In the output of a Gaussian calculation of frequencies, I can see a table listing all the 'forces' of all the atoms. It looks like this:

Center Number Atomic Number Forces (Hartrees/Bohr) X Y Z

  1        6          -0.001094502   -0.000768130   -0.006919164
  2        6           0.000528733   -0.000637636    0.005949765
  3        6           0.000424174   -0.000366750   -0.002904382
  4        6          -0.002619381   -0.000001885    0.005496979


What are these 'forces'? Are they related to those vibrations? But during vibrations, the force on each atom by other atoms is always changing, so why is only one constant 'force' value assigned to each atom?

  • $\begingroup$ I am not sure, but those might be related to the Hessian matrix, the partial second derivatives, which is diagonalized in a frequency calculation. Note that $F=m\frac{\mathrm{d}^2}{\mathrm{d}x^2}$ in 1D. $\endgroup$
    – Feodoran
    Jan 21, 2018 at 12:43
  • $\begingroup$ Did you optimize the geometry before running the frequency calculation? Frequency calculations don't have any meaning if they aren't taken at a stationary point of the potential energy surface. $\endgroup$
    – Tyberius
    Jan 21, 2018 at 16:59
  • $\begingroup$ @Feodoran But the forces are not calculated from that formula, right? The forces are calculated from the electronic structures of the atoms, regardless of the masses of the atoms, right? $\endgroup$
    – OhLook
    Jan 21, 2018 at 19:36
  • $\begingroup$ I always took them as force constant (spring on my mind) $\endgroup$
    – Alchimista
    Jan 21, 2018 at 20:09
  • $\begingroup$ @Alchimista But they're not. Look at their unit - Hartree/Bohr. Obviously they are just forces. $\endgroup$
    – OhLook
    Jan 21, 2018 at 21:26

1 Answer 1


The forces you have listed are the derivatives of the energy with respect to a displacement of a particular atom in each Cartesian direction. These forces are useful for finding stationary points on the potential energy surface, which occur when the force on each atom is zero. Computationally, it's not possible or worthwhile to try to get all the forces to be exactly zero, so stationary points are determined by a combination of criteria involving the forces and the displacement between optimization steps.

The vibrations in the system are related to the second derivatives of the energy with respect to the atomic coordinates, which are contained in the Hessian matrix.

  • $\begingroup$ Thank you for your helpful answer, but I still have a question: these forces are calculated based on the atoms' electronic structures and the distances between atoms, and these calculations don't involve masses of atoms, correct? $\endgroup$
    – OhLook
    Jan 21, 2018 at 21:30
  • $\begingroup$ @Ath they shouldn't depend on the masses explicitly. See this other question that references how to obtain forces from the Hartee Fock equation or the referenced source (Szabo and Ostlund's Modern Quantum Chemistry) for the full formulation. $\endgroup$
    – Tyberius
    Jan 21, 2018 at 22:00
  • $\begingroup$ You said '...stationary points on the potential energy surface, which occur when the force on each atom is zero', but I don't think the stationary points means forces are zero. A stationary point is a minimum energy in an area, which means it has smaller forces than other points in this area, but not necessarily zero $\endgroup$
    – OhLook
    Feb 9, 2018 at 22:13
  • $\begingroup$ @Uh-Oh by definition, a stationary point of a function is one for which the derivative (in this case, the force) is zero. Now true, in implementations we never manage to completely zero out all the forces so we set some minimum RMS force criteria, etc, but that is more an issue of our implementation and approximations than with how you would find a true stationary point. $\endgroup$
    – Tyberius
    Feb 9, 2018 at 23:20
  • 2
    $\begingroup$ @Tyberius conceivably those forces could also be used to run AIMD, not just geometry optimizations. Also, Uh-Oh , the force constant of a harmonic vibrational mode should not be confused with forces. $\endgroup$ Feb 12, 2018 at 20:06

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.