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# System prepended metadata

title: Chapter 8 -- Homework
tags: [Quantum Physics II]

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# Chapter 8 -- Homework
###### tags: `Quantum Physics II`
### Problem 8.1
#### (a)
The potential is $V(x)=\alpha |x|$.

The trial wave function is asked to be $$\psi(x)=Ae^{-bx^2},\quad b>0,$$ where $A$ is determined by normalization: $$\displaystyle1=|A|^2\int^\infty_{-\infty}e^{-2bx^2}dx=|A|^2\sqrt{\frac{\pi}{2b}}\implies A=\left(\frac{2b}{\pi}\right)^{1/4}.$$


The expectation value of kinetic energy is $$\require{cancel}\displaystyle\begin{align}\langle\hat{T}\rangle&=\int^\infty_{-\infty}\psi^*(x)\dfrac{-\hbar^2}{2m}\dfrac{d^2}{dx^2}\psi(x)dx\\&=\dfrac{-\hbar^2}{2m}|A|^2\int^\infty_{-\infty}e^{-bx^2}\dfrac{d^2}{dx^2}e^{-bx^2}dx\\&=\dfrac{\hbar^2b}{m}\sqrt{\dfrac{2b}{\pi}}\int^\infty_{-\infty}(-2bx+1)e^{-bx^2}dx\\&=\dfrac{\hbar^2b}{m}\sqrt{\dfrac{2b}{\pi}}\left[\small\int^\infty_{-\infty}e^{-bx^2}d(-bx^2)+\int^\infty_{-\infty}e^{-bx^2}dx\right]\\&=\dfrac{\hbar^2b}{m}\sqrt{\dfrac{2b}{\pi}}\Big[\cancelto{0}{\left. e^{-bx^2}\right|^\infty_{-\infty}}+\underbrace{\small\int^\infty_{-\infty}e^{-bx^2}dx}_{\sqrt{\pi/b}}\ \Big]\\&=\dfrac{\hbar^2b}{m}\sqrt{\dfrac{2b}{\pi}}\sqrt{\dfrac{\pi}{b}}=\dfrac{\hbar^2b}{2m}.\end{align}$$

The expectation value of potential energy is $$\displaystyle\begin{align}\langle\hat{V}\rangle&=\int^\infty_{-\infty}\psi^*(x)\alpha|x|\psi(x)dx=\alpha A^2\int^\infty_{-\infty}|x|e^{-2bx^2}dx\\&=2\alpha A^2\int^\infty_0 xe^{-2bx^2}dx=-\dfrac{\alpha A^2}{2b}\int^\infty_0e^{-2bx^2}d(-2bx^2)\\&=-\dfrac{\alpha A^2}{2b}\left[e^{-2bx^2}\right]^\infty_0=\dfrac{\alpha A^2}{2b}=\dfrac{\alpha}{2b}\sqrt{\dfrac{2b}{\pi}}=\dfrac{\alpha}{\sqrt{2b\pi}}.\end{align}$$

Hence, the expectation value of Hamiltonian is $$\langle\hat{H}\rangle=\langle\hat{T}\rangle+\langle\hat{V}\rangle=\dfrac{\hbar^2b}{2m}+\dfrac{\alpha}{\sqrt{2b\pi}}.$$

The variational principle tells us that $$\dfrac{\hbar^2b}{2m}+\dfrac{\alpha}{\sqrt{2b\pi}}\geq E_\text{gs},\qquad\forall\ b>0$$

To minimize $\langle\hat{H}\rangle$, we equate its first derivative (w.r.t $b$) with zero: $$\dfrac{d}{db}\langle\hat{H}\rangle=\dfrac{d}{db}\left(\dfrac{\hbar^2}{2m}b+\dfrac{\alpha}{\sqrt{2\pi}}b^{-1/2}\right)=\dfrac{\hbar^2}{2m}-\dfrac{\alpha}{2\sqrt{2\pi}}b^{-3/2}=0,$$ and there follows $$b=\left(\dfrac{\hbar^2}{2m}\dfrac{2\sqrt{2\pi}}{\alpha}\right)^{-2/3}=\left(\dfrac{m\alpha}{\sqrt{2\pi}\hbar^2}\right)^{2/3}.$$

Therefore, the lowest upper bound of ground state energy is $$\begin{align}\langle\hat{H}\rangle_{\min}&=\dfrac{\hbar^2}{2m}\left(\dfrac{m\alpha}{\sqrt{2\pi}\hbar^2}\right)^{2/3}+\dfrac{\alpha}{\sqrt{2\pi}}\left(\dfrac{m\alpha}{\sqrt{2\pi}\hbar^2}\right)^{-1/3}\\&=\dfrac{\hbar^2}{2m}\left(\dfrac{m^2\alpha^2}{2\pi\hbar^4}\right)^{1/3}+\dfrac{\alpha}{\sqrt{2\pi}}\left(\dfrac{\sqrt{2\pi}\hbar^2}{m\alpha}\right)^{1/3}\\&=\dfrac{1}{2}\left(\dfrac{\hbar^2\alpha^2}{2\pi m}\right)^{1/3}+\left(\dfrac{\hbar^2\alpha^2}{2\pi m}\right)^{1/3}=\dfrac{3}{2}\left(\dfrac{\hbar^2\alpha^2}{2\pi m}\right)^{1/3}.\tag*{$\blacksquare$}\end{align}$$


#### (b)
The potential is $V(x)=\alpha x^4$.

In this part, the same trial wave function will be used, so $\langle\hat{T}\rangle$ and $A$ are the same as those in part **(a)**.

The expectation value of potential energy is $$\displaystyle\begin{align}\langle\hat{V}\rangle&=\int^\infty_{-\infty}\psi^*(x)\alpha x^4\psi(x)dx=\alpha A^2\int^\infty_{-\infty}x^4e^{-2bx^2}dx\\&=\alpha\sqrt{\dfrac{2b}{\pi}}\left[\frac{3}{4}\sqrt{\pi}(2b)^{-5/2}\right]=\dfrac{3\alpha}{16b^2},\end{align}$$ in which we used the following techniques: $$\begin{align}\textstyle\int^\infty_{-\infty} e^{-px^2}dx&=\textstyle\sqrt{\frac{\pi}{p}},\\
\textstyle\frac{d}{dp}\int^\infty_{-\infty} e^{-px^2}dx&=\textstyle\int^\infty_{-\infty}(-x^2)e^{-px^2}dx=-\frac{1}{2}\sqrt{\pi}p^{-3/2},\\\textstyle\frac{d^2}{dp^2}\int^\infty_{-\infty} e^{-px^2}dx&=\textstyle\int^\infty_{-\infty} x^4e^{-px^2}dx=\frac{3}{4}\sqrt{\pi}p^{-5/2},\end{align}$$

Hence, the expectation value of Hamiltonian is $$\langle\hat{H}\rangle=\langle\hat{T}\rangle+\langle\hat{V}\rangle=\dfrac{\hbar^2b}{2m}+\dfrac{3\alpha}{16b^2}.$$

Minimizing $$\dfrac{d}{db}\langle\hat{H}\rangle=\dfrac{d}{db}\left(\dfrac{\hbar^2}{2m}b+\dfrac{3\alpha}{16b^2}\right)=\dfrac{\hbar^2}{2m}-\dfrac{3\alpha}{8b^3}=0$$  yields $b=\left(3m\alpha/4\hbar^2\right)^{1/3}$ and thus we have $$\begin{align}\langle\hat{H}\rangle_{\min}&=\dfrac{\hbar^2}{2m}\left(\dfrac{3m\alpha}{4\hbar^2}\right)^{1/3}+\dfrac{3\alpha}{16}\left(\dfrac{16\hbar^4}{9m^2\alpha^2}\right)^{1/3}\\&=\dfrac{1}{2}\left(\dfrac{3\hbar^4\alpha}{4m^2}\right)^{1/3}+\dfrac{1}{4}\left(\dfrac{3\hbar^4\alpha}{4m^2}\right)^{1/3}\\&=\dfrac{3}{4}\left(\dfrac{3\hbar^4\alpha}{4m^2}\right)^{1/3}.\tag*{$\blacksquare$}\end{align}$$

### Problem 8.8
Assume that the trial wave function is given by Eq.(8.28), $$\psi_1(\mathbf{r}_1,\mathbf{r}_2)\equiv\dfrac{Z^3}{\pi a^3}e^{-Z(r_1+r_2)/a},$$ where:
- $Z$ is effective nuclear charge and is *lower* than actual nuclear charge $Z_0$ due to shielding effect of electrons with each other;
- $a\equiv 4\pi \epsilon_0\hbar^2/m_e e^2$ is Bohr radius; and
- $\mathbf{r}_1$, $\mathbf{r}_2$ are respective positions of two electrons. 

Following Eq.(8.29),  the expectation value of $\hat{H}$ is $$\begin{align}\langle\hat{H}\rangle&=\left<-\tfrac{\hbar^2}{2m}\left(\nabla_1^2+\nabla_2^2\right)-\tfrac{e^2}{4\pi\epsilon_0}\left(\tfrac{Z_0}{r_1}+\tfrac{Z_0}{r_2}\right)\right>+\left<\tfrac{e^2}{4\pi\epsilon_0}\left(\tfrac{Z-Z_0}{r_1}+\tfrac{Z-Z_0}{r_2}+\tfrac{1}{\left|\mathbf{r}_1-\mathbf{r}_2\right|}\right)\right>\\&=2\left<-\frac{\hbar^2}{2m}\nabla^2-\frac{e^2}{4\pi\epsilon_0}\frac{Z}{r}\right>+2\frac{e^2}{4\pi\epsilon_0}(Z-Z_0)\left<\frac{1}{r}\right>+\frac{e^2}{4\pi\epsilon_0}\left<\frac{1}{\left|\mathbf{r}_1-\mathbf{r}_2\right|}\right>,\end{align}$$ in which:
- the 1st expectation value is just the ground state energy of single-electron atom, which, in terms of $a$, is $$E_1=-\dfrac{1}{2a}\dfrac{Ze^2}{4\pi\epsilon_0};$$
- the 2nd expectation value is given in (7.56) or (8.31), $$\left<\frac{1}{r}\right>=\dfrac{Z}{a};$$
- the 3rd expectation value is given in Eq. (8.32), $$\left<\frac{1}{\left|\mathbf{r}_1-\mathbf{r}_2\right|}\right>=-\dfrac{5Z}{4}E_1,$$ which is evaluated through Eqs. (8.21)-(8.26), with $Z_0=2$ replaced by $Z$.

Thus we have $$\langle\hat{H}\rangle=\left[2Z^2-4Z(Z-Z_0)-\tfrac{5}{4}Z\right]E_1=\left[-2Z^2+\left(4Z_0-\tfrac{5}{4}\right)Z\right]E_1,$$ similar to Eq. (8.33).

Minimizing $\langle\hat{H}\rangle$, $$\dfrac{d}{dZ}\langle\hat{H}\rangle=\left[-4Z+\left(4Z_0-\tfrac{5}{4}\right)\right]E_0=0,$$ we find the effective nuclear charge is $$Z=Z_0-\dfrac{5}{16}.\tag{1}$$

Finally, the best upper bound is $$\begin{align}\langle\hat{H}\rangle_{\min}&=\left[-2\left(Z_0-\tfrac{5}{16}\right)^2+\left(4Z_0-\tfrac{5}{4}\right)\left(Z_0-\tfrac{5}{16}\right)\right]E_1\\&=2\left(Z_0-\frac{5}{16}\right)^2E_1=2Z^2E_1.\tag{2}\end{align}$$

1. For hydrogen anion $\text{H}^-$, plugging in $Z_0=1$ to Eqs. (1) and (2), we have $$Z=1-\dfrac{5}{16}=\boxed{\dfrac{11}{16}},\\ \langle\hat{H}\rangle_{\min}=2\left(\frac{11}{16}\right)^2E_1=\frac{121}{128}\times(-13.6\text{ eV})=\boxed{-12.9\text{ eV}}>-13.6\text{ eV}.$$
2. For lithium cation $\text{Li}^+$, plugging in $Z_0=3$ to Eqs. (1) and (2), we have $$Z=3-\dfrac{5}{16}=\boxed{\dfrac{43}{16}}, \\\langle\hat{H}\rangle_{\min}=2\left(\frac{43}{16}\right)^2E_1=\frac{1849}{128}\times(-13.6\text{ eV})=\boxed{-196\text{ eV}}.$$

### Problem 8.9

The **direct** integral is defined by $$D\equiv a\left<\psi_0(r)\middle|\dfrac{1}{r'}\middle|\psi_0(r)\right>,$$ and the **exchange** integral is defined by $$X\equiv a\left<\psi_0(r)\middle|\dfrac{1}{r}\middle|\psi_0(r')\right>.$$

The trial wave function is (Eq. (8.37)) $$\psi_0(\mathbf{r})=\dfrac{1}{\sqrt{\pi a^3}}e^{-r/a}$$

First $$\begin{align}D&=a\left<\psi_0(r)\middle|\dfrac{1}{r'}\middle|\psi_0(r)\right>=a\left<\psi_0(r')\middle|\dfrac{1}{r}\middle|\psi_0(r')\right>\\&=a\int\left(\tfrac{1}{\sqrt{\pi a^3}}e^{-r'/a}\right)\frac{1}{r}\left(\tfrac{1}{\sqrt{\pi a^3}}e^{-r'/a}\right)d^3\mathbf{r}\\&=\dfrac{1}{\pi a^2}\iiint\dfrac{e^{-2r'/a}}{r}r^2\sin\theta drd\theta d\phi\end{align}$$

$r'=\sqrt{r^2+R^2-2rR\cos\theta}$

Then we only need to consider $\mathbf{r}$: $$\begin{align}D&=\dfrac{1}{\pi a^2}\iiint e^{-2\sqrt{r^2+R^2-2rR\cos\theta}/a}r\sin\theta drd\theta d\phi\\&=\dfrac{1}{\pi a^2}\left(\small\int_0^{2\pi}d\phi\right)\left[\small\int_0^\infty \left(\int_0^\pi e^{-2\sqrt{r^2+R^2-2rR\cos\theta}/a}\sin\theta d\theta\right)rdr\right]\end{align}$$

Let $y\equiv \sqrt{r^2+R^2-2rR\cos\theta}$, so that $d(y^2)=2ydy=2rR\sin\theta d\theta$, and then $$\require{cancel}\begin{align}D&=\dfrac{2\pi}{\pi a^2}\left[\int_0^\infty \left(\small\int_{|r-R|}^{r+R}\frac{ e^{-2y/a}}{\cancel{r}R}ydy\right)\cancel{r}dr\right]
\\&=\dfrac{2}{ a^2R}\left[\int_0^\infty\left(\small \left.-\frac{a}{2}e^{-2y/a}y\right|_{y=|r-R|}^{r+R}+\int_{|r-R|}^{r+R}\frac{a}{2}e^{-2y/a}dy\right)dr\right]
\\&=\dfrac{2}{ a^2R}\left\{\int_0^\infty\left.\small\left[-\frac{a}{2}\left(y+\frac{a}{2}\right)e^{-2y/a}\right]_{y=|r-R|}^{r+R}\right.dr\right\}
\\&=-\dfrac{1}{ aR}\left\{\int_0^\infty\left.\small\left[\left(r+R+\frac{a}{2}\right)e^{-2(r+R)/a}-\left(|r-R|+\frac{a}{2}\right)e^{-2|r-R|/a}\right]\right.dr\right\}
\\&=-\dfrac{1}{ aR}\left[\int_0^\infty\left(r+R+\frac{a}{2}\right)e^{-2(r+R)/a}dr-\int_R^\infty\left(r-R+\frac{a}{2}\right)e^{-2(r-R)/a}dr\right.\\&\hspace{4.75em}\left.-\int_0^R\left(-r+R+\frac{a}{2}\right)e^{2(r-R)/a}dr\right]
\\&=-\dfrac{e^{-2R/a}}{aR}\left\{\small\left[\int_0^\infty\left(r+R+\frac{a}{2}\right)e^{-2r/a}dr\right]-\left[e^{4R/a}\int_R^\infty\left(r-R+\frac{a}{2}\right)e^{-2r/a}dr\right.\right]
\\&\hspace{7.5em}\left.\small-\left[\int_0^R\left(-r+R+\frac{a}{2}\right)e^{2r/a}dr\right]\right\}
\\&=-\dfrac{e^{-2R/a}}{aR}\left\{\small\left[\frac{-a}{2}\left(R+\frac{a}{2}\right)e^{-2r/a}\right]_{r=0}^\infty+\left[\frac{-a}{2}\left(r+\frac{a}{2}\right)e^{-2r/a}\right]_{r=0}^\infty\right.
\\&\hspace{7.5em}\left.\small-\;e^{4R/a}\left[\frac{-a}{2}\left(-R+\frac{a}{2}\right)e^{-2r/a}\right]_{r=R}^\infty-e^{4R/a}\left[\frac{-a}{2}\left(r+\frac{a}{2}\right)e^{-2r/a}\right]_{r=R}^\infty\right.
\\&\hspace{7.5em}\left.\small-\left[\frac{a}{2}\left(R+\frac{a}{2}\right)e^{2r/a}\right]_{r=0}^R-\left[\frac{-a}{2}\left(r-\frac{a}{2}\right)e^{2r/a}\right]_{r=0}^R\right\}
\\&=-\dfrac{e^{-2R/a}}{aR}\left\{\left[\frac{a}{2}\left(R+\frac{a}{2}\right)\right]+\left[\frac{a}{2}\left(0+\frac{a}{2}\right)\right]\right.
\\&\hspace{7.5em}\left.-\;e^{4R/a}\left[\frac{a}{2}\left(\cancel{-R}+\frac{a}{2}\right)e^{-2R/a}\right]-e^{4R/a}\left[\frac{a}{2}\left(\cancel{R}+\frac{a}{2}\right)e^{-2R/a}\right]\right.
\\&\hspace{7.5em}\left.-\left[\frac{a}{2}\left(\cancel{R}+\frac{a}{2}\right)e^{2R/a}\right]+\left[\frac{a}{2}\left(R+\frac{a}{2}\right)\right]\right.
\\&\hspace{7.5em}\left.\small-\left[\frac{-a}{2}\left(\cancel{R}-\frac{a}{2}\right)e^{2R/a}\right]+\left[\frac{-a}{2}\left(0-\frac{a}{2}\right)\right]\right\}
\\&=-\dfrac{e^{-2R/a}}{aR}\left[\left(\dfrac{aR}{2}+\dfrac{a^2}{4}\right)+\dfrac{a^2}{4}-e^{2R/a}\dfrac{a^2}{4}-e^{2R/a}\dfrac{a^2}{4}\right.\\&\hspace{7.5em}-\left.e^{2R/a}\dfrac{a^2}{4}+\left(\dfrac{aR}{2}+\dfrac{a^2}{4}\right)-e^{2R/a}\dfrac{a^2}{4}+\dfrac{a^2}{4}\right]\\&=-\dfrac{e^{-2R/a}}{aR}\left(aR+a^2-a^2e^{2R/a}\right)\\&=\dfrac{a}{R}-\left(1+\dfrac{a}{R}\right)e^{-2R/a}.\tag*{$\blacksquare$}\end{align}$$

### Problem 8.10
$\psi=A[\psi_0(r)-\psi_0(r')]$

$\displaystyle 1=\int|\psi|^2d^3r=|A|^2\left[\cancelto{0}{\int\left|\psi_0(r)\right|^2d^3r}+\cancelto{0}{\int\left|\psi_0(r')\right|^2d^3r}-2\int\psi_0(r)\psi_0(r')d^3r\right]$

Define $I\equiv \left<\psi_0(r)\middle|\psi_0(r')\right>=\dfrac{1}{\pi a^3}\int e^{-(r+r')/a}d^3r$
$|A|^2=\dfrac{1}{2(1-I)}$

$\langle\hat{H}\rangle=\left(1+2\dfrac{D-X}{1-I}\right)E_1$

The potential energy associated with the proton-proton repulsion: $$V_{pp}=\dfrac{e^2}{4\pi\epsilon_0R}=-\dfrac{2a}{R}E_1.$$

The total energy of the system, in unit of $-E_1$ and as a funtion of $x\equiv R/a$ is $$\begin{align}F_-(x)&=\dfrac{E_{\text{tot}}}{-E_1}\\&=\dfrac{1}{-E_1}\left(V_{pp}+E_{\text{gs}}\right)\leq\dfrac{1}{-E_1}\left(V_{pp}+\langle\hat{H}\rangle_\min\right)=\dfrac{2a}{R}-1-2\dfrac{D-X}{1-I}\\&=\dfrac{2}{x}-1-2\dfrac{\tfrac{1}{x}-\left(1+\tfrac{1}{x}\right)e^{-2x}-(1+x)e^{-x}}{1-\left[1+x+(x^2/3)\right]e^{-x}}\\&=-1+\dfrac{2}{x}\dfrac{\left(1+\tfrac{1}{x}\right)e^{-2x}+\left(-1+\frac{2x^2}{3}\right)e^{-x}}{1-\left[1+x+(x^2/3)\right]e^{-x}}.\tag{$\blacksquare$}\end{align}$$

![](https://i.imgur.com/q1865oL.png)
- solid: $F_+(x)$
- dashed: $F_-(x)$
No minimum is found in $F_-(x)$.
