# Keeping the electronic configuration in the ground state is expensive in Ab-initio MD?

In E. Sandré & A. Pasturel's An Introduction to Ab-Initio Molecular Dynamics Schemes

If using small time steps (typically in the order of a few tenth of femto-second), the changes in the electronic configuration is small enough so that convergence of the electronic configuration is easily achieved using a few conjugate gradient steps. However, the computational cost of dynamical simulation appears to be mostly due to keeping the electronic configuration in the ground state.

I don't understand this. DFT and HF are methods that can only give electron ground state. Since it is either DFT or HF that is used in Ab-initio MD simulations, then what does this excited state is referring to?

My understanding:

First of all, the number of electron orbitals calculated by either DFT or HF approach equals the number of electrons in the system, so we get no electronic excited state.

Under Born-Oppenheimer approximation, while varying the atomic configuration, the electrons are expected to stay in their ground state. Since the validity of BO approximation dictates no abrupt change in electronic energy.

So, the reason of 'Keeping the electronic configuration in the ground state is expensive' is that when the atomic configuration changes in MD simulation, it is hard to make the electronic energy difference small.

Is my understanding correct?

• Thank you! I still have two confusions: 1. why H-F theorem is only valid for the electronic ground state? From its derivation, any eigenstate of $\hat H$ is okay and excited states are also eigenstates of $\hat H$. 2. what does it really mean that electrons are unlikely in their ground state during simulation? Since DFT and HF only give us electron ground state. – meTchaikovsky Oct 3 '18 at 10:56
• 1- I think it is because you use variational principle which assumes that you find the energy of the ground state. 2- For a configuration of nuclei you find the ground state electronic structure but when you move the nuclei there is no guarantee that the electrons will be in the ground state. Imagine that you calculate bond breaking for $H_2$ molecule. As you move H atoms far from each other at some point you should "leave" the ground state of an $H_2$ molecule. Yes, DFT will always result a ground state but the it will be wrong. – physicopath Oct 3 '18 at 11:02