This question: How do electron configuration microstates map to term symbols?
And the discussion of multielectron effects here: Quantum Computing and Animal Navigation
Inspired me to try to understand quantum mechanics of multi-electron atoms better. As it is aparrently difficult to solve anything in closed form, I found a numeric multi-electron quantum mechanics calculation package that is free for students! (Called GAMESS) My goal is to understand the basics at least at the level of the 'molecular orbital' picture, and be able to calculate some energies.

From the discussion in the other question and reading wikipedia I think I understand the basic idea behind Hartree Fock and the wavefunction made from a Slater determinant. And I was able to use the program to calculate the orbital energies and ground state energy for a ${}^1S_{0}$ term symbol atom like Neon. These make sense to me as each orbital filled with two electrons: $$ \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow} \ \underline{\uparrow \downarrow} \ \underline{\uparrow \downarrow} \\ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}$$

My main question is regarding atoms with open shells:
How is the wavefunction represented in the orbital picture for open shell atoms?

For example Lithium with term symbol ${}^2S_{1/2}$ $$ \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \ } & \underline{\quad \ } \ \underline{\quad \ } \ \underline{\quad \ } \\ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}$$

versus Boron with term symbol ${}^2P_{1/2}$

$$ \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow } & \underline{\uparrow \ } \ \underline{\quad \ } \ \underline{\quad \ } \\ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array}$$

It seems like these should be described exactly the same way in molecular orbitals: one open shell below which all the remaining shells are closed. However this apparently isn't the case. The Lithium atom is a "doublet" and can be described by ROHF with GAMESS, while the Boron case is apparently not just a doublet since somehow the degeneracy needs to be taken into account (does this 'kind of doublet' have a name besides the term symbol?) although this GAMESS document says they are both 'technically ROHF' states ( http://haucke-jena.de/gamess/gam_grund_1/gvb_ref.htm ).

Question "part 2":
Is this trying to say we need more than one slater determinant for these degenerate states?

That is, something like: $$ \frac{1}{\sqrt{2}} \left[ \left( \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow } & \underline{\uparrow \ } \ \underline{\quad \ } \ \underline{\quad \ } \\ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array} \right) + \left( \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow } & \underline{\quad \ } \ \underline{\quad \ } \ \underline{\uparrow \ } \\ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array} \right) \right] $$

So this would be called a "multi configuration" state I think (MCSCF?), but the quantum calculation documentation (previous link) says MCSCF isn't necessary for either of these states. So are those two "pieces" of the wavefunction somehow not different slater determinants?

So I'm very confused and would appreciate any help. Please clearly explain the terminology you use in an answer, as the difference between RHF, ROHF, GVB, MCSCF, etc. seem to be a large part of my confusion here.

Some example calculation results for an Oxygen atom.
energy = -74.8002167288 Hartrees --- triplet (ROHF with MULT=3)
energy = -74.6780126893 Hartrees --- singlet (ROHF with MULT=1)
energy = -74.6780126893 Hartrees --- singlet (RHF), should be same as ROHF MULT=1
energy = -74.7986255288 Hartrees --- ??? (GVB with setting specified by GAMESS doc for ${}^3P_2$ term symbol)

The GVB output also says:


                      1          2          3          4          5

                  2.000000   2.000000   1.333333   1.333333   1.333333

Which seems to me like a uniform mixing of $$ \left( \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow } & \underline{\uparrow \downarrow } \ \underline{\uparrow \ } \ \underline{\uparrow \ } \ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array} \right),

\left( \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow } & \underline{\uparrow \ } \ \underline{\uparrow \downarrow } \ \underline{\uparrow \ } \\ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array} \right),

\left( \begin{array}{ccc} \underline{\uparrow \downarrow} & \underline{\uparrow \downarrow } & \underline{\uparrow \ } \ \underline{\uparrow \ } \ \underline{\uparrow \downarrow } \\ \underset{1s}{\ } & \underset{2s}{\ } & \underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{2p} \end{array} \right) $$

But how can that center one be a ${}^3P_2$ state? Shouldn't it be a ${}^3S_1$ state? This all seems very conflicting. Especially that trying to include the effects of the degeneracy actually raised the energy instead of lowering it.

Here's the GAMESS input I used if someone else wants to play with it:

! p**4 3-P state
F(1)= 1.0 0.66666666666667
ALPHA(1)= 2.0 1.33333333333333 0.83333333333333
BETA(1)= -1.0 -0.66666666666667 -0.50000000000000 $END
 Oxygen atom
 O           8.0   0.0000000000   0.0000000000   0.0000000000

EDIT: My latex is kludgy. I had trouble making things line up. If someone could edit this so I can see the proper way to do things, that would be great.


1 Answer 1


I am not so certain about the terminology because I do not use electronic structure software, but since nobody has answered so far, I will try to see if my answer clarifies something.

In a closed-shell configuration you know that a single Slater determinant built from atomic orbitals (or for that matter, any function with angular dependence in terms of spherical harmonics) will be an eigenstate of $\hat{L}^2$ and $\hat{S}^2$ with quantum numbers L = S = 0. Additionally in this case there is no spin-orbit coupling, so that J = 0 as well. Therefore the angular momentum properties of this wave function will be well defined.

In the simplest ab initio molecular orbital calculation you obtain your atomic orbitals solving the restricted Hartree-Fock equations (RHF). Why restricted? Because if you have n electrons and the dimension of your basis (whatever STO, GTO or more advance basis, assuming you expand your atomic orbitals in a basis) is L > n/2, you populate the first n/2 orbitals and assume each orbital is doubly occupied by two electrons with spins $\uparrow$ and $\downarrow$. You could use the unrestricted Hartree-Fock (UHF) method, populating the lowest n spin-orbitals, but then the density of electrons with spin $\uparrow$ will be in general different than the density of electrons with spin $\downarrow$. Thus, although the energy of your system will be lower (some restrictions in the variational procedure are lifted) $S$ will not be well defined.

In general, for an open shell configuration the angular momentum will not be well defined by a single Slater determinant wave function. How do you create the proper wave function? You can use several strategies.

The more general strategy is to use a Multi-Configurational Self Consistent Field (MCSCF) approach, in which you generate several Slater determinants with the L orbitals, in which both the orbitals and the coefficients multiplying the Slater determinants are variational parameters that you have to determine consistently within the MCSCF approach.

A simpler method including correlation (among the many that exists) is the Configurational Interaction (CI) in which you optimize only once the orbitals and then generate the Slater determinants and then at the end of the day obtain the coefficients of these Slater determinants by diagonalizing the Hamiltonian.

The previous methods are called method with dynamic correlation or beyond the HF approach. More simply, you can use the RHF scheme and after obtaining your n/2 populated orbitals force the symmetry by suitable projector operators that generate the other Slater determinants needed so that your overall wave function is an eigenstate of $\hat{L}^2$ and $\hat{S}^2$ (and $\hat{L}_z$ and $\hat{S}_z$). This is good only in the so-called L-S coupling scheme when spin-orbit coupling is assumed small (small relativistic corrections). But of course you can later diagonalize the Hamiltonian with the spin-orbit coupling terms to resolve the fine-structure of the atom. I believe this is what GAMESS calls the ROHF method but I am not sure. To be sure you have to look at the manual.

More commonly, you can use the UHF method and after obtaining you n populated spin-orbitals you force the right symmetry by applying projector operators to filter the "wrong" components that give an incorrect spin symmetry. This method is often called the PHF or Projected Hartree-Fock. Finally, you can also use the RHF equations but start from a number of Slater determinants with the coefficients fixed by the projector operator (that is, they are no longer variational parameters) and then variationally optimize the orbitals for this set of determinants. The equations become more difficult and I believe this method, called EHF or Extended Hartree-Fock is not often applied. If you want to work with several Slater determinants you typically use CI or MCSCF.

Now going back to your examples.

In Lithium you have two microstates for the ground state that you can write assigning the 4 quantum numbers $n_j$, $l_j$, $m_j$ and $m_{s,j}$ for each electron $j$, assuming the usual Aufbau principle. However these two states are degenerate. Therefore you can write two Slater determinants which will give you the same ground state energy. In the absence of an external magnetic field, any of these Slater determinants (or whatever linear combination of the two) is a good description of your ground state. It will be an eigenstate of the four angular momentum operators (or of $\hat{J}^2$ and $\hat{J}_z$ too).

In your second example, Boron, you can write 6 microstates, because you can put your last electron in $|2,1,0,\pm 1/2 \rangle$ or $|2,1,\pm 1,\pm 1/2 \rangle$ where in the nice Dirac nomenclature the third quantum number corresponds to $m_j$ and the last to $m_{s,j}$. Although all the Slater determinants that you can built with these microstates assignments are degenerate eigenstates with well defined $\hat{L}^2$, $\hat{L}_z$, $\hat{S}^2$ and $\hat{S}_z$, the degeneracy will be broken when you include the spin-orbit coupling (fine structure). Then some states will be written in terms of a combination of several Slater determinants.

In your last example, Oxigen, you have two holes in the outer (open) shell. If you use the ROHF method (assuming I got the terminology right) then it suffices to use the Clebsch-Gordon coefficients to know the right combination of Slater determinants that determines well defined values of $\hat{L}^2$, $\hat{L}_z$, $\hat{S}^2$ and $\hat{S}_z$ for the set of 4 lowest-energy populated orbitals found by the RHF method.


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