D infinity h symmetry
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Hi Jonas,
CRYSTAL does not implement point groups with infinite-order rotations, so \(D_{\infty h}\) cannot be used directly. For linear molecules like CO₂, the practical approach is to approximate the symmetry using a finite-order rotation group.
In this case, \(D_{4h}\) (i.e., 24) is a good option in CRYSTAL. It preserves the key degeneracies of linear molecules, including the doubly degenerate bending modes of CO₂. Using a lower-symmetry group like \(D_{2h}\) would artificially lift these degeneracies.
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If you fear that a reduced symmetry might influence the thermodynamic results, you can simply run the calculation without imposing any symmetry at all. For a 3-atom system like CO₂ the computational cost is so minimal that you can run both symmetry-free and \(D_{4h}\) calculations. Comparing the two results should give you an indication of whether the imposed symmetry has any impact on the quantities you are interested in.
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