\psi\right)\\
&=-\nabla\cdot(c\psi^{\dagger}\mathbf{\alpha}\psi)=-\nabla\cdot\mathbf{j}
\end{aligned}
$$
Finally, we get the continuity equation
$$
\dfrac{\partial P}{\partial t}+\nabla\cdot\mathbf{j}=0
$$
$$
~\tag*{$\blacksquare$}
$$
## 20.2 Electromagnetic Interaction of the Dirac Paritlce
**Exercise 20.2.1** Derive Eq. (20.2.16).

**Solution.** To compute the $\boldsymbol{\pi}\times\boldsymbol{\pi}$, we could compute the $i$-th component first:
$$
\begin{aligned}
(\boldsymbol{\pi}\times\boldsymbol{\pi})_{i}&=\varepsilon_{ijk}\pi_{j}\pi_{k}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}+\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}+\dfrac{1}{2}\varepsilon_{ikj}\pi_{k}\pi_{j}\\
&=\dfrac{1}{2}\varepsilon_{ijk}\pi_{j}\pi_{k}-\dfrac{1}{2}\varepsilon_{ijk}\pi_{k}\pi_{j}\\
&=\dfrac{1}{2}\varepsilon_{ijk}[\pi_{j},\pi_{k}]
\end{aligned}
$$
Therefore, we need to compute the commutative relation between the mechanical momentum operator components, by definition, it is
$$
\begin{aligned}
[\pi_{j},\pi_{k}]&=\left[P_{j}-\dfrac{q}{c}A_{j},P_{k}-\dfrac{q}{c}A_{k}\right]\\
&=[P_{j},P_{k}]-\dfrac{q}{c}[P_{j},A_{k}]-\dfrac{q}{c}[A_{j},P_{k}]+\left(\dfrac{q}{c}\right)^{2}[A_{j},A_{k}]
\end{aligned}
$$
Now let's compute the commutators:
$$
\begin{aligned}
[P_{j},P_{k}]\psi&=(P_{j}P_{k}-P_{k}P_{j})\psi=P_{j}\left(-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{k}}\right)-P_{k}\left(-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{j}}\right)\\
&=(-\mathrm{i}\hbar)^{2}\dfrac{\partial}{\partial x_{j}}\left(\dfrac{\partial\psi}{\partial x_{k}}\right)-(-\mathrm{i}\hbar)^{2}\dfrac{\partial}{\partial x_{k}}\left(\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\hbar^{2}\dfrac{\partial^{2}\psi}{\partial x_{j}\partial x_{k}}+\hbar^{2}\dfrac{\partial^{2}\psi}{\partial x_{k}\partial x_{j}}=0
\end{aligned}
$$
$$
\begin{aligned}
[P_{j},A_{k}]\psi&=(P_{j}A_{k}-A_{k}P_{j})\psi=P_{j}(A_{k}\psi)-A_{k}(P_{j}\psi)\\
&=-\mathrm{i}\hbar\dfrac{\partial}{\partial x_{j}}(A_{k}\psi)-A_{k}\left(-\mathrm{i}\hbar\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\mathrm{i}\hbar\left(\dfrac{\partial A_{k}}{\partial x_{j}}\psi+A_{k}\dfrac{\partial\psi}{\partial x_{j}}\right)-A_{k}\left(-\mathrm{i}\hbar\dfrac{\partial\psi}{\partial x_{j}}\right)\\
&=-\mathrm{i}\hbar\dfrac{\partial A_{k}}{\partial x_{j}}\psi
\end{aligned}
$$
$$
\begin{aligned}
[A_{j},P_{k}]\psi&=(A_{j}P_{k}-P_{k}A_{j})\psi=A_{j}(P_{k}\psi)-P_{k}(A_{j}\psi)\\
&=A_{j}\left(-\mathrm{i}\hbar\dfrac{\partial \psi}{\partial x_{k}}\right)-(-\mathrm{i}\hbar)\dfrac{\partial}{\partial x_{k}}(A_{j}\psi)\\
&=A_{j}\left(-\mathrm{i}\hbar\dfrac{\partial \psi}{\partial x_{k}}\right)+\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\psi+\mathrm{i}\hbar A_{j}\dfrac{\partial\psi}{\partial x_{k}}\\
&=\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\psi
\end{aligned}
$$
$$
[A_{j},A_{k}]\psi=(A_{j}A_{k}-A_{k}A_{j})\psi=(A_{j}A_{k}-A_{j}A_{k})\psi=0
$$
Therefore,
$$
\begin{aligned}
(\boldsymbol{\pi}\times\boldsymbol{\pi})_{i}&=\dfrac{1}{2}\varepsilon_{ijk}[\pi_{j},\pi_{k}]\\
&=-\dfrac{q}{c}\left(-\mathrm{i}\hbar\dfrac{\partial A_{k}}{\partial x_{j}}+\mathrm{i}\hbar\dfrac{\partial A_{j}}{\partial x_{k}}\right)\\
&=\dfrac{\mathrm{i}q\hbar}{c}\left(\dfrac{\partial A_{k}}{\partial x_{j}}-\dfrac{\partial A_{j}}{\partial x_{k}}\right)\\
&=\dfrac{\mathrm{i}q\hbar}{c}\varepsilon_{ijk}\dfrac{\partial A_{k}}{\partial x_{j}}\\
&=\dfrac{\mathrm{i}q\hbar}{c}B_{i}
\end{aligned}
$$
Thus we get Eq. (20.2.16):
$$
\boldsymbol{\pi}\times\boldsymbol{\pi}=\dfrac{\mathrm{i}q\hbar}{c}\mathbf{B}
$$
$$
~\tag*{$\blacksquare$}
$$

**Exercise 20.2.2** Solve for the exact levels of the Dirac particle in a uniform magnetic field $\mathbf{B}=B_{0}\mathbf{k}$. Assume $\mathbf{A}=(B_{0}/2)(-y\mathbf{i}+x\mathbf{j})$. Consult Exercise 12.3.8. (Write the equation for $\chi$.)
## 20.3 More on Relativistic Quantum Mechanics

§ QM-20 Dirac方程

§ QM-20 The Dirac Equation

# 第19章 散射理论

## 19.1 引言

## 19.2 一维散射的回顾与概述

## 19.3 Born 近似（含时描述）

**习题 19.3.1** 证明

$$
\sigma_{\text{Yukawa}}=16 \pi r_0^2\left(\frac{g \mu r_0}{\hbar^2}\right)^2 \frac{1}{1+4 k^2 r_0^2}
$$

其中 $r_0=1 \mu_0$ 是作用程。将 $\sigma$ 与这一作用程对应的几何截面进行比较。



**习题 19.3.2** (1) 证明如果
$V(r)=-V_0 \theta\left(r_0-r\right)$，
那么

$$
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=4 r_0^2\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \frac{\left(\sin q r_0-q r_0 \cos q r_0\right)^2}{\left(q r_0\right)^6}
$$

(2) 证明当
$k r_0 \rightarrow 0$
时，散射趋于各向同性，并且

$$
\sigma \cong \frac{16 \pi r_0^2}{9}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2
$$

**习题 19.3.3** 证明对于 Gaussian 势
$V(r)=V_0 e^{-r^2 / r_0^3}$，
有

$$
\begin{aligned}
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} & =\frac{\pi r_0^2}{4}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \mathrm{e}^{-q^2 r_0^2 / 2} \\
\sigma & =\frac{\pi^2}{2 k^2}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2\left(1-\mathrm{e}^{-2 k^2 r_0^2}\right)
\end{aligned}
$$

[提示：由于
$q^2=2 k^2(1-\cos \theta)$，
因此
$\mathrm{d}(\cos \theta)=-\mathrm{d}(q^2) / 2 k^2$。]

**习题 19.3.4** 对 Gaussian 势验证上面的说法。

## 19.4 再论 Born 近似（时间无关近似）

**习题 19.4.1** 推导不等式 (19.4.44)。

## 19.5 分波展开

**习题 19.5.1** 证明：一个动能为
$100-\mathrm{MeV}$
的中子入射到一个固定原子核上时，
$l_{\max} \cong 2$。
[提示：核力的作用程约为一个 Fermi
$=10^{-5} \text{\AA}$。
另外，
$\hbar c \cong 200 \mathrm{MeV}~\mathrm{F}$
是核物理中更有用的估算口诀。]

**习题 19.5.2** 推导方程 (19.5.18)，并补全导向光学定理，即方程 (19.5.21) 的缺失步骤。

**习题 19.5.3** (1) 证明对于硬球，当
$k\rightarrow0$
时，
$\sigma_0 \rightarrow 4 \pi r_0^2$。

(2) 再考虑另一极限：
$k r_0$
非常大。由方程 (19.5.27) 以及 $j_l$ 和 $n_l$ 的渐近形式证明

$$
\sin ^2 \delta_l \xrightarrow[k r_0 \rightarrow \infty]{ } \sin ^2\left(k r_0-l \pi / 2\right)
$$

因此

$$
\begin{aligned}
\sigma=\sum_{l=0}^{l_{\max }=k r_0} \sigma_l & \cong \frac{4 \pi}{k^2} \int_0^{k r_0}(2 l) \sin ^2 \delta_l \mathrm{d} l \\
& \cong 2 \pi r_0^2
\end{aligned}
$$

这里我们把对 $l$ 的求和近似为积分，把
$2l+1$
近似为
$2l$，
并把振荡函数
$\sin^2\delta$
用其平均值
$1/2$
代替。

**习题 19.5.4** 证明深度为 $V_0$、作用程为 $r_0$ 的方势阱，其 $s$ 波相移为

$$
\delta_0=-k r_0+\tan ^{-1}\left(\frac{k}{k^{\prime}} \tan k^{\prime} r_0\right)
$$

其中 $k^{\prime}$ 和 $k$ 分别是势阱内部和外部的波数。

当 $k$ 很小时，$k r_0$ 是一个很小的数，我们将其忽略。现在考察随着势阱深度变化，也就是改变 $k^{\prime}$ 时，$\delta_0$ 会发生什么。证明每当

$$
k^{\prime} \simeq k_n^{\prime}=(2 n+1) \pi / 2 r_0
$$

时，$\delta_0$ 都具有方程 (19.5.30) 给出的共振形式，其中

$$
\Gamma / 2=\hbar^2 k_n / \mu r_0
$$

而 $k_n$ 是当
$k^{\prime}=k_n^{\prime}$
时 $k$ 的取值。

从一个浅到不足以支持任何束缚态的势阱开始，证明
$k_1^{\prime}$
对应于势阱刚好产生第一个零能束缚态。（见习题 12.6.9。）
（注意：零能束缚态对应于 $k=0$。）
随着势阱继续加深，这个能级向下移动，而很快在
$k_2^{\prime}$
处又会形成另一个零能束缚态，如此继续。

**习题 19.5.5** 证明即使一个势会吸收粒子，我们仍然可以用

$$
S_l(k)=\eta_l(k) \mathrm{e}^{2\mathrm{i} \delta_l}
$$

来描述它，其中
$\eta(<1)$
称为非弹性因子。

(1) 通过考察概率流，证明

$$
\begin{aligned}
\sigma_{\text{inel}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left[1-\eta_l^2\right] \\
\sigma_{\text{el}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left(1+\eta_l^2-2 \eta_l \cos 2 \delta_l\right)
\end{aligned}
$$

并且仍然有

$$
\sigma_{\mathrm{tot}}=\frac{4 \pi}{k} \operatorname{Im} f(0)
$$

(2) 考虑一个“黑盘”，它对
$r\leqslant r_0$
的所有粒子都完全吸收，而在更远处不起作用。将其理想化为：
对于
$l\leqslant k r_0$，
$\eta=0$；
对于
$l>k r_0$，
$\eta=1$、$\delta=0$。

证明
$\sigma_{\text{el}}=\sigma_{\text{inel}} \simeq \pi r_0^2$。
把求和换成积分，并假设
$k r_0 \gg 1$。
（见习题 19.5.3。）
为什么只要存在
$\sigma_{\text{inel}}$，
总会同时伴随一个
$\sigma_{\text{el}}$？

**习题 19.5.6** （光学定理）

(1) 证明入射波与散射波之间干涉所产生的流密度，其径向分量为

$$
j_r^{\mathrm{int}} \underset{r \rightarrow \infty}{\sim}\left(\frac{\hbar k}{\mu}\right) \frac{1}{r} \operatorname{Im}\left[\mathrm{i} \mathrm{e}^{\mathrm{i} k r(\cos \theta-1)} f^*(\theta) \cos \theta+\mathrm{i} \mathrm{e}^{\mathrm{i} k r(1-\cos \theta)} f(\theta)\right]
$$

(2) 说明只要
$\theta\neq0$，
由于
$r\rightarrow\infty$，
$j_r^{\text{int}}$
在任意小立体角上的平均值都为零。
[假设 $f(\theta)$ 是光滑函数。]

(3) 将
$j_r^{\text{int}}$
在前向方向的一个很小圆锥内积分，并证明（见提示）

$$
\int_{\text{forward cone}} j_r^{\text{int}} r^2 \mathrm{d}\Omega=-\left(\frac{\hbar k}{\mu}\right) \frac{4 \pi}{k} \operatorname{Im} f(0)
$$

因此，如果我们对靶后方区域中的总流进行积分，会发现干涉项（它只在紧邻前向方向、即靶后方显著）会造成粒子数的亏损，从而形成一个“阴影”。阴影区域中每秒缺失的粒子总数由上面的积分通量给出。把这一损失与入射通量
$\hbar k/\mu$
和截面 $\sigma$ 的乘积相等，就重新得到光学定理。

[提示：由于 $\theta$ 很小，可根据需要取
$\sin\theta\approx\theta$，
$\cos\theta=1$
或
$1-\theta^2/2$。
在计算 $\theta$ 积分的上限时，使用第 1 章中引入的思想：当一个函数的宗量趋于无穷时，如果该函数持续振荡，则其极限可以用平均值来处理。]

## 19.6 两粒子散射

**习题 19.6.1** (1) 从方程 (19.6.16) 和 (19.6.17) 出发，证明
$\mathrm{d}\sigma/\mathrm{d}\Omega$
与
$\mathrm{d}\sigma/\mathrm{d}\Omega_L$
之间的关系为

$$
\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega_L}\right|_{\theta_0}=\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}\right|_{2 \theta_0} 4 \cos \theta_0
$$

(2) 仅利用能量守恒和动量守恒，证明
$\theta_L \leqslant \pi / 2$。

(3) 对不等质量粒子散射，证明

$$
\tan \theta_L=\frac{\sin \theta}{\cos \theta+\left(m_1 / m_2\right)}
$$

其中 $m_2$ 是靶粒子的质量。

**习题 19.6.2** 利用方程 (19.3.16) 中的
$f_c(\theta)$
推导方程 (19.6.21)。

**习题 19.6.3** 假设
$f=f(\theta)$，
证明对于处于三重态的费米子，

$$
(\mathrm{d}\sigma/\mathrm{d}\Omega)_{\pi/2}=0.
$$

# Chapter 19 Scattering Theory
## 19.1 Introduction
## 19.2 Recapitulation of One-Dimensional Scattering and Overview
## 19.3 The Born Approximation (Time-Dependent Description)
**Exercise 19.3.1** Show that
$$
\sigma_{\text{Yukawa}}=16 \pi r_0^2\left(\frac{g \mu r_0}{\hbar^2}\right)^2 \frac{1}{1+4 k^2 r_0^2}
$$
where $r_0=1 \mu_0$ is the range. Compare $\sigma$ to the geometrical cross section associated with this range.



**Exercise 19.3.2** (1) Show that if $V(r)=-V_0 \theta\left(r_0-r\right)$,
$$
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=4 r_0^2\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \frac{\left(\sin q r_0-q r_0 \cos q r_0\right)^2}{\left(q r_0\right)^6}
$$

(2) Show that as $k r_0 \rightarrow 0$, the scattering becomes isotropic and
$$
\sigma \cong \frac{16 \pi r_0^2}{9}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2
$$

**Exercise 19.3.3** Show that for the Gaussian potential, $V(r)=V_0 e^{-r^2 / r_0^3}$,
$$
\begin{aligned}
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} & =\frac{\pi r_0^2}{4}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \mathrm{e}^{-q^2 r_0^2 / 2} \\
\sigma & =\frac{\pi^2}{2 k^2}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2\left(1-\mathrm{e}^{-2 k^2 r_0^2}\right)
\end{aligned}
$$
[Hint: Since $q^2=2 k^2(1-\cos \theta)$, $\mathrm{d}(\cos \theta)=-\mathrm{d}(q^2) / 2 k^2$.]

**Exercise 19.3.4** Verify the above claim for the Gaussian potential.

## 19.4 Born Again (The Time-Independent Approximation)
**Exercise 19.4.1** Derive the inequality (19.4.44).
## 19.5 The Partial Wave Expansion
**Exercise 19.5.1** Show that for a $100-\mathrm{MeV}$ (kinetic energy) neutron incident on a fixed nucleus, $l_{\max} \cong 2$. [Hint: The range of the nuclear force is roughly a Fermi$=10^{-5} \text{\AA}$. Also $\hbar c \cong 200 \mathrm{MeV}~\mathrm{F}$ is a more useful mnemonic for nuclear physics.]

**Exercise 19.5.2** Derive Eq. (19.5.18) and provide the missing steps leading to the optical theorem, Eq. (19.5.21).

**Exercise 19.5.3** (1) Show that $\sigma_0 \rightarrow 4 \pi r_0^2$ for a hard sphere as $k \rightarrow 0$.

(2) Consider the other extreme of $k r_0$ very large. From Eq. (19.5.27) and the asymptotic forms of $j_l$ and $n_l$ show that
$$
\sin ^2 \delta_l \xrightarrow[k r_0 \rightarrow \infty]{ } \sin ^2\left(k r_0-l \pi / 2\right)
$$
so that
$$
\begin{aligned}
\sigma=\sum_{l=0}^{l_{\max }=k r_0} \sigma_l & \cong \frac{4 \pi}{k^2} \int_0^{k r_0}(2 l) \sin ^2 \delta_l \mathrm{d} l \\
& \cong 2 \pi r_0^2
\end{aligned}
$$
if we approximate the sum over $l$ by an integral, $2 l+1$ by $2 l$, and the oscillating function $\sin ^2 \delta$ by its mean value of $1 / 2$.

**Exercise 19.5.4** Show that the $s$-wave phase shift for a square well of depth $V_0$ and range $r_0$ is
$$
\delta_0=-k r_0+\tan ^{-1}\left(\frac{k}{k^{\prime}} \tan k^{\prime} r_0\right)
$$
where $k^{\prime}$ and $k$ are the wave numbers inside and outside the well. For $k$ small, $k r_0$ is some small number and we ignore it. Let us see what happens to $\delta_0$ as we vary the depth of the well, i.e., change $k^{\prime}$. Show that whenever $k^{\prime} \simeq k_n^{\prime}=(2 n+1) \pi / 2 r_0$, $\delta_0$ takes on the resonant form Eq. (19.5.30) with $\Gamma / 2=\hbar^2 k_n / \mu r_0$, where $k_n$ is the value of $k$ when $k^{\prime}=k_n^{\prime}$. Starting with a well that is too shallow to have any bound state, show $k_1^{\prime}$ corresponds to the well developing its first bound state, at zero energy. (See Exercise 12.6.9.) (Note: A zero-energy bound state corresponds to $k=0$.) As the well is deepened further, this level moves down, and soon, at $k_2^{\prime}$, another zero-energy bound state is formed, and so on.

**Exercise 19.5.5** Show that even if a potential absorbs particles, we can describe it by
$$
S_l(k)=\eta_l(k) \mathrm{e}^{2\mathrm{i} \delta_l}
$$
where $\eta(<1)$, is called the inelasticity factor.

(1) By considering probability currents, show that
$$
\begin{aligned}
\sigma_{\text{inel}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left[1-\eta_l^2\right] \\
\sigma_{\mathrm{el}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left(1+\eta_l^2-2 \eta_l \cos 2 \delta_l\right)
\end{aligned}
$$
and that once again
$$
\sigma_{\mathrm{tot}}=\frac{4 \pi}{k} \operatorname{Im} f(0)
$$

(2) Consider a “black disk” which absorbs everything for $r \leqslant r_0$ and is ineffective beyond. Idealize it by $\eta=0$ for $l \leqslant k r_0$; $\eta=1$, $\delta=0$ for $l>k r_0$. Show that $\sigma_{\text{el}}=\sigma_{\text{inel}} \simeq \pi r_0^2$. Replace the sum by an integral and assume $k r_0 \gg 1$. (See Exercise 19.5.3.) Why is $\sigma_{\text{inel}}$ always accompanied by $\sigma_{\mathrm{el}}$?

**Exercise 19.5.6** (The Optical Theorem) (1) Show that the radial component of the current density due to interference between the incident and scattered waves is
$$
j_r^{\mathrm{int}} \underset{r \rightarrow \infty}{\sim}\left(\frac{\hbar k}{\mu}\right) \frac{1}{r} \operatorname{Im}\left[\mathrm{i} \mathrm{e}^{\mathrm{i} k r(\cos \theta-1)} f^*(\theta) \cos \theta+\mathrm{i} \mathrm{e}^{\mathrm{i} k r(1-\cos \theta)} f(\theta)\right]
$$

(2) Argue that as long as $\theta \neq 0$, the average of $j_r^{\text {int }}$ over any small solid angle is zero because $r \rightarrow \infty$. [Assume $f(\theta)$ is a smooth function.]

(3) Integrate $j_r^{\text {int }}$ over a tiny cone in the forward direction and show that (see hint)
$$
\int_{\text{forward cone}} j_r^{\text{int}} r^2 \mathrm{d}\Omega=-\left(\frac{\hbar k}{\mu}\right) \frac{4 \pi}{k} \operatorname{Im} f(0)
$$
Thus, if we integrate the total current in the region behind the target, we find that the interference term (important only in the near-forward direction, behind the target) produces a depletion of particles, casting a “shadow”. The total number of particles (per second) missing in the shadow region is given by the above expression for the integrated flux. Equating this loss to the product of the incident flux $\hbar k / \mu$ and the cross section $\sigma$, we regain the optical theorem. [Hint: Since $\theta$ is small, set $\sin \theta \approx \theta$, $\cos \theta=1$ or $1-\theta^2 / 2$ using the judgment. In evaluating the upper limit in the $\theta$ integration, use the idea introduced in Chapter 1, namely, that the limit of a function that oscillates as its argument approaches infinity is equal to its average value.]
## 19.6 Two-Particle Scattering
**Exercise 19.6.1** (1) Starting with Eqs. (19.6.16) and (19.6.17), show that the relation between $\mathrm{d}\sigma/\mathrm{d}\Omega$ and $\mathrm{d}\sigma/\mathrm{d}\Omega_L$ is
$$
\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega_L}\right|_{\theta_0}=\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}\right|_{2 \theta_0} 4 \cos \theta_0
$$

(2) Show that $\theta_L \leqslant \pi / 2$ by using just energy and momentum conservation.

(3) For unequal mass scattering, show that
$$
\tan \theta_L=\frac{\sin \theta}{\cos \theta+\left(m_1 / m_2\right)}
$$
where $m_2$ is the target mass.

**Exercise 19.6.2** Derive Eq. (19.6.21) using Eq. (19.3.16) for $f_c(\theta)$.

**Exercise 19.6.3** Assuming $f=f(\theta)$ show that $(\mathrm{d}\sigma/\mathrm{d}\Omega)_{\pi/2}=0$ for fermions in the triplet state.