Question

Suppose we used a minus sign in our trial wave function (Equation $\mathbf{7}\mathbf{.}\mathbf{37}\mathbf{)}\mathbf{:}\mathbf{\psi }\mathbf{=}\mathbf{A}\mathbf{[}{\mathbf{\psi }}_{\mathbf{0}}\mathbf{\left(}{\mathbf{r}}_{\mathbf{1}}\mathbf{\right)}\mathbf{-}{\mathbf{\psi }}_{\mathbf{0}}\mathbf{\left(}{\mathbf{r}}_{\mathbf{2}}\mathbf{\right)}\mathbf{]}$): Without doing any new integrals, find $\mathbf{F}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ (the analog to Equation ) for this case, and construct the graph. Show that there is no evidence of bonding. (Since the variational principle only gives an upper bound, this doesn't prove that bonding cannot occur for such a state, but it certainly doesn't look promising). Comment: Actually, any function of the form $\mathbf{\psi }\mathbf{=}\mathbf{A}\mathbf{\left[}{\mathbf{\psi }}_{\mathbf{0}}\mathbf{\right(}{\mathbf{r}}_{\mathbf{1}}\mathbf{)}\mathbf{+}{\mathbf{e}}^{\mathbf{i}}{\mathbf{\varphi \psi }}_{\mathbf{0}}\mathbf{(}{\mathbf{r}}_{\mathbf{2}}\mathbf{\left)}\mathbf{\right]}$has the desired property that the electron is equally likely to be associated with either proton. However, since the Hamiltonian (Equation 7.35 ) is invariant under the interchange $\mathit{P}\mathbf{:}{\mathit{r}}_{\mathbf{1}}\mathbf{\leftrightarrow }{\mathit{r}}_{\mathbf{2}}$ , its eigen functions can be chosen to be simultaneously eigen functions of P . The plus sign (Equation 7.37) goes with the eigenvalue +1 , and the minus sign (Equation 7.52 ) with the eigenvalue -1 ; nothing is to be gained by considering the ostensibly more general case (Equation 7.53), though you're welcome to try it, if you're interested.

Short answer

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

Step-by-step solution

The two equations

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 ${\psi }_0\left(r_1\right)$overlaps ${\psi }_0\left(r_2\right)$.

Equation 7.44 gives the expectation value of H in the trial state$\psi$ ,

role="math" localid="1658377951717" $⟨\mathrm{H}⟩={\mathrm{E}}_1-2|\mathrm{A}{|}^2\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0}\left[⟨{\mathrm{\psi }}_0\left({\mathrm{r}}_1\right)\left|\frac{1}{{\mathrm{r}}_2}\right|/{\mathrm{r}}_2\left|{\mathrm{\psi }}_0\right({\mathrm{r}}_1)⟩+⟨{\mathrm{\psi }}_0({\mathrm{r}}_1\left)\left|\frac{1}{{\mathrm{r}}_1}\right|{\mathrm{\psi }}_0\right({\mathrm{r}}_2)⟩\right]$

Definition of integral

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.

The value of  F(x)

Consider that,

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

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" $$\begin{gathered} |\mathrm{A}{|}^2=\frac{1}{2(1+\mathrm{I})} \\ ⟨\mathrm{H}⟩={\mathrm{E}}_1-2|\mathrm{A}{|}^2\frac{{\mathrm{e}}^2}{4{\mathrm{\pi ϵ}}_0}\left[⟨{\mathrm{\psi }}_0\left({\mathrm{r}}_1\right)\left|\frac{1}{{\mathrm{r}}_2}\right|/{\mathrm{r}}_2\left|{\mathrm{\psi }}_0\right({\mathrm{r}}_1)⟩+⟨{\mathrm{\psi }}_0({\mathrm{r}}_1\left)\left|\frac{1}{{\mathrm{r}}_1}\right|{\mathrm{\psi }}_0\right({\mathrm{r}}_2)⟩\right] \end{gathered}$$

The Equation 7.49 now becomes:

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

Adding repulsion

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.