# 上车后准备工作
1. **调整座椅** （座椅是否调节合适直接会影响到【刹车】踏板的控制）

---

2. **调节后视镜** （后视镜是否调节直接会影响到所有需要通过后视镜看的点位）
- 把旋钮拨到L调节左后视镜，调整车身占后视镜中$\dfrac{1}{3}$。
- 先往上调节保证能看到左后门把手，再玩下调，让左后门把手顶着后视镜的上边缘。
- 注意确保能看到左后轮。

类似地，调节右后视镜
- 把旋钮拨到R调节右后视镜，调整车身占后视镜中$\dfrac{1}{3}$。
- 先往上调节保证能看到右后门把手，再玩下调，让右后门把手顶着后视镜的上边缘。

如果后视镜调不到位置可以通过座椅的调节和坐姿的变化去改变。

---

3. 车辆行驶前不要着急踩刹车挂前进档，一定要检查下**手刹**是否送到底，**安全带**是否佩戴，检查**车门**是否完全关闭。仪表盘上都有显示。

---

4. 慢抬刹车 确认所有准备工作都已经好了就可以挂挡然后慢慢地松刹车

---

进入项目前容易扣分的地方：
- 手刹未松到底
- 未按规定使用安全带
- 车门未完全关闭



# 一、倒车入库

**考试操作要求：** 从道路一端控制线（两个前轮触地点再控制线以外），倒入车库停车，再前进出库向另一端控制线行驶，待两个前轮触地点均行驶过控制线后，倒入车库停车，前进驶出车库，回到起始点。考试过程中，车辆进退途中不得停车。项目完成时间不得超过210秒。

![reverse](https://xumin.net/static/img/08d5d0294382fbc0f089e79b4478d6f1.webp)

1. 进项目的时候方向盘一定要对好领线 **（2-3虚线中心）**，边走边修正方向尽量让车直着进入项目，车头偏左会导致肩膀点到的过早，车头偏右会导致肩膀点到的晚。车身偏左倒库时容易压左边线，车身偏右会导致进库的时候夹角不够，导致不知道如何调整或压库角和边线。
2. 入库后如果车身不正一定要及时调整车身。
3. 右倒库做完出库时一定要等左后轮出库角了再打方向，打早了会导致左倒库进库的时候夹角不够从而导致不知道怎么调整或压库角和边线，打晚了倒库时容易压右边线。
4. 左倒库出库的时候不用等后轮出库即将到库角的时候就可以打方向（具体情况根据车身与边线距离而定）。

**考试考核标准：**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. 不按规定路线、顺序行驶</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. 车身出线</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. 倒库不入</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">4. 未将两前轮触地点驶过控制线</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">5. 时间超过210秒</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">6. 中途停车超过2秒</td>
    <td class="tg-0pky">扣5分/次</td>
  </tr>
</tbody>
</table>

# 二、侧方停车
**考试操作要求：** 车辆在库前方一次倒车入库，中途不得停车，车轮不触轧车道边线，车身不触碰库位边线。再前进向左前方出库，出库前应开启左转向灯，出库过程中车轮不触轧车道边线，车身不触碰库位边线，出库后关闭转向灯。项目完成时间不得超过90秒。

![side parking](https://xumin.net/static/img/ac7b50eb73ff77edcdecd2b525aa95b3.webp)

1. 进项目的时候一定要对领线 **（对前方红颜色警示墩）**，边走边修正方向尽量让车直着进入项目，车身偏左进库时容易压左边线，（控制车身右侧与边线30-50公分之间）。
2. 停车后一定要看后视镜车身和边线是平齐，车头往左偏会导致进库的时候提前消失，这个时候等库角消失，即刻把方向打满，如果车头往右偏倒车的时候往右打点方向把车稍微往正的调一下，防止打方向的时候车身太靠左导致压左线。
3. 进库看库底角的时候，如果库角看到时候已经过了（回正晚了）最后一把往左打满的时候不要等后轮压线，即将压上的时候就打满把车往左拉一下，防止进库后压右边线。
4. 出库的时候先打灯再去换档，等参照点到线了再去动方向，车身正了及时回正方向，防止压右边线。

**注意：** 大众车的转向灯有两个挡位，要使劲一些才能打到位。

**考试考核标准：**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. 不按规定路线、顺序行驶</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. 车辆在入库停止后，车身出线</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. 项目完成时间超过90秒</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">4. 出库时不使用或错误使用转向灯</td>
    <td class="tg-0pky">扣10分</td>
  </tr>
  <tr>
    <td class="tg-0pky">5. 行驶中车身触碰库位边线</td>
    <td class="tg-0pky">扣10分/次</td>
  </tr>
  <tr>
    <td class="tg-0pky">6. 行驶中车轮触轧库位边线</td>
    <td class="tg-0pky">扣10分/次</td>
  </tr>
  <tr>
    <td class="tg-0pky">7. 行驶中车轮触轧车道边缘线</td>
    <td class="tg-0pky">扣10分/次</td>
  </tr>
  <tr>
    <td class="tg-0pky">8. 中途停车超过2秒</td>
    <td class="tg-0pky">扣5分/次</td>
  </tr>
</tbody>
</table>

# 三、曲线行驶

**考试操作要求：** 车辆从弯道的一端前进驶入，从另一端驶出。行驶中转向、速度平稳。中途不得停车，车轮不得碰轧车道边线。

![curve](https://xumin.net/static/img/58f80126dad6e00d895ef8988e135374.webp)

1. **进弯**：进弯的时候身体挺直让目光能看到引擎盖左右端的最远点，进弯前调节车身让车头卡在两条线的中间，进弯进好了这样后面就很感觉很顺畅。
2. **左弯**：到点左打一圈方向然后$90^{\circ}$修整法让点位贴着右边线行驶（大众车无需$+90^{\circ}$）。
3. 两弯交汇处，到点回正方向。
4. **右弯**，到点右打一圈方向然后$90^{\circ}$修整法让点位贴着左边线行驶。

**考试考核标准：**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. 不按规定路线、顺序行驶</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. 车道轧道路边缘线</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. 中途停车时间超过2秒</td>
    <td class="tg-0pky">扣5分/次</td>
  </tr>
</tbody>
</table>

# 四、直角转弯

**考试操作要求：** 驾驶车辆按规定的线路行驶，由左向右直角转弯，一次通过，中途不得停车，车轮不得碰轧车道边线。转弯前，应开启转向灯，完成转弯后，关闭转向灯。

![quarter_turn](https://xumin.net/static/img/977fde53c813f0c9922c2dd92142ce7e.webp)

1. 进项目的时候一定要对好领线，边走边修正方向尽量让车直着进入项目，车头偏左会导致参照点到的早，车头偏右会导致参照点到的晚。
2. 过库角的时候一定要注意看左后轮和库角的距离，标准是20公分，偏差小不用管如果距离库角特别近甚至压上了就证明点偏早了后面稍微晚2-3秒再打，如果距离过宽甚至于压上了右边线则证明点偏晚了后面稍后早2-3秒打。

**考试考核标准：**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. 不按规定路线、顺序行驶</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. 车道轧道路边缘线</td>
    <td class="tg-0pky">不合格</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. 转弯前不使用或错误使用转向灯</td>
    <td class="tg-0pky">扣10分/次</td>
  </tr>
  <tr>
    <td class="tg-0pky">4. 转弯后不关闭转向灯</td>
    <td class="tg-0pky">扣10分/次</td>
  </tr>
  <tr>
    <td class="tg-0pky">5. 中途停车时间超过2秒</td>
    <td class="tg-0pky">扣5分/次</td>
  </tr>
</tbody>
</table>

# Preparations After Getting into the Car

1. **Adjust the seat**
   Whether the seat is adjusted properly directly affects your control of the **brake pedal**.

---

2. **Adjust the rear-view mirrors**
   Proper mirror adjustment directly affects all reference points that need to be observed through the mirrors.

* Turn the knob to `L` to adjust the left rear-view mirror. Adjust it so that the vehicle body occupies about $\dfrac{1}{3}$ of the mirror.
* First adjust the mirror upward until you can see the left rear door handle, then adjust it downward so that the left rear door handle touches the upper edge of the mirror.
* Make sure that the left rear wheel is visible.

Similarly, adjust the right rear-view mirror:

* Turn the knob to `R` to adjust the right rear-view mirror. Adjust it so that the vehicle body occupies about $\dfrac{1}{3}$ of the mirror.
* First adjust the mirror upward until you can see the right rear door handle, then adjust it downward so that the right rear door handle touches the upper edge of the mirror.

If the mirror cannot be adjusted to the desired position, you can compensate by adjusting the seat position and your sitting posture.

---

3. Before the vehicle starts moving, do not rush to press the brake and shift into Drive. Be sure to check whether the **parking brake** has been fully released, whether the **seat belt** is fastened, and whether the **doors** are completely closed. These are all indicated on the instrument panel.

---

4. Slowly release the brake. Once you have confirmed that all preparations are complete, shift into gear and slowly release the brake pedal.

---

Common points where marks may be deducted before entering an exam item:

* Parking brake not fully released
* Seat belt not fastened as required
* Door not completely closed



# I. Reverse Parking into a Garage

**Exam operation requirements:** Starting from the control line at one end of the road, with the contact points of both front wheels outside the control line, reverse into the garage and stop. Then drive forward out of the garage toward the control line at the other end. After the contact points of both front wheels have passed the control line, reverse into the garage and stop again. Then drive forward out of the garage and return to the starting point. During the exam, the vehicle must not stop while moving forward or backward. The entire item must be completed within 210 seconds.

![reverse](https://xumin.net/static/img/08d5d0294382fbc0f089e79b4478d6f1.webp)

1. When entering the item, make sure the steering wheel is aligned correctly with the guide line **(the center between the 2nd and 3rd dashed lines)**. Make small steering corrections while moving and try to enter the item with the vehicle straight. If the front of the vehicle is too far left, the shoulder reference point will arrive too early. If the front is too far right, the shoulder reference point will arrive too late. If the vehicle body is too far left, it is easy to cross the left boundary line while reversing into the garage. If the vehicle body is too far right, the entry angle may be insufficient, making it difficult to adjust and possibly causing the vehicle to hit the garage corner or boundary line.

2. After entering the garage, if the vehicle body is not straight, adjust it immediately.

3. After completing the right-side reverse parking and driving out of the garage, wait until the left rear wheel has passed the garage corner before turning the steering wheel. Turning too early may result in an insufficient angle when performing the left-side reverse parking, making adjustment difficult or causing the vehicle to hit the garage corner or boundary line. Turning too late may cause the vehicle to cross the right boundary line while reversing.

4. When driving out after the left-side reverse parking, you do not need to wait until the rear wheel has completely left the garage. You may begin turning when the rear wheel is about to reach the garage corner. The exact timing depends on the distance between the vehicle body and the boundary line.

**Exam evaluation standards:**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. Failing to follow the prescribed route or sequence</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. Vehicle body crosses the boundary line</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. Failing to reverse fully into the garage</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">4. Failing to drive both front-wheel contact points past the control line</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">5. Completion time exceeds 210 seconds</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">6. Stopping for more than 2 seconds during the maneuver</td>
    <td class="tg-0pky">Deduct 5 points each time</td>
  </tr>
</tbody>
</table>

# II. Parallel Parking

**Exam operation requirements:** Reverse into the parking space in one continuous maneuver from in front of the space, without stopping midway. The wheels must not touch or cross the lane boundary line, and the vehicle body must not touch the parking-space boundary line. Then drive forward and exit toward the front left. Before exiting, turn on the left turn signal. During the exit, the wheels must not touch or cross the lane boundary line, and the vehicle body must not touch the parking-space boundary line. Turn off the turn signal after leaving the space. The entire item must be completed within 90 seconds.

![side parking](https://xumin.net/static/img/ac7b50eb73ff77edcdecd2b525aa95b3.webp)

1. When entering the item, align with the guide line **(aim at the red warning bollard ahead)**. Make small steering corrections while moving and try to enter with the vehicle straight. If the vehicle body is too far left, it is easy to cross the left boundary line while reversing into the space. Keep the right side of the vehicle approximately 30–50 cm from the boundary line.

2. After stopping, check in the mirror whether the vehicle body is parallel to the boundary line. If the front of the vehicle points to the left, the garage corner will disappear too early when reversing in. In this case, as soon as the garage corner disappears, turn the steering wheel fully. If the front points to the right, turn slightly to the right while reversing to straighten the vehicle a little, preventing the vehicle body from being too far left when turning and crossing the left line.

3. When observing the rear corner of the parking space during entry, if the corner has already passed when you see it, meaning the steering wheel was straightened too late, then on the final full-left turn, do not wait until the rear wheel touches the line. Turn fully just before it reaches the line to pull the vehicle slightly left and prevent crossing the right boundary line after entering the space.

4. When exiting, turn on the signal first and then shift gear. Wait until the reference point reaches the line before turning the steering wheel. Straighten the steering wheel promptly once the vehicle body is straight to avoid crossing the right boundary line.

**Note:** Volkswagen turn-signal levers have two detent positions. You need to push the lever firmly enough for it to engage fully.

**Exam evaluation standards:**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. Failing to follow the prescribed route or sequence</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. Vehicle body extends outside the parking space after stopping</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. Completion time exceeds 90 seconds</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">4. Failing to use the turn signal or using it incorrectly when exiting</td>
    <td class="tg-0pky">Deduct 10 points</td>
  </tr>
  <tr>
    <td class="tg-0pky">5. Vehicle body touches the parking-space boundary line while moving</td>
    <td class="tg-0pky">Deduct 10 points each time</td>
  </tr>
  <tr>
    <td class="tg-0pky">6. Wheels touch or cross the parking-space boundary line while moving</td>
    <td class="tg-0pky">Deduct 10 points each time</td>
  </tr>
  <tr>
    <td class="tg-0pky">7. Wheels touch or cross the lane-edge line while moving</td>
    <td class="tg-0pky">Deduct 10 points each time</td>
  </tr>
  <tr>
    <td class="tg-0pky">8. Stopping for more than 2 seconds during the maneuver</td>
    <td class="tg-0pky">Deduct 5 points each time</td>
  </tr>
</tbody>
</table>

# III. Curve Driving

**Exam operation requirements:** Drive forward into the curved road from one end and exit from the other end. Steering and speed should remain smooth and steady. Do not stop midway, and the wheels must not touch or cross the lane boundary lines.

![curve](https://xumin.net/static/img/58f80126dad6e00d895ef8988e135374.webp)

1. **Entering the curve:** Sit upright so that you can see the farthest visible points on both the left and right edges of the hood. Before entering the curve, adjust the vehicle so that the front is centered between the two boundary lines. If you enter the curve correctly, the rest of the maneuver will feel much smoother.

2. **Left curve:** At the reference point, turn the steering wheel one full turn to the left, then use the $90^{\circ}$ correction method to keep the reference point close to the right boundary line. Volkswagen vehicles do not require the additional $+90^{\circ}$ adjustment.

3. At the transition point between the two curves, straighten the steering wheel.

4. **Right curve:** At the reference point, turn the steering wheel one full turn to the right, then use the $90^{\circ}$ correction method to keep the reference point close to the left boundary line.

**Exam evaluation standards:**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. Failing to follow the prescribed route or sequence</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. Wheels cross the road-edge line</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. Stopping for more than 2 seconds during the maneuver</td>
    <td class="tg-0pky">Deduct 5 points each time</td>
  </tr>
</tbody>
</table>

# IV. Right-Angle Turn

**Exam operation requirements:** Drive along the prescribed route and make a right-angle turn from left to right in one continuous maneuver. Do not stop midway, and the wheels must not touch or cross the lane boundary lines. Before turning, turn on the turn signal. After completing the turn, turn off the turn signal.

![quarter\_turn](https://xumin.net/static/img/977fde53c813f0c9922c2dd92142ce7e.webp)

1. When entering the item, align correctly with the guide line. Make small steering corrections while moving and try to enter with the vehicle straight. If the front of the vehicle is too far left, the reference point will arrive too early. If the front is too far right, the reference point will arrive too late.

2. When passing the garage corner, pay close attention to the distance between the left rear wheel and the corner. The standard distance is 20 cm. If the deviation is small, no adjustment is needed. If the wheel is very close to the corner or even crosses it, the turning point was too early, so next time delay the steering input by about 2–3 seconds. If the distance is too large or the vehicle crosses the right boundary line, the turning point was too late, so next time turn about 2–3 seconds earlier.

**Exam evaluation standards:**

<table class="tg">
<tbody>
  <tr>
    <td class="tg-0pky">1. Failing to follow the prescribed route or sequence</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">2. Wheels cross the road-edge line</td>
    <td class="tg-0pky">Fail</td>
  </tr>
  <tr>
    <td class="tg-0pky">3. Failing to use the turn signal or using it incorrectly before the turn</td>
    <td class="tg-0pky">Deduct 10 points each time</td>
  </tr>
  <tr>
    <td class="tg-0pky">4. Failing to turn off the turn signal after the turn</td>
    <td class="tg-0pky">Deduct 10 points each time</td>
  </tr>
  <tr>
    <td class="tg-0pky">5. Stopping for more than 2 seconds during the maneuver</td>
    <td class="tg-0pky">Deduct 5 points each time</td>
  </tr>
</tbody>
</table>

C2驾照：科目二

C2 Driver’s License: Subject 2

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

上车后准备工作

附录

A.1 矩阵求逆

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

第20章 Dirac 方程

20.1 自由粒子的 Dirac 方程

第21章路径积分 II