Take, for example, $\ce{CO2}$. It is pretty clear that in the absence of rotation, the energy of this system depends only on the two bond lengths and the bond angle.

So if I wanted to build a potential energy surface for this molecule that was valid near and far from equilibrium, it should depend on these coordinates. But how do I rectify this with the fact that $\ce{CO2}$ has four vibrational modes? As we learn in undergraduate chemistry, it has two stretches and two degenerate bends.

If I build a Hamiltonian in the three valence coordinates, could it possibly describe the ground state energy correctly? That is, would it contain the zero-point energy for all four modes? Would it be able to find the right low-lying excited states of the bending mode(s)?

EDIT: Wildcat has brought up the point of symmetry and point groups below, and perhaps that can help get to the heart of my question. When we say that $\ce{CO2}$ is a linear molecule, that means that the lowest energy configuration has a bond angle of $\pi$. But in no way do I want to restrict the symmetry of the molecule. I want to work with a potential surface that is valid for symmetric and asymmetric geometries.

But here is where I get confused. Let's say I did restrict it to be linear. I would then have a two-dimensional potential, a function of the two bond lengths. Of the seven normal modes, I only keep two: the symmetric stretch and anti-symmetric stretch. The ground state energy would be the sum of the zero-point energy of these two modes. So therefore the number of modes is equal to the number of coordinates I used to write my Hamiltonian.

If I further constrict the symmetry, and impose the $D_{\infty h}$ point group, then the potential is now a function of one parameter. I will now only have one vibrational mode, the symmetric stretch, and the ground state energy is the zero-point energy of this mode. So, again, the number of modes is equal to the number of coordinates I use to write my Hamiltonian.

But what if the symmetry were not restricted in any way? I can write the potential as a function three internal coordinates, be they bond lengths and angle, othogonormal rectilinear coordinates, jacobi coordinates, etc. This potential, and the Hamiltonian I construct with it, should find the same modes described above, as well as the bending modes. But will it find both bending modes? I don't see why it should. Based on the two reduced-dimensional examples above, I would expect the number of modes to equal the number of coordinates. I would expect therefore, not to find the right ground-state energy, etc.

Specifically, I am concerned about the accuracy of dynamical simulations performed using this three-dimensional potential.

Likewise, what about a tetratomic liner molecule like $\ce{C2H2}$? I would use three bond lengths, two bond angles, and one dihedral angle, for a total of 6 coordinates - but it has 7 vibrational modes. Would I model the dynamics properly?

  • $\begingroup$ So, yes, you'll get the "correct" vibrational eigenstates. In quotes because the very separation of nuclear motion into vibrational and rotational is approximate, and thus, if you include only vibrational contribution (ZPE) into the total energy for both linear and non-linear configurations, that would be "not quite right" since you have different number of vibrational degrees of freedom in these cases. Better include corrections from all three types of nuclear motion: translational, rotational, and vibrational. $\endgroup$
    – Wildcat
    Jun 24, 2015 at 10:00
  • $\begingroup$ No, from you last edit, I see that you do not understand the matter, so I repeat one more time: the number of vibrational degrees of freedom is always either $3N-5$ for a linear configuration or $3N-6$ for a non-linear one, while the number of structural parameters is always $3N-6$. So, for $\ce{CO2}$ molecule you always have $3*3-6=3$ structural parameters and either $3*3-5=4$ or $3*3-6=3$ vibrational modes (for a linear configuration and a non-linear configuration respectively). $\endgroup$
    – Wildcat
    Jun 24, 2015 at 10:10

3 Answers 3


The Cartesian Hessian ($\mathbf{H}_C$) can be obtained by from the Hessian computed in internal coordinates ($\mathbf{H}_i$) through a non-linear transformation $$\mathbf{H}_C = \mathbf{B}^T\mathbf{H}_i\mathbf{B}$$ where $B_{ij}=\frac{\partial R_i}{x_j}$ and $R_i$ and $x_i$ is an internal and Cartesian coordinate, respectively.

In the case of the degenerate bending mode you can obtain the components of $\mathbf{H}_C$ that you need from $\frac{\partial^2 E}{\partial \theta^2}$, where $\theta$ is the bond angle, through suitable choices of $B_{ij}$'s.

So the Hessian generated in the three internal coordinates has all the information you need.


It is pretty clear that in the absence of rotation, the energy of this system depends only on the two bond lengths and the bond angle.

Yes, that is true, if you do not impose any restriction on geometry (such as, for instance, the linearity), the $\ce{CO2}$ molecule will indeed have $3N-6=3*3-6=3$ internal degrees of freedom and the same number of independent structural parameters (as which you can naturally choose two bond lengths and the angle between them). However, the point is that when you start to impose some restrictions on the geometry (symmetry), the number of internal degrees of freedom can increase by one (if you require your molecule to be linear), while the number of independent structural parameters will stay the same (though, few of them will be fixed and not varying anymore).

As a result the following three quantities which you mixed up are usually different:

  • the number of internal degrees of freedom: $3N-6$ (for a non-linear molecule) or $3N-5$ (for a linear molecule);
  • the number of independent structural parameters: always $3N-6$;
  • the number of varying (i.e. not-fixed) independent structural parameters: depends on the point group.

Thus, for instance, if you require the $\ce{CO2}$ molecule to be linear the number of internal degrees of freedom will be $3N-5=3*3-5=4$. It does not mean though that you need 4 independent structural parameters to describe it, 3 is still enough. And besides, by requiring the $\ce{CO2}$ to be linear you make it to belong to $\mathrm{C_{\infty v}}$ point group which fixes one of the parameters, the angle between bonds. So, while you have 4 internal degrees of freedom, you still need just 3 structural parameters to describe the molecule one of which is fixed in addition to that. So, if you want to scan PES, you have to vary just two of them: two bond lengths.

You can decrease the number of varying structural parameters even further: to just 1, namely, just the bond length, by making an additional (but quite reasonable) assumption that two $\ce{C-O}$ bonds in the $\ce{CO2}$ molecule have the same length, i.e. that the $\ce{CO2}$ molecule belongs to $\mathrm{D_{\infty h}}$ point group. Note that the number of internal degrees of freedom and the number of independent structural parameters will still be 4 and 3 respectively.

  • $\begingroup$ Thanks for pointing that out. I've edited the question to clarify and remove symmetry from the discussion. How would you answer for carbonyl sulfide? $\endgroup$ Jun 24, 2015 at 8:28
  • $\begingroup$ @largeboulder, if you want to scan PES for the carbonyl sulfide molecule and you want the molecule to be linear, you will need to vary just 2 geometrical parameters (two bond lengths). But you'll still have $3N−5=4$ internal degrees of freedom and 3 geometrical parameters exactly as in the case when you require $\ce{CO2}$ molecule to be just linear. $\endgroup$
    – Wildcat
    Jun 24, 2015 at 9:13
  • $\begingroup$ it isn't that I am requiring the molecule to be linear - it is that the minimum of the potential is at a linear geometry. But what I want to do is construct a potential that is valid near and far from the equilibrium, thus disregarding all symmetry. So if I construct a potential as a function of 3 internal coordinates (bond lengths and angles, rectilinear, jacobi, etc.) and then I use this potential to find the eigenstates (both low-energy and high-energy), is that sufficient? $\endgroup$ Jun 24, 2015 at 9:20
  • $\begingroup$ @largeboulder, in principle, you can vary all three parameters for $\ce{CO2}$ molecule, but I'm pretty sure the global minimum on the resulting PES will be the same as on PES obtained by constraining $\ce{CO2}$ molecule to $\mathrm{D_{∞h}}$ point group and varying just the bond length. The two PESes will, of course, be different, but the global minimum should be the same. $\endgroup$
    – Wildcat
    Jun 24, 2015 at 9:26
  • $\begingroup$ @largeboulder, but I still could not figure out your last question. Do you want to know will there be difference in ZPE at this equilibrium geometry with and without taking the symmetry of molecule into account? No, there should not be any difference: same number of vibrational modes contributing the same amount of energy. I'm pretty sure about that. $\endgroup$
    – Wildcat
    Jun 24, 2015 at 9:33

You need to incorporate all $3N-5$ modes (both bends and both stretches) in order to accurately calculate the ZPE even if, per Jan's answer, you can neglect one of the bending modes in constructing an accurate model of the PES. Even though the bends appear identical in mind's eye, both of their (degenerate) ZPE's must be included in order for the ground state calculated by the model to be accurate.

The way I think about it is, imagine a $\ce{CO2}$ molecule is oriented along the $z$-axis, with the carbon at the origin. If you only allow one bending mode, say with both oxygens translating toward $+x$ and the carbon toward $-x$, you have to hold the orientation of that motion fixed and your system will be unable to accommodate vibrational movement of any atoms in the $y$-direction. Accurate evaluation of the ZPE requires a model that allows bending along both axes.

The maddening thing about vibrating linear molecules in general, though, is that most linear combinations of a certain pair of these degenerate bending motions actually contain a rotational component. Using $\ce{CO2}$ as an example and ignoring anharmonicity, if the bending vibrations are $90^\circ$ out of phase with one another, the oxygen atoms gyrate around the $\ce{O=C=O}$ axis just like they would in a rotating molecule with a bent equilibrium geometry (imagine the orange dot is $\ce{C}$ and the green dot is a projection of the positions of the two $\ce{O}$'s):

Unit circle animation


Of course, the radius of the circle of projection of the motion of the oxygens will be quite small, I believe less than a degree, but it still is a rotation about the molecular axis.

Phase angles other than $90^\circ$ result in some combination of a bending motion and a rotation, where only in the $\theta=0^\circ$ case does the rotational aspect vanish:


For any linear molecule with three or more atoms, the degenerate bends involving the three atoms at each terminus of the molecule are the pair that can generate rotation. For $\ce{CO2}$ that means the entire molecule; for acetylene each vibration involves the two carbons and one of the hydrogens; etc. EDIT: The more I think about it, the more confident I am that every pair of degenerate bends in a linear molecule actually has this capability to generate rotation. It's just that any bends not centered at the central atom (if it exists) will cause the molecule to precess, as well as rotate, due to the asymmetry of the two 'arms' of the bending motion.

So, to some extent, I would actually argue your question is ill-posed: Linear molecules are always rotating! They're just rotating in a manner inconsistent with the standard rigid-rotor assumption, and (I think?) at energies much higher than those one would expect to see in conventional rotational spectroscopy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.