TIL that, when following the Slater(-Condon) rules, when calculating the energy's of the wavefunction of an atom/molecule, in using full CI expansion, the energy in the coverage between the single excited Slater Determinant and a higher excited Slater determinant (given as <phi(a,b;r,s) | H | phi(a;r)>) doesn't equal 0, because of some serious shenanigans. See below. To understand this, it has to be noted that normally orbitals are orthonormal to each other, making normally the coverage between two spinorbitals is given by epsilon * KroneckerDelta(a,r). (KroneckerDelta means: if a=r, KD = 1; else KD = 0) Epsilon is the own value of the operator H (which is the Hamiltonian, of which the own values are the energy of a system Fock operator. Ladies and gentlemen: Quantum Chemistry. Wish me luck for the coming finals. EDIT: oopsie-daisy, made quite some errors there. The <..|..|..> thingie calculates to an integral containing spinorbitals and some sort of Fock Operator. when one of the SD's is the SD resembling the unexcited wavefunction, a.k.a. the one calculated through the Hartree-Fock-method, and the other a single-excited SD(electron from orbital a to r), the equation ends up to be <fi(a)|F|fi®>.
Because spinorbitals fi(a),...,fi®,... are own functions of the Fock Operator F, under that operator the fi's are orthonormal, making the equation equal to epsilon * KD(a,r), which in that case is 0 (because a is not r) However, when you pick a singly-excited SD and a doubly-exicted SD: phi(a,d;r,s) (just read: none of them is the Hartree-Fock SD, and the difference in excitement is 1), the equation ends up in <fi(a)|F'|fi®>. F' is a Fock operator, similar, but not the same as F. Due to this, the spinorbitals aren't own functions anymore, which makes them under F' non-orthogonal, which causes coverage, hence it's not 0. Apologies for the glaring error. I should study some more (that's punishment enough)