2
$\begingroup$

I am currently going through Molecular Modelling: Principles and Applications by Andrew R. Leach (2nd Edition).

The Hamiltonian operator for an electron is given in equation (2.125) as

$$\mathscr{H}^{core} = -\frac{1}{2}\nabla_1^2 - \sum_{A = 1}^{M} \frac{Z_A}{r_{1A}}$$

Where the first term is the kinetic energy of the electron and the second term is its attraction to each nucleus.

This term contributes an energy term to the Fock matrix (page 57) given by

$$ H_{\mu\nu}^{core} = \int d\nu_1 \phi_{\mu}(1) \mathscr{H}^{core} \phi_{\nu}(1)$$

Where $\phi_{\mu}(1)$ and $\phi_{\nu}(1)$ are specific orbitals occupied by the electron.

Taking the example of the HeH$^{+}$ system [a two electron system] on page 62, there are two basis functions $1s_{A}$ centered on the helium atom and $1s_{B}$ centered on the hydrogen atom.

Given this basis the $H_{\mu\nu}^{core}$ terms for this system would be:

  1. $ H_{1s_{A} 1s_{A}}^{core} = \int d\nu_1 \phi_{1s_{A}}(1) \mathscr{H}^{core} \phi_{1s_{A}}(1)$
  2. $ H_{1s_{B} 1s_{B}}^{core} = \int d\nu_1 \phi_{1s_{B}}(1) \mathscr{H}^{core} \phi_{1s_{B}}(1) $
  3. $ H_{1s_{A} 1s_{B}}^{core} = H_{1s_{B} 1s_{A}}^{core} =\int d\nu_1 \phi_{1s_{A}}(1) \mathscr{H}^{core} \phi_{1s_{B}}(1) $

The first two terms make sense to me as they are simply the energy of the electron (kinetic + attaction to each nucleus) in a given orbital $1s_{A}$ or $1s_{B}$. That is if the electron is to be found in orbital $1s_{A}$ then it contributes $H_{1s_{A} 1s_{A}}^{core}$ units of energy to the Fock matrix.

I don't know how to interpret the third term with involves both $1s_{A}$ and $1s_{B}$ inside the integral. To put it colloquially, where does the energy in the third term come from?

Edit:

Could it be that the third term $ H_{1s_{A} 1s_{B}}^{core} = H_{1s_{B} 1s_{A}}^{core} =\int d\nu_1 \phi_{1s_{A}}(1) \mathscr{H}^{core} \phi_{1s_{B}}(1) $ indicates the probability of a transition of the electron between $1s_{A}$ and $1s_{B}$?

I saw the below answer and thought of this. https://physics.stackexchange.com/a/680191/219488

$\endgroup$
1
  • $\begingroup$ Imo @quantumwave's answer in the link you provided is just fine. $\endgroup$ Commented Feb 23, 2023 at 18:53

0