Point group of harmonic oscillator

Question

The book Molecular Quantum Mechanics by Atkins and Friedman [1] says the point group of a harmonic oscillator is $C_\mathrm{s},$ composed by the identity operator $E$ and a reflection $\sigma_\mathrm{h}$.

But I don't understand why the rotation of $\pu{{\pi} rad}$ 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 $C_2,$ composed by $E$ and a rotation $C_2?$

Reference

1. Atkins, P. W.; Friedman, R. Molecular Quantum Mechanics, 5th ed.; Oxford University Press: Oxford ; New York, 2011. ISBN 978-0-19-954142-3.

Answer 1

Maybe, go one step further and consider what the difference between $C_\mathrm{s}$ and $C_2$ really is?

Here are the character tables for both point groups:

$$\begin{array}{c|cc|cc} \hline C_\mathrm{s} (= C_\mathrm{h}) & E & \sigma_\mathrm{h} & & \\ \hline \mathrm{A'} & 1 & 1 & x,y, R_z & x^2, y^2, z^2, xy \\ \mathrm{A''} & 1 & -1 & z, R_x, R_y & xz, yz \\ \hline \end{array}$$

$$\begin{array}{c|cc|cc}\hline C_2 & E & C_2 & & \\ \hline \mathrm{A} & 1 & 1 & z, R_z & x^2, y^2, z^2, xy \\ \mathrm{B} & 1 & -1 & x, y, R_x, R_y & xz, yz \\ \hline \end{array}$$

The terms to the right describe rotations and translations in 3D molecules, and the harmonic oscillator isn't a 3D molecule, so these are irrelevant. Once we strip that away, we're left with the conclusion that except for the choice of labels, $C_\mathrm{s}$ and $C_2$ are really equivalent. In group theory terminology they are isomorphic.

• There are two symmetry operations, the identity operation $(E)$ and another one $(C_2$ or $\sigma_\mathrm{h})$
• There is one irreducible representation which is totally symmetric $(\mathrm{A'}$ or $\mathrm{A})$
• There is one irreducible representation which is asymmetric with respect to the non-identity operation $(\mathrm{A''}$ or $\mathrm{B})$

Sure, the labels are not the same, but we can only draw a clear distinction between the labels in 3D anyway. In the case of the harmonic oscillator there is only one non-trivial symmetry element and that doesn't change whether we call it $\sigma_\mathrm{h}$ or $C_2$. (This echoes the point in Hans Wurst's comment.)

So, it's entirely up to you whether you want to call the point group $C_\mathrm{s}$ or $C_2$. The maths works out the same in the end — as it has to.

Answer 2

As @Hans Wurst already mentioned, the authors are referring to 1D simple harmonic oscillator and the symmetry of the potential and wavefunction.

As a chemist the first think that comes to our mind when we think of a 1D harmonic oscillator is a diatomic molecule $(C_{\infty\mathrm v}$ or $D_{\infty\mathrm h}$. But it can also describe an atom adsorbed to a (flat) surface $(C_\mathrm{s}$ symmetry?).

However, I think the authors are referring to the symmetry of the harmonic oscillators potential and wavefunctions irrespective of the system to which they are applied. In this sense, you only can define two elements of symmetry, the identity and the symmetry around the origin, a mirror plane, thus you could say the oscillator belongs to the $C_\mathrm{s}$ symmetry.

This is what I understand from the book [1, p. 158]:

5.17 Symmetry and degeneracy

We have already mentioned (in Section 2.13) that the presence of degeneracy is a consequence of the symmetry of a system. We are now in a position to discuss this relation. To do so, we note that the hamiltonian of a system must be invariant under every operation of the relevant point group:

$$(RH) = H \tag{5.40}$$

A qualitative interpretation of eqn 5.40 is that the hamiltonian is the operator for the energy, and energy does not change under a symmetry operation. An example is the hamiltonian for the harmonic oscillator: the kinetic energy operator is proportional to $\mathrm d^2/\mathrm dx^2$ and the potential energy operator is proportional to $x^2$. Both terms are invariant under the replacement of $x$ by $−x,$ and so the hamiltonian spans the totally symmetric irreducible representation of the point group $C_\mathrm{s}$. Because $H$ is invariant under a similarity transformation of the group (that is, any symmetry operation leaves it unchanged), we can write

$$RHR^{−1} = H$$

Multiplication from the right by $R$ gives $RH = HR$, so we can conclude that symmetry operations must commute with the hamiltonian.

Reference

1. Atkins, P. W.; Friedman, R. Molecular Quantum Mechanics, 5th ed.; Oxford University Press: Oxford ; New York, 2011. ISBN 978-0-19-954142-3.