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?


Here is what the whole paragraph states:

  1. Use Hellmann-Feynman theorem to calculate forces.
  2. Hellmann-Feynman theorem is only valid for electronic ground state.
  3. During a dynamical simulation nuclei move, therefore, there is no guarantee that the system will be in electronic ground state. In other words the electronic structure calculated by ground state methods (such as DFT) will be wrong.
  4. If you want to have the system in ground state you need to update the electronic structure for each femtosecond, that is, in this short time scale from one times step to the other you can assume that the system is in ground state. Therefore you can safely use DFT.
  5. Now, because you need to perform a DFT (SCF) calculation in each femtosecond a dynamical simulation becomes very expensive. Assuming that you need to simulate the system, say, for nanoseconds.
  • $\begingroup$ 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. $\endgroup$ – meTchaikovsky Oct 3 '18 at 10:56
  • $\begingroup$ 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. $\endgroup$ – physicopath Oct 3 '18 at 11:02
  • $\begingroup$ Here is a quote from wikipedia entry for HF method. "The variational theorem states that for a time-independent Hamiltonian operator, any trial wave function will have an energy expectation value that is greater than or equal to the true ground state wave function corresponding to the given Hamiltonian. Because of this, the Hartree–Fock energy is an upper bound to the true ground state energy of a given molecule." $\endgroup$ – physicopath Oct 3 '18 at 11:07
  • $\begingroup$ Updated my answer, I hope it includes an answer to your second question in the comment. $\endgroup$ – physicopath Oct 3 '18 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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