Why Hamiltonian and L^2 Don't Commute in Linear Molecules

[filippo]

Lambda is the quantum number for the total orbital angular momentum of the
electrons about the internuclear axis. Unlike in atoms, the cylindrical
symmetry created by the strong electric field of the nuclei in a linear
molecule destroys the relationship [H ,L^2] = 0.

Can anyone tell me why the hamiltonian and L^2 don't commute anymore?

 

[tim_lou]

Moving to spherical coordinates, any Hamiltonian can be written as
$$H=-\frac{\hbar^2 p^2_r}{2m}+\frac{L^2}{2mr^2}+V(r, \theta, \phi)$$

p^2_r depends only on r. The L^2 terms contain derivatives in θ and φ. So, if V is a function of θ and φ, [L^2, V] will involve derivatives with respect to θ and φ. You can solve the differential equation [L^2, V]=0, and find that V can only depend on r.

So if there isn't spherical symmetry, H won't commute with L^2.