Usually vibrations (harmonic), derived from harmonic approximations are bigger than for experimental values. The question is, how does this depend on the size of the basis set. Many papers give that as a statement in their introduction without any explanation.

My idea would be that, if I use variational methods like DFT, then my energy will always be bigger than the real energy, as long as I use a finite basis set. And also increasing the set gives better results in DFT. The harmonic frequencies I determine computationally are derived from the Hessian matrix (or better the force constants for the harmonic oscillator) and this matrix is the second derivative of the energy, which is much too big for small basis sets.

So can I say that, because a smaller basis set will give me a bigger energy, the derived frequency will have to be bigger as well?

  • 1
    $\begingroup$ I'm curious: Why do you think that there's a correlation between the second derivative and the absolute value of a minimum? $\endgroup$
    – DSVA
    Jul 15, 2017 at 0:20
  • $\begingroup$ Thinking about it like that...well yeah you are right. But then I still don't understand why there is a dependency on the basis set size. $\endgroup$ Jul 15, 2017 at 18:57

2 Answers 2


No, you cannot say that larger energies will necessarily give larger frequencies.

Consider, for instance, a hypothetical (and purely fictional) method of determining the electronic energy of a system for a given geometry. Call this set of nuclear coordinates (geometry) $\mathbf{G}$, and our hypothetical method is a function which acts on these coordinates to produce an energy, i.e. $$f(\mathbf{G})=E$$

This method, however, is a very special kind of method. It is one which gives an absolutely unreliable energy. It gives you a number that has absolutely no basis in reality and can't be trusted at all. But, one thing we know about this method is that the error, which is truly horrendous and unpredictable, is constant! From this fact, it follows that this method predicts the correct frequencies, the experimental ones. That is, at the minimum of the potential predicted by our hypothetical method and the actual minimum, the difference between our energy, $E_0$, and the true energy, $\mathcal{E_0}$, is, $$E_0-\mathcal{E_0}=\epsilon$$ or, $$f(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon$$ For all real methods, however, we would need to write the above equation as, $$h(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon({\mathbf{G}})$$

The point of all this being that the hypothetical method I described was not variational or anything else you might desire out of a quantum chemical method. It only predicted energies for which the error term was constant. Whereas the real method, $h(\mathbf{G})$, predicts energies for which the error term is a function of the geometry $\mathbf{G}$.

The concise way of saying this is that frequencies depend only on the curvature of the potential well and not on the depth of the potential well.

With that established, we have answered the question that smaller basis sets, with their larger energies, do not necessarily give higher frequencies.

For instance, if you look at Sinha and Wilson's paper,[1] you will learn of something very empirical which computational chemists do sometimes that bothers me a lot but is illustrative for our purposes. In that paper, a big set of 42 organic molecules is taken, and the harmonic frequencies are calculated at a bunch of different methods and basis sets, then these frequencies are compared with the experimental values. A least squares fit is then performed to come up with scaling values to map the harmonic frequency onto the experimental frequency. As you say, these constants are numbers less than one, say 0.95. The reason these constants are less than one is what bothers me about doing this. Experimental values are anharmonic frequencies. We don't live in a harmonic world. Anharmonicity shifts the frequency down, by lowering one of the potential walls to accommodate bond-breaking, hence we need to scale our harmonic frequencies downwards. So we're basically comparing apples with oranges and there's no way to be sure that this will work in all cases.

More to your question, in this paper, the correction factors do not increase monotonically with basis set size as you suggested (this is the same as frequencies decreasing with basis set size). I won't reproduce their numbers here, because I don't know if that's frowned upon, but for both MP2 and B3LYP, the cc-pVnZ and aug-cc-pVnZ (n=D,T,Q) basis sets have T < D < Q for the scaling factor. So, the triple-zeta basis set invariably predicts a lower energy, but predicts a higher frequency.

Even more interestingly, if one only considers low-frequency modes, which are particularly poorly described by normal modes (these are torsions usually), the ordering becomes Q < T < D for the scaling factors. That's really a failure of the harmonic model more than anything, but it illustrates the point well that lowering the energy really doesn't guarantee you anything in terms of frequencies.

To address your comment, the reason there is any dependence on basis set size is because the error away from the equilibrium structure will change with basis set size, and hence the curvature of the potential changes which affects the frequency.

[1] Pankaj Sinha, Angela K. Wilson, J. Phys. Chem. A, 2004, 108 (42), 9213-9217. DOI: 10.1021/jp048233q

  • 3
    $\begingroup$ As a comment on your anharmonicity concerns: One quite often observes that anharmonic frequency corrections (eg from VSCF) lower the high and increase the low frequencies. I am not entirely sure how that happens exactly but as a matter of fact, that type of mismatch is also what you observe when you compare scaled computed harmonic frequencies with exptl. ones. $\endgroup$ Jul 21, 2017 at 6:42
  • 2
    $\begingroup$ @R_Berger this makes sense though because if the low frequency is a torsion or an umbrella mode, the largest correction will come from the coefficient on the x^4 term of a Taylor expansion. Whereas the high frequencies are Morse-like which has a first-order anharmonicity that decreases the frequency. Very good point though. I hope I didn't imply anything to the contrary. $\endgroup$
    – jheindel
    Jul 21, 2017 at 15:12
  • $\begingroup$ Not as far as I can see. $\endgroup$ Jul 22, 2017 at 20:35

smaller basis set would give a bigger energy that is sum of BO electron energy,but the frequency is derived from this energy

  • $\begingroup$ This has the makings of a decent answer, but could you expand on it a bit, please? $\endgroup$
    – jonsca
    Aug 10, 2017 at 2:02

Your Answer

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

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