# How to derive the one-electron schroedinger equation?

When you want to obtain a ground state for a given electronic system, you usually assume your state to be representable by a Slater determinant of molecular orbitals (MO). I understand how with the application of the variational principle you start working from the average value of the total Hamiltonian and you conclude that you can find those MO as the eigenstate of what is defined as a "one electron Hamiltonian".

Now, when doing electron dynamics, I have seen that many people start from the Schroedinger equation applied to those MO using that one electron Hamiltonian. However, I have never seen explained how to get there from the original equation that has to be applied to the full system (the Slater determinant and the real Hamiltonian of the system). I tried doing something similar to the variational procedure but the starting points seem to be very different. Maybe I'm missing something, but this does not seem like a trivial step to me.

• Then you can do this. (i) Assume your many body wave function ($|\Psi\rangle$) is a Slater determinant made up of orthonormal orbitals. (ii) Define a functional : $E(t)= \langle \Psi| i\hbar\frac{\partial}{\partial t} - \hat{H}(t) |\Psi\rangle$ ($\hat{H}(t)$ is time dependent many body Hamiltonian). (iii) Extremize $E(t)$ (Dirac-Frenkel variational principle) with respect to all orbitals with constraint that they remain orthonormal at any instant of time. (iii) After a bit of manipulations you will have TDHF equations. – Sunyam May 16 '18 at 10:54