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I am not talking about zero-point energy.

I wish to understand how the energy of a system is calculated relative to a zero baseline. This is best explained with an example.

From DFT, the energy could be:

Isolated H = -13.6 eV,
H2 = -31.8 eV

What are these numbers related to? What would be a system at 0 eV? Or what is the zero baseline from which these number are calculated from?

I have also noticed that force-field energies tend to be positive. What is the zero baseline here?

In summary:

  • What is the DFT baseline?
  • What is the Forcefield baseline?
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Start by looking at the Hamiltonian for a molecular system \begin{equation} \label{eq:coulomb_hamiltonian} \hat{H} = - \sum\limits_{α=1}^{ν} \frac{1}{2 m_{α}} \nabla_{α}^{2} - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2} - \sum\limits_{α=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{α}}{r_{αi}} + \sum\limits_{α=1}^{ν} \sum\limits_{β > α} \frac{Z_{α} Z_{β}}{r_{αβ}} + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, , \end{equation} where $m_{α}$ is the rest mass of nucleus $α$ and $Z_{α}$ is its atomic number, $r_{αi} = |\vec{r}_{α} - \vec{r}_{i}|$ is the distance between nucleus $α$ and electron $i$, $r_{αβ} = |\vec{r}_{α} - \vec{r}_{β}|$ is the distance between two nuclei $α$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.

Now, thinking classically, the total energy is equal to zero when all the terms in the Hamiltonian are equal to zero, so that the zero of the total energy corresponds to the case when, (again) classically speaking, particles are infinitely far away from each other and not moving. Obviously, the total energy of any stable molecular system should be lower than that, i.e. negative, for otherwise the system will be unstable with respect to dissociation towards particles. That is the way it works for any method of computational chemistry that solves the Schrödinger equation or its relativistic counterpart is some way or another (HF, post-HF, DFF, etc.).

But force field methods and energies are completely different beasts, because the meaning and the zero of energy in this realm depends on the functional form of potential energy and parameter set used in a particular force field. So, in molecular mechanics in general absolute energies have no meaning, only relative energies (energy differences) should be used for any kind of analysis.

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  • $\begingroup$ Shouldn't an isolated H be about -13.6 eV relative to the zero state? $\endgroup$ Commented Jan 14, 2020 at 15:32
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    $\begingroup$ @IvanNeretin, yes, it should be close to -0.5 Hartree in calculations. And I have no idea why OP's number is so far away from that value. $\endgroup$
    – Wildcat
    Commented Jan 14, 2020 at 15:47
  • $\begingroup$ @Wildcat. I apologise for the original numbers. These were chosen at random as I had no computational software, I have now updated to reflect computational values. $\endgroup$
    – Wychh
    Commented Jan 14, 2020 at 16:18

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