Download Atomic Term Symbols

Atomic Term Symbols.
In freshman chemistry you speak of the electronic configuration of an atom, e.g.
1s 2 2s 2 2 p 2 for the Carbon atom. There are 6 2p orbitals for carbon that we can label
p1 , p0 , p−1 , p1 , p0 , p−1 , 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 porbital implies l = 1 . The number of Slater determinants or so-called microconfigurations is the number of ways you can distribute 2 electrons over 6 spin-orbitals
⎛6⎞
6! 6 ⋅ 5
where the order doesn’t matter. This is ⎜ ⎟ =
=
= 15 . Let us denote the Slater
2
⎝ 2 ⎠ 2!4!
determinants as Φ k , k = 1..15 . This means that the 1s 2 2s 2 2 p 2 electronic configuration for
Carbon really indicates 15 different quantum states! These states would all have the same
1
energy, if not for the electron-electron repulsion operator ∑ in the Hamiltonian (same
i < j rij
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
Ψα = ∑ Φ k ckα
k =1
If we substitute this in the Schrödinger equation we get
∑ Hˆ Φ k ckα = ∑ Φ k ckα Eα
k
k
To get a well defined set of equations we integrate on both sides against Φ m , m = 1..15 ,
hence
∑
k
Φ m Hˆ Φ k ckα = ∑ Φ m Φ k ckα Eα
k
= ∑ δ mk ckα Eα
∑
k
H mk ckα = cmα Eα ∀m = 1...15
k
where we used the orthonormality of the Slater determinants. This is a so-called matrix
eigenvalue equation. It looks just like the Schrödinger equation HC = C E , 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
H mk = ∫∫∫ d 1d 2...d 6 Φ*m (1..6) Hˆ Φ k (1..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
= ∑ lˆz (i ) operator by summing the m values (-1, 0
determine the eigenvalue of the Lˆtotal
z
i
or 1 here) of the occupied orbitals in the Slater determinant. Likewise you can determine
the eigenvalue of the Sˆztotal = ∑ sˆz (i ) , by summing the ms values (-1/2, 1/2) for the
i
occupied orbitals in a determinant. The eigenstates of the Hamiltonian can be shown to
eigenfunctions of Lˆ2,total , Lˆtotal
, Sˆ 2,total , Sˆztotal (this is strictly valid only if we neglect the
z
rather minor effect of spin-orbit coupling, a relativistic effect). Therefore, every
eigenstate can be labelled as L, M , S , M S , M = − L...L, M S = − S ...S . The energy only
depends on the L, S quantum numbers, not on M , M S . Hence each E ( L, S ) energy level
has a degeneracy of (2 L + 1)(2 S + 1) . It is customary to use the letter-labels S, P, D, F, G
for total angular momentum L = 0,1, 2,3, 4 in analogy with the use of s, p, d , f , g orbitals
corresponding to the orbital angular momentum l = 0,1, 2,3, 4,... The spin quantum
number is indicated by the spin-degeneracy (2 S + 1) . The so-called atomic term symbol
(without spin-orbit coupling) indicates a particular energy eigenvalue and is indicated as
(2 S +1)
L , e.g. 1D, 3 P, 2 S , that have degeneracy 5, 9 and 2 respectively.
The story is not quite complete yet, because the true Hamiltonian also includes the spinorbit coupling or L ⋅ S coupling. As we have discussed for the hydrogen atom in some
detail, the operators Lˆ2 , Sˆ 2 , Jˆ 2 , Jˆ z , all commute, and they also commute with the
Hamiltonian, including spin-orbit coupling. The operators Lˆz , Sˆz do not commute with
the spin-orbit term in Ĥ , and should then not be used to label the states. We can
construct common eigenfunctions, which we can give quantum numbers n, L, S , J , M J .
The splitting due to spin-orbit coupling can be well approximated by making
eigenfunctions of the Jˆ 2 , Jˆ z operators from linear combinations of states in the (2 S +1) 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 S +1) L multiplet range
from J = L + S ... L − S . In each J-multiplet there are (2J+1) states having M J = − 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 S +1) LJ . As an example the 3 P level (S=1,L=1) spits into
J = 2,1,0 levels denoted 3 P2 , 3 P1 , 3 P0 , 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 D5/ 2 , 2 D3/ 2 . 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 M S → S value and of these the
highest M L → L value. Only the open shells are important here. For Carbon this yields
p1 p0 → M S = S = 1, M = L = 1 . This state is a member of the 3 P 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 M L → L value and of these the
determinant with the highest M S → S value. For carbon this yields
p1 p1 , M = L = 2, M S = S = 0 , 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 P2 (5),3 P1 (3),3 P0 (1), 1D2 (5), 1S0 (1) , 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.
⎛ 6 ⎞ 6! 6 ⋅ 5 ⋅ 4
1. Number of microstates ⎜ ⎟ =
=
= 20
3!
⎝ 3 ⎠ 3!3!
2. Highest M S = S determinant: p1 p0 p−1 → 4 S This is the lowest energy level (4 states)
3. Highest M L = L determinant: p1 p1 p0 → 2 D . This is the first excited energy level (10
states).
4. We have 14 out of 20 states accounted for. The remaining multiplet is 2 P with 6
states.
5. Including spin-orbit coupling we get the term symbols:
4
S3/ 2 (4), 2 D5/ 2 (6), 2 D3/ 2 (4), 2 P3/ 2 (4),2 P1/ 2 (2) , 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.