Two spinless particles of mass m_{1} and m_{2} 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** = γ**L** 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 L^{2}/(2I) where I is its moment of inertia,

I = m_{1}r_{1}^{2}+ m_{2}r_{2}^{2}= m'r^{2}, m' = m_{1}m_{2}/(m_{1}+ m_{2}).

The eigenfunctions of H are the eigenfunctions of L^{2}.

These are the spherical harmonics, Y_{lm}(θ,φ) = <**r**/r|l,m>.

The eigenvalues are E_{l}= l(l + 1)ħ^{2}/(2I). Each energy eigenvalue is 2l + 1 fold degenerate.

(b) The Hamiltonian is H = L^{2}/(2I) -**μ∙B**= L^{2}/(2I) – γBL_{z}.

The eigenvalues are E_{l}= l(l + 1)ħ^{2}/(2I) – γBmħ, where m takes on values from –l to +l in integer steps. The degeneracy of the eigenvalues is removed.

Two particles of mass m_{1} and m_{2} 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 L^{2}/(2I) where I is its moment of inertia,

I = m_{1}r_{1}^{2}+ m_{2}r_{2}^{2}= μr^{2}, μ = m_{1}m_{2}/(m_{1}+ m_{2}).

(b) The eigenfunctions of H are the eigenfunctions of L^{2}.

These are the spherical harmonics, Y_{lm}(θ,φ) = <**r**|l,m>.

The eigenvalues are E_{l}= l(l + 1)ħ^{2}/(2I).

It is customary to set B = ħ/(4πI), and write E_{l}= Bhl(l + 1).

The separation between adjacent levels is E_{l}- E_{l-1}= Bh[l(l + 1) - (l - 1)l] = 2Bhl.

The separation increases linear with l. Each energy eigenvalue is 2l + 1 fold degenerate.

Two atoms of masses m_{1} and m_{2} 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 ^{13}C^{16}O compare to
that of ^{12}C^{16}O?

Solution:

- Concepts:

The rigid rotator - Reasoning:

We model the system as a rigid rotator. - Details of the calculation:

Rotation about the CM: Eψ = L^{2}ψ/2I. I = m_{1}r_{1}^{2}+ m_{2}r_{2}^{2}. r_{1}+ r_{2}= r. m_{1}r_{1}= m_{2}r_{2}.

r_{1}= m_{2}r/(m_{1}+ m_{2}). r_{2}= m_{1}r/(m_{1}+ m_{2}). I = r^{2}m_{1}m_{2}/(m_{1}+ m_{2}) = μr^{2}.

μ = m_{1}m_{2}/(m_{1}+ m_{2}) = reduced mass.

L^{2}ψ = l(l + 1)ħ^{2}ψ, l = 0, 1, 2, … .

E = l(l + 1)ħ^{2}(m_{1}+ m_{2})/(2m_{1}m_{2}r^{2}).Ground state: E

_{0}= 0.

First excited state: E_{1}= ħ^{2}(m_{1}+ m_{2})/(m_{1}m_{2}r^{2}).^{13}C^{16}O: E_{1}= (29/208) ħ^{2}/r^{2}= 0.139 ħ^{2}/r^{2}.

^{12}C^{16}O: E_{1}= (28/192) ħ^{2}/r^{2}= 0.146 ħ^{2}/r^{2}.