InProblem 7.7we found that the trial wave function with shielding (Equation 7.27), which worked well for helium, is inadequate to confirm the existence of a bound state for the negative hydrogen ion.

${\mathit{\psi }}_{\mathbf{1}}\mathbf{(}{\mathit{r}}_{\mathbf{1}}\mathbf{,}{\mathit{r}}_{\mathbf{2}}\mathbf{)}\mathbf{\equiv }\frac{{\mathbf{z}}^{\mathbf{3}}}{\mathbf{\pi }{\mathbf{a}}^{\mathbf{3}}}{\mathit{e}}^{\mathbf{-}\mathbf{z}\mathbf{(}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}\mathbf{)}\mathbf{/}\mathbf{a}}$ (7.27)

Chandrasekhar used a trial wave function of the form

$\mathit{\psi }\mathbf{(}{\mathit{r}}_{\mathbf{1}}\mathbf{,}{\mathit{r}}_{\mathbf{2}}\mathbf{)}\mathbf{\equiv }\mathit{A}\mathbf{[}{\mathbf{\psi }}_{\mathbf{1}}\left(r_1\right){\mathbf{\psi }}_{\mathbf{2}}\left(r_2\right)\mathbf{+}{\mathbf{\psi }}_{\mathbf{2}}\left(r_1\right){\mathbf{\psi }}_{\mathbf{1}}\left(r_2\right)\mathbf{]}$ (7.62).

Where

${\mathit{\psi }}_{\mathbf{1}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{\equiv }\sqrt{\frac{{\mathbf{z}}_{\mathbf{1}}^{\mathbf{3}}}{{\mathbf{\pi a}}^{\mathbf{3}}}}{\mathit{e}}^{\mathbf{-}{\mathbf{z}}_{\mathbf{1}}\mathbf{/}\mathbf{a}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{,}$${\mathit{\psi }}_{\mathbf{1}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{\equiv }\sqrt{\frac{{\mathbf{z}}_{\mathbf{2}}^{\mathbf{3}}}{\mathbf{\pi }{\mathbf{a}}^{\mathbf{3}}}}{\mathit{e}}^{\mathbf{-}{\mathbf{z}}_{\mathbf{2}}\mathbf{/}\mathbf{a}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{,}$(7.63)

In effect, he allowed two different shielding factors, suggesting that one electron is relatively close to the nucleus, and the other is farther out. (Because electrons are identical particles, the spatial wave function must be symmetries with respect to interchange. The spin state—which is irrelevant to the calculation—is evidently anti symmetric.) Show that by astute choice of the adjustable parameters ${\mathit{Z}}_{\mathbf{1}}$and ${\mathit{Z}}_{\mathbf{2}}$you can get$<H>$less than -13.6.

Answer: :$<H>\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{1}}}{{\mathbf{x}}^{\mathbf{6}}\mathbf{+}{\mathbf{y}}^{\mathbf{6}}}\left(-x^8+2x^7+\frac{1}{2}{x}^6{y}^2-\frac{1}{2}{x}^5{y}^2-\frac{1}{8}{x}^3{y}^4+\frac{11}{8}x{y}^6-\frac{1}{2}{y}^8\right)\mathbf{.}$

Where$\mathit{x}\mathbf{\equiv }{\mathit{z}}_{\mathbf{1}}\mathbf{+}{\mathit{z}}_{\mathbf{2}}$.$\mathit{y}\mathbf{\equiv }\mathbf{2}\sqrt{{\mathbf{z}}_{\mathbf{1}}{\mathbf{z}}_{\mathbf{2}}}$Chandrasekhar used${\mathit{Z}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{039}$

(Since this is larger than 1, the motivating interpretation as an effective nuclear charge cannot be sustained, but never mindit’s still an acceptable trial wave function) and${\mathit{Z}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{283}$.

Total wave function is:

$\mathit{\psi }\mathbf{(}{\overrightarrow{\mathbf{r}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{r}}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathit{A}\left[{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right){\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\right]\mathbf{,}$

Where, ${\mathit{\psi }}_{\mathbf{1}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{\equiv }\sqrt{\frac{{\mathbf{z}}_{\mathbf{1}}^{\mathbf{3}}}{{\mathbf{\pi a}}^{\mathbf{3}}}}{\mathit{e}}^{\mathbf{-}{\mathbf{z}}_{\mathbf{1}}\mathbf{/}\mathbf{a}}\mathbf{}\mathbf{}\mathbf{,}{\mathit{\psi }}_{\mathbf{2}}\mathbf{\left(}\mathit{r}\mathbf{\right)}\mathbf{\equiv }\sqrt{\frac{{\mathbf{z}}_{\mathbf{2}}^{\mathbf{3}}}{{\mathbf{\pi a}}^{\mathbf{3}}}}{\mathit{e}}^{\mathbf{-}{\mathbf{z}}_{\mathbf{2}}\mathbf{r}\mathbf{/}\mathbf{a}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}$

The total wave function when residual forces are neglected is simply the product of the one quasi-particle proton and the one-quasi-particle neutron wave function.

Here the astute choice of adjustable parameters

${\left|A\right|}^2(1+2S^2+1)=1$

where

$$\begin{gathered} S=\int {\psi }_1\left(r\right){\psi }_2\left(r\right)d^{3}r=\frac{\sqrt{{\left(Z_1{Z}_2\right)}^3}}{{\mathrm{\pi a}}^3}\int e^{-(z_1+z_2)r/a}4{\mathrm{\pi r}}^2\mathrm{dr} \\ =\frac{4}{{\mathrm{a}}^3}{\left(\frac{\mathrm{y}}{2}\right)}^3\left[\frac{2{\mathrm{a}}^3}{{({\mathrm{z}}_1+{\mathrm{z}}_2)}^3}\right]={\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^3. \\ {\mathrm{A}}^2=\frac{1}{2\left[1+{(\mathrm{y}/\mathrm{x})}^6\right]}. \end{gathered}$$

$$\begin{gathered} H=-\frac{{\sout{h}}^2}{2m}\left({\nabla }_1^2+{\nabla }_2^2\right)-\frac{e^2}{4\mathrm{\pi }{\dot{\mathrm{o}}}_0}\left(\frac{1}{r_1}+\frac{1}{r_2}\right)+\frac{e^2}{4\mathrm{\pi }{\dot{\mathrm{o}}}_0}\frac{1}{\left|r_1+r_2\right|}, \\ H\psi =A\left\{\left[-\frac{{\sout{h}}^2}{2m}\left({\nabla }_1^2+{\nabla }_2^2\right)-\frac{e^2}{4\mathrm{\pi }{\dot{\mathrm{o}}}_0}\left(\frac{z_1}{r_1}+\frac{z_2}{r_2}\right)\right]{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)+\left[-\frac{{\sout{h}}^2}{2m}\left({\nabla }_1^2+{\nabla }_2^2\right)-\frac{e^2}{4\mathrm{\pi }{\dot{\mathrm{o}}}_0}\left(\frac{z_1}{r_1}+\frac{z_2}{r_2}\right)\right]{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\right\} \\ +A\frac{e^2}{4\mathrm{\pi }{\dot{\mathrm{o}}}_0}\left\{\left[\frac{Z_1-1}{r_1}+\frac{Z_2-1}{r_2}\right]{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)+\left[\frac{Z_1-1}{r_1}+\frac{Z_2-1}{r_2}\right]{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right)\right\}+V_{ee}\psi . \end{gathered}$$

where$V_{ee}\equiv \frac{e^2}{4\mathrm{\pi }{\dot{\mathrm{o}}}_0}\frac{1}{\left|r_1-r_2\right|}.$

The term in first curly brackets is $\left(z_1^2+z_2^2\right)E_1{\psi }_1\left(r_1\right){\psi }_2\left(r_2\right)+\left(z_2^2+z_1^2\right)E_1{\psi }_2\left(r_1\right){\psi }_1\left(r_2\right),$ so

Last term is equal to:

localid="1658396571051"

Instead $r_1,r_2$,we put r and last expression will simplify:

localid="1658404037903"

Now we need to calculate

First and last term are the same, so we have :

$=2\left(\frac{e^2}{4\mathrm{\pi }{\dot{\mathrm{o}}}_0}\right)A^2(B+C),$

localid="1658488118584"

$the{r}_{2}integralis\int e^{-2Z_2{r}_2/a}\frac{1}{\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos {\theta }_2}}{d}^3{r}_2$

$$\begin{gathered} =\frac{{\mathrm{\pi a}}^3}{Z_2^3{r}_1}\left[1-\left(1+\frac{Z_2{r}_1}{a}\right)e^{-2Z_2{r}_1/a}\right]\left(\mathrm{Eq}.8.25,\mathrm{but}\mathrm{with}\mathrm{a}\to \frac{2}{Z_2}a\right) \\ B=\frac{Z_1^3{Z}_2^3}{{\left({\mathrm{\pi a}}^3\right)}^2}\frac{\left({\mathrm{\pi a}}^3\right)}{Z_2^3}4\mathrm{\pi }{\int }_0^{\infty }{\mathrm{e}}^{-2{\mathrm{Z}}_1{\mathrm{r}}_1/\mathrm{a}}\frac{1}{{\mathrm{r}}_1}\left[1-\left(1+\frac{{\mathrm{Z}}_2{\mathrm{r}}_1}{\mathrm{a}}\right){\mathrm{e}}^{-2({\mathrm{Z}}_1+{\mathrm{Z}}_2){\mathrm{r}}_1/\mathrm{a}}\right]{\mathrm{r}}_1^2{\mathrm{dr}}_1 \end{gathered}$$

$$\begin{gathered} =\frac{4Z_1^3}{a^3}{\int }_0^{\infty }\left[r_1{e}^{-2Z_1{r}_1/a}-r_1{e}^{-2(Z_1+Z_2)r_1/a}-\frac{Z_2}{a}{r}_1^2{e}^{-2(Z_1+Z_2)r_1/a}\right]d{r}_1. \\ =\frac{4Z_1^3}{a^3}\left[{\left(\frac{a}{2Z_1}\right)}^2-{\left(\frac{a}{2(Z_1+Z_2}\right)}^2-\frac{Z_2}{a}2{\left(\frac{a}{2(Z_1+Z_2}\right)}^3\right]=\frac{Z_1^3}{a}\left(\frac{1}{Z_1^2}-\frac{1}{{(Z_1+Z_2)}^2}-\frac{Z_2}{{(Z_1+Z_2)}^3}\right) \end{gathered}$$

$$\begin{gathered} =\frac{Z_1{Z}_2}{a(Z_1+Z_2)}\left[1+\frac{Z_1{Z}_2}{(Z_1+Z_2)}\right]=\frac{y_2}{4ax}\left(1+\frac{y^2}{4x^2}\right) \\ C=\frac{Z_1{Z}_2}{{\left({\mathrm{\pi a}}^3\right)}^2}{\int }_{}^{}{e}^{-Z_1{r}_1/a}{e}^{-Z_2{r}_2/a}{e}^{-Z_2{r}_1/a}{e}^{-Z_1{r}_2/a}\frac{1}{\left|r_1-r_2\right|}{d}^3{r}_1{d}^3{r}_2 \\ =\frac{{\left(Z_1{Z}_2\right)}^3}{{\left({\mathrm{\pi a}}^3\right)}^2}{\int }_{}^{}{e}^{-\left(z_1+z_2\right)\left(r_1+r_2\right)/a}\frac{1}{\left|r_1-r_2\right|}{d}^3{r}_1{d}^3{r}_2 \end{gathered}$$

The integral is the same as, only with a $\to \frac{4}{Z_1+Z_2}a.$

Comparing Eqs. 7.20 and 7.25, we see that the integral itself was

localid="1658492734438" $$\begin{gathered} =\frac{E_1}{\left(x_6+y_6\right)}\left\{\left(x^2-\frac{1}{2}{y}^2\right)\left(x^6+y^6\right)-2x^2\left[x^2-\frac{1}{2}{y}^2-x+\frac{1}{2}\frac{y^6}{x^4}-\frac{y^6}{x^5}+\frac{y^2}{4x}+\frac{y^4}{16x^3}+\frac{5y^6}{16x^5}\right]\right\}. \\ =\frac{E_1}{\left(x_6+y_6\right)}\left(x^8+x^2{y}^6-\frac{1}{2}{x}^6{y}^2-\frac{1}{2}{y}^8-2x^8+x^6{y}^2+2x^7-x^2{y}^6+2x{y}^6--\frac{1}{2}{x}^5{y}^2-\frac{1}{8}{x}^3{y}^4-\frac{5}{8}x{y}^6\right). \\ =\frac{E_1}{\left(x_6+y_6\right)}\left(-x^8+2x^7+\frac{1}{2}{x}^6{y}^2-\frac{1}{2}{x}^5{y}^2-\frac{1}{8}{x}^3{y}^4+\frac{11}{8}x{y}^6-\frac{1}{2}{y}^8\right). \end{gathered}$$

Mathematical finds the minimum

At this point,, which is less than -13.6 eV but not by much.