5.73 Lecture #30
30 - 1
Matrix Elements of Many-Electron Wavefunctions
Last time:
1 /2
 ℜ 
ν = −
noninteger principal quantum number

 En , 
f (ν , r) 
solutions to Schröd. Eq. outside sphere of radius r0

g (ν , r)
l
l
l
need both f and g to satisfy boundary condition for E < 0 as r → ∞
ν=n−µ
l
πµ is phase shift of f ( ν, r )
l
l
infinite set of integer-spaced ν-values that satisfy r → ∞ boundary
condition
Wave emerges from core with ν-independent phase. Core transforms
wave with correct r → 0 limiting behavior into one that exits imaginary
sphere of radius r0, which contains the core region, with πµl phase shift.
Core sampled by set of different l’s.
Today:
Wavefunctions and Energy States of many-electron atoms
1.
orbitals → configurations → L - S states
2.
electrons are Fermions → ψ must be “ antisymmetrized”: KEY PROBLEM
3.
Slater determinants are antisymmetric wrt all e −i , e −j permutations
A.
Normalization
B.
Matrix Elements of one - e – Operators: e.g. HSO =
∑
i
C.
next few
lectures
4.
Matrix Elements of two - e – Operators: e.g. He =
(a very bad “perturbation”)
H eff in terms of ε n ,
l
∑
( )
a ri
li
⋅ si
e 2 rij
i> j
F k ,G k , ζ n parameters
123
orbital Slater-Condon
1/ rij
energy
l
spinorbit
Page 31-9 is an example of what we will be able to do.
* Interpretable trends: Periodic Table
* Atomic energy levels: mysterious code — no atom-to-atom relationships
evident without magic decoder ring.
updated September 19,
5.73 Lecture #30
30 - 2
Many-electron H
( )
2
e
H = ∑ h(i) + ∑
+ ∑ a ri l i ⋅ s i
rij
i=1
i > j= 1
i
12
4 4
3 1
4444244443
(0)
N
N
H
H (1)
sum of hydrogenic
1- e – terms: −
Z 2ℜ
n2
(unshielded orbital energies)
= −ε n
How do we set up matrix representation of this H?
H(0) defines basis set (complete, orthonormal, …)
(0)
H
=
h(i) ↔ ψ = ∏
N
N
∑
i =1
i =1
spatial
part
spin
part
[ φ (r ) sm (i) ]
i
i
s
“spin-orbital”
the φi(ri)’s could be
hydrogenic or
shielded-core
Rydberg-like orbitals.
[h(1) + h(2)]φ (1)φ (2) = [h(1)φ(1)]φ(2) + [h(2)φ(2)]φ(1)
a constant with respect to h(1).
H as sum, E as sum, ψ as product
Electronic Configuration: list of orbital occupancies
e.g. C 1s22s22p2 six e–
not sufficient to specify state of system
several L , S terms arise from this configuration: e.g. p2 →1D, 3P , 1S
r
r
r
r
L ≡ ∑ li
S ≡ ∑ si
i
we know that
i
2
L , Lz , S
2
e
, S z commute with h(i) +
2
rij
so we can use these to block diagonalize H.
Note that although
r
li
does not commute with e 2 rij , this is not
a problem for s2i and siz because h(i) and rij do not involve spin.
rij destroys l i but not L!
updated September 19,
5.73 Lecture #30
30 - 3
How do we get eigenstates of L2 , L z , S2 , S z ?
either:
I.
Method of ML, MS boxes
Advanced Inorganic
Which L–S terms exist, not the specific linear combinations
of spin-orbital products that correspond to these terms.
II.
Angular momentum coupling techniques
3-j
ladders plus orthogonality
projection operators
We will return to this problem and approach it both ways.
One rigorous symmetry must be imposed:
Pauli Exclusion Principle: electrons are Fermions and therefore any acceptable
wavefunction must be antisymmetric with respect to permutation of ANY pair of e–
electrons
1, 2 = u1(1) u2 ( 2)
e.g.
orbitals
P12 1, 2 = u1( 2) u2 (1) ≡ 2,1
e – are indistinguishable, ∴ [ H , Pij ] = 0
∴ all ψ ' s must belong to + or – eigenvalue of Pij (note that Pij2 = I)
+
–
[
Boson
Fermion
ψ ± = 2−1/2 1, 2 ± 2,1
[
(
integer spin)
(1/2 integer spin)
]
ψ+
bosons
ψ–
fermions
]
P12ψ = 2−1/2 2,1 ± 1, 2 = ±ψ ±
updated September 19,
5.73 Lecture #30
30 - 4
generalize to 3 e–? 3! combinations needed! Horrible
ψ’s have N! terms (each a product of N spin-orbitals)
matrix elements have (N!)2 additive terms!
TRICK! Slater Determinants
e–
u1(1) u2 (1)
= u1(1)u2 ( 2) − u1( 2)u2 (1)
u1( 2) u2 ( 2)
–
orbital
row is label for e
column is label for spin-orbital
you show that 3 × 3 Slater determinant gives 6 additive product terms
Determinants N × N
*
*
*
*
N! terms in expansion of determinant
determinant changes sign upon permutation of ANY two
rows [e–’s] or columns [spin-orbitals]
determinant is zero if any two rows or columns are
identical.
determinant may be uniquely specified by main diagonal
MUST SPECIFY IN ADVANCE A STANDARD ORDER
IN WHICH THE SPIN-ORBITALS ARE TO BE LISTED
ALONG MAIN DIAGONAL
e.g.
sα, sβ, p1α, p1β, p0α, p0β, p - 1α, p - 1β, …
[or for pN, suppress p in notation: 1α1β0α ∅ ML = 2, MS = 1/2]
Need a fancy notation to demonstrate how Slater determinants are to be
manipulated in evaluating matrix elements. This notation is to be forgotten as
soon as it has served its immediate purpose here.
updated September 19,
5.73 Lecture #30
30 - 5
# of binary permutations
ψN
−1/2
p
(
)
(
)
= N!
∑ −1 ℘[ u1(1) … uN ( N ) ]
℘
N! different ℘’s
℘ is ONE prescription for rearranging the orbitals from the initially
specified order
℘ is product of several Pij’s or, more useful for proving theorems, a product
of N factors Pi which tell whether the i-th electron is to be left in the i-th
spin-orbital or transferrred to some unspecified spin-orbital
[
]
N
℘ u1(1) … uN ( N ) = ∏
Piui (i )
i =1
A. Normalization
Verify that ( N!)
−1 / 2
is correct normalization factor
ψ N ψ N = (N!)−1 ∑
℘,℘′
(−1)p + p′℘[ u1 (1) … uN (N) ]℘′[ u1 (1) … uN (N) ].
Now rearrange into products of one - e – overlap integrals,
N
ψ N ψ N = (N!)−1 ∑
℘,℘′
(−1)p + p′ ∏
i= 1
Piu i (i) Pi′u i (i) .
ui are orthonormal
u(i) u(j) has no meaning because bra and ket must be associated with same e–
The only nonzero legal terms in
∑ are those where EACH Pi = Pi′
℘℘′
otherwise there will be AT LEAST 2 mis - matched 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-orbital
quantum numbers do not match.)
updated September 19,
5.73 Lecture #30
30 - 6
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 #30
30 - 7
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 #30
30 - 8
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 ]
next time G(i,j)
updated September 19,