For a given molecule, we can determine the normal modes of vibration and the corresponding vibrational frequencies by calculating the molecular Hessian of second-order energy derivatives, and finding the eigenvectors/eigenvalues. However, I've also seen approaches to finding the normal modes which rely only on the molecule's equilibrium geometry, and some group-theoretic tools.

In general, is it possible to precisely determine a molecules normal modes using only its equilibrium geometry, or does this type of analysis only give a qualitative understanding? Is it "easier" to find the normal modes of vibration, as opposed to finding the normal modes and the corresponding frequencies through diagonalizing the Hessian?

  • 1
    $\begingroup$ I think you are right - the group theoretic approach is only qualitative, and allows you to separate the normal modes by different symmetry. Also, if your molecule has no symmetry elements (e.g. C1 point group), then I don't think group theory would be very useful. $\endgroup$
    – S R Maiti
    Aug 2, 2022 at 11:24

1 Answer 1


This is an interesting question. The Hessian diagonalization one is general, however. Group theory is helpful when dealing with small symmetric molecules.

Although we have monograph for this topic, I can provide a concrete example.

Say the molecule H2O. We know that it has three vibrational modes, one for bending, and one for symmetric stretching and another for antisymmetric. From the point group point of view, the former two are symmetric ($A_1$), and the last one is antisymmetric ($B_2$).

If we use the group theory trick, which should work if, according to the orthogonality of normal modes, only one mode belongs to an irreducible representation, otherwise modes in the same irreducible rep will mix with each other, which cannot be separated with only group theory. We also know that Hessian relies on the underlying quantum chemistry (QC) computation.

So if we compute the frequencies of H2O, but with different QC method/basis set, we should have the displacements in the normal mode of antisymmetric stretching remain exactly the same. However, because the symmetric stretching and bending are both of $A_1$, the displacement vectors of those modes will change with the choice of the QC method/basis set.

Note that when doing this test, we should NOT perform the geometry optimization, which ensure that the "equilibrium" geometry is the same (although not, optimized or not will only raise a warning in most QC programs).

Here are the results from Molpro. HF/6-31g:

   Normal Modes

                                1 A1        2 A1        3 B2
   Wavenumbers [cm-1]         1826.82     3645.93     3770.62
   Intensities [km/mol]        104.40        1.26       33.92
   Intensities [relative]      100.00        1.21       32.49
             OX1             -0.00000     0.00000    -0.00000
             OY1             -0.00000    -0.00000    -0.06913
             OZ1             -0.07127    -0.04375    -0.00000
             HX2              0.00000     0.00000     0.00000
             HY2             -0.36845     0.60026     0.54867
             HZ2              0.56568     0.34722     0.39634
             HX3             -0.00000    -0.00000     0.00000
             HY3              0.36845    -0.60026     0.54867
             HZ3              0.56568     0.34722    -0.39634


   Normal Modes

                                1 A1        2 A1        3 B2
   Wavenumbers [cm-1]         1834.52     3613.18     3733.07
   Intensities [km/mol]        108.14        0.08       26.20
   Intensities [relative]      100.00        0.07       24.23
             OX1              0.00000    -0.00000     0.00000
             OY1             -0.00000    -0.00000    -0.06913
             OZ1             -0.07144    -0.04347     0.00000
             HX2              0.00000    -0.00000    -0.00000
             HY2             -0.36610     0.60169     0.54867
             HZ2              0.56703     0.34501     0.39634
             HX3             -0.00000     0.00000    -0.00000
             HY3              0.36610    -0.60169     0.54867
             HZ3              0.56703     0.34501    -0.39634

You will find our theory is correct.


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