Consider a two-particle rigid rotor. Ignoring the masses of electrons and the vibration of nuclei, a diatomic molecule may be taken as a rigid rotator. Let the masses of atoms be \(m_1\) and \(m_2\). It is assumed that the rotor involves no change in centre of gravity and bond length (\(r\)).
Let the center of gravity be at the origin of cartesian co-ordinates and let the distance \(m_1\) from the center of mass be \(r_1\) and distance of \(m_2\) be \(r_1\), then-
\[ m_1 r_1 = m_2 r_2, \quad r_1 + r_2 = r \]
\[ m_1(r-r_2) = m_2 r_2 \]
\[ m_1 r - m_1 r_2 = m_2 r_2 \]
\[ m_1 r_2 + m_2 r_2 = m_1 r \]
\[ (m_1 + m_2) r_2 = m_1 r \]
From this relation:
\[ r_1 = \frac{m_2 r}{m_1 + m_2}, \quad r_2 = \frac{m_1 r}{m_1 + m_2} \]
Moment of Inertia
Moment of Inertia (\(I\)) of a rotating body about the center of mass is given by:
\[ I = m_1 r_1^2 + m_2 r_2^2 \]
\[ I = \frac{m_1 m_2^2 r^2}{(m_1 + m_2)^2} + \frac{m_2 m_1^2 r^2}{(m_1 + m_2)^2} \]
\[ I = \frac{m_1 m_2 r^2 (m_1 + m_2)}{(m_1 + m_2)^2} = \frac{m_1 m_2 r^2}{m_1 + m_2} = \mu r^2 \]
where reduced mass is:
\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]
Kinetic Energy of Rotor
The kinetic energy of rotor is given by:
\[ T = \tfrac{1}{2} m_1 v_1^2 + \tfrac{1}{2} m_2 v_2^2 \]
\[ T = \tfrac{1}{2} m_1 \omega^2 r_1^2 + \tfrac{1}{2} m_2 \omega^2 r_2^2 \quad [v = \omega r] \]
\[ T = \tfrac{1}{2} \omega^2 (m_1 r_1^2 + m_2 r_2^2) = \tfrac{1}{2} I \omega^2 \]
\[ T = \frac{L^2}{2I} = \frac{L^2}{2 \mu r^2} \]
Where \(\omega\) and \(L\) are angular velocity and angular momentum respectively. Since the bond distance (\(r\)) is fixed, the potential energy is taken as zero. Hence, the rigid rotator has only kinetic energy.
Hamiltonian Operator
In quantum mechanics, we know that:
\[ \vec{L} = \vec{r} \times \vec{p} = \vec{r} \times \frac{h}{2 \pi i} \nabla \]
The Hamiltonian operator (\(\hat{H}\)) will contain only kinetic energy operator:
\[ \hat{H} = \hat{T} = \frac {(\hat{L^2})}{2 \mu r^2} = \frac{h^2 \nabla^2}{8 \pi^2 \mu} \]
Schrödinger Equation
The Schrödinger equation in operator form is given by:
\[ \hat{H} \psi = E \psi \]
So, the Schrödinger wave equation for a rigid rotator becomes:
\[ - \frac{h^2}{8 \pi^2 \mu} \nabla^2 \psi = E \psi \]