# Electron coordinates in QM programs

I find it somewhat unusual that one does not need to define electron coordinates in an electronic structure calculation. Taking Hartree Fock for example, the Fock operator includes the one-electron operator that depends on electron coordinate. However, the matrix representation of the Fock operator in the basis of $$\{\phi_{\mu}\}$$ is

$$F_{\mu v}=\int d \mathbf{r}_{1} \phi_{\mu}^{*}(1) f(1) \phi_{v}(1)$$

So because the entries in the Fock matrix are all integrals over all space $$\int_{r_{1}=0}^{\infty} d \mathbf{r}_{1}$$, it doesn't need the electron coordinate. Is that correct?

• Electron is not a thing that has coordinates. Jan 5 at 18:16
• Yes, very much so Jan 5 at 18:16
• @IvanNeretin I'm very much aware of the quantum treatment of electron as a wave rather than a particle with definite coordinates. I'm more concerned with the numerical implementation of electronic structure method!
– Jung
Jan 5 at 18:46

The answer is yes and no. Electronic coordinates are needed when calculating the Fock matrix elements $$F_{rs}$$, which consist of one-electron integrals and two-electron integrals. Once you get the Fock matrix elements, then the Hatree-Fock-Roothaan equations are solved by matrix algebra through diagonalization without using electron coordinates. Therefore, electron coordinates are used explicitly in the step of Fock matrix elements calculations but not in the step of matrix diagonalization.

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Think about the integration as solving the Schrödinger (or Fock, or whatever) equation for all coordinates in space at the same time.

The whole point of this is to obtain a wavefunction which yields the energy eigenvalue to your operator no matter what electron coordinates you plug in. In other terms,

$$\hat H \Psi ( \mathbf{x} ) = E \Psi ( \mathbf{x} )$$

for any $$\mathbf{x}$$.

So, once you have your final wavefunction, nothing stops you from "putting your electrons in specific locations", but that's not really all that interesting because we know that this will still yield the same energy eigenvalue - because that's how we constructed the wavefunction.