# Why I can't use a single Slater determinant to describe the ground state of open-shell atoms and molecules?

If we have a closed-shell atom or molecule (in the Born-Oppenheimer aproximation), and adopt the central field aproximation, we can write the ground state wave function of the system using a Slater determinant, constructed using the spin-orbitals that are the solutions of the equations

$$(h_{i} + V_{i}) \varphi_{i} = \varepsilon_{i} \varphi_{i}$$

with the lowests energies, where $$h_{i}$$ is the sum of the kinect energy and interaction with the nucleo(s) of the electrons, and $$V$$ is the potential. However, I read that if the system is open-shell, in general we can't write the ground state wave function using only a single Slater determinant, we have to take linear combinations of Slater determinants. Can someone explaim why this happpen, and indicate a reference? I would like to understand it.

You are looking for multi-reference cases.$$^1$$ Another term usually applied is "strong correlation". See What exactly is meant by 'multi-configurational' and 'multireference'?.

Instead of trying to answer myself on what those cases might be, let me quote Frank Jensen (Introduction to Computational Chemistry, 2nd ed., Wiley, 2007, p. 154):

Consider again the ozone molecule with the two resonance structures shown in Figure 4.9 [which shows a) a zwitterion with a positive charge on the central oxygen b) a triplet configuration with single electrons on terminal oxygens]. Each type of resonance structure essentially translates into a different determinant. If more than one non-equivalent resonance structure is important, this means that the wave function cannot be described even qualitatively correctly at the RHF single-determinant level (benzene, for example, has two equivalent cyclo hexatriene resonance structures, and is adequately described by an RHF wave function). A UHF wave function allows some biradical characters, with the disadvantages discussed [earlier in the text]. Alternatively, a second restricted type [Configurational State Function (CSF)] (consisting of two determinants) with two singly occupied MOs may be included in the wave function. The simplest [multi-configuration SCF] for ozone contains two configurations (often denoted TCSCF), with the optimum MOs and configurational weights determined by the variational principle. The CSFs entering an MCSCF expansion are pure spin states, and MCSCF wave functions therefore do not suffer from the problem of spin contamination.

In some cases, a single determinant may give you an incomplete picture. This is more often the case for triplet or open-shell species, but open-shell singlet cases can also demand a multi-determinant treatment.

Note that multi-reference treatments may also be necessary to obtain accurate quantitative results if the physical ground and excited states are close in energy.

$$^1$$ The "reference" is due to the use of the Hartree-Fock determinant as the reference state for configuration interaction (CI) post-SCF calculations. There are systems where only bad results can be achieved when using a single reference in a truncated CI. (In full CI, you have every possible reference included.)

• I very much agree with your answer. I'd like to add that some open-shell cases can be correctly described by a single determinante (e.g. with ROHF). On the other hand, unrestricted methods use two SD already, but that is quite different from what you have described here. – Martin - マーチン Jan 5 at 12:19
• @Martin-マーチン I often encounter this claim that "unrestricted methods (effectively) use two Slater determinants" but I haven't really found a rigorous proof of this. Do you have any good references? – eimrek May 5 at 11:24
• @eimrek There is no rigorous proof, because rigorously speaking it's not true. However, it is almost true. See for example the Wikipedia Page on UHF, the Pople-Nesbet-Berthier eq: $$\mathbf{F}^\alpha\mathbf{C}^\alpha=\mathbf{S}\mathbf{C}^\alpha\mathbf{\epsilon}^\alpha,\quad\mathbf{F}^\beta\mathbf{C}^\beta=\mathbf{S}\mathbf{C}^\beta\mathbf{\epsilon}^\beta$$ Looks independent enough, but $\mathbf{F}^{\alpha,\beta}$ depend on their own solutions and on each other, so they are coupled. So that's more of a qualitative statement. – Martin - マーチン May 5 at 12:00