LCAO Method: H$_2^+$ Molecule

The H$_2^+$ molecule can be represented by the following diagram:

where $A$ and $B$ represent two hydrogen nuclei and $e$ represents the electron. The Hamiltonian of the system is,R1(376)

$$\hat{H}= -\frac{\hbar^2}{2M_A}\nabla_{R_A}^2 -\frac{\hbar^2}{2M_B}\nabla_{R_B}^2 +H_{el},$$ (54)

where,

$$\hat{H}_{el} =-\frac{\hbar^2}{2m}\nabla_r^2 -\frac{e^2}{r_A} -\frac{e^2}{r_B} + \frac{e^2}{R_{AB}}.$$ (55)

This is another three-body Hamiltonian, similar to the Helium atom Hamiltonian, where instead of having two electrons and one nucleus we have two nuclei and one electron. In order to compute the eigenstates, we assume that the kinetic energy of the nuclei can be neglected when compared to the other terms in the Hamiltonian (Born-Oppenheimer approximation). The electronic energy is computed at various internuclear distances $R_{AB}$, by considering that the term $\frac{e^2}{R_{AB}}$, in Eq. (55) is a constant factor parametrized by $R_{AB}$. (In practice, this constant factor is ignored when solving the eigenvalue problem, since it can be added at the end of the calculation).

According to the linear combination of atomic orbitals (LCAO) method, a convenient trial state for $H_2^+$ can be written as follows,

$$\mid \Psi > = C_A \mid \phi_A > + C_B \mid \phi_B >, \$$   (compare this equation with Eq. (21))

where $\mid \phi_A >$, and $\mid \phi_B >$, are 1S atomic orbitals of atoms A and B, respectively.

According to the variational theorem, the optimum coefficients $C_A$ and $C_B$ can be found by minimizing the expectation value of the energy,

$$<E> =\frac{<\psi\vert\hat{H'}_{el}\vert\psi>}{<\psi\vert\psi>} = ... ...+2C_A C_B H_{AB} + C_B^2 H_{BB}}{C_A^2 S_{AA} +2C_A C_B S_{AB} + C_B^2 S_{BB}},$$

with respect to $C_A$ and $C_B$. Here, $H_{jk}= <\phi_j\vert\hat{H'}_{el}\vert \phi_k >$, $S_{jk}=<\phi_j \vert\phi_k>$, and

$$\hat{H'}_{el} = -\frac{\hbar^2}{2m} \nabla_r^2 -\frac{e^2}{r_A} -\frac{e^2}{r_B}.$$

Exercise 52: Show that the condition,

$\Bigg(\frac{\partial <E>}{\partial C_A } \Bigg)_{C_B} =0$ implies $C_A (H_{AA} -<E>) +C_B(H_{AB} - S_{AB} <E>) =0,$ and

$\Bigg (\frac{\partial <E>}{\partial C_B } \Bigg)_{C_A} =0$ implies $C_A (H_{AB} - S_{AB} <E>) +C_B(H_{BB} - <E>) =0$, when $<\phi_j \vert \phi_j> = 1.$

These equations are called secular equations and have a nontrivial solution (i.e., a solution different from the trivial solution $C_A=0$, $C_B=0$), when the determinant of the expansion coefficients vanishes, i.e.,

$$\left \vert \begin{matrix}H_{AA}-<E> & H_{AB}-S_{AB}<E> \\ H_{BA}-S_{BA}<E> & H_{BB}-<E> \\ \end{matrix} \right \vert = 0,$$

This determinant is called the secular determinant.

Since $\mid \phi_A >$ and $\mid \phi_B >$ are 1S orbitals, $H_{AA}=H_{BB}$, and $S_{AB}=S_{BA}=S$. Therefore,

$$(H_{AA}-<E>)^2 -(H_{AB}-S<E>)^2 =0,$$

and

$$E_{\pm}=\frac{H_{AA} \pm H_{AB}}{1 \pm S}.$$

Substituting $<E>_+$ in the secular equations we obtain,

$${C_A}_{\pm} = \pm {C_B}_{\pm}.$$

Therefore,

$\psi_+ = {C_A}_+ (\phi_A+ \phi_B)$, where ${C_A}_+ = \frac{1}{\sqrt{2+2S}}$,

$\psi_- = {C_A}_- (\phi_A- \phi_B)$, where ${C_A}_- = \frac{1}{\sqrt{2-2S}}.$

The strategy followed in this section for solving the eigenvalue problem of $H_2^+$ can be summarized as follows:

1. Expand the solution $\mid \Psi >$ according to a linear combination of atomic orbitals (LCAO).

2. Obtain a set of $n$ secular equations according to the variational approach.

3. Solve the secular determinant by finding the roots of the characteristic equation, a polynomial of degree $n$ in $E$.

4. Substitute each root into the secular equations and find the eigenvectors (e.g., the expansion coefficients in the LCAO) that correspond to such root.

The energies $<E >_{\pm}$ are functions of $H_{AA}$, $H_{AB}$ and $S$. The integral $H_{AA}$ is defined as the sum of the energy of an electron in a 1S orbital and the attractive energy of the other nucleus:

$$H_{AA} = \int d \tau \phi_A^* \left[-\frac{\hbar^2}{2 m} \nabla_r... ...{r_B} \right] \phi_A = E_{1S}(H) - \int d \tau \phi_A^* \frac{e^2}{r_B} \phi_A.$$ (56)

As the nuclei $A$ and $B$ are brought closer together, the second term in Eq. (56) (i.e., the term $\int d \tau \phi_A^* \frac{e^2}{r_B} \phi_A$) tends to make the energy of $H_2^+$ more negative, increasing the stability of the molecule. The term $\frac{e^2}{R_{AB}}$ is responsible for the repulsion between nuclei and increases monotonically as the two nuclei get closer together, counteracting the stabilization caused by $-\frac{e^2}{r_B}$. Therefore, the sum $H_{AA}+\frac{e^2}{R_{AB}}$ is not responsible for the stabilization of the system as the nuclei are brought closer together.

The integral $H_{AB}$ defined as follows,

$$H_{AB} = \int d \tau \phi_A^* \left( -\frac{\hbar^2}{2m} \nabla_r^2 -\frac{e^2}{r_A} -\frac{e^2}{r_B} \right) \phi_B,$$ (57)

is called resonance integral and takes into account the fact that the electron is not restricted to any of the two $1S$ atomic orbitals, but it can rather be exchanged between the two orbitals.

At large values of $R_{AB}$, the resonance integral $H_{AB}$ goes to zero. Decreasing $R_{AB}$, $H_{AB}$ becomes more negative and stabilizes the molecule relative to the asymptotically separated atoms. The eigenvalues $<E >_{\pm}$ can be represented as a function of $R_{AB}$ by the following diagram:

Note that $<E>_+$ is lower than $<E>_-$ because $H_{AA}$ and $H_{AB}$ are negative.

In analogy to the variational approach implemented to study the Helium atom, one could further improve the variational solution of $H_2^+$ by optimizing the exponents $\xi$ (e.g., effective nuclear charges) in the functions that represent $\phi_A$ and $\phi_B$,

$$\phi_{A/B} = \frac{(\frac{\xi}{2a})^{3/2}}{\sqrt{\pi}} e^{- \frac{\xi r_{A/B}}{2a}}.$$ (58)

Such variational correction of the effective nuclear charge is known as scaling.

Exercise 53: According to the quantum mechanical description of $H_2^+$ explain:

(1) Why do molecules form? What is a chemical bond?

(2) Consider state $\psi_+= (2+2S)^{-1/2}(\chi_A +\chi_B)$ where nucleus $A$ is at $R_A=(\frac{R}{2},0,0)$ and nucleus $B$ is at $R_B=(-\frac{R}{2},0,0)$. Compute $\psi^* \psi$ at the coordinate (0,0,0), and compare such probability density to the sum of probability amplitudes due to $\phi_A$ and $\phi_B$.