3
$\begingroup$

The electronic Hamiltonian in the canonical orbital basis has the following form: $$\tag{1} H_e = h_0 + \sum_{ij=1}^{N} h_{ij} \sum_{\sigma \in \{\uparrow,\downarrow\}} a_{i\sigma}^\dagger a_{j\sigma} + \frac{1}{2}\sum_{ijkl=1}^{N} g_{ijkl} \sum_{\sigma\tau \in \{\uparrow,\downarrow\}} a_{i\sigma}^\dagger a_{j\tau}^\dagger a_{k\tau}a_{l\sigma}, $$ in which $i,j,k,l$ index over the spatial orbitals. If one wishes to work in an active space $S \subset \{1,\ldots,N\}$, this defines a new Hamiltonian:

$$\tag{2} H_e = \tilde{h}_0 + \sum_{ij\in S} \tilde{h}_{ij} \sum_{\sigma \in \{\uparrow,\downarrow\}} a_{i\sigma}^\dagger a_{j\sigma} + \frac{1}{2}\sum_{ijkl \in S} \tilde{g}_{ijkl} \sum_{\sigma\tau \in \{\uparrow,\downarrow\}} a_{i\sigma}^\dagger a_{j\tau}^\dagger a_{k\tau}a_{l\sigma}. $$

My question is, how does one derive the new coefficients $\tilde{h}_0, \tilde{h}_{ij}, \tilde{g}_{ijkl}$ given the core and virtual orbitals? There is a lot of software for doing this automatically, but I was unable to find a derivation for the formulas used (the closest I found was this which only contains formulas but no derivation).

$\endgroup$
1
  • $\begingroup$ +1. Have you looked at some of the original papers about the CASSCF method, such as this one by Bjorn Roos et al., which was published in 1980? $\endgroup$ Commented Mar 16 at 0:34

1 Answer 1

2
$\begingroup$

When you have a set of inactive orbitals, you just replace the bare one-electron operators in the active space with the Fock operator, which contains the Coulomb and exchange contributions from the inactive orbitals. If you want to dig out the mathematical formulation, the purple bible is usually the place to start from.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .