Question

In Chem 465, we learn that the Hamiltonian of a rigid rotor is H=BJ2 where B is the rotational constant and J is the rotational angular momentum. The rotational energies are: E=BJ(J+1) 5 where J=0,1,2… is the rotational angular momentum quantum number. Consider a perturbation: 𝐻𝐻’ = 1 2 𝜆𝜆�𝐽𝐽̂+ + 𝐽𝐽̂−�. where λ is a constant. The operators 𝐽𝐽̂+ and 𝐽𝐽̂− have the following effects on the spherical harmonics 𝑌𝑌𝐽𝐽 𝑚𝑚𝐽𝐽 : 𝐽𝐽̂−𝑌𝑌𝐽𝐽 𝑚𝑚𝐽𝐽 = 𝐽𝐽𝑌𝑌𝐽𝐽−1 𝑚𝑚𝐽𝐽−1 , 𝐽𝐽̂−𝑌𝑌0 0 = 0 𝐽𝐽̂+𝑌𝑌𝐽𝐽 𝑚𝑚𝐽𝐽 = (𝐽𝐽 + 1)𝑌𝑌𝐽𝐽+1 𝑚𝑚𝐽𝐽+1 (1) [5 pts] Prove that the first-order perturbation vanishes based on symmetry arguments. (2) [10 pts] For the ground rotational state (J=0), compute the second-order contribution to its energy due to the perturbation by the J=1 state. (3) [3 pts] For the ground rotational state (J=0), prove that the second-order contribution to its energy due to the perturbation by the J=2 state vanishes

Answer 1

Selection rules only permit transitions between consecutive rotational levels: ΔJ=J±1, and require the molecule to contain a permanent dipole moment. Due to the dipole requirement, molecules such as HF and HCl have pure rotational spectra and molecules such as H2 and N2 are rotationally inactive.