The H $_2^+$ Ion

The simplest molecule we can work with is the 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 $\psi_A=\sqrt{1\over{\pi a^3_0}}e^{-r_{1A}/a_0}$ is g.s. around nucleus A. $\psi_A$ and $\psi_B$ are not orthogonal; there is overlap. We must compute the normalization constant.

$$\bgroup\color{black}{1\over{C^2_\pm}}=\left<\psi_A\pm\psi_B\v... ...i_B\right>=2\pm 2\int d^r\psi_A\psi_B\equiv 2 \pm 2 S(R)\egroup$$

where

$$\bgroup\color{black}S(R)=\left(1 + {R\over{a_0}} + {R^2\over{3 a^2_0}}\right)e^{-R/a_0}\egroup$$

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

We can compute the energy of these states.

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

$$\bgroup\color{black} \left<\psi_A\vert H_0\vert\psi_A\right> = E_1 + {e^2\over R}\left(1+{R\over{a_0}}\right)e^{-2R/a_0}\egroup$$

$$\bgroup\color{black} \left<\psi_A\vert H_0\vert\psi_B\right> ... ... -{e^2\over{a_0}}\left(1+{R\over{a_0}}\right)e^{-R/a_0}\egroup$$

$$\bgroup\color{black}\left<H_0\right>= {E_1+{e^2\over R}\left(... ...{a_0}}\left(1+R/a_0\right)e^{-R/a_0} \over {1\pm S(R)}}\egroup$$

$$\bgroup\color{black}\left<H_0\right>_{\pm}=E_1 {1-(2/y)(1+y)... ...y}-2(1+y)e^{-y}\right] \over{ 1\pm (1+y+y^2/3)e^{-y}}} \egroup$$

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 shows the energies expected from our calculation for the space symmetric and antisymmetric states as well as the result of a more complete calculation as a function of the distance between the protons.

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.