## States arising from various calculations

Each MECI calculation invoked by use of the keywords C.I.=n and C.I.=(n,m) normally gives rise to states of quantized spins. When C.I. is used without any other modifying keywords, the number of states shown in the Tables 1 and 2 will be obtained. These numbers of spin states will be obtained irrespective of the chemical nature of the system.

 Table 1: States arising from C.I.=n No. of M.O.s Number of States Arising from in MECI Odd Electron Systems Even Electron Systems Doublets Quartets Sextets Singlets Triplets Quintets 1 1 1 2 2 3 1 3 8 1 6 3 4 20 4 20 15 1 5 75 24 1 50 45 5

 Table 2: Number of states arising from C.I.=(n,m) Number of States Arising from Keywords Odd-electron systems Even-electron systems C.I.=(2,1) 2 = (2!/(0!2!))*(2!/(1!1!)) 4 C.I.=(3,1) 9 9 C.I.=(3,2) 3 9 C.I.=(4,1) 24 16 C.I.=(5,1) 50 25 C.I.=(8,1) 224 = (8!/(6!2!))*(8!/(7!1!)) 64 C.I.=(8,4) 3920 = (8!/(3!5!))*(8!/(4!4!)) 4900 = (8!/(4!4!))*(8!/(4!4!))

If the C.I. is not performed correctly then the symmetry features of the system will not be correctly identified. Two systems, methane and a transition metal atom are used to verify that the correct symmetry features are produced.

### Calculation of the States for methane, CH4

The following data-set is provided to allow the quantities in this discussion to be reproduced::

``` vectors c.i.=8 meci
Methane

H     0.00000000 +0    0.0000000 +0    0.0000000 +0     0     0     0
C     1.08523001 +0    0.0000000 +0    0.0000000 +0     1     0     0
H     1.08523001 +0  109.4712210 +0    0.0000000 +0     2     1     0
H     1.08523001 +0  109.4712210 +0 -120.0000000 +0     2     1     3
H     1.08523001 +0  109.4712210 +0  120.0000000 +0     2     1     3```

Methane provides a good test of the configuration interaction calculation.  If the C.I. is not performed correctly, the symmetry features of the system will not be correctly identified.

Using the standard semiempirical basis set, there are eight atomic orbitals, 2s, 2px, 2py, and 2pz on carbon, and a 1s on each of the four hydrogen atoms.  In the molecule these form eight molecular orbitals that have symmetries, in the order of increasing energy: 1a1, 1t2, 2t2, and 2a1.  Methane is a small molecule, only five atoms and 8 molecular orbitals, so evaluation of the various terms in the C.I. calculation is very fast.  What is not fast is calculating the states formed when C.I.=8 or C.I.=(8,4) is used.

Both configuration interaction calculations are the same - all eight M.O.s and all eight electrons are used.  So the active space consists of eight M.O.s, and permutation of the eight electrons, four of α and four of β spin, in these M.O.s result in (8!*(4!)-2)2 = 4900 microstates or configurations.  Diagonalization of the resulting configuration interaction matrix is slow because of the computational effort involved.  By default, the spin-states and point-group symmetry representations of the resulting states are evaluated; these are shown in the following Table:

 Table 3 Number of States for each representation for Methane Spin-state Spin A1 A2 E T1 T2 # States (Ms = 0) # States (all Ms) SINGLET 0 104 60 152 196 236 1764 1764 TRIPLET 1 97 97 191 296 296 2352 7056 QUINTET 2 31 31 65 88 88 720 3600 SEPTET 3 7 0 4 4 12 63 441 NONET 4 1 0 0 0 0 1 9 Totals: 240 188 412 584 632 4900 12870

#### Interpretation of these quantities

Each state has a set of quantum numbers.  In this system, as in all systems that have non-infinite point-groups, these are given the symbols of the type "nSRMs." In other types of systems, such as the infinite groups, space groups, and magnetic groups, other symbols are used, but these are not relevant here and can be ignored.  The quantum number n is 1 for the lowest-energy state of symmetry SRMs, 2 for the next one, and so on.  S is the spin-state of the state; this can be 0, 1, 2, 3, or 4 here.  Each spin-state has (2S + 1) magnetic components, all of the same energy, so in chemistry these are commonly referred to as Singlet, Triplet, Quintet, Septet, and Nonet.  Note: by convention, quantum numbers for atomic and molecular orbitals use lower-case letters, quantum numbers for entire systems, such as a wave-function for methane, use upper-case letters.  That convention will be used here. When discussing individual states, the standard convention is to specify the spin-state in terms of the degeneracy of the state, i.e., the number of magnetic components, so the nonet state A1 would be written as 9A1, and for a specific component of spin, as the subscript suffix, e.g. 9A14. The quantum number Ms refers to the magnetic component of spin, and is most simply defined as half the difference in the number of α and β electrons in each state.  Because the number of α and β electrons are the same in all microstates used here, all states have a Ms of zero, i.e.,  Ms = 0.  From here on, for convenience the Ms quantum number will be dropped unless needed. Finally R is the irreducible representation of the point-group symmetry of the state. A1 and A2 are non-degenerate, that is, they are used for states that have exactly one component, E is a degenerate state having two components, and T1 and T2 are degenerate states that have three components. For convenience, the degeneracy of a specific irreducible representation can be written DR.

All states are orthogonal to all other states. This orthogonality can be intrinsic, thus all spin-states are automatically orthogonal, or result from diagonalization, thus all states of A1 symmetry are expressed by orthonormal eigenfunctions.

Given that there are 4900 states in this system, the addition of the individual states must add to 4900.  This addition is made complicated by the degeneracy of the various individual states.  Because of this, the simple addition 104 + 60 + etc. + 97 + 97 + 191 + etc. is replaced by the sum of the product of the degeneracy of each state times the frequency, F, of occurrence of that state.  So the simple sum is replaced by (DR)F, thus the number of Singlet states is given by: 1×104 + 1×60 + 2×152 + 3×196 + 3×236 = 1764, and the number of Triplet states is 1×97 + 1×97 + 2×191 + 3×296 + 3×296 = 2352. The sum of all these states is 4900, as expected.

This calculation was simplified by only using Ms = 0. If that constraint is removed, so that eight electrons are permuted among 16 molecular orbitals, i.e., the number of α and β molecular orbitals, then the number of states would rise to 16!*(8!)-2 = 12870.  Calculating all these states in one job would be very difficult to perform, as all 12870 microstates would need to be supplied by keyword MICROS, and diagonalization of such a large matrix would take a long time, as would the symmetry analysis.  Fortunately, states of different Ms are automatically orthogonal, so the set of 12870 microstates can be separated into nine sets where all microstates in a set have the same Ms. These sets could be represented by Ms = -4, -3, -2, -1, 0, +1, +2, +3, and +4. Because of time-reversal symmetry the states resulting from Ms = -n are equivalent to those for Ms = +n; this reduces the number of sets from nine to five, vis: Ms = 0,  1,  2,  3, and 4.  The set with Ms = 0 has already been run, so all that remains is to run the sets Ms = 1,  2  3, and  4. Editing the data-set for methane is straightforward; to select all states that have Ms = 1 use the keyword MS=1.

As expected, the set of states for Ms = 4 consists of only one state, the nonet 9A1, there being only one way to put eight α electrons into eight molecular orbitals.

For the set of states with Ms = 3 there are eight ways to put seven α electrons in eight molecular orbitals and eight ways to put one β electron in eight molecular orbitals; this results in eight times eight or 64 microstates and therefore 64 states.  The number and symmetries of these states are given by the sum of the entries in SEPTET and NONET in Table 3. All the states are SEPTET except one of the A1 states which is  9A13; this particular state is part of the nine-fold degenerate  manifold 9A14:-4. Similarly, the sets of states in Ms = 2 and Ms = 1 can be understood as the TRIPLET and SINGLET states in Table 3 plus the entries in the next lower line. That many states are degenerate can be verified by examining the associated energy level for the various values of Ms in a state, they should all be identical. Movement within a state from one Ms to another can be accomplished by using the step operator, otherwise known as the shift or ladder operator, but caution is advised when working with these operators - before starting, some practice sessions are recommended, for example showing the Ŝ+|β> = α and that Ŝ-|αα> = 2(αβ +βα).

This description can be concluded by explaining the entries in the column "# States."  Starting at the bottom, with the number of states for the spin-state NONET, 1, the number of states is given by (8!/(8!*0!))2 = 1. On the next line up, the SEPTET line, the number of states is (8!/(7!*1!))2 -  (8!/(8!*0!))2 = 63, and, to generalize, the number of entries is given by (8!/((8-n)!*n!))2 -  (8!/((8-(n-1))!*(n-1)!))2 , where "n" is the spin, i.e. ½( |number of α electrons - number of β electrons|).  The first of these two terms is obviously the component of spin, Ms, the second can be understood as excluding all states that have a higher total spin.

#### Multiplication of the Irreducible Representations in point-group Td

For users who want to explore the symmetry relationships within point-group Td a bit further, the following multiplication table might be useful.

A1 x X = X
E × A2 = E
T1 × A2 = T2
T2 × A2 = T1
E × E = A1 + A2 + E
E × T1 = T1 + T2 = E × T2
T1 × T2 = A2 + E + T1 + T2
T1 × T1 = A1 + E + T1 + T2 = T2 × T2

### Calculation of the symmetries of States of transition metal atoms and ions   (Works with Version 21.186 and higher)

Transition metal atom states are an amazingly rich field for symmetry analysis.  Unlike the quantum numbers for electrons that are very limited, state quantum numbers span a wide range and provide a wealth of detailed information on the atomic structure.  A chromium atom is used in the following discussion.  This element was chosen as the representative for all transition metals; if you'd prefer to use a different element, feel free to do so, but be aware that all the quantum numbers would need to be re-calculated.

Transition metal atoms use nine atomic orbitals, in order of increasing energy these are a set of five d-orbitals, one s-orbital, and three p-orbitals. In the interest of generality, all nine atomic orbitals will be used in the following discussions.  This provides a fairly complete set of systems and allows some of the unusual features of state functions to be examined. For other work the five d-orbitals could be used on their own, in which case the analysis would be much simpler. In order for the symmetry properties of states to be explored, all the math involved in constructing the configuration interaction matrix and all the analysis of the symmetries of the states must be correct.  Put another way, an error in either the math of the C.I. or in the analysis of the state eigenfunctions would prevent the symmetry from being identified.  Put still another way, the symmetry analysis provides strong evidence that the underlying atomic theory is correct.

Chemists are very familiar with the shapes of atomic orbitals and there is a strong temptation for them to assign physical meaning to these shapes.  Physicists, on the other hand, are more comfortable using complex atomic orbitals that lend themselves to being labeled with the magnetic quantum number ms. So for example the five d-orbitals are assigned the magnetic quantum numbers of +2, +1, 0, -1, and -2.  Both descriptions are valid, and in MOPAC the underlying theory used in generating the states uses chemist's ideas of atomic orbitals, but the symmetry analysis can most easily be done using the physicist's model of quantized magnetic moments. So for convenience, the physicist's model will be used from here on; this allows the nine atomic orbitals to be assigned the following magnetic quantum numbers:

 Magnetic Quantum Numbers of Atomic Orbitals s p p p d d d d d 0 +1 0 -1 +2 +1 0 -1 -2

In addition to the magnetic quantum number, each electron has a spin quantum number of +½ or -½, these correspond to the chemist's model of of α and β electrons. Each atomic orbital can hold zero, one or two electrons.

With this information, the process of populating the atomic orbitals can be started.  A given configuration of electrons in a set of atomic orbitals is called a microstate. If there are no electrons present, there is only one possible microstate - the empty state.  If one electron is present, it can be in any one of the nine atomic orbitals, so there are nine microstates.  For two electrons, one of spin +½ and one of spin -½, there are 81 microstates.  If both electrons have the same spin, say +½, then this number drops to 36.  Because electron spins add, the component of spin momentum of these microstates is +1. A second set of 36 microstates exists when the spin of both electrons is -½ and the total spin momentum is -1.  So for two electrons in nine atomic orbitals, the total number of microstates possible is 81 + 36 + 36 = 153.

The following table shows the number of states for all possible numbers of electrons, and the formulae for calculating these numbers. In the table "No. of States" is the number of microstates with the same component of spin  momentum.  "Total no. of States" is the addition of the individual sets of states with the same component of orbital momentum. Sets of microstates with a negative component of orbital momentum are not shown because they have the same values as the sets with a positive Ms, but these should be included in the sum.

 No. Electrons Ms Formula No. of States Ms Formula No. of States Ms Formula No. of States Ms Formula No. of States Ms Formula No. of States Total no. of States No. Electrons 0 0 (9!/(0!9!))2 1 1 0 1 ½ (9!/(1!8!))×(9!/(0!9!)) 9 18 1 2 0 (9!/(1!8!))2 81 1 (9!/(2!7!))×(9!/(0!9!)) 36 153 2 3 ½ (9!/(2!7!))×(9!/(1!8!)) 324 1½ (9!/(3!6!))×(9!/(0!9!)) 84 816 3 4 0 (9!/(2!7!))2 1296 1 (9!/(3!6!))×(9!/(1!8!)) 756 2 (9!/(4!5!))×(9!/(0!9!)) 126 3060 4 5 ½ (9!/(3!6!))×(9!/(2!7!)) 3024 1½ (9!/(4!5!))×(9!/(1!8!)) 1134 2½ (9!/(5!4!))×(9!/(0!9!)) 126 8568 5 6 0 (9!/(3!6!))2 7056 1 (9!/(4!5!))×(9!/(2!7!)) 4536 2 (9!/(5!4!))×(9!/(1!8!)) 1134 3 (9!/(6!3!))×(9!/(0!9!)) 84 18564 6 7 ½ (9!/(4!5!))×(9!/(3!6!)) 10584 1½ (9!/(5!4!))×(9!/(2!7!)) 4536 2½ (9!/(6!3!))×(9!/(1!8!)) 756 3½ (9!/(7!2!))×(9!/(0!9!)) 36 31824 7 8 0 (9!/(4!5!))2 15876 1 (9!/(5!4!))×(9!/(3!6!)) 10584 2 (9!/(6!3!))×(9!/(2!7!)) 3024 3 (9!/(7!2!))×(9!/(1!8!)) 324 4 (9!/(8!1!))×(9!/(0!9!)) 9 43758 8 9 ½ (9!/(5!4!))×(9!/(4!5!)) 15876 1½ (9!/(6!3!))×(9!/(3!6!)) 7056 2½ (9!/(7!2!))×(9!/(2!7!)) 1296 3½ (9!/(8!1!))×(9!/(1!8!)) 81 4½ (9!/(9!0!))×(9!/(0!9!)) 1 48620 9 10 0 (9!/(5!4!))2 15876 1 (9!/(6!3!))×(9!/(4!5!)) 10584 2 (9!/(7!2!))×(9!/(3!6!)) 3024 3 (9!/(8!1!))×(9!/(2!7!)) 324 4 (9!/(9!0!))×(9!/(1!8!)) 9 43758 10 11 ½ (9!/(6!3!))×(9!/(5!4!)) 10584 1½ (9!/(7!2!))×(9!/(4!5!)) 4536 2½ (9!/(8!1!))×(9!/(3!6!)) 756 3½ (9!/(9!0!))×(9!/(2!7!)) 36 31824 11 12 0 (9!/(6!3!))2 7056 1 (9!/(7!2!))×(9!/(5!4!)) 4536 2 (9!/(8!1!))×(9!/(4!5!)) 1134 3 (9!/(9!0!))×(9!/(3!6!)) 84 18564 12 13 ½ (9!/(7!2!))×(9!/(6!3!)) 3024 1½ (9!/(8!1!))×(9!/(5!4!)) 1134 2½ (9!/(9!0!))×(9!/(4!5!)) 126 8568 13 14 0 (9!/(7!2!))2 1296 1 (9!/(8!1!))×(9!/(6!3!)) 756 2 (9!/(9!0!))×(9!/(5!4!)) 126 3060 14 15 ½ (9!/(8!1!))×(9!/(7!2!)) 324 1½ (9!/(9!0!))×(9!/(6!3!)) 84 816 15 16 0 (9!/(8!1!))2 81 1 (9!/(9!0!))×(9!/(7!2!)) 36 153 16 17 ½ (9!/(9!0!))×(9!/(8!1!)) 9 18 17 18 0 (9!/(9!0!))2 1 1 18

If, instead of using nine atomic orbitals where each orbital can hold up to two electrons, 18 atomic orbitals are used where the first nine are of  α spin and the second nine are of β spin, then the total number of states can be calculated more rapidly using the expression 18!/(n!(18-n)!) where "n" is the number of electrons. This symmetry relationship allows the calculation of the total number of states to be evaluated in two different ways.

Three microstates are unique in that they do not interact with any other microstates; these are the zero and 18 electron configurations, and the nine-electron configuration when the magnetic component of spin is 4½. Only the nine-electron microstate is interesting, one of the other two is empty, the other is full. In the nine-electron system all the electrons are spin-up, so the total spin is 4½, and the resulting unique state has a spin-degeneracy of 10, hence it is a Decet.

In the configuration interaction calculation each microstate (configuration) interacts with all the microstates in the set. Solving the sets where the component of spin is 0 or ½ is straightforward when the number of electrons is in the range 0 - 5 or 13 - 18.  When the number of electrons is in the range 6 - 12 the computational effort to generate and manipulate the state wavefunctions rises rapidly, and for the nine-electron system several CPU days of effort are required. Any attempt to solve the state wavefunctions using the 18 atomic orbital sets should be regarded as a wasted effort.  First, the number of permutations rises rapidly as shown in the "Total no. of States" column, so that the nine-electron system would likely require a few CPU months to run.  Second, because states of different component of spin do not interact, verification that the states that differ only by component of spin are degenerate is straightforward.  Third, the total spin of a state is readily calculable, and can be used in constructing the state symbol. Finally, the sets of states for 0, 1, and 2 electrons for the various Ms shown in the table can easily be shown to be equivalent to the sets of states for the 18 atomic orbital permutations.  If, however, these points do not dissuade, it is possible to model 18 atomic orbital sets by using keyword MICROS.

### Results

The following tables show the number of states in terms of their orbital symmetry and their spin.  In the column headed "(2L + 1)" are the orbital degeneracies of the terms, and the various entries in the row headed "M = (2S + 1):" gives the spin degeneracies of the terms.  Orbital angular momentum is indicated by the letters S, P, D, F, G, H, I, K, and L; these map onto the quantum numbers L = 0, 1, 2, 3, 4, 5, 6, 7, and 8. Lower-case letters "g" and "u" indicate whether the state is symmetric (gerade) with respect to inversion or antisymmetric (ungerade). If a state function is ungerade, it changes sign on inversion, i.e., when the Cartesian coordinates x, y, and z are inverted to give -x, -y, and -z.  Column "No. of Terms" gives the sum of the product of the orbital term and its degeneracy.  So in the two-electron system there are three singlet and one triplet D(g) states, so there would be (3 + 1)*5 = 20 terms. This should be equal to the number of states in the column headed "No. of Terms" in the table.  In fact, it is correct for almost all, over 99%, of the sets of electrons.

 1 electron (2L + 1) DOUBLET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 2 Harmonic S(g) 1 1 1 2 P(g) 3 0 0 0 D(g) 5 1 5 10 F(g) 7 0 0 0 G(g) 9 0 0 0 H(g) 11 0 0 0 I(g) 13 0 0 0 K(g) 15 0 0 0 L(g) 17 0 0 0 S(u) 1 0 0 0 P(u) 3 1 3 6 D(u) 5 0 0 0 F(u) 7 0 0 0 G(u) 9 0 0 0 H(u) 11 0 0 0 I(u) 13 0 0 0 K(u) 15 0 0 0 L(u) 17 0 0 0 Totals: 9 18

 2 electrons (2L + 1) SINGLET TRIPLET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 1 3 Harmonic S(g) 1 3 0 3 3 P(g) 3 0 2 6 18 D(g) 5 3 1 20 30 F(g) 7 0 1 7 21 G(g) 9 1 0 9 9 H(g) 11 0 0 0 0 I(g) 13 0 0 0 0 K(g) 15 0 0 0 0 L(g) 17 0 0 0 0 S(u) 1 0 0 0 0 P(u) 3 2 2 12 24 D(u) 5 1 1 10 20 F(u) 7 1 1 14 28 G(u) 9 0 0 0 0 H(u) 11 0 0 0 0 I(u) 13 0 0 0 0 K(u) 15 0 0 0 0 L(u) 17 0 0 0 0 Totals: 81 153

 3 electrons (2L + 1) DOUBLET QUARTET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 2 4 Harmonic S(g) 1 3 0 3 3 P(g) 3 5 4 27 51 D(g) 5 8 1 45 55 F(g) 7 4 3 49 91 G(g) 9 3 0 27 27 H(g) 11 2 0 11 11 I(g) 13 0 0 0 0 K(g) 15 0 0 0 0 L(g) 17 0 0 0 0 S(u) 1 1 2 3 7 P(u) 3 7 2 27 39 D(u) 5 6 3 45 75 F(u) 7 5 2 49 77 G(u) 9 2 1 27 45 H(u) 11 1 0 11 11 I(u) 13 0 0 0 0 K(u) 15 0 0 0 0 L(u) 17 0 0 0 0 Totals: 324 492

 4 electrons (2L + 1) SINGLET TRIPLET QUINTET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 1 3 5 Harmonic S(g) 1 9 2 1 12 20 P(g) 3 5 16 3 72 204 D(g) 5 16 12 4 160 360 F(g) 7 7 14 3 168 448 G(g) 9 9 6 1 144 288 H(g) 11 2 4 0 66 154 I(g) 13 2 0 0 26 26 K(g) 15 0 0 0 0 0 L(g) 17 0 0 0 0 0 S(u) 1 3 6 3 12 36 P(u) 3 10 12 2 72 168 D(u) 5 11 16 5 160 420 F(u) 7 10 12 2 168 392 G(u) 9 6 8 2 144 360 H(u) 11 3 3 0 66 132 I(u) 13 1 1 0 26 52 K(u) 15 0 0 0 0 0 L(u) 17 0 0 0 0 0 Totals: 1296 3060

 5 electrons (2L + 1) DOUBLET QUARTET SEXTET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 2 4 6 Harmonic S(g) 1 14 4 3 21 62 P(g) 3 26 21 2 147 444 D(g) 5 39 20 5 320 9440 F(g) 7 31 21 2 387 1106 G(g) 9 25 11 2 342 954 H(g) 11 12 6 0 198 528 I(g) 13 6 1 0 91 208 K(g) 15 1 0 0 15 30 L(g) 17 0 0 0 0 0 S(u) 1 10 10 1 21 66 P(u) 3 30 16 3 147 426 D(u) 5 35 25 4 320 970 F(u) 7 33 18 3 387 1092 G(u) 9 23 14 1 342 972 H(u) 11 13 5 0 198 506 I(u) 13 5 2 0 91 234 K(u) 15 1 0 0 15 30 L(u) 17 0 0 0 0 0 Totals: 3024 8568

 6 electrons (2L + 1) SINGLET TRIPLET QUINTET SEPTET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 1 3 5 7 Harmonic S(g) 1 24 13 9 2 48 122 P(g) 3 22 53 17 2 282 840 D(g) 5 50 57 26 3 680 1860 F(g) 7 34 61 19 2 812 2282 G(g) 9 37 40 14 1 828 2106 H(g) 11 15 26 5 0 506 1298 I(g) 13 13 8 2 0 312 650 K(g) 15 2 3 0 0 75 165 L(g) 17 1 0 0 0 17 17 S(u) 1 10 17 7 0 34 96 P(u) 3 34 49 19 4 318 912 D(u) 5 38 61 24 1 620 1740 F(u) 7 42 59 20 3 868 2380 G(u) 9 29 42 13 0 756 1980 H(u) 11 19 25 6 0 550 1364 I(u) 13 9 10 1 0 260 572 K(u) 15 3 3 0 0 90 180 L(u) 17 0 0 0 0 0 0 Totals: 7056 18564

 7 electrons (2L + 1) DOUBLET QUARTET SEXTET OCTET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 2 4 6 8 Harmonic S(g) 1 32 15 9 0 56 178 P(g) 3 67 58 12 2 417 1362 D(g) 5 101 65 20 1 935 2950 F(g) 7 91 68 13 1 1211 3780 G(g) 9 77 45 10 0 1188 3546 H(g) 11 45 28 3 0 836 2420 I(g) 13 25 10 1 0 468 1248 K(g) 15 8 3 0 0 165 420 L(g) 17 2 0 0 0 34 68 S(u) 1 22 21 3 0 46 146 P(u) 3 76 53 17 2 444 1446 D(u) 5 91 70 15 1 885 2800 F(u) 7 97 65 16 1 1253 3906 G(u) 9 71 48 7 0 1134 3384 H(u) 11 49 27 4 0 880 2530 I(u) 13 22 11 0 0 429 1144 K(u) 15 9 3 0 0 180 450 L(u) 17 1 0 0 0 17 34 Totals: 10578 31812

 8 electrons (2L + 1) SINGLET TRIPLET QUINTET SEPTET NONET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 1 3 5 7 9 Harmonic S(g) 1 36 25 18 3 0 82 222 P(g) 3 44 99 40 8 1 576 1818 D(g) 5 89 114 55 9 0 1335 3845 F(g) 7 68 125 47 7 0 1729 5089 G(g) 9 73 88 35 3 0 1791 4797 H(g) 11 38 61 17 1 0 1287 3443 I(g) 13 27 26 7 0 0 780 1820 K(g) 15 7 11 1 0 0 285 675 L(g) 17 4 1 0 0 0 85 119 S(u) 1 18 35 18 2 1 74 236 P(u) 3 59 90 40 9 0 594 1776 D(u) 5 74 122 55 8 1 1300 3900 F(u) 7 79 119 47 7 0 1764 5040 G(u) 9 62 94 35 3 0 1746 4860 H(u) 11 43 59 17 1 0 1320 3432 I(u) 13 22 29 7 0 0 754 1872 K(u) 15 9 10 1 0 0 300 660 L(u) 17 2 2 0 0 0 68 136 Totals: 15870 43740

 9 electrons (2L + 1) DOUBLET QUARTET SEXTET OCTET DECET No. of Terms Terms times   Spin Degeneracy M = (2S +1): 2 4 6 8 10 Harmonic S(g) 1 40 22 12 0 0 74 240 P(g) 3 92 82 20 4 0 594 1992 D(g) 5 136 92 30 2 0 1300 4180 F(g) 7 128 100 22 2 0 1764 5628 G(g) 9 110 68 16 0 0 1746 5292 H(g) 11 70 44 6 0 0 1320 3872 I(g) 13 38 18 2 0 0 754 2080 K(g) 15 14 6 0 0 0 300 780 L(g) 17 4 0 0 0 0 68 136 S(u) 1 34 36 9 2 1 82 292 P(u) 3 98 70 22 2 0 576 1872 D(u) 5 130 103 28 4 0 1325 4360 F(u) 7 131 92 23 1 0 1729 5432 G(u) 9 108 76 15 1 0 1800 5562 H(u) 11 73 41 6 0 0 1320 3806 I(u) 13 36 22 2 0 0 780 2236 K(u) 15 13 5 0 0 0 270 690 L(u) 17 4 1 0 0 0 85 204 Totals: 15887 48654

Notes:  The totals for the number of terms and terms time spin degeneracy for the 7, 8, and 9 electron systems are incorrect.  This was not due to a fault in the configuration interaction calculation or to a fault in the symmetry analysis, instead it was due to perfect degeneracy of various states.  Degeneracy of this type is a consequence of the limited C.I., i.e., the use of one "s", three "p", and five "d" atomic orbitals, and because the states were degenerate, the state eigenvectors could not be resolved into their component states.

If space-group symmetry is not used, by adding keyword NOSYM, then the total numbers of terms are correct, and the perfect degeneracy can be seen in the eigenvalues.

Interesting states:
(A) The 9-electron states 6I(g) and 6I(u) have the highest degeneracy for this set of atomic orbitals, 6 × 13 = 78-fold degenerate.
(B) The 9-electron state 10S(u) has the highest spin, 4½, and is totally symmetric to rotation, but is antisymmetric to inversion.
(C) The states L(g) and L(u) have the highest angular momentum, L = 8, in which the highest positive component of momentum is achieved when the atomic orbital occupancy is two electrons, one of
α and one of  β spin, in each atomic orbital that has a positive magnetic quantum number, ml, i.e.:

 s p +1 p 0 p -1 d +2 d +1 d 0 d -1 d -2 0 2 0 0 2 2 0 0 0
(D) All three sets of electrons, 7, 8, and 9, result in large C.I.s that can take up to four CPU days to run.