The rigid rotator

Problem:

Two spinless particles of mass m1 and m2 are separated by a fixed distance r.  Their center of mass is fixed at the origin of the coordinate system and they are free to rotate about their center of mass.  (The system is a "rigid rotator".)
(a)  Find the eigenvalues and eigenfunctions of the Hamiltonian.  What is the degeneracy of the eigenvalues?
(b)  Assume the system has a magnetic moment μ = γL and is placed in a uniform magnetic field B = Bk.  What are the eigenvalues of H now?  What is the degeneracy of the eigenvalues?

Solution:

• Concepts:
The eigenfunctions of the orbital angular momentum operator.
• Reasoning:
We can write the Hamiltonian for the motion about the center of mass as that of a fictitious particle of reduced mass m' moving in a central potential.  We solve the Schroedinger equation in spherical coordinates.
• Details of the calculation:
(a)  The Hamiltonian of the rotator is L2/(2I) where I is its moment of inertia,
I = m1r12 + m2r22 = m'r2,  m' = m1m2/(m1 + m2).
The eigenfunctions of H are the eigenfunctions of L2.
These are the spherical harmonics, Ylm(θ,φ) = <r/r|l,m>.
The eigenvalues are El = l(l + 1)ħ2/(2I).  Each energy eigenvalue is 2l + 1 fold degenerate.
(b)  The Hamiltonian is H = L2/(2I) - μ∙B = L2/(2I) – γBLz.
The eigenvalues are El = l(l + 1)ħ2/(2I) – γBmħ, where m takes on values from –l to +l in integer steps.  The degeneracy of the eigenvalues is removed.

Problem:

Two particles of mass m1 and m2 are separated by a fixed distance r.  Their center of mass is fixed at the origin of the coordinate system and they are free to rotate about their center of mass.  (The system is a "rigid rotator".)
(a)  Write down the Hamiltonian of the system.
(b)  Find the eigenvalues and eigenfunctions of this Hamiltonian.  What is the separation between adjacent levels?  What is the degeneracy of the eigenvalues?

Solution:

• Concepts:
The time-independent Schroedinger equation in three dimensions, the eigenfunctions of the orbital angular momentum operator.
• Reasoning:
We can write the Hamiltonian for the motion about the center of mass as that of a fictitious particle of reduced mass m moving in a central potential.  We solve the Schroedinger equation in spherical coordinates.
• Details of the calculation:
(a)  The Hamiltonian of the rotator is L2/(2I) where I is its moment of inertia,
I = m1r12 + m2r22 = μr2,  μ = m1m2/(m1 + m2).
(b) The eigenfunctions of H are the eigenfunctions of L2.
These are the spherical harmonics, Ylm(θ,φ) = <r|l,m>.
The eigenvalues are El = l(l + 1)ħ2/(2I).
It is customary to set  B = ħ/(4πI), and write El = Bhl(l + 1).
The separation between adjacent levels is  El - El-1 = Bh[l(l + 1) - (l - 1)l] = 2Bhl.
The separation increases linear with l.  Each energy eigenvalue is 2l + 1 fold degenerate.

Problem:

Two atoms of masses m1 and m2 are bound together in a diatomic molecule.  The separation of their nuclei is r.  What are the rotational kinetic energy levels of the molecule?  How does the energy of the first excited state of 13C16O compare to that of 12C16O?

Solution:

• Concepts:
The rigid rotator
• Reasoning:
We model the system as a rigid rotator.
• Details of the calculation:
Rotation about the CM:  Eψ = L2ψ/2I.  I = m1r12 + m2 r22.  r1 + r2 = r.  m1r1 = m2 r2.
r1 = m2r/(m1 + m2).  r2 = m1r/(m1 + m2).  I = r2m1m2/(m1 + m2) = μr2.
μ = m1m2/(m1 + m2)  = reduced mass.
L2ψ = l(l + 1)ħ2ψ,  l = 0, 1, 2, … .
E = l(l + 1)ħ2(m1 + m2)/(2m1m2r2).

Ground state:                E0 = 0.
First excited state:        E1 = ħ2(m1 + m2)/(m1m2r2).

13C16O:  E1 = (29/208) ħ2/r2 = 0.139 ħ2/r2.
12C16O:  E1 = (28/192) ħ2/r2 = 0.146 ħ2/r2.