# 7.3: The Electronic Ground State


We now try to investigate the lowest electronic levels of H$$^+_2$$. First we look for symmetries, and note that, since $${\bf r}_1 = {\bf r} + {\bf R}/2$$ and $${\bf r}_2 = {\bf r} − {\bf R}/2$$, the electronic Hamiltonian is invariant under the parity operation $${\bf r} \rightarrow −{\bf r}$$. If $$\hat{\mathcal{P}}$$ denotes the parity operator, then

$[\hat{\mathcal{P}} , \hat{H} ] = 0 \nonumber$

These are commuting operators, so they have can have the same eigenfunctions. These eigenfunctions are called gerade if the parity is even and ungerade if the parity is odd:

$\hat{\mathcal{P}} U^g_j ({\bf r, R}) = U^g_j ({\bf r, R}), \quad \hat{\mathcal{P}} U^u_j ({\bf r, R}) = −U^u_j ({\bf r, R}) \nonumber$

Now think about wave functions. If $$R$$ is large, the system separates into a hydrogen atom and a proton (two degenerate states). The hydrogen atom has a large spacing between levels, so we use degenerate perturbation theory with $$1s$$ levels only. Quite generally, this procedure of taking linear combinations of atomic orbitals is known as the LCAO method. Note that this basis set is normalised, but neither complete nor orthogonal.

Since there must be solutions which are eigenfunctions of the parity operator, we take normalised linear combinations of gerade or ungerade symmetry of $$1s$$ orbitals:

$\psi^g = [u_{1s}(r_1) + u_{1s}(r_2)]/ \sqrt{2} \quad \text{ and } \quad \psi^u = [u_{1s}(r_1) − u_{1s}(r_2)]/ \sqrt{2} \nonumber$

We calculate the expectation value of the electronic Hamiltonian using these LCAO molecular wavefunctions:

$E^{g, u}(\mathbf{R})=\int \psi^{g, u *}(\mathbf{r}, \mathbf{R}) \hat{H} \psi^{g, u}(\mathbf{r}, \mathbf{R}) \mathrm{d}^{3} r=\langle u_{1 s}\left(r_{1}\right)|\hat{H}| u_{1 s}\left(r_{1}\right)\rangle \pm\langle u_{1 s}\left(r_{1}\right)|\hat{H}| u_{1 s}\left(r_{2}\right)\rangle \nonumber$

where + and - correspond to $$u$$ and $$g$$ respectively, giving $$E^g ({\bf R})$$ and $$E^u ({\bf R})$$ for each value of $${\bf R}$$;

The evaluation of the integrals is complicated, but the results have the form:

$E^{g}(\mathbf{R})=E_{1 s}+\frac{e^{2}}{\left(4 \pi \epsilon_{0}\right) R} \times \frac{\left(1+R / a_{0}\right) \exp \left(-2 R / a_{0}\right)+\left[1-(2 / 3)\left(R / a_{0}\right)^{2}\right] \exp \left(-R / a_{0}\right)}{1+\left[1+\left(R / a_{0}\right)+(1 / 3)\left(R / a_{0}\right)^{2}\right] \exp \left(-R / a_{0}\right)} \nonumber$

and

$E^{u}(\mathbf{R})=E_{1 s}+\frac{e^{2}}{\left(4 \pi \epsilon_{0}\right) R} \times \frac{\left(1+R / a_{0}\right) \exp \left(-2 R / a_{0}\right)-\left[1-(2 / 3)\left(R / a_{0}\right)^{2}\right] \exp \left(-R / a_{0}\right)}{1-\left[1+\left(R / a_{0}\right)+(1 / 3)\left(R / a_{0}\right)^{2}\right] \exp \left(-R / a_{0}\right)} \nonumber$

where $$a_0$$ is the Bohr radius and $$E_{1s}$$ is the ground-state energy of atomic hydrogen.

The two curves $$E^g − E_{1s}$$ and $$E^u − E_{1s}$$ are plotted as a function of $$R$$. Note that the curve which corresponds to the symmetric (gerade) orbital exhibits a minimum at $$R = R_0$$, where $$R_0/a_0 \simeq 2.5$$, corresponding to $$E^g − E_{1s} = −1.77$$ eV. Since this is an upper bound on the ground-state energy, this implies that there is a stable bound state, a molecular ion. The curve represents an effective attraction between the two protons. By contrast, the curve corresponding to the ungerade orbital has no minimum, so that a H$$^+_2$$ ion in this state will dissociate into a proton and a hydrogen atom. If we think of the protons being attracted by the electron and repelled by each other, the symmetrical state should be the more tightly bound because the electron spends more of its time between the protons, where it attracts both of them. This is an example of covalent bonding.

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