# Chapter 10 -- Homework
###### tags: `Quantum Physics II`
B05202054 何信佑
https://hackmd.io/@ulynx/QP2_Ch10_Hw
## Problem 10.1
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**Rutherford scattering**. An incident particle of charge $q_1$ and kinetic energy $E$ scatters off a heavy stationary particle of charge $q_2$.
- **(a)** Derive the formula relating the impact parameter to the scattering angle. *Answer:* $b=(q_1q_2/8\pi\epsilon_0 E)\cot(\theta/2)$.
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### Solution of (a)
1. For a two-body central-force problem with conservative field, we have
- conservation of mechanical energy: $$E=\dfrac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right)+V(r)=\text{constant},\tag{1}$$ where $V(r)=\dfrac{q_1q_2}{4\pi\epsilon_0}\dfrac{1}{r}$ is the Coulomb potential energy. For simplicity, we let $\kappa=\dfrac{q_1q_2}{4\pi\epsilon_0}$.
- conservation of angular momentum: $$\ell=mr^2\dot\phi=\text{constant},\tag{2}$$ which implies $\dot\phi=\ell/mr^2$.

Putting $\dot\phi$ into Eq. (1) and solve for $\dot{r}^2$, we obtain $$\dot{r}^2=\dfrac{2}{m}\left[E-V(r)\right]-\dfrac{\ell^2}{m^2r^2}\tag{3}$$
Let $u=1/r$; thus $$\require{cancel}\dot r=\dfrac{dr}{dt}=\dfrac{dr}{du}\dfrac{du}{d\phi}\dfrac{d\phi}{dt}=-\dfrac{1}{u^2}\dot\phi\dfrac{du}{d\phi}=-\dfrac{1}{\cancel{u^2}}\dfrac{\ell}{m\cancel{r^2}}\dfrac{du}{d\phi}=-\dfrac{\ell}{m}\dfrac{du}{d\phi}.$$ Then (3) becomes $$\begin{align}&\hspace{3em}\left(-\dfrac{\ell}{m}\dfrac{du}{d\phi}\right)^2=\dfrac{2}{m}\left(E-\kappa u\right)-\dfrac{\ell^2u^2}{m^2}\\&\implies \dfrac{du}{d\phi}=\sqrt{\dfrac{2m}{\ell^2}\left(E-\kappa u\right)-u^2}\\&\implies d\phi=\dfrac{du}{\sqrt{\frac{2m}{\ell^2}\left(E-\kappa u\right)-u^2}}.\tag{4}\end{align}$$

2. We should integrate Eq. (4)
- from $u=0$ ($r=\infty$) and $\phi=0$
- to $u=u_\max$ ($r=r_\min$) and $\phi=\phi_0$.

The result is $$\displaystyle \phi_0=\int^{u_\max}_0 \dfrac{du}{\sqrt{\frac{2m}{\ell^2}\left(E-\kappa u\right)-u^2}}.\tag{5}$$ From the graph we see $\phi_0+\phi_0+\theta=\pi$, and hence Eq. (5) is $$ \theta=\pi-2\phi_0=\pi-2\int^{u_\max}_0 \frac{du}{\sqrt{\frac{2m}{\ell^2}\left(E-\kappa u\right)-u^2}}.\tag{6}$$

3. Let $u_1$ and $u_2$, with $u_1<u_2$, be the two roots of $\frac{2m}{\ell^2}\left(E-\kappa u\right)-u^2=0$ or $$u^2+\frac{2m\kappa}{\ell^2}u-\frac{2mE}{\ell^2}=0;\tag{7}$$ i.e. $$\frac{2m}{\ell^2}\left(E-\kappa u\right)-u^2=-(u-u_1)(u-u_2).$$

But (Eq. (4)) $$\dfrac{du}{d\phi}=\sqrt{\frac{2m}{\ell^2}\left(E-\kappa u\right)-u^2}=\sqrt{(u-u_1)(u_2-u)}$$ entails that $u_\max$ is the larger root of Eq. (7), i.e. $u_2=u_\max$.

Use the integral formula $$\displaystyle \int_a^b\frac{du}{\sqrt{(u-u_1)(u_2-u)}}=\left.-\sin^{-1}\left(\frac{-2u+u_2+u_1}{u_2-u_1}\right)\right|_a^b$$ to carry out Eq. (6): $$\displaystyle \begin{align}\theta&=\pi-2\int^{u_\max}_0 \dfrac{du}{\sqrt{(u-u_1)(u_\max-u)}}\\&=\pi+2\left[\sin^{-1}\left(\frac{-2u+u_\max+u_1}{u_\max-u_1}\right)\right]_0^{u_\max}\\&=\pi+2\left[\sin^{-1}(-1)-\sin^{-1}\left(\frac{u_\max+u_1}{u_\max-u_1}\right)\right]\\&=\pi+2\left[-\dfrac{\pi}{2}-\sin^{-1}\left(\frac{u_\max+u_1}{u_\max-u_1}\right)\right]\\&=-2\sin^{-1}\left(\frac{u_\max+u_1}{u_\max-u_1}\right)\tag{8}\end{align}$$

4. Solving Eq. (7), we have $$u_\max=-\dfrac{m\kappa}{\ell^2}+\sqrt{\left(\dfrac{m\kappa}{\ell^2}\right)^2+\dfrac{2mE}{\ell^2}}\\u_1=-\dfrac{m\kappa}{\ell^2}-\sqrt{\left(\dfrac{m\kappa}{\ell^2}\right)^2+\dfrac{2mE}{\ell^2}}$$ and $$\begin{align}\dfrac{u_\max+u_1}{u_\max-u_1}&=\dfrac{-\frac{m\kappa}{\ell^2}}{\sqrt{\left(\frac{m\kappa}{\ell^2}\right)^2+\frac{2mE}{\ell^2}}}=\dfrac{-1}{\sqrt{1+\frac{2mE}{\ell^2}\left(\frac{\ell^2}{m\kappa}\right)^2}}\\&=\dfrac{-1}{\sqrt{1+\frac{2E\ell^2}{m\kappa^2}}}=\dfrac{-1}{\sqrt{1+\left(2bE/\kappa\right)^2}}\tag{9}\end{align}$$ where we used $E=\frac{1}{2}mv^2$, $\ell=mvb$, where $b$ is the impact parameter, so that $$\dfrac{2E\ell^2}{m\kappa^2}=\dfrac{\cancel{m}v^2m^2v^2b^2}{\cancel{m}\kappa^2}=\dfrac{\left(mv^2\right)^2b^2}{\kappa^2}=\dfrac{(2E)^2b^2}{\kappa^2}=\left(\dfrac{2bE}{\kappa}\right)^2$$

5. Finally, we put Eq. (9) into Eq. (8): $$\begin{align}&\theta=-2\sin^{-1}\left(\dfrac{u_\max+u_1}{u_\max-u_1}\right)=-2\sin^{-1}\dfrac{-1}{\sqrt{1+\left(2bE/\kappa\right)^2}}\\&\implies -\dfrac{\theta}{2}=\sin^{-1}\dfrac{-1}{\sqrt{1+\left(2bE/\kappa\right)^2}}\\&\implies \sin\left(-\dfrac{\theta}{2}\right)=\dfrac{-1}{\sqrt{1+\left(2bE/\kappa\right)^2}}\\&\implies \left(\dfrac{2bE}{\kappa}\right)^2=\dfrac{1}{\sin^2(\theta/2)}-1=\dfrac{1-\sin^2(\theta/2)}{\sin^2(\theta/2)}=\cot^2(\theta/2)\\&\implies \boxed{b=\dfrac{\kappa}{2E}\cot\left(\dfrac{\theta}{2}\right)=\dfrac{q_1q_2}{8\pi\epsilon_0 E}\cot\left(\dfrac{\theta}{2}\right)}\tag{10}\end{align}$$

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- **(b)** Determine the differential scattering cross-section. *Answer:* $$D(\theta)=\left[\dfrac{q_1q_2}{16\pi\epsilon_0E\sin^2(\theta/2)}\right]^2.\tag{10.11}$$
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### Solution of (b)
Following Eq. (10), we calculate the differential cross-section (Equation 10.4) $$\begin{align}D(\theta)&=\dfrac{b}{\sin\theta}\left|\dfrac{db}{d\theta}\right|\\&=\dfrac{1}{\sin\theta}\dfrac{\kappa}{2E}\cot\left(\dfrac{\theta}{2}\right)\left|\dfrac{\kappa}{2E}\dfrac{d}{d\theta}\cot\left(\frac{\theta}{2}\right)\right|\\&=\left(\dfrac{\kappa}{2E}\right)^2\dfrac{1}{2\sin(\frac{\theta}{2})\cancel{\cos(\frac{\theta}{2})}}\dfrac{\cancel{\cos\frac{\theta}{2}}}{\sin\frac{\theta}{2}}\left|\dfrac{d}{d\theta}\cot\left(\frac{\theta}{2}\right)\right|\\&=\left(\dfrac{\kappa}{2E}\right)^2\dfrac{1}{2\sin^2(\theta/2)}\left|-\dfrac{1}{2}\csc^2\left(\frac{\theta}{2}\right)\right|\\&=\left[\dfrac{\kappa}{4E\sin^2(\theta/2)}\right]^2=\boxed{\left[\dfrac{q_1q_2}{16\pi\epsilon_0E\sin^2(\theta/2)}\right]^2}.\tag*{$\blacksquare$}\end{align}$$

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- **(c\)** Show that the total cross-section for Rutherford scattering is *infinite*.
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### Solution of (c\)
The total cross-section is the integral of $D(\theta)$ over all solid angles: (Equation 10.7) $$\begin{align}\sigma&=\int D(\theta)d\Omega=\iint \left[\tfrac{q_1q_2}{16\pi\epsilon_0E\sin^2(\theta/2)}\right]^2\sin(\theta)d\theta d\phi\\&=\left(\dfrac{q_1q_2}{16\pi\epsilon_0E}\right)^2\underbrace{\left(\small\int_0^{2\pi}d\phi\right)}_{=\,2\pi}\underbrace{\left(\small\int_0^\pi\dfrac{\sin\theta }{\sin^4(\theta/2)}d\theta\right)}_{\equiv\,I}.\tag{11}\end{align}$$ But the integral $I$ doesn't converge because when $\theta$ is small (we can assume $0<\theta<\epsilon\ll\pi$, where $\epsilon$ is a very small), by approximation $\sin\theta\approx\theta$ and $\sin(\theta/2)\approx\theta/2$ we see that $$\begin{align}&\quad\int_0^\pi\dfrac{\sin\theta }{\sin^4(\theta/2)}d\theta>\int_0^\epsilon\frac{\sin\theta }{\sin^4(\theta/2)}d\theta\approx\int_0^\epsilon\frac{\theta }{(\theta/2)^4}d\theta=16\int_0^\epsilon\frac{d\theta}{\theta^3}\\&=16\left.\left(-\dfrac{1}{2\theta^2}\right)\right|_0^\epsilon=8\left(-\dfrac{1}{\epsilon^2}+\lim_{\theta\to0}\dfrac{1}{\theta^2}\right)= +\infty.\tag*{$\blacksquare$}\end{align}$$
## Problem 10.4
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Consider the case of low-energy scattering from a spherical delta-function shell: $$V(r)=\alpha\delta (r-a),$$ wjere $\alpha$ and $a$ are constants. Calculate the scattering amplitude, $f(\theta)$, the differential cross-section, $D(\theta)$, and the total cross-section, $\sigma$. Assume $ka\ll 1$, so that only the $\ell=0$ term contributes significantly. (To simplify matters, throw out all $\ell\neq 0$ terms right from the start.) The main problem, of course, is to determine $C_0$. Express your answer in terms of the dimensionless quantitu $\beta\equiv 2ma\alpha/\hbar^2$. *Answer:* $\sigma=4\pi a^2\beta^2/(1+\beta)^2$.
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## Problem 10.7
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Find the $S$-wave ($\ell=0$) partial wave phase shift $\delta_0(k)$ for scattering from a delta-function shell (Problem 10.4). Assume that the radial wave function $u(r)$ goes to $0$ as $r\to 0$. *Answer:* $$-\cot^{-1}\left[\cot(ka)+\dfrac{ka}{\beta\sin^2(ka)}\right],\quad\text{where}~~\beta\equiv\dfrac{2m\alpha a}{\hbar^2}$$
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### Solution
1. Within $r<a$, the potential energy $V=0$, and the radial wave funciton of $S$-wave, $u(r)=rR(r)$, satisfies $$\dfrac{d^2u}{dr^2}+k^2u=0\quad \left(r<a,~k=\tfrac{\sqrt{2mE}}{\hbar}\right),$$ whose solution is $$u(r)=A\sin(kr+\delta_0').\tag{1}$$
$\because R(r)=\dfrac{u(r)}{r}$ is finite as $r\to 0$, $\delta_0'=0$, $\therefore~u(r)=A\sin(kr)\quad (r<a)$
2. Within $r<a$, the potential energy $V=0$, and the radial wave funciton is $$u(r)=C\sin(kr+\delta_0)\quad(r>a).\tag{2}$$

3. Due to continuity of the wave function at $r=a$, we require $$C\sin(ka+\delta_0)=A\sin(ka).\tag{3}$$ Besides, from the condition $$\left.\dfrac{du}{dr}\right|_{r=a^-}-\left.\dfrac{du}{dr}\right|_{r=a^+}=\dfrac{2m\alpha}{\hbar^2}u(a)$$ we obtain $$\begin{align}Ck\cos(ka+\delta_0)-Ak\cos(ka)&=\dfrac{2m\alpha}{\hbar^2}A\sin(ka)\\\implies C\cos(ka+\delta_0)&=\left[\dfrac{2m\alpha}{\hbar^2k}\sin(ka)+\cos(ka)\right]A\tag{4}\end{align}$$

4. Eq. (4), divided by Eq. (3), yields $$\begin{align}&\hspace{3em}\dfrac{\cos(ka+\delta_0)}{\sin(ka+\delta_0)}=\dfrac{2m\alpha}{\hbar^2k}+\dfrac{\cos(ka)}{\sin(ka)}\\&\implies\dfrac{\cos(ka)\cos(\delta_0)-\sin(ka)\sin(\delta_0)}{\sin(ka)\cos(\delta_0)+\cos(ka)\sin(\delta_0)}=\dfrac{2m\alpha}{\hbar^2k}+\dfrac{\cos(ka)}{\sin(ka)}\\&\implies \cancel{\cos(ka)\cos(\delta_0)}-\sin(ka)\sin(\delta_0)\\&\hspace{4em}=\left[\sin(ka)\cos(\delta_0)+\cos(ka)\sin(\delta_0)\right]\left[\dfrac{2m\alpha}{\hbar^2k}+\dfrac{\cos(ka)}{\sin(ka)}\right]\\&\hspace{4em}=\left[\sin(ka)\cos(\delta_0)+\cos(ka)\sin(\delta_0)\right]\dfrac{2m\alpha}{\hbar^2k}\\&\hspace{6em}+\cancel{\cos(ka)\cos(\delta_0)}+\cos(ka)\sin(\delta_0)\dfrac{\cos(ka)}{\sin(ka)}\\&\implies -\sin(ka)\cos(\delta_0)\dfrac{2m\alpha}{\hbar^2k}\\&\hspace{4em}=\sin(ka)\sin(\delta_0)+\cos(ka)\sin(\delta_0)\left[\dfrac{2m\alpha}{\hbar^2k}+\dfrac{\cos(ka)}{\sin(ka)}\right]\\&\implies \cos(\delta_0)\dfrac{2m\alpha}{\hbar^2k}\\&\hspace{4em}=-\sin(\delta_0)-\cot(ka)\sin(\delta_0)\left[\dfrac{2m\alpha}{\hbar^2k}+\cot(ka)\right]\\&\implies \cot(\delta_0)=\dfrac{\hbar^2k}{2m\alpha}\left\{-1-\cot(ka)\left[\dfrac{2m\alpha}{\hbar^2k}+\cot(ka)\right]\right\}\\&\hspace{5.7em}=-\dfrac{\hbar^2k}{2m\alpha}\left[1+\cot^2(ka)\right]-\cot(ka)\\&\hspace{5.7em}=-\dfrac{\hbar^2k}{2m\alpha}\dfrac{1}{\sin^2(ka)}-\cot(ka)\\&\implies \delta_0=\cot^{-1}\left[-\dfrac{\hbar^2k}{2m\alpha}\dfrac{1}{\sin^2(ka)}-\cot(ka)\right]\\&\hspace{3.6em}=-\cot^{-1}\left[\cot(ka)+\dfrac{\hbar^2k}{2m\alpha}\dfrac{1}{\sin^2(ka)}\right].\tag{5}\end{align}$$
5. If we use $\beta\equiv\dfrac{2m\alpha a}{\hbar^2}$, then Eq. (5) becomes $$\delta_0=\boxed{-\cot^{-1}\left[\cot(ka)+\dfrac{ka}{\beta}\dfrac{1}{\sin^2(ka)}\right]}.\tag*{$\blacksquare$}$$
## Problem 10.8
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Check that Equation 10.65 $$G(\mathbf{r})=-\dfrac{e^{ikr}}{4\pi r}$$ satisfies Equation 10.52 $$\left(\nabla^2+k^2\right)G(\mathbf{r})=\delta^3(\mathbf{r}),$$ by direct substitution. *Hint:* $\nabla^2(1/r)=-4\pi\delta ^3(\mathbf{r})$.
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### Solution
$\boldsymbol\nabla G(\mathbf{r})=\boldsymbol\nabla\left(-\dfrac{e^{ikr}}{4\pi r}\right)=-\dfrac{1}{4\pi}\left[\dfrac{ik}{r}\boldsymbol\nabla(r) +\boldsymbol\nabla\left(\dfrac{1}{r}\right)\right]e^{ikr}\tag{1}$

$\begin{align}\nabla^2G(\mathbf{r})&=\boldsymbol\nabla\boldsymbol\cdot\boldsymbol\nabla G(\mathbf{r})=-\dfrac{1}{4\pi}\boldsymbol\nabla\boldsymbol\cdot\left\{\left[\small\dfrac{ik}{r}\boldsymbol\nabla(r)+\boldsymbol\nabla\left(\dfrac{1}{r}\right)\right]e^{ikr}\right\}\\&=-\dfrac{1}{4\pi}\left\{\left[\small ik\boldsymbol\nabla\left(\dfrac{1}{r}\right)\boldsymbol\cdot\boldsymbol\nabla(r)+\dfrac{ik}{r}\nabla^2(r)+\nabla^2\left(\dfrac{1}{r}\right)\right]e^{ikr}\right.\\&\hspace{3em}+\left.\left[\small\dfrac{ik}{r}\boldsymbol\nabla(r)+\boldsymbol\nabla\left(\dfrac{1}{r}\right)\right]\boldsymbol\cdot ik e^{ikr}\boldsymbol\nabla (r)\right\}\\&=-\dfrac{1}{4\pi}\left[\small2ik\boldsymbol\nabla\left(\dfrac{1}{r}\right)\boldsymbol\cdot\boldsymbol\nabla(r)+\dfrac{ik}{r}\nabla^2(r)+\nabla^2\left(\dfrac{1}{r}\right)+\dfrac{(ik)^2}{r}\left(\boldsymbol\nabla r\right)^2\right]e^{ikr}\tag{2}\end{align}$

Using $$\boldsymbol\nabla r=\dfrac{\mathbf{r}}{r},\quad\boldsymbol\nabla\left(\dfrac{1}{r}\right)=-\dfrac{\mathbf{r}}{r^3},\quad\nabla^2 r=\dfrac{2}{r},\quad\nabla^2\left(\dfrac{1}{r}\right)=-4\pi\delta^3(\mathbf{r})$$ in Eq. (2), we obtain $$\begin{align}\nabla^2 G(\mathbf{r})&=-\dfrac{1}{4\pi}\left[2ik\left(-\dfrac{\mathbf{r}}{r^3}\right)\boldsymbol\cdot\dfrac{\mathbf{r}}{r}+\dfrac{ik}{r}\dfrac{2}{r}-4\pi\delta^3(\mathbf{r})-\dfrac{k^2}{r}\dfrac{\mathbf{r}}{r}\boldsymbol\cdot\dfrac{\mathbf{r}}{r}\right]e^{ikr}\\&=-\dfrac{1}{4\pi}\left[-\cancel{\dfrac{2ik}{r^2}}+\cancel{\dfrac{2ik}{r^2}}-4\pi\delta^3(\mathbf{r})-\dfrac{k^2}{r}\right]e^{ikr}\\&=\underbrace{-\dfrac{e^{ikr}}{4\pi r}}_{G(\mathbf{r})}\left[-4\pi r\delta^3(\mathbf{r})-k^2\right]\\\implies&\nabla^2G(\mathbf{r})+k^2G(\mathbf{r})=e^{ikr}\delta^3(\mathbf{r}).\tag{3}\end{align}$$ Because for any integrable function $f(r)$ and positive number $R$, $$4\pi\int^R_0f(r)e^{ikr}\delta^3(\mathbf{r})dr=4\pi\int^R_0f(r)\delta^3(\mathbf{r})dr=f(\mathbf{0})$$ must holds, we use $e^{ikr}\delta^3(\mathbf{r})=\delta^3(\mathbf{r})$ in Eq. (3) to reach $$\boxed{\left(\nabla^2+k^2\right)G(\mathbf{r})=\delta^3(\mathbf{r})}.\tag*{$\blacksquare$}$$

## Problem 10.9
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Show that the ground state of hydrogen (Equation 4.80) satisfies the integral form of the Schrödinger equation, for the appropriate $V$ and $E$ (note that $E$ is *negative*, so $k=i\kappa$, where $\kappa\equiv\sqrt{-2mE}/\hbar$).
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### Solution
The ground state wave funciotn of hydrogen is (Equation 4.80) $$\psi(\mathbf{r})=\dfrac{1}{\sqrt{\pi a^3}}e^{-r/a},\tag{1}$$ where $$a=\dfrac{4\pi\epsilon_0\hbar^2}{me^2}\tag{2}$$ is the Bohr radius.

The potential energy of hydrogen is $$V(\mathbf{r})=-\dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r}=-\dfrac{\hbar^2}{ma}\dfrac{1}{r}\tag{3}$$

The integral form of the Schrödinger equation (Equation 10.67) is $$\psi(\mathbf{r})=\psi(\mathbf{r}_0)-\dfrac{m}{2\pi\hbar^2}\int\dfrac{e^{ik|\mathbf{r}-\mathbf{r}_0|}}{|\mathbf{r}-\mathbf{r}_0|}V(\mathbf{r}_0)\psi(\mathbf{r}_0)d^3r_0.\tag{4}$$ The ground state energy of hydrogen is (from Equation 4.77) $$E=-\dfrac{m}{2\hbar^2}\left(\dfrac{e^2}{4\pi\epsilon_0}\right)^2=-\dfrac{m}{2\hbar^2}\left(\dfrac{\hbar^2}{ma}\right)^2=-\dfrac{\hbar^2}{2ma^2},$$ and thus $$k=\dfrac{\sqrt{\small 2mE}}{\hbar}=i\dfrac{\sqrt{\small -2mE}}{\hbar}=i\dfrac{\sqrt{\small \hbar^2/a^2}}{\hbar}=\frac{i}{a}\tag{5}$$

Without an incident wave function , $\psi(\mathbf{r}_0)=0$. After substituting Eqs. (1), (3) and (5) into the right-hand side of Eq. (4), there follows $$\begin{align}&\quad 0-\dfrac{m}{2\pi\hbar^2}\int\dfrac{e^{-|\mathbf{r}-\mathbf{r}_0|/a}}{|\mathbf{r}-\mathbf{r}_0|}V(\mathbf{r}_0)\psi(\mathbf{r}_0)d^3r_0\\&=-\dfrac{m}{2\pi\hbar^2}\int\dfrac{e^{-|\mathbf{r}-\mathbf{r}_0|/a}}{|\mathbf{r}-\mathbf{r}_0|}\left(-\dfrac{\hbar^2}{ma}\dfrac{1}{r_0}\right)\left(\dfrac{e^{-r_0/a}}{\sqrt{\pi a^3}}\right)d^3r_0\\&=\dfrac{\cancel{m}}{2\pi\cancel{\hbar^2}}\dfrac{\cancel{\hbar^2}}{\cancel{m}a}\dfrac{1}{\sqrt{\pi a^3}}\int\dfrac{e^{-|\mathbf{r}-\mathbf{r}_0|/a}}{|\mathbf{r}-\mathbf{r}_0|}\dfrac{e^{-r_0/a}}{r_0}d^3r_0\\&=\dfrac{1}{2\pi a\sqrt{\pi a^3}}\int\dfrac{e^{-|\mathbf{r}-\mathbf{r}_0|/a}}{|\mathbf{r}-\mathbf{r}_0|}\dfrac{e^{-r_0/a}}{r_0}d^3r_0\\&=\dfrac{1}{2\pi a\sqrt{\pi a^3}}\iiint\frac{\exp\left(-\frac{1}{a}\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}\right)}{\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}}\dfrac{e^{-r_0/a}}{r_0}r_0^2\sin\theta dr_0d\theta d\phi\\&=\dfrac{1}{2\pi a\sqrt{\pi a^3}}\left(\small\int_0^{2\pi} d\phi\right)\iint\frac{\exp\left(-\frac{1}{a}\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}\right)}{\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}}r_0e^{-r_0/a}\sin\theta dr_0d\theta \\&=\dfrac{1}{\cancel{2\pi} a\sqrt{\pi a^3}}\cancel{2\pi}\int_0^{+\infty} \left[\small r_0e^{-r_0/a}\int_0^\pi\tfrac{\exp\left(-\frac{1}{a}\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}\right)}{\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}}\sin\theta d\theta\right] dr_0 \\&=\dfrac{1}{a\sqrt{\pi a^3}}\int_0^{+\infty} \left[ r_0e^{-r_0/a}I(r_0)\right] dr_0,\tag{6}\end{align}$$ where the radial part integral $$\begin{align}I(r_0)&\equiv \int_0^\pi\frac{\exp\left(-\frac{1}{a}\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}\right)}{\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}}\sin\theta d\theta\\&=\int^{r+r_0}_{|r-r_0|} \dfrac{e^{-u/a}}{rr_0}du=\dfrac{1}{rr_0}\left[-ae^{-u/a}\right]^{r+r_0}_{u=|r-r_0|}\\&=\dfrac{-a}{rr_0}\left(e^{-(r+r_0)/a}-e^{-|r-r_0|/a}\right)\tag{7}\end{align}$$ can be carried out by transformation $u=\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}$ and $du=\frac{rr_0\sin\theta d\theta}{\sqrt{\small r^2+r_0^2-2rr_0\cos\theta}}$.

And then we use Eq. (7) to continue the evaluation of Eq. (6): $$\begin{align}&\quad\dfrac{1}{\bcancel{a}\sqrt{\pi a^3}}\int_0^{+\infty} \left[ \cancel{r_0}e^{-r_0/a}\dfrac{-\bcancel{a}}{r\cancel{r_0}}\left(e^{-(r+r_0)/a}-e^{-|r-r_0|/a}\right)\right] dr_0\\&=-\dfrac{1}{r}\dfrac{1}{\sqrt{\pi a^3}}\left[\small\int_0^{+\infty} e^{-r_0/a}e^{-(r+r_0)/a}dr_0-\int_0^r e^{-r_0/a}e^{-(r-r_0)/a} dr_0-\int_r^{+\infty}e^{-r_0/a} e^{(r-r_0)/a} dr_0\right]\\&=-\dfrac{1}{r}\dfrac{1}{\sqrt{\pi a^3}}\left[\small e^{-r/a} \int_0^{+\infty}e^{-2r_0/a}dr_0-e^{-r/a}\int_0^r dr_0-e^{r/a} \int_r^{+\infty}e^{-2r_0/a} dr_0\right]\\&=-\dfrac{1}{r}\dfrac{1}{\sqrt{\pi a^3}}\left[\small \bcancel{\frac{a}{2}e^{-r/a}}-re^{-r/a}-\bcancel{\frac{a}{2}e^{r/a} e^{-2r/a}}\right]\\&=-\dfrac{1}{\cancel{r}}\dfrac{1}{\sqrt{\pi a^3}}(-\cancel{r}e^{-r/a})=\dfrac{1}{\sqrt{\pi a^3}}e^{-r/a}=\psi(\mathbf{r}),\tag{8}\end{align}$$ which ends up match the left-hand side of Eq. (6). Therefore, Eq. (1) satisfies Eq. (4). $\blacksquare$

## Problem 10.12
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Calculate the total cross-section for scattering from a Yukawa potential, in the Born approximation. Express your answer as a function of $E$.
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### Solution
The **scattering amplitude** of Yukawa potential is already given by Equation 10.91 in Example 10.5: $$f(\theta)=-\dfrac{2m\beta}{\hbar^2(\mu^2+\kappa^2)},\tag{1}$$ where $\mu$ is a constant (appearing in Equation 10.90) and $\kappa=2k\sin(\theta/2)$ is anglular dependent. The differential cross-section is the square of modulus of scattering amplitude (Equation 10.14): $$D(\theta)=|f(\theta)|^2=\left(\dfrac{2m\beta}{\hbar^2}\right)^2\dfrac{1}{(\mu^2+\kappa^2)^2},\tag{2}$$ and the total cross-section is the integral of $D(\theta)$ over all solid angles (Equation 10.7): $$\begin{align}\sigma&=\int D(\theta)d\Omega=\left(\dfrac{2m\beta}{\hbar^2}\right)^2\iint \dfrac{1}{(\mu^2+\kappa^2)^2}\sin\theta d\theta d\phi\\&=\left(\dfrac{2m\beta}{\hbar^2}\right)^2\underbrace{\left(\small\int_0^{2\pi} d\phi\right)}_{=\,2\pi}\underbrace{\small\int_0^\pi \dfrac{\sin\theta d\theta}{\left[\mu^2+4k^2\sin(\theta/2)\right]^2} }_{\equiv\,I}\tag{3}\end{align}$$

The integral over polar angle is $$\begin{align}I&=\int_0^\pi \dfrac{\sin\theta d\theta}{\left[\mu^2+4k^2\sin(\theta/2)\right]^2}=\int_0^\pi \dfrac{2\sin(\theta/2)\cos(\theta/2)}{\left[\mu^2+4k^2\sin(\theta/2)\right]^2}d\theta\\&=\dfrac{1}{2k^2}\int_0^{4k^2}\dfrac{du}{\left(\mu^2+u\right)^2}=\dfrac{1}{2k^2}\left.\left(-\dfrac{1}{\mu^2+u}\right)\right|_0^{4k^2}\\&=\dfrac{-1}{2k^2}\left(\dfrac{1}{\mu^2+4k^2}-\dfrac{1}{\mu^2}\right)=\dfrac{-1}{2\cancel{k^2}}\dfrac{-4\cancel{k^2}}{\mu^2\left(\mu^2+4k^2\right)}=\dfrac{2}{\mu^2\left(\mu^2+4k^2\right)},\tag{4}\end{align}$$ which is done by substitution $u=4k^2\sin(\theta/2)$ and $du=4k^2\sin(\theta/2)\cos(\theta/2)d\theta$.

Hence, Eq. (3) becomes $$\sigma=2\pi\left(\dfrac{2m\beta}{\hbar^2}\right)^2\dfrac{2}{\mu^2\left(\mu^2+4k^2\right)}=\left(\dfrac{4m\beta}{\mu\hbar^2}\right)^2\dfrac{\pi}{\mu^2+4k^2};$$ with $k^2=2mE/\hbar^2$, we can write $$\sigma=\left(\dfrac{4m\beta}{\mu\hbar^2}\right)^2\dfrac{\pi}{\mu^2+8mE/\hbar^2}=\boxed{\left(\dfrac{4m\beta}{\mu\hbar}\right)^2\dfrac{\pi}{\hbar^2\mu^2+8mE}}.\tag*{$\blacksquare$}$$
## Problem 10.19
::: warning
Prove the **optical theorem**, which relates the cross-sectionto the imaginary part of the forward scattering amplitude: $$\sigma=\dfrac{4\pi}{k}\text{Im}[ f(0)]\tag{10.104}$$
*Hint:* Use Equations 10.47 and 10.48.
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### Solution
Equation 10.47 gives the forward scattering amplitude: $$f(\theta)=\dfrac{1}{k}\sum_{\ell=0}^{+\infty}(2\ell+1)e^{i\delta_\ell}\sin(\delta_\ell)P_\ell(\cos\theta),\tag{1}$$ where $P_\ell$ is the $\ell$-th Legendre polynomial. Putting $\theta=0$ into Eq. (1), we have $$f(0)=\dfrac{1}{k}\sum_{\ell=0}^{+\infty}(2\ell+1)e^{i\delta_\ell}\sin(\delta_\ell)\cancelto{1}{P_\ell(1)},$$ whose imaginary part is $$\text{Im}[f(0)]=\dfrac{1}{k}\sum_{\ell=0}^{+\infty}(2\ell+1)\sin^2(\delta_\ell).\tag{2}$$

From Equation 10.48, the total cross-section is $$\sigma=\dfrac{4\pi}{k^2}\sum_{\ell=0}^{+\infty}(2\ell+1)\sin^2(\delta_\ell).\tag{3}$$

Comparing Eqs. (2) and (3), we conclude that $$\boxed{\sigma=\dfrac{4\pi}{k}\text{Im}[ f(0)]}.\tag*{$\blacksquare$}$$

## Problem 10.23