Chapter 7, Problem 7.56P

(a)

To determine

The first-order correction to the ground state, ${\psi }_0\left(r_1,r_2\right)={\psi }_{100}\left(r_1\right){\psi }_{100}\left(r_2\right)$.

Answer to Problem 7.56P

The first-order correction to the ground state, ${\psi }_0\left(r_1,r_2\right)={\psi }_{100}\left(r_1\right){\psi }_{100}\left(r_2\right)$ is $34\text{ eV}$.

Explanation of Solution

Write the expression for the first-order correction to the ground state

    $E_0^1=\langle {\psi }_0|\frac{e^2}{4\pi {\epsilon }_{0}r}|{\psi }_0\rangle$

Solving the above equation by multiply and divide by $a$

$E_0^1=\frac{e^2}{4\pi {\epsilon }_{0}a}\langle {\psi }_0|\frac{a}{r}|{\psi }_0\rangle$

Where, $r=\left|r_1-r_2\right|$. From the result of Problem 5.15, the matrix element $\langle \frac{1}{\left|r_1-r_2\right|}\rangle =\frac{5}{4a}$.

The above equation becomes,

$\begin{array}{c}{E}_0^1=\frac{e^2}{4\pi {\epsilon }_{0}a}\frac{5}{4}\\ =\frac{5}{2}\frac{e^2}{8\pi {\epsilon }_{0}a}\\ =\frac{5}{2}\left|E_1\right|\end{array}$

Substitute $13.6\text{ eV}$ for $\left|E_1\right|$ in the above equation

$\begin{array}{c}{E}_0^1=\frac{5}{2}\left(13.6\text{ eV}\right)\\ =34\text{ eV}\end{array}$

Conclusion:

Thus, the first-order correction to the ground state, ${\psi }_0\left(r_1,r_2\right)={\psi }_{100}\left(r_1\right){\psi }_{100}\left(r_2\right)$ is $34\text{ eV}$.

(b)

To determine

Show that $E_{\pm }^1=\frac{1}{2}\left(K\pm J\right)$, where $K\equiv 2{\displaystyle \int {\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)H\text{'}{\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)d^3{r}_1{d}^3{r}_2}$ and $J\equiv 2{\displaystyle \int {\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)H\text{'}{\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_2\right)d^3{r}_1{d}^3{r}_2}$. And evaluate the value of the two integrals and compare the results with Figure 5.2.

Answer to Problem 7.56P

This has been proved that $E_{\pm }^1=\frac{1}{2}\left(K\pm J\right)$, where $\begin{array}{l}K\equiv 2{\displaystyle \int {\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)H\text{'}{\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)d^3{r}_1{d}^3{r}_2},\text{and}\\ J\equiv 2{\displaystyle \int {\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)H\text{'}{\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_2\right)d^3{r}_1{d}^3{r}_2}\end{array}$.

The value of the two integrals, $K=\frac{136}{81}\left|E_1\right|$ and $J=\frac{128}{729}\left|E_1\right|$. The calculated energies are in agreement with Figure 5.2.

Explanation of Solution

The degenerate first-excited states are

${\psi }_{\pm }\left(r_1,r_2\right)=\frac{1}{\sqrt{2}}\left[{\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)\pm {\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_2\right)\right]$

The Hamiltonian is invariant when the $r_1$ and $r_2$ interchanges, and these two states are even and odd under that interchange so they won’t be mixed by the perturbation. Therefore, degenerate perturbation theory can be used.

$\begin{array}{c}{E}_{\pm }^1=\langle {\psi }_{\pm }|\frac{e^2}{4\pi {\epsilon }_{0}r}|{\psi }_{\pm }\rangle \\ =\frac{1}{2}{\displaystyle \iint [{\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)\frac{e^2}{4\pi {\epsilon }_{0}r}{\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)}\\ \pm {\psi }_{100}^{*}\left(r_1\right){\psi }_{200}^{*}\left(r_2\right)\frac{e^2}{4\pi {\epsilon }_{0}r}{\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_2\right)\\ \pm {\psi }_{200}^{*}\left(r_1\right){\psi }_{100}^{*}\left(r_2\right)\frac{e^2}{4\pi {\epsilon }_{0}r}{\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)\\ \pm {\psi }_{200}^{*}\left(r_1\right){\psi }_{100}^{*}\left(r_2\right)\frac{e^2}{4\pi {\epsilon }_{0}r}{\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_2\right)]d^3{r}_1{d}^3{r}_2\end{array}$

Dropping the complex conjugation as the orbitals are real. And interchange $r_1$ and $r_2$ in the third and fourth terms. Therefore, the above equation becomes,

$E_{\pm }^1=\frac{1}{2}\left(K\pm J\right)$        (I)

Here,

$K=2{\displaystyle \int {\left|{\psi }_{200}\left(r_2\right)\right|}^2}{\displaystyle \int {\left|{\psi }_{100}\left(r_1\right)\right|}^2\frac{e^2}{4\pi {\epsilon }_{0}r}d{r}_{1}d{r}_2}$        (II)

Where,

$\begin{array}{c}{V}_1={\displaystyle \int {\left|{\psi }_{100}\left(r_1\right)\right|}^2\frac{e^2}{4\pi {\epsilon }_{0}r}d{r}_{1}d{r}_2}\\ =\frac{e^2}{4\pi {\epsilon }_{0}a}{\displaystyle \int \frac{8}{\pi a^3}{e}^{-4r_1/a}\frac{a}{\left|r_1-r_2\right|}d{r}_1}\\ =2\left|E_1\right|\frac{16}{a^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-4r_1/a}}{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{a\sin {\theta }_1}{\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos {\theta }_1}}d{\theta }_1{r}_1^{2}d{r}_1}\end{array}$        (III)

Where,

$Z_1={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{a\sin {\theta }_1}{\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos {\theta }_1}}d{\theta }_1}$

Let $u=\cos \theta$

$\begin{array}{c}{Z}_1={\displaystyle \underset{-1}{\overset{1}{\int }}\frac{a}{\sqrt{r_1^2+r_2^2-2r_1{r}_{2}u}}du}\\ =-{\frac{a}{r_1{r}_2}\sqrt{r_1^2+r_2^2-2r_1,r_{2}u}|}_{-1}^1\\ =\frac{a}{r_1{r}_2}\left(r_1+r_2-\left|r_1-r_2\right|\right)\\ =\{\begin{array}{cc}2a/r_1& r_1>r_2\\ 2a/r_2& r_1<r_2\end{array}\end{array}$

Substituting the above equation in equation (III)

$\begin{array}{c}{V}_1=2\left|E_1\right|\frac{32}{a^3}\left[{\displaystyle \underset{0}{\overset{r_2}{\int }}{e}^{-4r_1/a}\frac{a}{r_2}{r}_1^{2}d{r}_1}+{\displaystyle \underset{r_2}{\overset{\infty }{\int }}{e}^{-4r_1/a}a{r}_{1}d{r}_1}\right]\\ =2\left|E_1\right|\frac{32}{a^3}{\left(\frac{a}{4}\right)}^3\left[\frac{a}{r_2}{\displaystyle \underset{0}{\overset{4r_2/a}{\int }}{z}^2{e}^{-z}dz}+4{\displaystyle \underset{4r_2/a}{\overset{0}{\int }}z{e}^{-z}dz}\right]\\ =\left|E_1\right|\left\{\frac{a}{r_2}\left[2-\left(2+\frac{8r_2}{a}+{\left(\frac{4r_2}{a}\right)}^2\right)e^{-4r_2/a}\right]+4\left(1+\frac{4r_2}{a}\right)e^{-4r_2/a}\right\}\\ =2\left|E_1\right|\left[\frac{a}{r_2}-\left(2+\frac{a}{r_2}\right)e^{-4r_2/a}\right]\end{array}$

Substituting the above equation in equation (II),

$\begin{array}{c}K=8\pi {\displaystyle \underset{0}{\overset{\infty }{\int }}{\left|{\psi }_{200}\left(r_2\right)\right|}^2{V}_1\left(r_2\right)r_2^{2}d{r}_2}\\ =8\pi \left(2\left|E_1\right|\right){\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{\pi a^3}{\left(1-\frac{r_2}{a}\right)}^2{e}^{-2r_2/a}\left[\frac{a}{r_2}-\left(2+\frac{a}{r_2}\right)e^{-4r_2/a}\right]r_2^{2}d{r}_2}\\ =\frac{16}{a^3}\left|E_1\right|{\displaystyle \underset{0}{\overset{\infty }{\int }}a{r}_2{\left(1-\frac{r_2}{a}\right)}^2{e}^{-2r_2/a}d{r}_2}-{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(2r_2^2+a{r}_2\right){\left(1-\frac{r_2}{a}\right)}^2{e}^{-6r_2/a}d{r}_2}\end{array}$

Using variable substitute, where $z=2r_2/a$ in the first integral and $z=6r_2/a$ in the second integral. The above equation becomes,

$\begin{array}{c}K=\frac{16}{a^3}\left|E_1\right|{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{a^{2}z}{2}{\left(1-\frac{z}{2}\right)}^2{e}^{-z}\frac{a}{2}dz}-{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\frac{1}{18}{a}^2{z}^2+\frac{1}{6}{a}^{2}z\right){\left(1-\frac{z}{6}\right)}^2{e}^{-z}\frac{a}{6}dz}\\ =2\left|E_1\right|{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\frac{16}{9}z-2z^2+\frac{14}{27}{z}^3-\frac{1}{486}{z}^4\right)e^{-z}dz}\\ =2\left|E_1\right|\left(\frac{16}{9}-4+\frac{42}{9}-\frac{4}{81}\right)\\ =\frac{136}{81}\left|E_1\right|\end{array}$        (IV)

Similarly solve for $J$

$J=2{\displaystyle \int {\psi }_{200}\left(r_2\right){\psi }_{100}\left(r_2\right)}{\displaystyle \int {\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_1\right)\frac{e^2}{4\pi {\epsilon }_{0}r}{d}^3{r}_1{d}^3{r}_2}$        (V)

Where,

$\begin{array}{c}{V}_2={\displaystyle \int {\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_1\right)\frac{e^2}{4\pi {\epsilon }_{0}r}{d}^3{r}_1}\\ =\frac{e^2}{4\pi {\epsilon }_{0}a}\frac{2\sqrt{2}}{\pi a^3}{\displaystyle \int \left(1-\frac{r_1}{a}\right)e^{-3r_1/a}\frac{a}{\left|r_1-r_2\right|}d{r}_1}\\ =2\left|E_1\right|\frac{2\sqrt{2}}{\pi a^3}\left(2\pi \right){\displaystyle \underset{0}{\overset{\infty }{\int }}\left(1-\frac{r_1}{a}\right)e^{-3r_1/a}{Z}_1\left(r_1\right)r_1^{2}d{r}_1}\end{array}$

$Z_1$ is solved while computing for $K$, Substitute $z=3r_1/a$

$\begin{array}{c}{V}_2=2\left|E_1\right|\frac{8\sqrt{2}}{a^3}\left[{\displaystyle \underset{0}{\overset{r_2}{\int }}\frac{a{r}_1^2}{r_2}\left(1-\frac{r_1}{a}\right)e^{-3r_1/a}d{r}_1}+{\displaystyle \underset{r_2}{\overset{\infty }{\int }}a{r}_1\left(1-\frac{r_1}{a}\right)e^{-3r_1/a}d{r}_1}\right]\\ =2\left|E_1\right|\frac{8\sqrt{2}}{a^3}\left[{\displaystyle \underset{0}{\overset{3r_2/a}{\int }}\frac{a^3{z}^2}{9r_2}\left(1-\frac{z}{3}\right)e^{-z}\frac{a}{3}dz}+{\displaystyle \underset{3r_2/a}{\overset{\infty }{\int }}\frac{a^{3}z}{3}\left(1-\frac{z}{3}\right)e^{-z}\frac{a}{3}dz}\right]\\ =2\left|E_1\right|\frac{8\sqrt{2}}{9}\left[\frac{a}{3r_2}{\displaystyle \underset{0}{\overset{3r_2/a}{\int }}\left(z^2-\frac{z^3}{3}\right)e^{-z}\frac{a}{3}dz}+{\displaystyle \underset{3r_2/a}{\overset{\infty }{\int }}\left(z-\frac{z^2}{3}\right)e^{-z}\frac{a}{3}dz}\right]\\ =2\left|E_1\right|\frac{8\sqrt{2}}{9}\left[\frac{a}{3r_2}\left(2-\left(2+\frac{6r_2}{a}+\frac{9r_2^2}{a^2}\right)e^{-3r_2/a}\right)-\left[2-\left(2+\frac{6r_2}{a}+\frac{9r_2^2}{a^2}+\frac{9r_2^3}{a^3}\right)e^{-3r_2/a}\right]\right]\\ +\left(1+\frac{3r_1}{a}\right)e^{-3r_2/a}-\frac{1}{3}\left(2+\frac{6r_2}{a}+\frac{9r_2^2}{a^2}\right)e^{-3r_2/a}\end{array}$

Solving further,

$V_2=\left|E_1\right|\frac{16\sqrt{2}}{27}\left(1+\frac{3r_2}{a}\right)e^{-3r_2/a}$

Substitute the above equation in equation (IV)

$\begin{array}{c}J=8\pi {\displaystyle \int {\psi }_{200}\left(r_2\right){\psi }_{100}\left(r_2\right)}{V}_2\left(r_2\right)r_2^{2}d{r}_2\\ =8\pi \left|E_1\right|\frac{16\sqrt{2}}{27}\frac{2\sqrt{2}}{\pi a^3}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(1-\frac{r_2}{a}\right)e^{-3r_2/a}\left(1+\frac{3r_2}{a}\right)e^{-3r_2/a}{r}_2^{2}d{r}_2}\\ =\frac{512}{27}\left|E_1\right|\frac{1}{a^3}\left(\frac{a^3}{6}\right){\displaystyle \underset{0}{\overset{\infty }{\int }}\left(1-\frac{z}{6}\right)\left(1+\frac{z}{2}\right)e^{-z}{z}^{2}dz}\\ =\frac{64}{729}\left|E_1\right|{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(z^2+\frac{z^3}{3}-\frac{z^4}{12}\right)e^{-z}dz}\end{array}$

Solving further,

$\begin{array}{c}J=\frac{64}{729}\left|E_1\right|\left(2+2-2\right)\\ =\frac{128}{729}\left|E_1\right|\end{array}$        (VI)

Substitute equation (IV) and (VI) in (I),

The energies are

$\begin{array}{c}{E}^0+\frac{1}{2}\left(K+J\right)=4E_1+4E_2-\frac{68}{81}{E}_1-\frac{64}{729}{E}_1\\ =4E_1+4\frac{E_1}{4}-\frac{68}{81}{E}_1-\frac{64}{729}{E}_1\\ =\frac{2969}{729}{E}_1\\ =-55.39\text{ eV}\end{array}$

Similarly for the symmetric spatial state of parahelium

$\begin{array}{c}{E}^0+\frac{1}{2}\left(K-J\right)=4E_1+4E_2-\frac{68}{81}{E}_1+\frac{64}{729}{E}_1\\ =4E_1+4\frac{E_1}{4}-\frac{68}{81}{E}_1+\frac{64}{729}{E}_1\\ =\frac{3097}{729}{E}_1\\ =-57.78\text{ eV}\end{array}$

The calculated value is in agreement with the results from Figure 5.1 by predicting that orthohelium is lower in energy.

Conclusion:

It has been proved that $E_{\pm }^1=\frac{1}{2}\left(K\pm J\right)$, where $K\equiv 2{\displaystyle \int {\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)H\text{'}{\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)d^3{r}_1{d}^3{r}_2}$ and $J\equiv 2{\displaystyle \int {\psi }_{100}\left(r_1\right){\psi }_{200}\left(r_2\right)H\text{'}{\psi }_{200}\left(r_1\right){\psi }_{100}\left(r_2\right)d^3{r}_1{d}^3{r}_2}$. And the value of the two integrals, $K=\frac{136}{81}\left|E_1\right|$ and $J=\frac{128}{729}\left|E_1\right|$. The calculated energies are in agreement with Figure 5.2.