# 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 $p_f a_0 / Z \hbar \gg 1$.

Chapter 18 Time-Dependent Perturbation Theory

# Chapter 17 Time-Independence Perturbation Theory
## 17.1 The Formalism
## 17.2 Some Examples
**Exercise 17.2.1** Consider $H^1=\lambda x^4$ for the oscillator problem.

(1) Show that
$$
E_n^1=\frac{3 \hbar^2 \lambda}{4 m^2 \omega^2}\left[1+2 n+2 n^2\right]
$$

(2) Argue that no matter how small $\lambda$ is, the perturbation expansion will break down for some large enough $n$. What is the physical reason?



**Exercise 17.2.2** Consider a spin-1/2 particle with gyromagnetic ratio $\gamma$ in a magnetic field $\mathbf{B}=B \mathbf{i}+B_0 \mathbf{k}$. Treating $B$ as a perturbation, calculate the first- and second-order shifts in energy and first-order shift in wave function for the ground state. Then compare the exact answers expanded to the corresponding orders.

**Exercise 17.2.3** In our study of the H atom, we assumed that the proton is a point charge $e$. This leads to the familiar Coulomb interaction $\left(-e^2/r\right)$ with the electron.

(1) Show that if the proton is a uniformly dense charge distribution of radius $R$, the interaction is
$$
\begin{aligned}
V(r) & =-\frac{3 e^2}{2 R}+\frac{e^2 r^2}{2 R^3}, & & r \leqslant R \\
& =-\frac{e^2}{r}, & & r>R
\end{aligned}
$$

(2) Calculate the first-order shift in the ground-state energy of hydrogen due to this modification. You may assume $\mathrm{e}^{-R / a_0} \simeq 1$. You should find $E^1=2 e^2 R^2 / 5 a_0^3$.

**Exercise 17.2.4** (1) Prove the Thomas-Reiche-Kuhn sum rule
$$
\sum_{n^{\prime}}\left(E_{n^{\prime}}-E_n\right)|\langle n^{\prime}\mid X\mid n\rangle|^ 2=\sum_{n^{\prime}}\left(E_{n^{\prime}}-E_n\right)\langle n\mid X\mid n^{\prime}\rangle\langle n^{\prime}\mid X\mid n\rangle=\frac{\hbar^2}{2 m}
$$
where $|n\rangle$ and $\left|n^{\prime}\right\rangle$ are eigenstates of $H=P^2 / 2 m+V(X)$. [Hint: Eliminate the $E_{n^{\prime}}-E_n$ factor in favor of $H$.]

(2) Test the sum rule on the $n$th state of the oscillator.

**Exercise 17.2.5** We have seen that if we neglect the repulsion $e^2 / r_{12}$ between the two electrons in the ground state of He , the energy is $-8 \mathrm{Ry}=-108.8~\mathrm{eV}$. Treating $e^2 / r_{12}$ as a perturbation, show that
$$
\langle 100,100| H^1|100,100\rangle=\frac{5}{2} \mathrm{Ry}
$$
so that $E_0^0+E_0^1=-5.5 . \mathrm{Ry}=-74.8~\mathrm{eV}$. Recall that the measured value is $-78.6~\mathrm{eV}$ and the variational estimate is $-77.5~\mathrm{eV}$. [Hint: $\left\langle H^1\right\rangle$ can be viewed as the interaction between two concentric, spherically symmetric exponentially falling charge distributions. Find the potential $\phi(r)$ due to one distribution and calculate the interaction energy between this potential and the other charge distribution.]

**Exercise 17.2.6** Verify Eq. (17.2.34).

**Exercise 17.2.7** For the oscillator, consider $H^1=-qfX$. Find an $\Omega$ that satisfies Eq. (17.2.38). Feed it into Eq. (17.2.39) for $E_n^2$ and compare with the earlier calculation.

**Exercise 17.2.8** Fill in the steps connecting Eqs. (17.2.41) and (17.2.43). Try to use symmetry arguments to reduce the labor involved in evaluating the integrals.
## 17.3 Degenerate Pertubation Theory
**Exercise 17.3.1** Use the dipole selection rules to show that $H^1$ has the above form and carry out the evaluation of $\Delta$.

**Exercise 17.3.2** Consider a spin-1 particle (with no orbital degrees of freedom). Let $H=A S_z^2+B\left(S_x^2-S_y^2\right)$, where $S_i$ are $3 \times 3$ spin matrices, and $A \gg B$. Treating the $B$ term as a perturbation, find the eigenstates of $H^0=A S_z^2$ that are stable under the perturbation. Calculate the energy shifts to first order in $B$. How are these related to the exact answers?

**Exercise 17.3.3** Consider the case where $H^0$ includes the Coulomb plus spin-orbit interaction and $H^1$ is the effect of a weak magnetic field $\mathbf{B}=B \mathbf{k}$. Using the appropriate basis, show that the first-order level shift is related to $j_z$ by
$$
E^1=\left(\frac{e B}{2 m c}\right)\left(1 \pm \frac{1}{2 l+1}\right) j_z, \quad j=l \pm 1 / 2
$$
Sketch the levels for the $n=2$ level assuming that $E^1 \ll E_{\text{f.s.}}^1$.

**Exercise 17.3.4** We discuss here some tricks for evaluating the expectation values of certain operators in the eigenstates of hydrogen.

(1) Suppose we want $\langle 1 / r\rangle_{nlm}$. Consider first $\langle\lambda / r\rangle$. We can interpret $\langle\lambda / r\rangle$ as the first order correction due to a perturbation $\lambda / r$. Now this problem can be solved exactly; we just replace $e^2$ by $e^2-\lambda$ everywhere. (Why?) So the exact energy, from Eq. (13.1.16) is $E(\lambda)=$ $-\left(e^2-\lambda\right)^2 m / 2 n^2 \hbar^2$. The first-order correction is the term linear in $\lambda$, that is, $E^1=m e^2 \lambda / n^2 \hbar^2=$ $\langle\lambda / r\rangle$, from which we get $\langle 1 / r\rangle=1 / n^2 a_0$, in agreement with Eq. (13.1.36). For later use, let us observe that as $E(\lambda)=E^0+E^1+\cdots=E(\lambda=0)+\lambda(\mathrm{d}E/\mathrm{d}\lambda)_{\lambda=0}+\cdots$, one way to extract $E^1$ from the exact answer is to calculate $\lambda(\mathrm{d}E/\mathrm{d}\lambda)_{\lambda=0}$.

(2) Consider now $\left\langle\lambda / r^2\right\rangle$. In this case, an exact solution is possible since the perturbation just modifies the centrifugal term as follows:
$$
\frac{\hbar^2 l(l+1)}{2 m r^2}+\frac{\lambda}{r^2}=\frac{\hbar^2 l^{\prime}\left(l^{\prime}+1\right)}{2 m r^2}\tag{17.3.23}
$$
where $l^{\prime}$ is a function of $\lambda$. Now the dependence of $E$ on $l^{\prime}(\lambda)$ is, from Eq. (13.1.14),
$$
E\left(l^{\prime}\right)=\frac{-m e^4}{2 \hbar^2\left(k+l^{\prime}+1\right)^2}=E(\lambda)=E^0+E^1+\cdots
$$
Show that
$$
\left\langle\frac{\lambda}{r^3}\right\rangle=E^1=\left.\lambda \frac{\mathrm{d}E}{\mathrm{d}\lambda}\right|_{\lambda=0}=\left(\frac{\mathrm{d}E}{\mathrm{d}l^{\prime}}\right)_{l^{\prime}=l} \cdot\left(\frac{\mathrm{d}l^{\prime}}{\mathrm{d} \lambda}\right)_{l^{\prime}=l} \cdot \lambda=\frac{\lambda}{n^3 a_0^2\left(l+\frac{1}{2}\right)}
$$
Canceling $\lambda$ on both sides, we get Eq. (17.3.11).

(3) Consider finally $\left\langle l / r^3\right\rangle$. Since there is no such term in the Coulomb Hamiltonian, we resort to another trick. Consider the radial momentum operator, $p_r=-i \hbar(\partial / \partial r+1 / r)$, in terms of which we may write the radial part of the Hamiltonian
$$
\left(\frac{-\hbar^2}{2 m}\right)\left(\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}\right)
$$
as $p_r^2 / 2 m$. (Verify this.) Using the fact that $\left\langle\left[H, p_r\right]\right\rangle=0$ in the energy eigenstates, and by explictly evaluating the commutator, show that
$$
\left\langle\frac{1}{r^3}\right\rangle=\frac{1}{a_0(l)(l+1)}\left\langle\frac{1}{r^2}\right\rangle
$$
combining which with the result from part (2) we get Eq. (17.3.20).

(4) Find the mean kinetic energy using the trick from part (1), this time rescaling the mass. Regain the virial theorem.

Chapter 17 Time-Independence Perturbation Theory

# Chapter 16 Variational and WKB Methods
## 16.1 The Variatonal Method
**Exercise 16.1.1** Try $\psi=\exp \left(-\alpha x^2\right)$ for $V=\dfrac{1}{2} m \omega^2 x^2$ and find $\alpha_0$ and $E\left(\alpha_0\right)$.



**Exercise 16.1.2** For a particle in a box that extends from $-a$ to $a$, try (within the box) $\psi=(x-a)(x+a)$ and calculate $E$. There is no parameter to vary, but you still get an upper bound. Compare it to the true energy, $E_0$. (Convince yourself that the singularities in $\psi^{\prime \prime}$ at $x= \pm a$ do not contribute to the energy.)

**Exercise 16.1.3** For the attractive delta function potential $V=-a V_0 \delta(x)$ use a Gaussian trial function. Calculate the upper bound on $E_0$ and compare it to the exact answer ($-m a^2 V_0^2/2\hbar^2$) (Exercise 5.2.3).

**Exercise 16.1.4** For the oscillator choose
$$
\begin{aligned}
\psi & =(x-a)^2(x+a)^2, & & |x| \leqslant a \\
& =0, & & |x|>a
\end{aligned}
$$
Calculate $E(a)$, minimize it and compare to $\hbar \omega / 2$.

**Exercise 16.1.5** Solve the variational problem for the $l=1$ states of the electron in a potential $V=-e^2/r$. In your trial function incorporate (i) correct behavior as $r \rightarrow 0$, appropriate to $l=1$, (ii) correct number of nodes to minimize energy, (iii) correct behavior of wave function as $r \rightarrow \infty$ in a Coulomb potential (i.e., exponential instead of Gaussian damping). Does it matter what $m$ you choose for $Y_l^m$ ? Comment on the relation of the energy bound you obtain to the exact answer.
## 16.2 The Wentzel-Kramers-Brillouin Method
**Exercise 16.2.1** Consider the function $W(E)$ introduced in Eq. (16.2.21a). Since $-E$ is the $t$ derivative of $S(t)$, it follows that $W(E)$, must be the Legendre transform of $S$. In this case $t$ must emerge as the derivative of $W(E)$ with respect to $E$. Verify that differentiation of the formula
$$
W(E)=\int_{x^{\prime}}^x \sqrt{2 m\left(E-V\left(x^{\prime \prime}\right)\right)} \mathrm{d} x^{\prime \prime}
$$
gives the time $t$ taken to go from start to finish at this preassigned energy.

**Exercise 16.2.2** Consider the free particle problem using the approach given above. Now that you know the wave functions explicitly, evaluate the integral in Eq. (16.2.18b) by contour integration to obtain
$$
U\left(x, x^{\prime}, t\right)=\theta\left(x-x^{\prime}\right) \mathrm{e}^{(\mathrm{i}/\hbar) \sqrt{2 m E}\left(x-x^{\prime}\right)}+\theta\left(x^{\prime}-x\right) \mathrm{e}^{(-\mathrm{i}/\hbar) \sqrt{2 m E}\left(x-x^{\prime}\right)}
$$
In doing the contour integrals, ask in which half-plane you can close the contour for a given sign of $x-x^{\prime}$. Compare the above result to the semiclassical result and make sure it all works out. Note that there is no need to form the combination $U\left(x, x^{\prime}, t\right)+U^*\left(x, x^{\prime}, t\right)$; we can calculate both the principal part and the delta function contributions explicitly because we know the $p$-dependence of the integrand explicitly. To see this more clearly, avoid the contour integral approach and use instead the formula for $(x \pm i \varepsilon)^{-1}$ given above. Evaluate the principal value integral and the contribution from the delta function and see how they add up to the result of contour integration. Although both routes are possible here, in the problem with $V \neq 0$ contour integration is not possible since the $p$-dependence of the wave function in the complex $p$ plane is not known. The advantage of the $U+U^*$ approach is that it only refers to quantities on the real axis and at just one energy.

**Exercise 16.2.3** Let us take a second look at our derivation. We worked quite hard to isolate the eigenfunctions at one energy: we formed the combination $U\left(x, x^{\prime}, z\right)+U^*\left(x, x^{\prime}, z\right)$ to get rid of the principal part and filter out the delta function. Now it is clear that if in Eq. (16.2.18a) we could integrate in the range $-\infty \leqslant t \leqslant \infty$, we would get the $\delta$ function we want. What kept us from doing that was the fact the $U(t)$ was constructed to be used for $t>0$. However, if we use the time evolution operator $\mathrm{e}^{-(\mathrm{i}/\hbar)Ht}$ for negative times, it will simply tell us what the system was doing at earlier times, assuming the same Hamiltonian. Thus we can make sense of the operator for negative times as well and define a transform that extends over all times. This is true in classical mechanics as well. For instance, if a stone is thrown straight up from a tall building and we ask when it will be at ground level, we get two answers, one with negative $t$ corresponding to the extrapolation of the given initial conditions to earlier times. Verify that $U\left(x, x^{\prime}, z\right)+U^*\left(x, x^{\prime}, z\right)$ is indeed the transform of $U(t)$ for all times by seeing what happens to $\langle x\mid \mathrm{e}^{-(\mathrm{i}/\hbar)Hf}\mid x^{\prime}\rangle$ under complex conjugation and the exchange $x \leftrightarrow x^{\prime}$. Likewise, ask what happens to $U_{\mathrm{cl}}$ when we transform it for all times. Now you will find that no matter what the sign of $x-x^{\prime}$ a single right moving trajectory can contribute-it is just that the time of the stationary point, $t^*$, will change sign with the sign of $x-x^{\prime}$. (The same goes for a left moving trajectory.)

**Exercise 16.2.4** Alpha particles of kinetic energy $4.2~\mathrm{MeV}$ come tunneling out of a nucleus of charge $Z=90$ (after emission). Assume that $V_0=0$, $x_0=10^{-12}~\mathrm{cm}$, and $V(x)$ is just Coulombic for $x \geqslant x_0$. (See Fig. 16.1) Estimate the mean lifetime. [Hint: Show $\gamma=\left(8 Z e^2 / \hbar v\right)\left[\cos ^{-1} y^{1 / 2}-y^{1 / 2}(1-y)^{1 / 2}\right]$, where $y=x_0 \mid x_e$. Show that $y \ll 1$ and use $\cos^{-1} y^{1/2} \simeq \dfrac{1}{2}\pi -y^{1/2}$ before calculating numbers.]

**Exercise 16.2.5** In 1974 two new particles called the $\psi$ and $\psi^{\prime}$ were discovered, with rest energies $3.1$ and $3.7~\mathrm{GeV}$, respectively ($1~\mathrm{GeV}=10^9~\mathrm{eV}$). These are believed to be nonrelativistic bound states of a “charmed” quark of mass $m=1.5~\mathrm{GeV}/c^2$ (i.e., $mc^2=1.5~\mathrm{GeV}$) and an antiquark of the same mass, in a linear potential $V(r)=V_0+kr$. By assuming that these are the $n=0$ and $n=1$ bound states of zero orbital angular momentum, calculate $V_0$ using the WKB formula. What do you predict for the rest mass of $\psi^{\prime \prime}$, the $n=2$ state? (The measured value is $\simeq 4.2~\mathrm{GeV}/c^2$.) [Hints: (1) Work with $\mathrm{GeV}$ instead of $\mathrm{eV}$. (2) There is no need to determine $k$ explicitly.]

**Exercise 16.2.6** Obtain Eq. (16.2.39) for the $\lambda x^4$ potential by the scaling trick.

**Exercise 16.2.7** Find the allowed levels of the harmonic oscillator by the WKB method.

**Exercise 16.2.8** Consider the $l=0$ radial equation for the Coulomb problem. Since $V(r)$ is singular at the turning point $r=0$, we can't use ($n+3/4$).

(1) Will the additive constant be more or less than $3/4$?

(2) By analyzing the exact equation near $r=0$, it can be shown that the constant equals $1$. Using this constant show that the WKB energy levels agree with the exact results.