张量积 I

# 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

Appendix

# 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

Chapter 21 Path Integrals——II

# 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

Chapter 20 The Dirac Equation

# 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.

张量积 I

1. 简介

Appendix

A.1 Matrix Inversion

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

Chapter 20 The Dirac Equation

20.1 The Free-Particle Dirac Equation

Chapter 19 Scattering Theory

19.1 Introduction

19.2 Recapitulation of One-Dimensional Scattering and Overview

19.3 The Born Approximation (Time-Dependent Description)