Atomic Term Symbols.

In freshman chemistry you speak of the electronic configuration of an atom, e.g.

2 2 21 2 2s s p for the Carbon atom. There are 6 2p orbitals for carbon that we can label

1 0 1 1 0 1, , , , ,p p p p p p− − , where the barred orbitals indicate β spin, while unbarred orbitals denote α spin. I am omitting the fact that it is the 2p (rather than 3p orbital) for convenience. The subscripts denote the m-quantum number (-1, 0 or 1) while the p-orbital implies 1l = . The number of Slater determinants or so-called micro-configurations is the number of ways you can distribute 2 electrons over 6 spin-orbitals

where the order doesn’t matter. This is 6 6! 6 5 152 2!4! 2⎛ ⎞ ⋅

= = =⎜ ⎟⎝ ⎠

. Let us denote the Slater

determinants as , 1..15k kΦ = . This means that the 2 2 21 2 2s s p electronic configuration for Carbon really indicates 15 different quantum states! These states would all have the same

energy, if not for the electron-electron repulsion operator 1i j ijr<∑ in the Hamiltonian (same

orbital energies). So this is a situation where inclusion of e-e repulsion qualitatively changes the picture. The Slater determinants themselves are not eigenfucntions of the Hamiltonian. However the eigenstates can be well approximated as linear combinations of Slater determinants. The simplest way to achieve a qualitatively correct picture would be to write each state as a linear combination of the 15 possible Slater determinants (for Carbon)

15

1k k

kcα α

=

Ψ = Φ∑

If we substitute this in the Schrödinger equation we get ˆ

k k k kk k

H c c Eα α αΦ = Φ∑ ∑

To get a well defined set of equations we integrate on both sides against , 1..15m mΦ = , hence

ˆ

1...15

m k k m k kk k

mk kk

mk k mk

H c c E

c E

H c c E m

α α α

α α

α α α

δ

Φ Φ = Φ Φ

=

= ∀ =

∑ ∑

∑

∑

where we used the orthonormality of the Slater determinants. This is a so-called matrix eigenvalue equation. It looks just like the Schrödinger equation E=HC C , but in a basis of Slater determinants. This approach is called Configuration Interaction, and it is really an application of the ‘linear variational principle’. The hardest part is the calculation of the matrix elements

* ˆ... (1..6) (1..6)mk m kH d d d H= Φ Φ∫∫∫ 1 2 6 (for Carbon with 6 electrons)

The 6*3=18 dimensional integrals can be reduced to so-called 1- and 2-particle integrals which are 3-dimensional respectively 6-dimensional integrals. The 6-dimensional integrals, so-called two-electron integrals are the hardest integrals we ever need in quantum chemistry (even for molecules), and they can be calculated efficiently on a computer. If we know all the integrals / Hamiltonian matrix elements we can diagonalize the matrix and we would get 15 eigenvalues and 15 orthonormal states from the matrix diagonalization problem. The Configuration Interaction method works for both molecules and atoms, and is conceptually simple. If one uses relatively small CI matrices with the purpose to resolve the degeneracies that would occur without electron interaction the method is often called CASSCF, since we also want to optimize the orbitals in the electron configuration (Complete Active Space Self-Consistent Field). CI is also used to account for the remainder of electron correlation, due to the Vee term in the Hamiltonian. Then one requires the inclusion of a very large number of determinants (1,000,000 or more), and other methods than CI, e.g. Coupled Cluster theory are widely used also. For atoms one can use spherical symmetry to predict the degeneracy of the various states. The procedure is fairly simple. The idea is that for each Slater determinant you can determine the eigenvalue of the ˆˆ ( )total

z zi

L l i=∑ operator by summing the m values (-1, 0

or 1 here) of the occupied orbitals in the Slater determinant. Likewise you can determine the eigenvalue of the ˆ ˆ ( )total

z zi

S s i=∑ , by summing the sm values (-1/2, 1/2) for the

occupied orbitals in a determinant. The eigenstates of the Hamiltonian can be shown to eigenfunctions of 2, 2,ˆ ˆˆ ˆ, , ,total total total total

z zL L S S (this is strictly valid only if we neglect the rather minor effect of spin-orbit coupling, a relativistic effect). Therefore, every eigenstate can be labelled as , , , , ... , ...S SL M S M M L L M S S= − = − . The energy only depends on the ,L S quantum numbers, not on , SM M . Hence each ( , )E L S energy level has a degeneracy of (2 1)(2 1)L S+ + . It is customary to use the letter-labels S, P, D, F, G for total angular momentum 0,1,2,3,4L = in analogy with the use of , , , ,s p d f g orbitals corresponding to the orbital angular momentum 0,1,2,3,4,...l = The spin quantum number is indicated by the spin-degeneracy (2 1)S + . The so-called atomic term symbol (without spin-orbit coupling) indicates a particular energy eigenvalue and is indicated as (2 1)S L+ , e.g. 1 3 2, ,D P S , that have degeneracy 5, 9 and 2 respectively. The story is not quite complete yet, because the true Hamiltonian also includes the spin-orbit coupling or L S⋅ coupling. As we have discussed for the hydrogen atom in some detail, the operators 2 2 2ˆˆ ˆ ˆ, , , zL S J J , all commute, and they also commute with the

Hamiltonian, including spin-orbit coupling. The operators ˆˆ ,z zL S do not commute with

the spin-orbit term in H , and should then not be used to label the states. We can construct common eigenfunctions, which we can give quantum numbers , , , , Jn L S J M . The splitting due to spin-orbit coupling can be well approximated by making eigenfunctions of the 2ˆ ˆ, zJ J operators from linear combinations of states in the (2 1)S L+ multiplet. In principle one could use the linear variational principle (i.e. CI). However, we can do it alternatively by exploiting the commutation relations. Let me just state the result. In general the J quantum numbers corresponding to the (2 1)S L+ multiplet range from ...J L S L S= + − . In each J-multiplet there are (2J+1) states having ...JM J J= − . In a magnetic field all of these energy levels would split. Without a magnetic field the (2J+1) levels in a given (L,S,J) multiplet are degenerate. We indicate such a multiplet by the so-called term symbol (2 1)S

JL+ . As an example the 3P level (S=1,L=1) spits into 2,1,0J = levels denoted 3 3 3

2 1 0, ,P P P , that each have (2J+1) members, e.g. 5, 3, 1 respectively, 9 states in total. A 2 D level (S=1/2, L=2), splits into J=5/2 (6 states) and J=3/2 (4 states), hence 10 states in total. The atomic term values are 2 2

5/ 2 3/ 2,D D . You may notice that the splittings discussed for the hydrogen atom follow this rule also. The energy splittings between the levels having the same L and S, but different J is very small, on the order of wave numbers (for light atoms). Levels that have different S and L values, even if they same electronic configuration can easily be split by an eV or more (~ 8000 cm-1). Hund’s rules predict the atomic term symbol of the ground state, and the energy ordering. 1. The state with the highest value of S has the lowest energy, and the energy increases with decreasing S. 2. For states with the same S, the state with the largest value of L has the lowest energy. Energy increases with decreasing L. 3. For states with the same S and L: if the shell is less than half filled the smallest value of J has lowest energy, otherwise the highest value of J is the most stable. In excersises/examination questions you are often asked to construct certain term values. I would ask you to do the following: 1. Determine the number of microstates. This is 15 for carbon. 2. Determine the Slater determinant of the highest SM S→ value and of these the highest LM L→ value. Only the open shells are important here. For Carbon this yields

1 0 1, 1Sp p M S M L→ = = = = . This state is a member of the 3P multiplet, which has 9 members in total. This is the ground state of Carbon (neglecting spin-orbit coupling). The highest possible S quatum number determines the ground state.

3. Determine the Slater determinant of the highest LM L→ value and of these the determinant with the highest SM S→ value. For carbon this yields

1 1 , 2, 0Sp p M L M S= = = = , i.e. the 1D multiplet, having 5 members that each have the same energy. The other members are typically linear combinations of Slater determinants. This is the first excited state of the atom. For atoms the first excited electronic state is on the order of one eV=8050 cm-1 higher in energy than the ground state. 4. In this fashion we have accounted for 14 out of a total of 15 possible states. The missing member is the 1S (1 state). In general one has to do a little more work to get the remaining term values. 5. Finally include spin-orbit coupling and list all term values, indicating the degeneracy of each level. For carbon: 3 3 3 1 1

2 1 0 2 0(5), (3), (1), (5), (1)P P P D S , in total we then have 5+3+1+5+1 = 15 microstates as required. Let us do one more example, the Nitrogen atom, which has 3 p electrons.

1. Number of microstates 6 6! 6 5 4 203 3!3! 3!⎛ ⎞ ⋅ ⋅

= = =⎜ ⎟⎝ ⎠

2. Highest SM S= determinant: 41 0 1p p p S− → This is the lowest energy level (4 states)

3. Highest LM L= determinant: 2

1 1 0p p p D→ . This is the first excited energy level (10 states). 4. We have 14 out of 20 states accounted for. The remaining multiplet is 2P with 6 states. 5. Including spin-orbit coupling we get the term symbols: 4 2 2 2 2

3/ 2 5/ 2 3/ 2 3/ 2 1/ 2(4), (6), (4), (4), (2)S D D P P , for a total of 20 states. In a magnetic field all of these level would have different energies, and this is an experimental way to determine that these energy levels really do exist! Another way to demonstrate the degeneracies is by calculating properties using statistical mechanics. The degeneracy of the states enters the formulas for entropy and free energy, or the partition function. We only get agreement with experiment if we include the proper degeneracies. C’est ca.