# 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.

Chapter 19 Scattering Theory

# Chapter 18 Time-Dependent Perturbation Theory
## 18.1 The Problem
## 18.2 First-Order Perturbation Theory
**Exercise 18.2.1** Show that if $H^1(t)=-e \mathscr{E} X /\left[1+(t / \tau)^2\right]$, then, to first order,
$$
P_{0 \rightarrow 1}=\frac{e^2 \mathscr{E}^2 \pi^2 \tau^2}{2 m \omega \hbar} \mathrm{e}^{-2 \omega \tau}
$$



**Exercise 18.2.2** A hydrogen atom is in the ground state at $t=-\infty$. An electric field $\mathbf{E}(t)=(\mathbf{k} \mathscr{E}) e^{-t^2 / \tau^2}$ is applied until $t=\infty$. Show that the probability that the atom ends up in any of the $n=2$ states is, to first order,
$$
P(n=2)=\left(\frac{e \mathscr{E}}{\hbar}\right)^2\left(\frac{2^{15} a_0^2}{3^{10}}\right) \pi \tau^2 e^{-\omega^2 \tau^2 / 2}
$$
where $\omega=\left(E_{2 l m}-E_{100}\right) / \hbar$. Does the answer depend on whether or not we incorporate spin in the picture?

**Exercise 18.2.3** Consider a particle in the ground state of a box of length $L$. Argue on semiclassical grounds that the natural time period associated with it is $T \simeq m L^2 / \hbar \pi$. If the box expands symmetrically to double its size in time $\tau \ll T$ what is the probability of catching the particle in the ground state of the new box? (See Exercise (5.2.1).)

**Exercise 18.2.4** In the $\beta$ decay $\mathrm{H}^3$ (two neutrons + one proton in the nucleus) $\rightarrow\left(\mathrm{He}^3\right)^{+}$ (two protons + one neuron in the nucleus), the emitted electron has a kinetic energy of $16~\mathrm{keV}$. Argue that the sudden approximation may be used to describe the response of an electron that is initially in the $1 s$ state of $\mathrm{H}^3$. Show that the amplitude for it to be in the ground state of $\left(\mathrm{He}^3\right)^{+}$is $16(2)^{1 / 2} / 27$. What is the probability for it to be in the state
$$
|n=16, l=3, m=0\rangle~\text{of}~\left(\mathrm{He}^3\right)^{+} ?
$$

**Exercise 18.2.5** An oscillator is in the ground state of $H=H^0+H^1$, where the time-independent perturbation $H^1$ is the linear potential ($-f x$). If at $t=0$, $H^1$ is abruptly turned off, show that the probability that the system is in the $n$th eigenstate of $H^0$ is given by the Poisson distribution
$$
P(n)=\frac{\mathrm{e}^{-\lambda} \lambda^n}{n!}, \quad \text { where } \quad \lambda=\frac{f^2}{2 m \omega^3 \hbar}
$$
[Hint: Use the formula
$$
\exp [A+B]=\exp [A] \exp [B] \exp \left[-\frac{1}{2}[A, B]\right]
$$
where $[A, B]$ is a $c$ number.]

**Exercise 18.2.6** Consider a system subject to a perturbation $H^1(t)=H^1 \delta(t)$. Show that if at $t=0^{-}$the system is in the state $\left|i^0\right\rangle$, the amplitude to be in a state $\left|f^0\right\rangle$ at $t=0^{+}$is, to first order,
$$
d_f=\frac{-\mathrm{i}}{\hbar}\left\langle f^0\right| H^1\left|i^0\right\rangle \quad(f \neq i)
$$
Notice that (1) the state of the system does change instantaneously; (2) Even though the perturbation is "infinite" at $t=0$, we can still use first-order perturbation theory if the “area under it” is small enough.
## 18.3 Higher Orders in Perturbation Theory
**Exercise 18.3.1** Derive Eq. (18.3.30).

**Exercise 18.3.2** In the paramagnetic resonance problem Exercise 14.4.3 we moved to a frame rotating in real space. Show that this is also equivalent to a Hilbert space rotation, but that it takes us neither to the interaction nor the Heisenberg picture, except at resonance. What picture is it at resonance? (If $\mathbf{B}=B_0 \mathbf{k}+B \cos \omega t \mathbf{i}-B \sin \omega t \mathbf{j}$, associate $B_0$ with $H_S^0$ and $B$ with $H_S^1$.)
## 18.4 A General Discussion of Electromagnetric Interactions
**Exercise 18.4.1** By taking the divergence of Eq. (18.4.5) show that the continuity equation must be obeyed if Maxwell's equations are to be mutually consistent.

**Exercise 18.4.2** Calculate $\mathbf{E}$ and $\mathbf{B}$ corresponding to $(\mathbf{A}, \phi)$ and $(\mathbf{A}^{\prime}, \phi^{\prime})$ using Eqs. (18.4.7) and (18.4.9) and verify the above claim.

**Exercise 18.4.3** Suppose we are given some A and $\phi$ that do not obey the Coulomb gauge conditions. Let us see how they can be transformed to the Coulomb gauge.

(1) Show that if we choose
$$
\Lambda(\mathbf{r}, t)=-c \int_{-\infty}^t \phi\left(\mathbf{r}, t^{\prime}\right) \mathrm{d} t^{\prime}
$$
and transform to ($\mathbf{A}^{\prime}, \phi^{\prime}$) then $\phi^{\prime}=0$. $\mathbf{A}^{\prime}$ is just $\mathbf{A}-\boldsymbol{\nabla} \Lambda$, with $\boldsymbol{\nabla} \cdot \mathbf{A}^{\prime}$ not necessarily zero.

(2) Show that if we gauge transform once more to ($\mathbf{A}^{\prime \prime}, \phi^{\prime \prime}$) via
$$
\Lambda^{\prime}=-\frac{1}{4 \pi} \int \frac{\nabla \cdot \mathbf{A}^{\prime}\left(\mathbf{r}^{\prime}, t\right) \mathrm{d}^3 \mathbf{r}^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}
$$
then $\boldsymbol{\nabla} \cdot \mathbf{A}^{\prime \prime}=0$. [Hint: Recall $\nabla^2\left(1 /\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)=-4 \pi \delta^3\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$.]

(3) Verify that $\phi^{\prime \prime}$ is also zero by using $\nabla \cdot \mathbf{E}=0$.

(4) Show that if we want to make any further gauge transformations within the Coulomb gauge, $\Lambda$ must be time independent and obey $\nabla^2 \Lambda=0$. If we demand that $|\mathbf{A}| \rightarrow 0$ at spatial infinity, $\mathbf{A}$ becomes unique.

**Exercise 18.4.4** (Proof of Gauge Invariance in the Schrödinger Approach)

(1) Write $H$ for a particle in the potentials $(A, \phi)$.

(2) Write down $H_{\Lambda}$, the Hamiltonian obtained by gauge transforming the potentials.

(3) Show that if $\psi(\mathbf{r}, t)$ is a solution to Schrödinger's equation with the Hamiltonian $H$, then $\psi_{\Lambda}(\mathbf{r}, t)$ given in Eq. (18.4.33) is the corresponding solution with $H \rightarrow H_{\Lambda}$.
## 18.5 Interaction of Atoms with Electromagnetic Radiation
**Exercise 18.5.1** (1) By going through the derivation, argue that we can take the $\mathrm{e}^{\mathrm{i}\mathbf{k} \cdot \mathbf{r}}$ factor into account exactly, by replacing $\mathbf{p}_f$ by $\mathbf{p}_f-\hbar \mathbf{k}$ in Eq. (18.5.19).

(2) Verify the claim made above about the electron momentum distribution.

**Exercise 18.5.2** (1) Estimate the photoelectric cross section when the ejected electron has a kinetic energy of $10~\mathrm{Ry}$. Compare it to the atom's geometric cross section $\simeq \pi a_0^2$.

(2) Show that if we consider photoemission from the $1 s$ state of a charge $Z$ atom, $\sigma \propto Z^5$, in the limit $p_f a_0 / Z \hbar \gg 1$.