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, Y_lm(θ,φ) = <r/r|l,m>.
  The eigenvalues are E_l = l(l + 1)ħ²/(2I).  Each energy eigenvalue is 2l + 1 fold degenerate.
  (b)  The Hamiltonian is H = L²/(2I) - μ·B = L²/(2I) - γBL_z.
  The eigenvalues are E_l = 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, Y_lm(θ,φ) = <r|l,m>.
  The eigenvalues are E_l = l(l + 1)ħ²/(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.

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.