Title: ETH_Sem5 / AdvQC / 14_Crucial_Problems_QC_Practice

Q: What is the spin state problem?
A: To be able to predict the correct energy gaps between states of different spin \(S\) is crucial in chemistry (e.g., two-state reactivity). Present-day DFT has difficulties to cope with this problem. Problem: In DFT, the spin-state of many complexes heavily depends on the chosen coefficients in the exchange-correlation function, e.g., the results strongly depend on exact exchange (from HF) admixture \(c_3\) in B3LYP.
\[\begin{aligned} E_\mathrm{xc}^\mathrm{B3LYP} &= c_3E_\mathrm{x}^\mathrm{ex.ex} + (1-c_3)E_\mathrm{x}^\mathrm{LSDA} \\ &\hphantom{=} + c_1E_\mathrm{x}^\mathrm{B88} + c_2E_\mathrm{c}^\mathrm{LYP} \\ &\hphantom{=}+ (1-c_2)E_\mathrm{c}^\mathrm{VWN} \end{aligned}\]
Exact exchange enters linearly in a functional, but the splitting is an energy difference evaluated for two different electron densities and two different electron-nucleus potentials (structures). The (almost) linear dependence of spin-state splittings on exact-exchange admixture is very common.

Q: How can the spin-density be calculated in DFT? What is meant by ``broken-symmetry'' and why is it useful?
A: The spin density can be calculated by:
\[Q(\pos) = \rho_\alpha(\pos) - \rho_\beta(\pos)\]
The orbital contributions determine degree of spin polarization. If the \(\alpha\) and \(\beta\) electrons are treated seperately their respective orbitals are not artificially forced to be degenerate (broken symmetry). Thus the resulting spin density will differ from the regular one!

Q: What can be the problem if one wants to calculate reference spin densities with CASSCF?
A: For example consider an Fe(\(3\mathrm{d}\)) nitroxyl complex. Defining the active orbital space:
- Minimal active space: Fe 3d- and both NO \(\pi^\star\)-orbitals \(\Rightarrow\) CAS(7,7)
- Consider also both NO \(\pi\)-orbitals \(\Rightarrow\) CAS(11,9)
- Additional shell of Fe d-orbitals (double-shell orbitals) is gradually included \(\Rightarrow\) CAS(11,11) to CAS(11,14)
- Include one ligand \(\sigma\)-orbital and the antibonding \(\sigma^\star\)-orbital upon the CAS(11,11) \(\Rightarrow\) CAS(13,13)
\(\Rightarrow\) Analyze convergence of spin density w.r.t. the dimension of the active orbital space
CAS(11,14): spin density;
all others: spin density differences to CAS(11,14)
As one can see, a stable CAS with all important orbitals is difficult to obtain!
Problem: Reference spin densities for very large CAS required but larger than CAS(18,18) is not feasible!