5.73 Lecture #34
34 - 1
e2/rij and Slater Sum Rule Method
LAST TIME:
1. L2,S2 method for setting up NLMLSMS⟩ many-electron basis
states in terms of linear combination
of Slater determinants
2
* ML = 0, MS = 0 block: L → L + L −
S2 → S + S −
* diagonalize S2 (singlets and triplets)
* diagonalize L2 in same basis that diagonalizes S2
[Recall: to get matrix elements of L2, first evaluate L2 ψ i
and then left multiply by ψ j
2. coupled representations njωls⟩ and NJLSMJ⟩
3. Projection operators: automatic projection of L2
eigenfunctions* remove unwanted L ″ part
* preserve normalization of wanted L′ part
* remove overlap factor
TODAY:
1. Slater Sum Rule Trick (trace invariance): MAIN IDEA OF LECTURE.
2
2. evaluate ∑ e rij matrix elements
(tedious, but good for you)
i> j
[2-e– operator, spatial coordinates only, scalar wrt J,L,S]
* multipole expansion of charge distribution due to “other electrons”
* matrix element selection rules for e2/rij in both Slater determinantal and
many-e– basis sets
* Gaunt Coefficients (ck) (tabulated) and Slater-Condon (Fk,Gk) Coulomb and
Exchange parameters. Because of sum rule, can evaluate mostly
ab r1 ab and ab r1 ba type matrix elements and never ab r1 cd
ij
ij
ij
type matrix elements.
3. Apply Sum Rule Method
4. Hund’s 1st and 2nd Rules
updated September 19,
5.73 Lecture #34
34 - 2
1. Slater’s Sum Rule Method
It is almost always possible to evaluate e2/rij matrix elements without
solving for all |LMLSMS⟩ basis states.
* trace of any Hermitian matrix, expressed in ANY representation, is
the sum of the eigenvalues of that matrix (thus invariant to unitary
transformation)
*
∑e
2
ˆ ˆ has
/ rij and every scalar operator with respect to Jˆ (or L,S)
i> j
nonzero matrix elements diagonal in J and MJ (or L and ML) and
independent of MJ or (ML,MS)
[W-E Theorem: J is a GENERIC ANGULAR MOMENTUM with
respect to which e2/rij is classified]
Recall from definition of r12, that e2/rij is a scalar operator with respect to
ˆ L,
ˆ Sˆ but not with respect to ji or li.
J,
Interelectronic Repulsion:
∑e
2
rij
i> j
* destroys orbital approximation ∅ $$ for electronic structure calculations
* “correlation energy,” “shielding”
r
r12
e1−
r
r1
e2−
r
r2
•
e1− at (r1 ,θ1 , φ1 )
e2− at (r2 ,θ 2 , φ2 )
r
r r
r12 = r2 − r1
scalar with
respect to J,
L, S, si but
not ji, l i
r122 = r12 − 2r1 ⋅ r2 + r22
r r 1/ 2
r12 = r + r − 2 r1 r2 cos(r1 , r2 )
[
2
1
2
2
]
updated September 19,
5.73 Lecture #34
34 - 3
r
expand r12−1 as power series in  < 
 r> 
where r< is smaller of r1 , r2
(integrals evaluated in 2 regions :
r1 < r2 , r2 < r1 )
lengthy algebra see Eyring, Walter, and Kimball “Quantum Chemistry”


pages 369 - 371 and, for relationship between Legendre 
polynomials and Y m θ ,φ , pages 52 - 59.



l
( )
2n-pole moment (n=0 monopole, n=1 dipole,…)
multipole
expansion
∞
1
=∑
rij n = 0
n
∑
m=−n
n
4
π
r

 < Y m θ ,φ Y m θ ,φ *
 2 n + 1 r>n +1 n ( i i ) n ( j j )
[
convergent
series
not principal
q.n.!
]
angular momenta
magnitude n, projection m
scalar product of 2 angular momenta, one
for i-th particle, one for j-th
* converts m to –m
n-pole charge distribution ∅ n-th rank tensor ∅ 2n+1 components
No dependence on s, so 1 / rij is scalar with respect to S, si , s j .
[Y
m
n
(θi , φ i ) =
θi , φ i
li
= n, m i = m
l
]
updated September 19,
5.73 Lecture #34
34 - 4
The reason for this rather complicated looking expansion is that it is
well suited for integrals over atomic orbitals which are expressed in
terms of ri, θi, φi, which are coordinates of the i-th e– with respect to the
center of symmetry (nucleus) rather than the other e–. It enables use of
AO basis states. Otherwise 1/rij integrals would be nightmares.
Selection rules for matrix elements:
not principal q. n.

 ∆l i ≤ n , ∆m i = m , ∆msi = 0
orbitals

 ∆l j ≤ n , ∆m j = −m , ∆ms j = 0
l
l
term in multipole expansion
triangle rule,
li
− l′i ≤ n ≤ l i + l′i
(steps of 2 because of parity)
overall: ∆L = 0, ∆S = 0, ∆ML = 0, ∆MS = 0, and indep. of ML, MS
Can use any ML, MS from box diagram.
It is also clear how to evaluate the angular factors of the atomic orbital matrix
elements using 3-j coefficients. Special tables of “Gaunt Coefficients” (also C&S
pages 178-179, Golding, page 41, see handout).
updated September 19,
5.73 Lecture #34
34 - 5
general 1/r12 matrix element (∆so = 0, 1, and 2 are possible)
e2±
ab 1 r12 cd =
1
cd
r12
ab
e1±
ab
−
1
dc
r12
ab
1
cd = δ ( ms a , msc )δ ( msb , ms d )δ ( m a + m b , m c + m
r12
l
**
1/r12 does not operate
on spin coodinates
**
l
l
∑
k{
=0
ck
(
m a , l cm
la
l
lc
e1–
tensor rank for
product of AOs
occupied by e± # 1
)c (
k
)×
1/r12 scalar with respect
to
Lˆ = ˆl + ˆl
12
∞
ld
1
m b , l dm
lb
l
2
ld
e2–
)×
GAUNT COEFFICIENTS —
ANGULAR FACTOR OF
INTEGRAL
must be same
as for # 2 for scalar
product of n -th rank
tensors
R k (nal a nbl b ncl c nd l d )
radial factor
c
k
(m,
l
l
l
′m ′
l
)
tabulated
triangle rule:
 2l ′ + 1 
≡

 2l + 1 
l
- l′ ≤ k ≤ l + l′
l
+ l′ + k = even
1 /2
k ′
A000
Amk ′− m′ ,m′ ,− m
ll
l l
l
l
l
l
Clebsch-Gordan coefficients
that result from integral
over product of 3 spherical
harmonics — one from
operator, two from orbitals
(from properties of A )
kll ′
000
( parity)
updated September 19,
5.73 Lecture #34
34 - 6
m′ 1 + m = m
e1− integral
restrictions on k and m:
l
l1
n1l 1m 1 Ykm n1′l ′1m ′ 1
l
l
triangle rule
for intraconfiguration matrix elements, Rk(abcd) takes on especially simple
form (because the same one or two orbitals appear in the bra and in the ket).
Rk ( ab, ab) ≡ F k ( a, b) 
“Slater − Condon” parameters
k
k
are reduced matrix elements dependent only on ,
R ( ab, ba) ≡ G ( a, b) (these
, , and not on any of the m quantum numbers.) All
la
lb l c ld
l
L-S states from one configuration are expressed in terms
of the same set of Fk, Gk parameters.
e2
ab
ab
r12
e2
J(a , b) ≡ ab ab =
r12
spins must match
or K term vanishes
= J( a, b) − δ ( ms a , msb )K( a, b)
EXCHANGE
DIRECT
∑ c k (l am , l am )c k (l bm , l bm
∞
k =0
la
la
ak
(
lb
m a , l bm
la
l
lb
) × F (n
k
lb
al a
, nb , l b
)
) [ a * (1)a(1)Opb
ˆ * (2)b(2)dτ dτ
]
∫∫
1
2
charge distributions
(
e2
K (a , b) ≡ ab ba = δ msa ,msb
r12
)∑ [ (
∞
k =0
ck
)] G (n
2
m a , l bm
la
l
lb
b k (l a m a , l b m
l
lb
k
al a
, nbl b
)
)
ˆ (2)b * (2)dτ dτ
[∫∫ a * (1)b(1)Opa
]
1
something else!
for special cases, such as nd 2 , nal a = nbl b and F k = G k
Now we are ready to use tables of c k (or, more conveniently, a k and b k )
to set up e 2 rij matrix
updated September 19,
2
5.73 Lecture #34
34 - 7
( recall 1I, 3 H, 1G, 3 F, 1D, 3 P, 1S)
Easy example: nf 2
I 60 = 3α 3β 

3
H 51 = 3α 2α 
1
these are the only L-S states represented
by a single Slater determinant —
extremes of ML,MS box diagram
since e2/rij is a scalar operator with respect to Lˆ , Sˆ , Jˆ , matrix elements are ML, MS,
MJ independent — so we can use any ML,MS component to evaluate the matrix
element — whichever is most convenient!
e1−
e2−
2
1
I
2
e 1
I = ∑ c k (33, 33)c k (33, 33) F k (nf , nf ) − δ (α , β )∑ [c k (33, 33)] G k (nf , nf )
m
r12
k = 0, 2, 4, 6
k
l
l
=
3
H
∑
[c (33, 33)] F (nf , nf )
∑
{[c (33,33)c (32,32)]F (nf ,nf ) − [c (33,32)] G (nf ,nf )}
k = 0, 2, 4, 6
e2 3
H =
r12
Fk(nf2)
=0
one spin α
2l
k = 0 ,2 ,4 ,6
2
k
k
other spin β
k
k
k
e1−
e2−
(a,a)
(b,b)
2
k
k
both spins α
F k (nf 2 )
(a,b)
Here is where everyone makes mistakes!
Use table of ck in Golding/C&S handout (C&S page 179).
Note that [1/7361•64]1/2 is implicit after the first entry for f2, k = 6.
k=0
2
4
6
ck(33,33)
1
–1/3
1/11
–[1/7361•64]1/2
ck(32,32)
1
0
–7/33
–[6/7361•64]1/2
ck(33,32)
0
+1/3
–301/2/33
–[7/7361•64]1/2
Dk
1
225
1089 = 332
7361•64
convenient
factor
updated September 19,
5.73 Lecture #34
34 - 8
Dk is a factor that simplifies the expressions. Each term has the form Fk/Dk. Call
this ratio Fk. Get simpler looking expressions when you replace Fk by DkFk (Dk
appears in denominators of ck as […/Dk]1/2 )
1
 1
 1  4 
 6
e2 1
1
I
I = F0 +   F2 + 
F +
F
 9
 121
 7361 ⋅ 64 
r12
= F0 + 25 F2 + 9 F4 + F6
3
Always have two
factors of ck. Thus Fk
gets divided by Dk to
yield Fk.
 1 
 1   −7 
2
30  4  −6 − 7  6
e2 3
H
H = F 0 +  −  (0) − (1 3)  F 2 +     −
F + 
F
r12
 7361 ⋅ 64 
 3 

 11  33  33 ⋅ 33 
= F0 −
−13
1 2
51
4
F −
F
F6
2
9
7361 ⋅ 64
(33)
= F0 − 25 F2 − 51F4 − 13 F6
A lot of book – keeping, but easy to learn how to use tables of ck , a k , b k , D k .
But it is much more work for f 3 than for f 2 .
SUM RULE METHOD:
Basic idea is that the sum of diagonal elements in the single Slater determinant basis
set within an ML, MS box is equal to the sum of the eigenvalues!
Look at M L = 3, M S = 1 box:
3α 0α and 2α 1α . This box generates
but trace is E ( 3 H ) + E ( 3 F ) and we already know E ( 3 H )!
So
3
H 31 and
3
F 31 ,
E( 1 I ) = 3α3β
E( 3 H ) = 3α 2α
E( 3 F ) = 3α 0α + 2α1α − E( 3 H )
E( 1G) = 3α1β + 3β1α + 2α 2β − E( 1 I ) − E( 3 H )
E( 1 D) = 3α − 1β + 3β − 1α + 2α 0β + 2β0α
+ 1α1β − E( 1 I ) − E( 1G) − E( 3 H ) − E( 3 F )
E( 3 P) = 3α − 2α + 2α − 1α + 1α 0α − E( 3 H ) − E( 3 F )
E( 1S ) = sum of seven
– sum of six E( 2 S +1 L)
updated September 19,
5.73 Lecture #34
34 - 9
This seems rather laborious, but it is much easier than:
generating each LML = L SMS = S⟩ as an explicit linear combination of Slater
determinants
* then calculating matrix elements of e2/rij, because there are many nonzero offdiagonal matrix elements between Slater determinants in the same ML,MS box.
*
2 2S+1
Here is the final result for the energies of all (nf)
( )
( )
L terms:
( )
E = E0 +E1 +E2
E
(0 )
= sum of orbital energies from h
(0 )
=−
Z 2ℜ
n2
Bare nucleus
hydrogenic orbital
energy — or partly
shielded by filled shells.
= εn
l
2
( )
SO
H2
E1 = e
+ 1
3
rij
123 next lecture
ready now
E
(2 )
(
= (intraconfigurational spin - orbit) + interconfigurational e 2 rij
shielded
by all filled
subshells
For nf2
CI
shielded
by same
subshell
I
2ε nf + F0 (nf 2 )
+ 25 F2 nf
3
H
2ε nf
+
F0
– 25 F2
– 51 F4
– 13 F6
1
G
2ε nf
+ F0
– 30 F2
+ 97 F4
+ 78 F6
3
F
2ε nf
+
F0
– 10 F2
– 33 F4
– 286 F6
1
D
2ε nf
+
F0
+ 19 F2
– 99 F4
+ 715 F6
3
P
2ε nf
+
F0
+ 45 F2
+ 33 F4
– 1287 F6
1
S
2ε nf
+
F0
+ 60 F2
+ 198 F4
+ 1716 F6
1
)
shielded-core
configurational
energy
(
2
)
(
+ 9 F4 nf
2
)
(
+ F6 nf
2
)
intraconfiguration L-S term splittings
(there is NO center of Gravity Rule for degeneracy weighted L-S terms)
updated September 19,
5.73 Lecture #34
34 - 10
Now it is easy to show that all Fk’s are > 0 and Fk >> Fk+2 etc. (roughly factor of 10
per step in k)
From this we get an empirical rule
Lowest E of all L–S terms is the one with
*
*
MAXIMUM S
of those with Maximum S, lowest is the one with MAXIMUM L
These are Hund’s first and second (of three) rules.
Note also that Hund’s rules do nothing about predicting the energy order of L-S terms
except for the identity of the single, lowest energy L-S term.
updated September 19,
5.73 Lecture #34
34 - 11
Nonlecture
There are several interesting problems also solved by this e2/rij formalism.
1. Energy splittings between and Slater determinantal characters of two or more L,S
terms of the same L and S that belong to the same L,S configuration
e.g. d 3 → two 2D terms
see pages 47 - 50 of Golding for 2 × 2 secular determinant for 2D of d 3
2. matrix elements of e2/rij between same–L,S terms that belong to two different
configurations
e.g.
nd 2
1
S,3P,1D,3F,1G
 1S,3P,1D,3F,1G 
ndn ′d  3 1 3 1 3 no Pauli restrictions
 S, P, D, F, G 
so there will be
1
S ~1S
3
1
P ~ 3P
D~1D
3
F ~ 3F
1
G ~1G
interconfigurational CI’s, and each of these 5 interaction matrix elements will
NOT be of the same magnitude.
updated September 19,