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.

 

[Entropia]

The reason for the lack of commutativity between the Hamiltonian and L^2 in linear molecules is due to the loss of cylindrical symmetry. In atoms, the cylindrical symmetry created by the nuclei's electric field allows for the conservation of the total orbital angular momentum of the electrons, represented by the quantum number lambda. However, in linear molecules, this symmetry is destroyed by the strong electric field of the nuclei along the internuclear axis. As a result, the total orbital angular momentum is no longer conserved, and the operators for the Hamiltonian and L^2 no longer commute. This means that the two quantities cannot be measured simultaneously with certainty, and there is some uncertainty in their values. This break in commutativity is a consequence of the loss of cylindrical symmetry in linear molecules and is an important factor to consider in studying their quantum properties.