91Ó°ÊÓ

Equation 7.43 gives the normalization factor as,

$|\mathrm{A}{|}^2=\frac{1}{2(1+\mathrm{I})}$where I is overlap integral which determines the amount by which ${\mathrm{ψ}}_0\left(r_1\right)$overlaps ${\mathrm{ψ}}_0\left(r_2\right)$.

Equation 7.44 gives the expectation value of H in the trial state$\mathrm{ψ}$ ,

role="math" localid="1658377951717" $⟨\mathrm{H}âŸ\copyright ={\mathrm{E}}_1-2|\mathrm{A}{|}^2\frac{{\mathrm{e}}^2}{4{\mathrm{Ï€Ï\mu }}_0}\left[⟨{\mathrm{ψ}}_0\left({\mathrm{r}}_1\right)\left|\frac{1}{{\mathrm{r}}_2}\right|/{\mathrm{r}}_2\left|{\mathrm{ψ}}_0\right({\mathrm{r}}_1)âŸ\copyright +⟨{\mathrm{ψ}}_0({\mathrm{r}}_1\left)\left|\frac{1}{{\mathrm{r}}_1}\right|{\mathrm{ψ}}_0\right({\mathrm{r}}_2)âŸ\copyright \right]$

An integral is a number that is assigned to a function in order to explain displacement, area, volume, and other notions that occur when infinitesimal data is combined.

Integration is the term used to describe the process of locating integrals.

Consider that,

$\mathrm{ψ}=A\left[{\mathrm{ψ}}_0\right(r_1)-{\mathrm{ψ}}_0(r_2\left)\right]$

Two things that change sign are - integral I from Equation 7.43 and sign inside the parenthesis in the Equation 7.44 . So,

role="math" localid="1658378005528" $$\begin{gathered} |\mathrm{A}{|}^2=\frac{1}{2(1+\mathrm{I})} \\ ⟨\mathrm{H}âŸ\copyright ={\mathrm{E}}_1-2|\mathrm{A}{|}^2\frac{{\mathrm{e}}^2}{4{\mathrm{Ï€Ï\mu }}_0}\left[⟨{\mathrm{ψ}}_0\left({\mathrm{r}}_1\right)\left|\frac{1}{{\mathrm{r}}_2}\right|/{\mathrm{r}}_2\left|{\mathrm{ψ}}_0\right({\mathrm{r}}_1)âŸ\copyright +⟨{\mathrm{ψ}}_0({\mathrm{r}}_1\left)\left|\frac{1}{{\mathrm{r}}_1}\right|{\mathrm{ψ}}_0\right({\mathrm{r}}_2)âŸ\copyright \right] \end{gathered}$$

The Equation 7.49 now becomes:

$⟨HâŸ\copyright =\left[1+2\frac{D-X}{1-I}\right]E_1$

Add proton-proton repulsion$V_{pp}$ in order to get total energy$F\left(x\right)$, where$\mathrm{x}=\mathrm{R}/\mathrm{a}$:

$$\begin{gathered} \mathrm{F}\left(\mathrm{x}\right)={\mathrm{E}}_1+\frac{3{\mathrm{E}}_1}{1-{\mathrm{e}}^{-\mathrm{x}}\left(1+\mathrm{x}+{\displaystyle \frac{{\mathrm{x}}^2}{3}}\right)}\left[\frac{1}{\mathrm{x}}-\left(1+\frac{1}{\mathrm{x}}\right){\mathrm{e}}^{-2\mathrm{x}}-\left(1+\mathrm{x}\right){\mathrm{e}}^{-\mathrm{x}}\right]-\frac{2}{\mathrm{x}}{\mathrm{E}}_1 \\ =-{\mathrm{E}}_1\left\{-1+\frac{2}{\mathrm{x}}-\frac{2}{\mathrm{x}}\frac{1-\left(1+\mathrm{x}\right){\mathrm{e}}^{-2\mathrm{x}}-\left(\mathrm{x}+{\mathrm{x}}^2\right){\mathrm{e}}^{-\mathrm{x}}}{1-{\mathrm{e}}^{-\mathrm{x}}\left(1+\mathrm{x}+{\displaystyle \frac{{\mathrm{x}}^2}{3}}\right)}\right\} \\ =-{\mathrm{E}}_1\left\{-1+\frac{2}{\mathrm{x}}\frac{1-{\mathrm{e}}^{-\mathrm{x}}\left(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{3}\right)-1+\left(1+\mathrm{x}\right){\mathrm{e}}^{-2\mathrm{x}}+\left(\mathrm{x}+{\mathrm{x}}^2\right){\mathrm{e}}^{-\mathrm{x}}}{1-{\mathrm{e}}^{-\mathrm{x}}\left(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{3}\right)}\right\} \\ =-{\mathrm{E}}_1\left\{-1+\frac{2}{\mathrm{x}}\frac{\left(1+\mathrm{x}\right){\mathrm{e}}^{-2\mathrm{x}}+\left({\displaystyle \frac{2{\mathrm{x}}^2}{3}}-1\right){\mathrm{e}}^{-\mathrm{x}}}{1-{\mathrm{e}}^{-\mathrm{x}}\left(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{3}\right)}\right\} \end{gathered}$$

In the graph below, this function is represented by a black line.

The Equation 7.51has a bound state, as shown by the blue line.