5.73 Lecture #31
31 - 1
MATRIX ELEMENTS OF F(i) AND G(i,j)
Last time:
orbitals → configurations → states (“terms”)
Fermions: Slater Determinants: Pauli Exclusion Principle
Notation for Slater Determinant: main diagonal .
TODAY:
1. SLATER DETERMINANTAL MATRIX ELEMENTS
A. Normalization
B. F(i) One - e – operator
e.g.
H SO = ∑ a(ri )l i ⋅ si
i
–
C. G(i, j) Two - e operator e.g.
H = ∑ e 2 rij
e
i> j
Recall: specify standard order (because Determinant changes sign upon binary
permutation)
Goal: make inconvenience of Slater determinants almost vanish — matrix
elements will be almost what you expect for simple non-antisymmetrized
products of spin-orbitals.
pages 31-2,3,4 are repeat of 30-6, 7,8
ψ N = ( N!)
A. Normalization:
−1 / 2
( −1) p℘[ u1 (1) … uN ( N )
∑
℘
]
verify that (N!)–1/2 is correct factor
ψ N ψ N = (N!) −1
∑
℘ ,℘′
[
] [
(−1)p + p′℘ u1 (1) … u N (N) ℘′ u1 (1) … u N (N)
]
rearrange into products of one - e - overlap integrals:
= (N!) −1
∑
℘ ,℘′
N
(−1)p + p′ ∏ Pi ui Pi′ui
i =1
*
ui are othonormal
*
u(i) u(j) has no meaning because bra and ket must be associated with SAME e–
only nonzero LEGAL terms in
∑
are those where EACH Pi = Pi′ otherwise get AT LEAST
℘,℘′
*
2 MISMATCHED bra - kets
ui ( k ) u j ( k ) … u j ( l ) ui ( l )
=0
=0
(Here the electron names match in each bra-ket but the spin-orbitals do not
match.)
Think of a one- or two-e– operator as a scheme for dealing with or “hiding” the
small number of mismatched spin-orbitals.
updated September 19,
5.73 Lecture #31
31 - 2
Thus it is necessary that ℘ = ℘′ , p = p′ , (−1) p + p ′ = +1
and ψ N ψ N
−1
(
)
= N! ∑
[
℘ u1 (1) u1 (1) … uN (N ) uN (N )
℘
=1
]
=1
each term in sum over ℘ gives + 1, but there are N possibilities for
P1 , N − 1 possibilities for P2
∴ N ! possibilities for sum over ℘
ψ N ψ N = ( N !)
Thus the assumed (N!)
–1/2
−1
∑ 1=1
℘
normalization factor is correct.
B. Matrix elements of one-electron operators
F=
∑
( )
i
ψA
e.g.
f ri
r
r
L = ∑ li
i
−1/ 2
p
(
)
(
)
≡ N!
∑ −1 ℘ a1 (1)
… a N (N )
℘
ψ B ≡ ( N !)
−1/ 2
p′
(
)
1
℘′ b1 (1)
−
∑
… bN (N )
℘′
ψ A Fψ B
−1
p + p′
(
)
(
)
= N! ∑
−1
℘[ a1(1) …] f ( ri )℘′[ b1(1) …]
i ,℘,℘′
−1
p + p′
(
)
(
)
= N! ∑
−1
[ P1a1(1) P1b1(1) ]
i ,℘,℘′
( )
…  Pi a i (i ) f ri Pi′bi (i )

[
 … P a ( N ) P′ b ( N )
N N
N N

]
Product of N orbital matrix element factors in each term of sum. Of these, N–1 are
orbital overlap integrals and only one involves the one-e– operator.
updated September 19,
5.73 Lecture #31
31 - 3
SELECTION RULE ΨA F ΨB = 0 if ψ A and ψ B differ by more than one spin - orbital
(at least one of the orbital overlap integrals would be zero)
two cases remain:
1. differ by one spin-orbital
ψ A = u1(1)… a k ( k)…uN ( N )  the mismatched orbitals

ψ B = u1(1)… bk ( k)…uN ( N )  are in the same position
use ui to denote common spin-orbitals
use ak, bk ≠ 0 to denote unique spin-orbitals
for this choice, all N Pi factors of each ℘ must be identical to all N factors of ℘′
additional requirement: ℘ must bring mismatched orbitals into i-th position
so that they match up with the f(ri) operator to give a k (i ) f ( ri ) bk (i )
ANY OTHER ARRANGEMENT GIVES
( )
a ( l ) bk ( l ) ui (i ) f ri ui (i ) = 0
1k4
4244
3 1442443
=0
≠0
(N – 1)! ways of arranging the e– in the other N – 1 matched orbitals and
there are N identical terms (in which the e– is in the privileged location) in
the sum over i
−1
ψ A F ψ B = ( N!) ( N − 1)! N ak f bk
updated September 19,
5.73 Lecture #31
31 - 4
If the order of spin-orbitals in ψA or ψB must be arranged away from the
standard order in order to match the positions of ak and bk, then we get an
additional factor of (–1)p where p is the number of binary permutations
p
(
)
= −1 ak f bk
ψA F ψB
i.e.
for difference
of one
spin-orbital
A = 12 5 7
B = 12 3 5 = − 12 5 3
ψ A Fψ B = − 7 F 3
2. ψA = ψB
Differ by zero spin-orbitals
ψA F ψB
−1
(
)
= N! ∑
i ,℘
[ P a (i ) f ( r ) P a (i ) ]
i i
i
all other factors are =1
[
N ! identical terms from sum over ℘ again ( N − 1)! N
ψ A Fψ B =
∑
]
( )
a i f ri a i
i
Normalization

* 1 − e − Operator F
*
Examples of f3:
i i
comes out almost the same as naive
expectation WITHOUT need for
antisymmetrization!
ψ = 3α1α − 2α
L z = h(3 + 1 − 2)
L z S z = h2
( 32 + 12 − 1)
J + 3α1α − 2α = L + 3α1α − 2α + S + 3α1α − 2α
=
[0 + 101/2 3α 2α − 2α
h
+ 101/2 3α1α − 1α + 0 + 0 + 0 ]
now G(i,j)
updated September 19,
5.73 Lecture #31
31 - 5
C. G(i,j) : 4 cases
1. differ by more than 2 spin-orbitals: Matrix Element → 0
2. differ by 2 spin-orbitals: one pair of nonzero matrix elements
3. differ by 1 spin-orbital: sum over pairs of nonzero matrix elements
4. expectation value : differ by 0 spin-orbitals: double sum over pairs of
matrix elements
1. is obvious — only way to make up for orbital mismatch is to hide the
mismatched orbitals in ⟨|g(i,j)|⟩ (rather than in an overlap integral). But one
can only hide 2-mis-matched pairs in, e.g.
ai a j g(i, j ) bi b j
ψ A G(i, j ) ψ b = 0
if ψA, ψB differ by more than 2 pairs of
spin-orbitals
2. differ by two pairs of spin-orbitals
ψ A = u1 (1) … a1 (i ) … a 2 ( j) … uN (N )
ψ B = (−1) p u1 (1) … b1 (i ) … b2 ( j) … uN (N )
permutations needed to put b1
and b2 in the i and j positions
−1
ψ A G ψ B = ( N!) ∑
i> j
[ P a (i )
i 1
∑
℘,℘′
( −1) p + p ′ [orthogonality integrals] ×
Pj a2 ( j ) g(i, j ) Pi′b1 (i ) Pj′b2 ( j )
]
*
are (N – 2)! ways of permuting the N – 2 matched uk functions that are
not filled with e– i and j. Moreover these permutations must involve
Pk = P′k (all k ≠ i,j).
*
also N(N – 1) identical terms in sum over i > j
updated September 19,
5.73 Lecture #31
31 - 6
Thus there are ( N − 2)!( N − 1) N = N! identical terms in ∑
∑ sums.
i > j ℘,℘′
But there are still two possibilities:
1. ℘ = ℘′ ∴ p = p′ and Pi = Pi′, Pj = Pj′
2. ℘ same as ℘′ except for i, j pair where
THIS MEANS WE
PUT THE j-th e– in
P′j where WE PUT
THE i-th e– in Pi
Pi = Pj′
(−1) p+ p′ = −1
Pj = Pi′
the 2 ℘’s differ by one binary permutation
[
THUS: ψ A G ψ B = ± a1 (1)a2 (2) g(1, 2) b1 (1)b2 (2) − a1 (1)a2 (2) g(1, 2) b2 (1)b1 (2)
For ψA,ψB different
by 2 spin-orbitals
# of permutations needed to make ψB match ψA
— no sign ambiguity if standard order is
initially specified
3. ψA,ψB differ by only one pair of spin-orbitals
You work this out
ψ A G ψ B = ±∑
n
(a matched with un,
un matched with b)
(a,b matched)
[ a(1)u (2) g(1, 2) b(1)u (2)
n
n
− a(1)un (2) g(1, 2) un (1)b(2)
]
not arbitrary
4. differ by zero spin-orbitals : expectation value
ψA GψA =
∑ [ u (1)u
n
m
(2) g(1, 2) un (1)um (2) − un (1)um (2) g(1, 2) um (1)un (2)
n>m
DIRECT
what we would expect
without antisymmetrization
EXCHANGE
unexpected: consequence
of antisymmetrization
ρ(n1 )g(1 ,2 )ρ(m2 )
updated September 19,
]
]
5.73 Lecture #31
31 - 7
The ONLY real surprise that results from the antisymmetrization requirement
for two-electron operators is one extra term (and some signs) that has no
counterpart if antisymmetrization had been ignored.
SUMMARY
* antisymmetrize → Slater determinants
* matrix elements are hardly more complicated than those of simple spinorbital products
•signs due to permutation [Standard order]
•extra terms in G(i,j)
Do some examples for p2
1.
2.
3.
What L,S terms belong to p2 (Lecture #32: method of crossing out microstates)
What is the correct linear combination of Slater determinants that corresponds to
a specific L-S term in either the JLSMJ⟩ or the LMLSMS⟩ basis set
•ladders plus orthogonality (Lecture #32)
•L2 and S2 matrices
•3-j coefficients
e 2 rij → F k (nl, n ′l ′), G k (nl, n′λ ′) Slater Condon parameters
relative energies of L - S terms expressed in terms of F k and G k ' s
4.
Matrix elements of HSO
• ζ (NLS)
• ζ (NLS) ↔ ζnl
• full HSO in terms of ζnl
updated September 19,
5.73 Lecture #31
31 - 8
EXAMPLES:
Slater : p2 1α1β →1D
M L = 2, M S = 0
Lz = 1α1β Lz 1α1β
=
Sz
[1 + 1] = 2h
h
1  1  
= h  +  −   = 0h
 2  2 
tricky! L2z = ∑
L2z
2
lz
i
i
+∑
–
The only 2 terms in
sum are 1,2 & 2,1
L2z =
h
2
[12 + 12 ] + ∑
i≠ j
1–e
operator
Sz2 =
h
2
2
2
i≠ j
Easier to do this by
applying L z = ∑ l zi
i
–
2–e
operator
twice
 1α1β l l 1α1β − 1α1β l l 1β1α 
zi zj
zi zj


from spin-mismatch
2
h
= 2 h + h [1 + 1 − 0 − 0] = 4
2
lz lz
i
j
2
as expected
1
1 1 

2 1
2
0
0
h −
+
+
−
−
−
 4 4 
 4 4
 = 0 h as expected
L2 − L2z = L2x + L2y =
(
(
1
L+ L− + L− L+
2
)
)
1
L+ L− + L− L+ + L2z
2
can you show L2 = h2 6 for 1α1β of p2 ?
L2 =
updated September 19,
5.73 Lecture #31
31 - 9
Patterns of Lowest-Lying States: “Aufbau” for adults!
Atom
C
N
O
lowest config.
1s22s22p2
1s22s22p3
1s22s22p4
1
4
1
L-S terms
S, 1D, 3P
3
Lowest Term
S, 2P, 2D
4
P0
(regular)
excitation
[lowest L-S-J
term of each
configuration]
characteristic
transition
(2p ↔2s)
3
D
3
S 1
5
P2
(inverted)
resultant
configuration
S, 3,1P, 3,1D
[5 S 2 ]
1,3
P
3
[ P0 ]
1,3
F, 1,3P
[3 F 2 ]
2s2p4
2s2p5
2s22p23s
2s22p33s
2s22p23d
2s22p33d
2,4
2,4
P, 2D, 2S
[4P5/2]
2,4
P, 2D, 2S
[4P1/2]
1,3
P
[ P2 ]
5,3
S, 3,1P, 3,1D
[5 S 2 ]
3
F, 2,4P, 2D 2G, 2S
[4F3/2]
5,3
D, 3,1F, 3,1P, 1,3G, 1,3S
[5D0]
“regular”
“inverted”
5/2
4
lowest L-S
states of the
two relevant
configuration
3
S3/2
(no fine structure)
Transitions to 2p← 2s 2s2p3
lowest
2
2p3s
3s
←
2p
2s
configurations
∆l = ±1
3d ← 2p 2s22p3d
5,3
S, 1D, 3P
S2
3
P 3/2
1/2
0
Pi 1
2
“inverted”
4
2
3
P 1
0
1/2
Pi 3/2
5/2
3
4
S
3/2
0
Pi 1
2
ζ2p(C) < ζ2p(N) < ζ2p(O)
updated September 19,