← Multielectron Atoms and Molecules

Molecular Rotations

Friday, March 18, 2022

Rotations in Quantum Mechanics

In classical physics, a rotor with rotational inertial II has rotational kinetic energy K=12Iω2=L2/2IK=\frac{1}{2}I\omega^2=L^2/2I, where L=IωL=I\omega is the rotational angular momentum. There is no obvious analogy in quantum mechanics, but we can imagine a particle of mass mm rotating on the end of a rod of length rr. However, in quantum mechanics it is not possible to have rotations in a fixed plane (due to the uncertainty relationship), so we instead imagine a particle confined to move in three dimensions on a sphere of radius rr.

The resulting energies take the form

EL=L(L+1)22I    L=0,1,2,3,...E_L=\frac{L\left(L+1\right)\hbar^2}{2I}~~~~L=0,1,2,3,...

where LL is the angular momentum quantum number. The rotational levels have energies 0,2(2/2I),6(2/2I),12(2/2I),...0, 2\left(\hbar^2/2I\right), 6\left(\hbar^2/2I\right), 12\left(\hbar^2/2I\right),... (note: not equally spaced).

We can then find the kinetic energy by replacing classical angular momentum with quantum angular momentum L=L(L+1)\left|\boldsymbol{\overrightarrow{L}}\right|=\sqrt{L\left(L+1\right)}\hbar. Note: LL is used to represent the angular momentum quantum number of a system whereas ll is used to represent the angular momentum quantum number of a particle.

Much like the vibrational states, there is a selection rule for rotational states too:

Rotational selection rule:    ΔL=±1\text{Rotational selection rule:}~~~~\Delta L=\pm 1

Rotating Diatomic Molecules

Imagine a diatomic molecule whose center of mass is at the origin (x1m1=x2m2x_1m_1=x_2m_2).

Rotating diatomic molecule

The rotational inertia is I=m1x12+m2x22I=m_1x_1^2+m_2x_2^2, or, using the reduced mass m=m1m2/(m1+m2)m=m_1m_2/\left(m_1+m_2\right) and equilibrium separation Req=x1+x2R_{\text{eq}}=x_1+x_2:


Plugging this into the energy states, we get


where BB is the rotational parameter, defined for diatomic molecules as


The energy absorbed or emitted must follow the selection rule and transition between adjacent levels, so

ΔE=EL+1EL=B(L+1)(L+2)BL(L+1)=2B(L+1)\Delta E=E_{L+1}-E_L=B\left(L+1\right)\left(L+2\right)-BL\left(L+1\right)=2B\left(L+1\right)

Here again we see that the spacing between energy levels depends on LL itself rather than being equally spaced as it is in vibrational transitions. Emitted photons can have energies 2B,4B,6B,...2B,4B,6B,....