Change numbering for Scattering from 5. to 6.

mkdocs.yml

@@ -38,13 +38,14 @@ nav:
     - 5.4. Berry phase: 'adiabatic_berry.md'
     - 5.5. Exercises: 'adiabatic_exercises.md'
   - 6. Scattering:
-    - 6.1. Classical analysis of scattering: 'ch51.md'
-    - 6.2. Quantum scattering: 'ch52.md'
-    - 6.3. The optical theorem: 'ch53.md'
-    - 6.4. Partial wave analysis: 'ch54.md'
-    - 6.5. Calculation of scattering cross sections with partial waves: 'ch55.md'
-    - 6.6. Summary: 'ch56.md'
-    - 6.7. Problems: 'ch57.md' 
+    - 6.1. Classical analysis of scattering: 'ch61.md'
+    - 6.2. Quantum scattering: 'ch62.md'
+    - 6.3. The optical theorem: 'ch63.md'
+    - 6.4. Partial wave analysis: 'ch64.md'
+    - 6.5. Calculation of scattering cross sections with partial waves: 'ch65.md'
+    - 6.6. Summary: 'ch66.md'
+    - 6.7. Problems: 'ch67.md'
+    - 6.8. Solutions: 'ch68.md'
   - Appendices:
     - Appendix A - Complex function theory survival guide: 'appA.md'
     - Appendix B - Green's functions in QM: 'appB.md'

src/ch51.md → src/ch61.md

-# 5. Scattering in classical and in quantum mechanics {#Chap:Scat}
+# 6. Scattering in classical and in quantum mechanics {#Chap:Scat}
 
 Scattering experiments are perhaps the most important tool for obtaining
 detailed information on the structure of matter, in particular the
@@ -42,7 +42,7 @@ freedom which are not included explicitly in the system (inclusion of
 these degrees of freedom would cause the energy to be conserved). In
 this chapter we shall consider elastic scattering.
 
-## 5.1. Classical analysis of scattering
+## 6.1. Classical analysis of scattering
 
 A well known problem in classical mechanics is that of the motion of two
 bodies attracting each other by a gravitational force whose value decays
@@ -70,7 +70,7 @@ supposed to be at rest. This might not always justified in a real
 experiment, but in a standard approach in classical mechanics, the full
 two-body problem is reduced to a one-body problem with with a reduced
 mass, which is the present case (in
-problem [5.1](./ch57.md) we will perform the same procedure for a
+problem [6.1](./ch67.md) we will perform the same procedure for a
 quantum two-body system). The incident particles interact with the
 scattering centre located at ${\bf r}=
 \pmb 0$ by the usual scalar two-point potential $V(r)$ which satisfies
@@ -96,7 +96,7 @@ interaction potential $V(r)$ which governs the interaction process.
 of the outcoming particle </figcaption>
 </figure>
 
-In figure [5.1](#Fig:Scat), the geometry of the process is shown. In addition
+In figure [6.1](#Fig:Scat), the geometry of the process is shown. In addition
 a small cone, spanned by the spherical polar angles $d\vartheta$ and
 $d\varphi$, is displayed. It is assumed here that the scattering takes
 place in a small neighbourhood of the scattering centre, and for the
@@ -107,7 +107,7 @@ scattering centre is given by
 $dA = R^2 \sin\vartheta d\vartheta d\varphi$. The quantity
 $\sin\vartheta d\vartheta d\varphi$ is called *spatial angle* and is
 usually denoted by $d\Omega$. This $d\Omega$ defines a cone like the one
-shown in figure [5.1](#Fig:Scat). Now consider the number of particles which will
+shown in figure [6.1](#Fig:Scat). Now consider the number of particles which will
 hit the detector within this small area per unit time. This number,
 divided by the total incident flux (see above) is called the
 *differential scattering cross section*, $d\sigma/
@@ -121,7 +121,7 @@ rotations around the $z$-axis, so the differential scattering cross
 section only depends on $\vartheta$. The only two relevant parameters of
 the incoming particle then are its velocity and its distance $b$ from
 the $z$-axis. This distance is called the *impact parameter* -- it is
-also shown in figure [5.1](#Fig:Scat).
+also shown in figure [6.1](#Fig:Scat).
 
 We first calculate the scattering angle $\vartheta$ as a function of the
 impact parameter $b$. We perform this calculation for the example of
@@ -132,14 +132,14 @@ given by specifying $r(t)$, $\vartheta(t)$. However, for the Kepler
 problem it is more convenient to focus on the *shape* of the orbitals,
 which is given as
 $$r = \lambda \frac{1+\epsilon}{\epsilon \cos(\vartheta-C) - 1}$$ with
-$$\tag{5.3}\label{Eq:KepHelp}
+$$\tag{6.3}\label{Eq:KepHelp}
 \epsilon = \sqrt{ 1 + \frac{2E \ell^2}{\mu A^2}};$$ this parameter is
 called *eccentricity* -- for a hyperbola, we have $|\epsilon|>1$. Here,
 $\ell$ is the angular momentum and $\mu$ the reduced mass. The
 integration constant $C$ reappears in the cosine because we have not
 chosen $\vartheta=0$ at the perihelion -- the closest approach occurs
 when the particle crosses the dashed line in
-figure [5.1](#Fig:Scat) which bisects the in- and outgoing particle
+figure [6.1](#Fig:Scat) which bisects the in- and outgoing particle
 direction.
 
 We know that for the incoming particles, for which $\vartheta=\pi$,
@@ -151,7 +151,7 @@ cosine is $C-\pi$, so that we find $$\vartheta_\infty - C = C - \pi,$$
 or $\vartheta_\infty = 2C-\pi$. The subscript $\infty$ indicates that
 this value corresponds to $t \rightarrow \infty$. From the last two
 equations we find the following relation between the scattering angle
-$\vartheta_\infty$ and $\epsilon$: $$\tag{5.6}\label{Eq:ThetaEps}
+$\vartheta_\infty$ and $\epsilon$: $$\tag{6.6}\label{Eq:ThetaEps}
  \sin(\vartheta_\infty/2) = \cos(\pi/2-\vartheta/2) = \cos(\pi - C) = 1/\epsilon.$$
 
 We want to know $\vartheta_\infty$ as a function of $b$ rather than
@@ -162,7 +162,7 @@ $$E = \frac{\mu}{2} v^2_{\text{inc}},$$ so that the impact parameter can
 be found as $$b = \frac{\ell}{\sqrt{2\mu E}}.$$ Using
 Eq. \eqref{Eq:KepHelp} and the fact that
 $\cot(x) = \sqrt{1-\sin^2(x)}/\sin(x)$, we can finally write
-\eqref{Eq:ThetaEps} in the form: $$\tag{5.10}\label{Eq:BTheta}
+\eqref{Eq:ThetaEps} in the form: $$\tag{6.10}\label{Eq:BTheta}
 \cot(\vartheta_\infty/2) = \sqrt{\epsilon^2 - 1} = \frac{2 E b}{|A|}.$$
 
 From the relation between $b$ and $\vartheta_\infty$ we can find the

src/ch52.md → src/ch62.md

-# 5.2. Quantum scattering
+# 6.2. Quantum scattering
 
 In quantum scattering, we know parts of the wave function describing the
 particles that are scattered off some other particles. Just as for
 classical scattering, we consider a two-particle collision, which we can
 reduce to a single particle problem in which an incident particle
 scatters off a static potential which is nonzero only near the origin --
-see problem [5.1](./ch57.md). Assuming that the potential vanishes indeed
+see problem [6.1](./ch67.md). Assuming that the potential vanishes indeed
 outside of a sphere with radius $r_{\text{max}}$ centered at the origin,
 we know the wave functions for the incident beam along the $z$-axis and
 the scattered waves. So, outside of the sphere, we have
@@ -52,7 +52,7 @@ Before calculating $G_0$ let us assume we have it at our disposal. We
 then may write the solution to the *full* Schrödinger equation,
 i.e. including the potential $V$, in terms of a solution $\phi({\bf r})$
 to the 'bare' Schrödinger equation, that is, the Schrödinger
-equation with potential $V\equiv 0$: $$\tag{5.14}\label{Eq:Dyson}
+equation with potential $V\equiv 0$: $$\tag{6.14}\label{Eq:Dyson}
 \psi({\bf r}) = \phi({\bf r}) + \int G_0({\bf r}, {\bf r}') V({\bf r}') \psi({\bf r}')\; d^3 r',$$
 which is an explicit form of the equation we met in
 chapter [Appendix B](./appB.md):
@@ -124,7 +124,7 @@ transform of the scattering potential.
 
 For future reference, we note that the exact expression for the
 scattering amplitude is given by
-$$f(\vartheta, \varphi) = -\frac{m}{2\pi \hslash^2} \int V({\bf r}') \tag{5.15}\label{ExactF}
+$$f(\vartheta, \varphi) = -\frac{m}{2\pi \hslash^2} \int V({\bf r}') \tag{6.15}\label{ExactF}
 e^{{\rm i}{\bf k}\cdot {\bf r}} \psi({\bf r}) \; d^3 r'.$$ where
 ${\bf k}$ is the outgoing wave with polar angles $\vartheta,\varphi$.
 This is the equation one would obtain from \eqref{Eq:Dyson}

src/ch53.md → src/ch63.md

-# 5.3. The optical theorem {#Sec:OptTheor}
+# 6.3. The optical theorem {#Sec:OptTheor}
 
 We conclude this chapter by deriving an important theorem which results
 from the conservation of matter. This is the *optical theorem*, which
@@ -76,7 +76,7 @@ Writing $\psi = \phi_{\text{in}} + \phi_{\text{out}}$, this can be
 rewritten as
 $$-k \sigma_{\text{tot}} = {\text{Im}} \int_V \left[ \phi^*_{\text{in}} \nabla^2 \left( \psi - \phi_{\text{in}}\right) - 
 \left( \psi - \phi_{\text{in}}\right)\nabla^2 \phi^*_{\text{in}} \right] d^3r = {\text{Im}} \int_V \left[ \phi^*_{\text{in}}  \nabla^2 \left( \psi \right) - 
- \psi\nabla^2 \left( \phi^*_{\text{in}} \right)\right] d^3r. \tag{5.16}\label{Eq:OptTheor}$$
+ \psi\nabla^2 \left( \phi^*_{\text{in}} \right)\right] d^3r. \tag{6.16}\label{Eq:OptTheor}$$
 Now we use that the full wave function satisfies the full Schrödinger
 equation $$\nabla^2 \psi = -k^2 \psi + \frac{2m V({\bf r}) }{\hslash^2} \psi,$$
 and the incoming wave satisfies the Schrödinger equation with potential
@@ -87,7 +87,7 @@ The first and the third term in the integral cancel and we have
 $$-k \sigma_{\text{tot}} = {\text{Im}} \int_V \phi^*_{\text{in}} \frac{2m V(r)}{\hslash^2} \psi \; d^3 r.$$
 The last term is recognised as the exact scattering amplitude for
 $\theta=0$ (up to a negative pre-factor) -- see
-Eq. [(5.15)](./ch52.md#mjx-eqn-ExactF).
+Eq. [(6.15)](./ch62.md#mjx-eqn-ExactF).
 
 Therefore, we find the *optical theorem*:
 $$\sigma_{\text{tot}} = {\text{Im}} \frac{4\pi f(\theta=0)}{k}.$$ If the

src/ch54.md → src/ch64.md

-# 5.4. Partial wave analysis
+# 6.4. Partial wave analysis
 
 The scattering process is described by the solutions of the
 single-particle Schrödinger equation involving the (reduced) mass $m$,
@@ -13,7 +13,7 @@ dimensions, which could be solved using the 'brute force' discretisation
 methods, but exploiting the spherical symmetry of the potential, we can
 solve the problem in another, more elegant, way which, moreover, works
 much faster on a computer. More specifically, in
-Section [5.5](./ch55.md) we shall establish a relation between the
+Section [6.5](./ch65.md) we shall establish a relation between the
 *phase shift* and the scattering cross sections. In this section, we
 shall restrict ourselves to a description of the concept of phase shift
 and describe how it can be obtained from the solutions of the radial
@@ -37,7 +37,7 @@ $$
 \frac{\hslash^2 l(l+1)}{2m r^2} \right] \right\}
     u_l (r) = 0 . \label{RadSchrod}$$
 
-Figure [5.2](#Fig:phase) shows the solution of the radial Schrödinger equation with $l=0$ for a
+Figure [6.2](#Fig:phase) shows the solution of the radial Schrödinger equation with $l=0$ for a
 square well potential for various well depths -- our discussion applies
 also to nonzero values of $l$. 
 
@@ -52,12 +52,12 @@ be written as a linear combination of the two independent solutions
 $j_l$ and $n_l$, the regular and irregular spherical Bessel functions.
 We write this linear combination in the particular form
 $$u_l(r>{r_{\text{max}}}) \propto kr \left[\cos\delta_l j_l(kr) - \sin\delta_l n_l(kr)
-\right]. \tag{5.17}\label{Eq:LinComb}$$ $\delta_l$ is determined via a matching
+\right]. \tag{6.17}\label{Eq:LinComb}$$ $\delta_l$ is determined via a matching
 procedure at the well boundary. The motivation for writing $u_l$ in this
 form follows from the asymptotic expansion for the spherical Bessel
 functions:
 
-$$\tag{5.18}\label{AsymBes}
+$$\tag{6.18}\label{AsymBes}
 \begin{align}
 kr j_l(kr) &\approx \sin(kr-l\pi/2)  \\ \nonumber
 kr n_l(kr) &\approx -\cos(kr-l\pi/2)  \\ \nonumber
@@ -78,13 +78,13 @@ The phase shift as a function of energy and $l$ contains all the
 information about the scattering properties of the potential. In
 particular, the phase shift enables us to calculate the scattering cross
 sections and this will be done in section
-[5.5](./ch55.md);
+[6.5](./ch65.md);
 here we simply quote the results. The differential cross section is
 given in terms of the phase shift by
 $$\frac{d\sigma}{d\Omega} = \frac{1}{k^2} \left| \sum_{l=0}^\infty (2l+1)
 e^{i\delta_l}
-       \sin(\delta_l) P_l (\cos \theta)\right|^2 \tag{5.19}\label{Eq:DiffCrSec}$$
-and for the total cross section we find $$\tag{5.20}\label{sigmatotdef}
+       \sin(\delta_l) P_l (\cos \theta)\right|^2 \tag{6.19}\label{Eq:DiffCrSec}$$
+and for the total cross section we find $$\tag{6.20}\label{sigmatotdef}
 \sigma_{{\text{tot}}} = 2 \pi \int d\theta \sin \theta
 \frac{d\sigma}{d\Omega}(\theta)
 =  \frac{4\pi}{k^2} \sum_{l=0}^\infty (2l+1) \sin^2 \delta_l.$$
@@ -161,7 +161,7 @@ injected at this energy are strongly scattered and this shows up as a
 peak in the total cross section.
 
 Such peaks can be seen
-Figure [5.4](#Fig:Sigma), which shows the total cross section as a
+Figure [6.4](#Fig:Sigma), which shows the total cross section as a
 function of the energy calculated with a program as described above. The
 peaks are due to $l=4$, $l=5$ and $l=6$ scattering, with energies
 increasing with $l$.
@@ -174,7 +174,7 @@ the range $\rho$ of the Lennard-Jones potential.
  </figcaption>
 </figure>
 
-Figure [5.5](#Fig:ExpRes) finally shows the experimental results for the
+Figure [6.5](#Fig:ExpRes) finally shows the experimental results for the
 total cross section for H--Kr. We see that the agreement is excellent.
 
 <figure markdown>
@@ -186,4 +186,4 @@ scattering of hydrogen atoms by krypton atoms as function of centre of mass ener
 </figure>
 
 You should be able now to reproduce the data of
-Figure [5.4](#Fig:Sigma) with your program.
+Figure [6.4](#Fig:Sigma) with your program.

src/ch55.md → src/ch65.md

-# 5.5. Calculation of scattering cross sections with partial waves {#Sec:ScatTheor}
+# 6.5. Calculation of scattering cross sections with partial waves {#Sec:ScatTheor}
 
-In this section we derive Eqs. [5.19](./ch54.md#mjx-eqn-Eq:DiffCrSec) and [5.20](./ch54.md#mjx-eqn-sigmatotdef). At a large distance from the scattering
+In this section we derive Eqs. [6.19](./ch64.md#mjx-eqn-Eq:DiffCrSec) and [6.20](./ch64.md#mjx-eqn-sigmatotdef). At a large distance from the scattering
 centre we can make an *Ansatz* for the wave function. This consists of
 the incoming beam and a scattered wave:
 $$\psi ({\bf r}) \sim e^{i\bf k \cdot r } + f(\theta) \frac{e^{ikr}}{r}. 
-\tag{5.22}\label{eq_psi}$$ $\theta$ is the angle between the incoming beam and the
+\tag{6.22}\label{eq_psi}$$ $\theta$ is the angle between the incoming beam and the
 line passing through ${\bf r}$ and the scattering centre. $f$ does not
 depend on the azimuthal angle $\varphi$ because the incoming wave has
 azimuthal symmetry, and the spherically symmetric potential will not
@@ -12,7 +12,7 @@ generate $m\neq 0$ contributions to the scattered wave. $f(\theta)$ is
 called the scattering amplitude. From the *Ansatz* it follows that the
 differential cross section is given directly by the square of this
 amplitude:
-$$\frac{d\sigma}{d\Omega} = |f( \theta ) |^2  \tag{5.23}\label{eq_sigma}$$ with
+$$\frac{d\sigma}{d\Omega} = |f( \theta ) |^2  \tag{6.23}\label{eq_sigma}$$ with
 the appropriate normalisation.
 
 Beyond ${r_{\text{max}}}$, the solution can also be written in the form
@@ -22,13 +22,13 @@ Y_l^m
 leaving out all $m\neq 0$ contributions because of the azimuthal
 symmetry:
 $$\psi({\bf r}) = \sum_{l=0}^{\infty} A_l \frac{u_{l}(r)}{r} P_l
-(\cos\theta)  \tag{5.23}\label{eq_psi3}$$ where we have used the fact that
+(\cos\theta)  \tag{6.23}\label{eq_psi3}$$ where we have used the fact that
 $Y^l_0(\theta, \phi)$ is proportional to $P_l[\cos(\theta)]$. Because
 the potential vanishes in the region $r>r_{\text {max}}$, the solution
 $u_{l}(r)/r$ is given by the linear combination of the regular and
 irregular spherical Bessel functions, and as we have seen this reduces
 for large $r$ to
-$$u_l (r) \approx  \sin(kr - \frac{l\pi}{2} + \delta_l ).\tag{5.25}\label{bess_approx}$$
+$$u_l (r) \approx  \sin(kr - \frac{l\pi}{2} + \delta_l ).\tag{6.25}\label{bess_approx}$$
 We want to derive the scattering amplitude $f(\theta)$ by equating the
 expressions \eqref{eq_psi} and \eqref{eq_psi3} for the wave function. For large $r$ we obtain,
 using \eqref{bess_approx}:
@@ -42,7 +42,7 @@ wave
 $$e^{i \bf k \cdot r} = \sum_{l=0}^\infty (2l+1) i^l j_l(kr) P_l (\cos
 \theta).$$ $f(\theta)$ can also be written as an expansion in Legendre
 polynomials:
-$$f(\theta) = \sum_{l=0}^\infty f_l P_l (\cos \theta) \tag{5.x}\label{f_exp},$$
+$$f(\theta) = \sum_{l=0}^\infty f_l P_l (\cos \theta) \tag{6.x}\label{f_exp},$$
 so that we obtain: 
 $$\begin{gathered}
 \sum_{l=0}^\infty A_l \left[ \frac{\sin(kr - l\pi/2 + \delta_l )}{kr}
@@ -50,13 +50,13 @@ $$\begin{gathered}
 \sum_{l=0}^\infty \left[ (2l+1) i^l j_l (kr) + f_l \frac{e^{ikr}}{r}
                        \right] P_l (\cos \theta).
 \end{gathered}$$
-If we substitute the [asymptotic form](./ch54.md#mjx-eqn-AsymBes) of $j_l$ in the right hand side, we find:
+If we substitute the [asymptotic form](./ch64.md#mjx-eqn-AsymBes) of $j_l$ in the right hand side, we find:
 $$
 \begin{gather}
 \sum_{l=0}^\infty A_l \left[ \frac{\sin(kr - l\pi/2 + \delta_l )}{kr}
 \right] P_l (\cos \theta) = 
 \\
-\frac{1}{r}\sum_{l=0}^\infty \left[ \frac{2l+1}{2ik} (-)^{l+1} e^{-ikr} + \left(f_l+ \frac{2l+1}{2ik} \right) e^{ikr} \right] P_l (\cos \theta) \tag{5.26}\label{equat}.
+\frac{1}{r}\sum_{l=0}^\infty \left[ \frac{2l+1}{2ik} (-)^{l+1} e^{-ikr} + \left(f_l+ \frac{2l+1}{2ik} \right) e^{ikr} \right] P_l (\cos \theta) \tag{6.26}\label{equat}.
 \end{gather}$$
 Both the left and the right hand side of
 \eqref{equat} contain
@@ -68,7 +68,7 @@ should both be equal on both sides in
 condition leads to 
 $$
 A_l = (2l+1) e^{i\delta_l} i^l$$ and
-$$f_l = \frac{2l+1}{k} e^{i \delta_l} \sin(\delta_l).\tag{5.27}\label{f_delta}$$
+$$f_l = \frac{2l+1}{k} e^{i \delta_l} \sin(\delta_l).\tag{6.27}\label{f_delta}$$
 
 Using \eqref{eq_sigma}, \eqref{f_exp}, and \eqref{f_delta}, we can write down an expression for the differential cross section in terms of the phase shifts $\delta_l$:
 $$

src/ch56.md → src/ch66.md

-# 5.6. Summary
+# 6.6. Summary
 
 In this chapter, we have analysed scattering of particles by a potential
 localised in a finite region around some point, which we take as the

src/ch57.md → src/ch67.md

-# 5.7. Problems
+# 6.7. Problems
 
-## 5.1. Two body problem
+## 6.1. Two body problem
 
 Show that for two identical
 particles, 1 and 2, with coordinates ${\bf r}_1$ and ${\bf r}_2$,
@@ -17,7 +17,7 @@ Also show that $\psi({\bf r}, {\bf R})$ can be written as
 $\phi({\bf r})\chi({\bf R})$. Find suitable eigen-equations for
 $\phi$ and $\chi$ (this is separation of variables).
 
-## 5.2.  
+## 6.2.  
 Consider the 1-D scattering problem illustrated in the figure below,
 with an arbitrary localised potential (without any particular
 spatial symmetry) in Region II, and $V(x)=0$ in Regions I and III.
@@ -69,7 +69,7 @@ $\{A,B,C,D\}=\{0,S_{\mathrm{LR}},S_{\mathrm{RR}},1\}$.
 6.  Are (c) and (d) necessary or simply sufficient conditions for
     $S$ to be unitary?
 
-## 5.3.  
+## 6.3.  
 In this problem, we consider scattering in 1 dimension for the
 potential in the following picture:
 
@@ -124,7 +124,7 @@ at the (horizontal) coordinate $r_s$.
 4.  If your program works correctly, you find several peaks. Give an
     explanation for those peaks.
 
-5.  Now imagine a potential as in the picture, but now with
+6.  Now imagine a potential as in the picture, but now with
     $b=\infty$. In that case, the problem is that of the square
     well. You can find the bound states of the well graphically by
     intersecting $\tan(kx)$ with the line $q$, where $k$ is the wave
@@ -137,19 +137,19 @@ at the (horizontal) coordinate $r_s$.
     investigate variations of the parameters $w$, $V_w$ and explain
     what you see
 
-## 5.4.  
+## 6.4.  
 Obtain the total cross-section for the potential given by
 $$V(r) = \begin{cases} -V_0 & {\text{ for }} r<a; \\
                         0 & {\text{ for }} r>a.
         \end{cases}$$ Explain the result you get when taking the low
 energy limit?
 
-## 5.5.
+## 6.5.
 Consider scattering off a spherical cage (such as a bucky ball),
 which is described by a $\delta$-function $$V(r) = g \delta (r-a).$$
 Find the scattering amplitude.
 
-## 5.6.
+## 6.6.
 In time-dependent perturbation theory, we can calculate the
 probability to move from some initial state to some final state as a
 result of a perturbation which was turned on for a finite time.
@@ -211,7 +211,7 @@ the explicit time-dependence of these coefficients.
     expressions, we assume the normalisation
     $$\left\langle{\bf k}| {\bf r}\right\rangle= \frac{e^{-{\rm i}{\bf k}\cdot {\bf r}}}{(2\pi)^{3/2}}.$$
 
-## 5.7.
+## 6.7.
 In this problem we consider a *Josephson junction*: two
 superconducting regions separated by a thin insulating layer. In
 each superconduting island, Cooper pairs of charge $2e$ and mass
@@ -252,14 +252,14 @@ this problem, we neglect the finite thickness of the layer.
     $$I = I_c \sin\delta(t),$$ where $I_c$ is some constant (it is
     the critical current) which you do not need to calculate.
 
-## 5.8.
+## 6.8.
 Calculate the differential cross section $d\sigma/d\Omega$ in the
 first Born approximation for a potential of the form
 $$V(r) = -V_0 e^{-r/R}.$$ *Hint:*
 $\int_0^\infty e^{-ar} r dr = 1/a^2.$ Note that $a$ may be complex
 in this expression!
 
-## 5.9.  *Scattering off a charge distribution*
+## 6.9.  *Scattering off a charge distribution*
 We consider scattering of particles with charge $e$ off a charge
 distribution $\rho({\bf r})$ located near the origin. The
 electrostatic potential felt by the incoming particles is given as
@@ -276,7 +276,7 @@ $$-\nabla^2 \varphi({\bf r}) = \rho({\bf r})/\epsilon_0.$$
 2.  Show that the scattering amplitude is given by
     $$f(\theta,\phi) = - \frac{me}{2\pi\hslash^2} \frac{\tilde{\rho}({\bf q})}{\epsilon_0 q^2}.$$
 
-## 5.10. 
+## 6.10. 
 In the Born approximation, the forward-scattering amplitude (the
 scattering angle $\vartheta = 0$) is real, yielding
 $\sigma_{\text{tot}} = 0$, and therefore it appears to be in

src/ch58.md → src/ch68.md

-# 5.8. Solutions
+# 6.8. Solutions
 
 Click :fontawesome-solid-wand-magic-sparkles: above to activate the Jupyter notebook mode and display the source code.
 
-## 5.3. Resonant scattering with a 1D square potential
+## 6.3. Resonant scattering with a 1D square potential
 
 In this problem, we consider scattering in 1 dimension for the
 potential in the following picture:
@@ -31,7 +31,7 @@ Consider a separation between a region with potential $V_\text{R}$
 on the right and $V_\text{L}$ on the left. The separation is located
 at the (horizontal) coordinate $r_s.$
 
-### 5.3.1
+### 6.3.1
 
 Give the scattering matrix which relates the expansion coefficients on the left of $r_s$ to those at the right (suppose those at the right are given). The matrix elements depend on $V_{\text{L}}$, $V_{\text{R}}$, $E$ and $r_s.$ The energy can be smaller or larger than $V_{\text{L}}$, $V_{\text{R}}$.
 
@@ -87,7 +87,7 @@ def make_mat(kl,kr,rs):
     return m11, m12, m21, m22
 ```
 
-### 5.3.2
+### 6.3.2
 
 If we want to find the transmission amplitude, we take on the
 very right exclusively a right-moving wave, with amplitude 1. By
@@ -125,7 +125,7 @@ nvec = np.ndarray((enernum, 2),dtype=complex)
 
 ```
 
-### 5.3.3. 
+### 6.3.3. 
 In the leftmost region, there is an incoming and a reflected
 wave. The transmission is defined as the amplitude of the
 outgoing wave in the rightmost region, provided that the