H$_2$ Molecule

The $H_2$ molecule can be represented by the following diagram:

The diagram includes two electrons, represented by $e_1$ and $e_2$, and two protons $A$ and $B$. The Hamiltonian of the system is,

$$\hat{H}= -\frac{\hbar^2}{2M_A} \nabla_{R_A}^2 -\frac{\hbar^2}{2M_B} \nabla_{R_B}^2 +\hat{H}_{el},$$

where

$$\hat{H}_{el}= -\frac{\hbar^2}{2m} \nabla_1^2 -\frac{e^2}{r_{A1}}-... ... -\frac{e^2}{r_{A2}}-\frac{e^2}{r_{B2}} +\frac{e^2}{r_{12}}+\frac{e^2}{R_{AB}}.$$

In analogy to the He atom, it is possible to identify one-electron Hamiltonians (i.e., associated with electrons 1 and 2),

$$H_2^+(1)=-\frac{\hbar^2}{2m} \nabla_1^2 -\frac{e^2}{r_{A1}}-\frac{e^2}{r_{B1}},$$

and,

$$H_2^+(2)=-\frac{\hbar^2}{2m} \nabla_2^2 -\frac{e^2}{r_{A2}}-\frac{e^2}{r_{B2}}.$$

A zeroth order solution is obtained by neglecting the repulsion between electrons. Since $\frac{e^2}{R_{AB}}$ contributes only with a constant value to the energy (e.g., a constant parametrized by $R_{AB}$), we can make use of the theorem of separation of variables and obtain the solution of the eigenvalue problem,

$$\hat{H} \mid \psi > = E \mid \psi >,$$

as the product

$$\mid \psi > = A \mid \Phi_1 > \mid \Phi_2 >,$$ (59)

where $\mid \Phi_1 >$ and $\mid \phi_2 >$ are eigenstates of the $H_2^+$ Hamiltonian and A is the anti-symmetrizing spin wave function,

$$A = \frac{1}{N\sqrt{2}} \left[ \alpha(1) \beta(2) -\beta(1) \alpha(2) \right ].$$

Note that the hydrogen molecule occupies the same place in the theory of molecular electronic structure as the helium atom in the theory of atomic electronic structure. Therefore, the correction due to electronic repulsion can be calculated according to first order perturbation theory as follows,

$$E = 2 E_{H_2^+}(R_{AB}) + < \psi \vert\frac{e^2}{r_{12}}\vert \psi> - \frac{e^2}{R_{AB}}.$$ (60)

Note that the last term discounts the repulsion between nuclei that has been over-counted.

The equilibrium distance, $R_{AB}^{(eq)}$, is obtained by minimizing $E$ with respect to $R_{AB}$. Substituting such value into Eq. (60), we obtain the minimum energy of the $H_2$ molecule.

The complete ground state of $H_2$ is described as follows,

$$\psi= \frac{1}{N\sqrt{2}} \left[ \alpha(1) \beta(2) -\beta(1) \al... ...eft[ 1S_A(1)1S_A(2) + 1S_A(1)1S_B(2) + 1S_B(1)1S_A(2) + 1S_B(1)1S_B(2)\right ],$$ (61)

where $N$ is a normalization factor, obtained by substituting $\mid \Phi_1 >$ and $\mid \Phi_2 >$ in Eq. (59), by the ground state wave function of $H_2^+$,

$$\Phi_j = \frac{1}{\sqrt{N}} \left [ 1S_A(j) + 1S_B(j) \right].$$

According to Eq. (61), the probability of finding both electrons close to nucleus A (i.e., the probability of finding the electronic configuration $H_A^-$ $H_B^+$), is determined by the square of the expansion coefficient associated with the term $1S_A(1)1S_A(2)$. Analogously, the probability of finding both electrons close to nucleus B is proportional to the square of the expansion coefficient associated with the term $1S_B(1)1S_B(2)$. Therefore, terms $1S_A(1)1S_A(2)$, $1S_B(1)1S_B(2)$ describe ionic configurations, while terms $1S_A(1)1S_B(2)$ and $1S_B(1)1S_A(2)$ describe covalent structures.

Unfortunately, the LCAO wavefunction, introduced by Eq. (61), predicts the same probability for ionic and covalent configurations, $H_A^+ H_B^-$, $H_A^-$ $H_B^+$, and $H_A H_B$, respectively. This is quite unsatisfactory since it is contrary to the chemical experience. The LCAO model predicts that upon dissociation half of the $H_2$ molecules break into ions $H^-$ and $H^+$. Contrary to such prediction, the $H_2$ molecule dissociates almost always into two hydrogen atoms.

Heitler-London(HL) Method:

The Heitler-London approach aims to correct the shortcomings of the LCAO description by neglecting the ionic terms altogether. Therefore, the HL wave function of $H_2$ includes only covalent terms as follows,

$$\psi_{HL}= \frac{1}{N' \sqrt{2}} \left[ \alpha(1) \beta(2) -\beta(1) \alpha(2) \right] \left[ 1S_A(1)1S_B(2) + 1S_B(1)1S_A(2) \right ].$$

This wave function gives a better description of the energy as a function of $R_{AB}$ and predicts the proper asymptotic behavior at large internuclear distances.

Exercise 54: Prove that, according to the HL approach,

$$E= \frac{J+K}{1+S^2},$$

with

$$J=<1S_A(1)1S_B(2)\vert H\vert 1S_A(1)1S_B(2)>,$$

and

$$K=<1S_A(1)1S_B(2)\vert H\vert 1S_B(1)1S_A(2)>.$$