Chapter 8, Problem 8.25P

To determine

Show that by astute choice of the adjustable parameters $Z_1$ and $Z_2$ , the expectation value of the Hamiltonian $\langle H\rangle$ is less than $-13.6\text{ eV}$.

Answer to Problem 8.25P

It has been prove that by astute choice of the adjustable parameters $Z_1$ and $Z_2$ , the expectation value of the Hamiltonian $\langle H\rangle$ is less than $-13.6\text{ eV}$.

Explanation of Solution

The Normalization condition

$\begin{array}{c}1={\displaystyle \int {\left|\psi \right|}^2{d}^3{r}_1{d}^3{r}_2}\\ ={\left|A\right|}^2\left[{\displaystyle \int {\psi }_1^2{d}^3{r}_1}{\displaystyle \int {\psi }_2^2{d}^3{r}_2}+2{\displaystyle \int {\psi }_1{\psi }_2{d}^3{r}_1}{\displaystyle \int {\psi }_1{\psi }_2{d}^3{r}_2}+{\displaystyle \int {\psi }_2^2{d}^3{r}_1}{\displaystyle \int {\psi }_1^2{d}^3{r}_2}\right]\\ ={\left|A\right|}^2\left(1+2S^2+1\right)\\ =2{\left|A\right|}^2\left(1+2S^2\right)\end{array}$        (I)

Solve for $S$

$\begin{array}{c}S\equiv {\displaystyle \int {\psi }_1\left(r\right){\psi }_2\left(r\right)d^{3}r}\\ =\frac{\sqrt{{\left(Z_1{Z}_2\right)}^3}}{\pi a^3}{\displaystyle \int e^{-\left(Z_1+Z_2\right)r/a}\left(4\pi r^2\right)dr}\\ =\frac{4}{a^3}{\left(\frac{y}{2}\right)}^3\left[\frac{2a^3}{{\left(Z_1+Z_2\right)}^3}\right]\\ ={\left(\frac{y}{x}\right)}^3\end{array}$

Substitute the above equation in equation (I) to find the value of $A^2$

$A^2=\frac{1}{2\left[1+{\left(\frac{y}{x}\right)}^6\right]}$        (II)

Write the Hamiltonian of the given system

$H=-\frac{{\hslash }^2}{2m}\left({\nabla }_1^2+{\nabla }_2^2\right)-\frac{e^2}{4\pi {\epsilon }_0}\left(\frac{1}{r_1}+\frac{1}{r_2}\right)+\frac{e^2}{4\pi {\epsilon }_0}\frac{1}{\left|r_1-r_2\right|}$

Then,

$\begin{array}{l}H\psi =A\left\{\begin{array}{l}\left[-\frac{{\hslash }^2}{2m}\left({\nabla }_1^2+{\nabla }_2^2\right)-\frac{e^2}{4\pi {\epsilon }_0}\left(\frac{Z_1}{r_1}+\frac{Z_2}{r_2}\right)\right]{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)\\ +\left[-\frac{{\hslash }^2}{2m}\left({\nabla }_1^2+{\nabla }_2^2\right)-\frac{e^2}{4\pi {\epsilon }_0}\left(\frac{Z_1}{r_1}+\frac{Z_2}{r_2}\right)\right]{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\end{array}\right\}\\ +A\frac{e^2}{4\pi {\epsilon }_0}\left\{\begin{array}{l}\left(\frac{Z_1-1}{r_1}+\frac{Z_2-1}{r_2}\right){\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)\\ +\left(\frac{Z_2-1}{r_1}+\frac{Z_1-1}{r_2}\right){\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\end{array}\right\}+V_{ee}\psi \end{array}$        (III)

Where, $V_{ee}=\frac{e^2}{4\pi {\epsilon }_0}\frac{1}{\left|r_1-r_2\right|}$.

The first term inside the bracket in equation (III) is

$\begin{array}{l}=\left\{\left(Z_1^2+Z_2^2\right)E_1{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)+\left(Z_2^2+Z_1^2\right)E_1{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\right\}\\ =\left\{\left(Z_1^2+Z_2^2\right)E_1\psi \right\}\end{array}$

Thus, equation (III) becomes

$H\psi =\left\{\left(Z_1^2+Z_2^2\right)E_1\psi \right\}+A\frac{e^2}{4\pi {\epsilon }_0}\left\{\begin{array}{l}\left(\frac{Z_1-1}{r_1}+\frac{Z_2-1}{r_2}\right){\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)\\ +\left(\frac{Z_2-1}{r_1}+\frac{Z_1-1}{r_2}\right){\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\end{array}\right\}+V_{ee}\psi$

From the above equation, the expectation value of $\langle H\rangle$ is

$\begin{array}{l}\langle H\rangle =\left(Z_1^2+Z_2^2\right)E_1+\langle V_{ee}\rangle +A^2\frac{e^2}{4\pi {\epsilon }_0}\\ \times \left\{\langle {\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)+{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)|\left(\begin{array}{l}\left(\frac{Z_1-1}{r_1}+\frac{Z_2-1}{r_2}\right)|{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)\rangle \\ +\left(\frac{Z_2-1}{r_1}+\frac{Z_1-1}{r_2}\right)|{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\rangle \end{array}\right)\right\}\end{array}$        (IV)

Solving the terms inside the bracket separately,

$\begin{array}{c}\left\{\right\}=\left(Z_1-1\right)\langle {\psi }_1\left(r_1\right)|\frac{1}{r_1}|{\psi }_1\left(r_1\right)\rangle +\left(Z_2-1\right)\langle {\psi }_2\left(r_2\right)|\frac{1}{r_2}|{\psi }_2\left(r_2\right)\rangle \\ +\left(Z_2-1\right)\langle {\psi }_1\left(r_1\right)|\frac{1}{r_1}|{\psi }_2\left(r_1\right)\rangle \langle {\psi }_2\left(r_2\right)|{\psi }_1\left(r_2\right)\rangle \\ +\left(Z_1-1\right)\langle {\psi }_1\left(r_1\right)|{\psi }_2\left(r_1\right)\rangle \langle {\psi }_2\left(r_2\right)|\frac{1}{r_1}|{\psi }_1\left(r_2\right)\rangle +\left(Z_1-1\right)\langle {\psi }_2\left(r_1\right)|\frac{1}{r_1}|{\psi }_1\left(r_1\right)\rangle \langle {\psi }_1\left(r_2\right)|{\psi }_2\left(r_2\right)\rangle \\ +\left(Z_2-1\right)\langle {\psi }_2\left(r_1\right)|{\psi }_1\left(r_1\right)\rangle \langle {\psi }_1\left(r_2\right)|\frac{1}{r_2}|{\psi }_2\left(r_2\right)\rangle +\left(Z_2-1\right)\langle {\psi }_2\left(r_1\right)|\frac{1}{r_1}|{\psi }_2\left(r_1\right)\rangle \\ +\left(Z_1-1\right)\langle {\psi }_1\left(r_2\right)|\frac{1}{r_2}|{\psi }_1\left(r_2\right)\rangle \\ =2\left(Z_1-1\right){\langle \frac{1}{r}\rangle }_1+2\left(Z_2-1\right){\langle \frac{1}{r}\rangle }_2+2\left(Z_1-1\right)\langle {\psi }_1|{\psi }_2\rangle \langle {\psi }_1|\frac{1}{r_1}|{\psi }_2\rangle +2\left(Z_2-1\right)\langle {\psi }_1|{\psi }_2\rangle \langle {\psi }_1|\frac{1}{r_1}|{\psi }_2\rangle \end{array}$

But,

${\langle \frac{1}{r}\rangle }_1=\langle {\psi }_1\left(r_1\right)|\frac{1}{r_1}|{\psi }_1\left(r_1\right)\rangle =\frac{Z_1}{a}$

${\langle \frac{1}{r}\rangle }_2=\langle {\psi }_2\left(r_2\right)|\frac{1}{r_2}|{\psi }_2\left(r_2\right)\rangle =\frac{Z_2}{a}$

Therefore, equation (IV) becomes,

$\begin{array}{l}\langle H\rangle =\left(Z_1^2+Z_2^2\right)E_1+\langle V_{ee}\rangle \\ +A^2\frac{e^2}{4\pi {\epsilon }_0}\left\{\frac{1}{a}\left(Z_1-1\right)Z_1+\frac{1}{a}\left(Z_2-1\right)Z_2+\left(Z_1+Z_2-2\right)\langle {\psi }_1|{\psi }_2\rangle \langle {\psi }_1|\frac{1}{r_1}|{\psi }_2\rangle \right\}\end{array}$

Solve the expectation value of $V_{ee}$

$\begin{array}{l}\langle V_{ee}\rangle =\frac{e^2}{4\pi {\epsilon }_0}\langle \psi |\frac{1}{\left|r_1-r_2\right|}|\psi \rangle \\ =\frac{e^2}{4\pi {\epsilon }_0}{A}^2\langle {\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)+{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)|\frac{1}{\left|r_1-r_2\right|}|{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)+{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\rangle \\ =\frac{e^2}{4\pi {\epsilon }_0}{A}^2\left[2\langle {\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)|\frac{1}{\left|r_1-r_2\right|}|{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)\rangle +2\langle {\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)|\frac{1}{\left|r_1-r_2\right|}|{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\rangle \right]\\ =\frac{2e^2}{4\pi {\epsilon }_0}{A}^2\left(B+C\right)\end{array}$

Where, $B=\langle {\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)|\frac{1}{\left|r_1-r_2\right|}|{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)\rangle$ and $C=\langle {\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)|\frac{1}{\left|r_1-r_2\right|}|{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\rangle$.

Solving for $B$,

$\begin{array}{c}B=\frac{Z_1^3{Z}_2^3}{{\left(\pi a^3\right)}^2}{\displaystyle \int e^{-2Z_1{r}_1/a}{e}^{-2Z_2{r}_2/a}\frac{1}{\left|r_1-r_2\right|}{d}^3{r}_1{d}^3{r}_2}\\ =\frac{Z_1^3{Z}_2^3}{{\left(\pi a^3\right)}^2}\frac{\pi a^3}{Z_2^3}\left(4\pi \right){\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-2Z_1{r}_1/a}\frac{1}{r_1}\left[1-\left(1+\frac{Z_2{r}_1}{a}\right)e^{-2Z_2{r}_1/a}\right]}{r}_1^{2}dr\\ =\frac{4Z_1^3}{a^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left[r_1{e}^{-2Z_1{r}_1/a}-r_1{e}^{-2\left(Z_1+Z_2\right)r_1/a}-\frac{Z_2}{a}{r}_1^2{e}^{-2\left(Z_1+Z_2\right)r_1/a}\right]}d{r}_1\\ =\frac{4Z_1^3}{a^3}\left[{\left(\frac{a}{2Z_1}\right)}^2-{\left(\frac{a}{2\left(Z_1+Z_2\right)}\right)}^2-\frac{Z_2}{a}2{\left(\frac{a}{2\left(Z_1+Z_2\right)}\right)}^3\right]\end{array}$

Solving further,

$\begin{array}{c}B=\frac{Z_1^3}{a}\left[\frac{1}{Z_1^2}-\frac{1}{{\left(Z_1+Z_2\right)}^2}-\frac{Z_2}{{\left(Z_1+Z_2\right)}^3}\right]\\ =\frac{Z_1{Z}_2}{a\left(Z_1+Z_2\right)}\left[1+\frac{Z_1{Z}_2}{{\left(Z_1+Z_2\right)}^2}\right]\\ =\frac{y^2}{4ax}\left[1+\frac{y^2}{4x^2}\right]\end{array}$

Solving for $C$

$\begin{array}{c}C=\frac{Z_1^3{Z}_2^3}{{\left(\pi a^3\right)}^2}{\displaystyle \int e^{-Z_1{r}_1/a}{e}^{-Z_2{r}_2/a}{e}^{-Z_1{r}_1/a}{e}^{-Z_2{r}_2/a}\frac{1}{\left|r_1-r_2\right|}{d}^3{r}_1{d}^3{r}_2}\\ =\frac{Z_1^3{Z}_2^3}{{\left(\pi a^3\right)}^2}\frac{\pi a^3}{Z_2^3}\left(4\pi \right){\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-\left(Z_1+Z_2\right)\left(r_1+r_2\right)/a}\frac{1}{\left|r_1-r_2\right|}{d}^3{r}_1{d}^3{r}_2}\\ =\frac{{\left(Z_1{Z}_2\right)}^3}{{\left(\pi a^3\right)}^2}\frac{5{\pi }^2}{256}\frac{4^5{a}^5}{{\left(Z_1+Z_2\right)}^5}\\ =\frac{20}{a}\frac{{\left(Z_1{Z}_2\right)}^3}{{\left(Z_1+Z_2\right)}^5}\end{array}$

Solving further,

$C=\frac{5}{16a}\frac{y^6}{x^5}$

Therefore, $\langle V_{ee}\rangle$

$\begin{array}{c}\langle V_{ee}\rangle =\frac{2e^2}{4\pi {\epsilon }_0}{A}^2\left(B+C\right)\\ =\frac{2e^2}{4\pi {\epsilon }_0}{A}^2\left(\frac{y^2}{4ax}\left[1+\frac{y^2}{4x^2}\right]+\frac{5}{16a}\frac{y^6}{x^5}\right)\\ =2A^2\left(-2E_1\right)\frac{y^2}{4x}\left(1+\frac{y^2}{4x^2}+\frac{5y^4}{4x^4}\right)\end{array}$

The expectation value of the Hamiltonian becomes,

$\langle H\rangle =E_1\left\{\begin{array}{l}{x}^2-\frac{1}{2}{y}^2-\frac{2}{\left[1+{\left(y/x\right)}^6\right]}\left[x^2-\frac{1}{2}{y}^2-x+\frac{1}{2}\left(x-2\right)\frac{y^6}{x^5}\right]\\ -\frac{2}{\left[1+{\left(y/x\right)}^6\right]}\frac{y^2}{4x}\left(1+\frac{y^2}{4x^2}+\frac{5y^4}{4x^4}\right)\end{array}\right\}$

Solving the above equation,

$\langle H\rangle =\frac{E_1}{\left(x^6+y^6\right)}\left(-x^8+2x^7+\frac{1}{2}{x}^6{y}^2-\frac{1}{2}{x}^5{y}^2-\frac{1}{2}{x}^5{y}^2-\frac{1}{8}{x}^3{y}^4+\frac{11}{8}x{y}^6-\frac{1}{2}{y}^8\right)$

At $x=1.32245$, $y=1.08505$ corresponding to $Z_1=1.0392$, $Z_2=0.2832$.

The expectation value,

$\begin{array}{c}{\langle H\rangle }_{\min }=1.0266E_1\\ =-13.962\text{ eV}\end{array}$

Hence proved.

Conclusion:

It has been prove that by astute choice of the adjustable parameters $Z_1$ and $Z_2$ , the expectation value of the Hamiltonian $\langle H\rangle$ is less than $-13.6\text{ eV}$.