# 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?

• I believe they are saying that when you take a small time step, not much can change so it shouldn't take long to converge the energy. This makes sense. The confusing part is how one can go from that to saying that it is hard to keep the electronic configuration in the ground state. So, I believe what they mean is that it is possible to accidentally get onto an excited surface during the time step. This seems like it would be system-dependent. The point is you have to expend time to verify you are still on the same surface somehow in addition to verifying the orbitals are converged. – jheindel Oct 5 '18 at 4:12
• They refer to approximate ground state (ie nit exact but close), not a specific excited state. Ie in spite the SCF requires only few steps, it is still the most expensive part of the calculation if we want to system wave function to be calculated with SCF in every single step. A common spread up is to let somehow a looser convergence on the electronic wave function, and calculate the forces without calculating every single time a well converged wavefunction. – Greg Mar 10 '19 at 15:34

Almost, that's the key here. The gradient will become more of an approximation as you move away from the nuclear coordinates for which the underlying SCF solution was originally calculated (and merely recalculate the gradient at new coordinates using old MO coefficients). One way of dealing with this problem is to take the old SCF solution as a starting point for a new SCF solution, which should converge in just a few steps. But those few SCF iterations can easily dominate the cost, given that a gradient calculation after obtaining the SCF solution typically has a computational cost on the order of 1-3 SCF iterations.$$^{1}$$
$$^{1}$$ A specialized quantum chemistry program, that would keep certain integrals and other intermediates across the different SCF procedures, would help somewhat with the computational cost at the expense of more on-disk storage and more uncertainty, since all those quantities implicitly depend on the nuclear coordinates.