Nuclear motion in diatomic molecules

Nuclear motion in diatomic molecules involves the rotation, vibration, and overall translation of nuclei, significantly impacting molecular behavior and spectroscopic transitions.

From eq57, the Schrodinger equation for nuclear motion in a diatomic-molecule is

$\biggr\[-\frac{\hbar^2}{2m_{\alpha}}\nabla^2_{\alpha}-\frac{\hbar^2}{2m_{\beta}}\nabla^2_{\beta}+U(R)\biggr\]\psi_N=E\psi_N\;\;\;\;\;\;\;\;60$

where $\alpha$ and $\beta$ refer to the nuclei and $R$ is the inter-nuclear distance.

The eigenfunctions and eigenvalues of eq60, which is a two-particle problem, are $E=E_{tr}+E_{int}$ and $\psi_N=\psi_{tr}\psi_{int}$ respectively, where the subscripts tr and int refer to translational motion and internal motion respectively. Since the operator $\frac{\hat{p}^2_{\mu}}{2\mu}$ in eq49 is equivalent to $-\frac{\hbar^2}{2\mu}\nabla^2$ (see this article for derivation), the Schrodinger equation for internal motion is

$\biggr\[-\frac{\hbar^2}{2\mu}\nabla^2+U(R)\biggr\]\psi_{int}=E_{int}\psi_{int}\;\;\;\;\;\;\;\;61$

The potential energy $U$ is spherically symmetric, as it depends on $R$ only. This implies that eq61 is a central force problem, with the following Schrodinger equation and eigenfunction:

$\biggr\[-\frac{\hbar^2}{2\mu}\biggr$\frac{2}{R}\frac{\partial}{\partial R}+\frac{\partial^2}{\partial R^2}\biggr$+\frac{\hat{L}^2}{2\mu R^2}+U(R)\biggr\]\psi_{int}=E_{int}\psi_{int}\;\;\;\;\;\;\;\;62$

$\psi_{int}=P(R)Y^M_J(\theta,\phi)\;\;\;\;\;\;\;\;63$

where $\hat{L}^2$ is the operator for the square of the magnitude of the orbital angular momentum of the molecule, $P(R)$ is a function of $R$ and $Y^M_J(\theta,\phi)$ is a spherical harmonic function with quantum numbers $J$ and $M$ .

The kinetic energy of internal motion is the sum of rotational kinetic energy and vibrational kinetic energy. This implies that the Hamiltonian of eq62 consists, in part, of two kinetic energy operators. If we assume that the diatomic molecule is a rigid rotor ( $R$ is a constant), whose rotational motion is independent of its vibrational motion, then there is no potential energy contribution in the Hamiltonian for the rotation of a rigid rotor because the bond length remains fixed at its equilibrium distance. In other words, rotation is characterised only by $\theta$ and $\phi$ , and the Hamiltonian for vibrational motion consists of both a kinetic energy operator and a potential energy function, as the vibrational motion of the molecule is described by a change in $R$ .

Since ${\biggr\[\frac{\hat{L}^2}{2\mu R^2}Y^M_J(\theta,\phi)\biggr\]P(R)=E_{rot}P(R)Y^M_J(\theta,\phi)}$ , where ${E_{rot}=\frac{J(J+1)\hbar^2}{2\mu R^2}}$ (see this article for derivation)

$\biggr\[-\frac{\hbar^2}{2\mu}\biggr$\frac{2}{R}\frac{\partial}{\partial R}+\frac{\partial^2}{\partial R^2}\biggr$+U(R)\biggr\]P(R)Y^M_J(\theta,\phi)=(E_{int}-E_{rot})P(R)Y^M_J(\theta,\phi)$

$Y^M_J(\theta,\phi)$ becomes a non-zero constant because the Hamiltonian in the above equation is a function of $R$ . So,

$\biggr\[-\frac{\hbar^2}{2\mu}\biggr$\frac{2}{R}\frac{\partial}{\partial R}+\frac{\partial^2}{\partial R^2}\biggr$+U(R)\biggr\]P(R)=(E_{int}-E_{rot})P(R)\;\;\;\;\;\;\;\;64$

To simplify eq64, we substitute ${P(R)=\frac{F(R)}{R}}$ , ${P'(R)=\frac{F'(R)}{R}-\frac{F(R)}{R^2}}$ and ${P''(R)=\frac{F''(R)}{R}-2\frac{F'(R)}{R^2}}+2\frac{F(R)}{R^3}$ in eq64 to give

$\biggr\[-\frac{\hbar^2}{2\mu}\frac{d^2}{dR^2}+U(R)\biggr\]F(R)=(E_{int}-E_{rot})F(R)\;\;\;\;\;\;\;\;65$

We can approximate $U(R)$ by expanding it in a Taylor series around the equilibrium internuclear distance $R=R_e$ :

$U(R)=U(R_e)+U'(R_e)(R-R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2+\cdots\;\;\;\;\;\;\;\;66$

At $R_e$ , the slope of $U(R)$ is zero, i.e. $U'(R_e)=0$ . For small vibrations, where $R$ is close to $R_e$ , we ignore all the higher power terms, giving ${U(R)=U(R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2}$ , which when substituted in eq65 gives

$\biggr\[-\frac{\hbar^2}{2\mu}\frac{d^2}{dR^2}+U(R_e)+\frac{1}{2}U''(R_e)(R-R_e)^2\biggr\]F(R)=(E_{int}-E_{rot})F(R)$

The molecule is in its most stable state when $R=R_e$ for a particular electronic configuration, with $U(R_e)$ being a minimum point on the potential energy curve. This implies that the electrons are in their lowest energy state for that configuration and $U(R_e)$ is called the equilibrium electronic energy $E_{elec}$ of the molecule.

Question

Elaborate on the concept of $E_{elec}$ .

Answer

$E_{elec}$ is the energy of the electrons that are interacting with the nuclei at their equilibrium positions. It is neither the purely electronic energy $E_{e}$ nor $E_{e}+\hat{V}_{NN}$ in eq56. It is also not a form of vibrational energy, which is characterised by a displacement of the nuclei from their equilibrium positions. Since $E_{elec}$ is a constant for a particular electronic state, it is subtracted out when we calculate vibrational transition energies for a given electronic state. This is why it is sometimes set to zero.

Letting $x=R-R_e$ and $k=U''(R_e)$ and therefore $F(R)=F(x+R_e)=S(x)$ , we have

$\biggr\[-\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2}+\frac{1}{2}kx^2\biggr\]S(x)=E_{vib}S(x)\;\;\;\;\;\;\;\;67$

where $E_{vib}=E_{int}-E_{rot}-E_{elec}$ .

Question

Show that ${\frac{d^2}{dR^2}F(R)=\frac{d^2}{dx^2}S(x)}$ .

Answer

According to the chain rule, ${\frac{dS(x)}{dx}=\frac{dF(R)}{dx}=\frac{dF(R)}{dR}\frac{dR}{dx}}$ . Since $x=R-R_e$ , we have ${\frac{dR}{dx}=1}$ and ${\frac{dS(x)}{dx}=\frac{dF(R)}{dR}}$ . So, ${\frac{d^2S(x)}{dx^2}=\frac{d}{dx}F'(R)=\frac{dF'(R)}{dR}\frac{dR}{dx}=\frac{d^2F(R)}{dR^2}}$ .

Eq67 is the Schrodinger equation for a one-dimensional harmonic oscillator. Assuming that $S(x)$ behaves the same as a wave function of the harmonic oscillator, it is given by eq46. $E_{vib}$ is then represented by eq50. Therefore, $E$ and $\psi_N$ in eq60 are