The $\mathrm{H}_2^+$ Ion

The simplest molecule we can work with is the $\mathrm{H}_2^+$ ion. It has two nuclei (A and B) sharing one electron (1).

$$\bgroup\color{black}H_0 = {p^2_e\over {2m}} - {e^2\over{r_{1A}}} - {e^2\over{r_{1B}}} + {e^2\over{R_{AB}}}\egroup$$

$R_{AB}$ is the distance between the two nuclei.

The lowest energy wavefunction can be thought of as a (anti)symmetric linear combination of an electron in the ground state near nucleus A and the ground state near nucleus B

$$\bgroup\color{black}\psi_\pm\left(\vec{r},\vec{R}\right)=C_\pm(R)\left[\psi_A\pm\psi_B\right]\egroup$$

where $\psi_A=\sqrt{1\over{\pi a^3_0}}e^{-r_{\scriptscriptstyle 1A}/a_0}$ is g.s. around nucleus A. $\psi_A$ and $\psi_B$ are not orthogonal; there is overlap. We must compute the normalization constant to estimate the energy.

$$\bgroup\color{black}{1\over{C^2_\pm}}=\left<\psi_A\pm\psi_B\v... ...\pm 2\langle\psi_A\vert\psi_B\rangle \equiv 2 \pm 2 S(R)\egroup$$

where

$$\bgroup\color{black}S(R)\equiv\langle\psi_A\vert\psi_B\rangle... ...1 + {R\over{a_0}} + {R^2\over{3 a^2_0}}\right)e^{-R/a_0}\egroup$$

These calculations are ``straightforward but tedious'' (Gasiorowicz).

We can now compute the energy of these states.

$$\begin{eqnarray*} \left<H_0\right>_\pm &=&{1\over {2[1\pm S(R)]}}\left<\psi_A\pm... ...> \pm \left<\psi_A\vert H_0\vert\psi_B\right> \over {1\pm S(R)}} \end{eqnarray*}$$

We can compute the integrals needed.

$$\begin{eqnarray*} \left<\psi_A\vert H_0\vert\psi_A\right> &=& E_1 + {e^2\over R}... ...ght)S(R) -{e^2\over{a_0}}\left(1+{R\over{a_0}}\right)e^{-R/a_0} \end{eqnarray*}$$

We have reused the calculation of $S(R)$ in the above. Now, we plug these in and rewrite things in terms of $y=R/a_0$, the distance between the atoms in units of the Bohr radius.

$$\begin{eqnarray*} \left<H_0\right>_\pm &=& {E_1+{e^2\over R}\left(1+R/a_0\right)... ...2/3)e^{-y}-2(1+y)e^{-y}\right] \over{ 1\pm (1+y+y^2/3)e^{-y}}} \end{eqnarray*}$$

The symmetric (bonding) state has a large probability for the electron to be found between nuclei. The antisymmetric (antibonding) state has a small probability there, and hence, a much larger energy.

The graph below shows the energies from our calculation for the space symmetric $(E_g)$ and antisymmetric $(E_u)$ states as well as the result of a more complete calculation (Exact $E_g$) as a function of the distance between the protons $R$. Our calculation for the symmetric state shows a minimum arount 1.3 Angstroms between the nuclei and a Binding Energy of 1.76 eV. We could get a better estimate by introduction some parameters in our trial wave function and using the variational method.

The antisymmetric state shows no minimum and never goes below -13.6 eV so there is no binding in this state.

By setting ${d\langle H\rangle\over dy}=0$, we can get the distance between atoms and the energy.

	Distance	Energy
Calculated	1.3 Å	-1.76 eV
Actual	1.06 Å	-2.8 eV

Its clear we would need to introduce some wfn. parameters to get good precision.