Question

The book Molecular Quantum Mechanics by Atkins and Friedman [1] says the point group of a harmonic oscillator is Cs,

composed by the identity operator E
and a reflection σh
.

But I don't understand why the rotation of πrad
about the y axis is not a symmetry operation in this case. It looks like this operation also leaves the hamiltonian invariant, since V(x)=V(−x) for a harmonic oscillator. So, why can't the point group in this case be C2, composed by E and a rotation C2?

Answer 1

The rotation of πrad about the y-axis is not a symmetry operation in the point group of a harmonic oscillator because it does not leave the Hamiltonian invariant.

Step-by-step explanation:

The point group of a harmonic oscillator is Cs, composed of the identity operator E and a reflection σh. The reason why the rotation of πrad about the y-axis is not a symmetry operation in this case is because it does not leave the Hamiltonian invariant.

While the potential energy function, V(x), of a harmonic oscillator is symmetric with respect to x, it is not symmetric with respect to rotation around the y-axis. Therefore, the point group in this case cannot be C2.