The \bgroup\color{black}$\mathrm{H}_2^+$\egroup Ion

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

\begin{displaymath}\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\end{displaymath}

\bgroup\color{black}$R_{AB}$\egroup 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


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

...\pm 2\langle\psi_A\vert\psi_B\rangle \equiv 2 \pm 2 S(R)\egroup\end{displaymath}


...1 + {R\over{a_0}} + {R^2\over{3 a^2_0}}\right)e^{-R/a_0}\egroup\end{displaymath}

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

We can now compute the energy of these states.

\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)}}

We can compute the integrals needed.

\left<\psi_A\vert H_0\vert\psi_A\right> &=& E_1 + {e^2\over R}...

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

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

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 \bgroup\color{black}$(E_g)$\egroup and antisymmetric \bgroup\color{black}$(E_u)$\egroup states as well as the result of a more complete calculation (Exact \bgroup\color{black}$E_g$\egroup) as a function of the distance between the protons \bgroup\color{black}$R$\egroup. 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 \bgroup\color{black}${d\langle H\rangle\over dy}=0$\egroup, 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.

Jim Branson 2013-04-22