Diagrammatic representation of RS perturbation theory
first used in QE by R. Feynman
Diagrammatic Perturbation Theory for Two States
• lowest eigenvalue of two state problem
Perturbation V is presented by dot.
Two zero order states |1> and |2>
are lines with arrows
Hole
Particle
Simple representation of matrix elements V
<label of line in | V | label of line out >
N-th order energy, N=1,..4 is product of n matrix elements V, each matrix
elements is dot with line going into it and line coming out of it => algebraic
representation by pictures containing N dot connected in some way.
Rules:
NOT ALLOWED
1)
N dot vertically
2) Connect all n
dots together with
continuous line
3) Do it in all possible
distinct ways. Two diagram
are equivalent if each and
every dot is connected to an
identical pair of dots
IDENTICAL
“rubber bands” glued to dots
Diagrams to 4th order
Only distinct diagrams
Arrows (1 is label
for down line, 2
for up line)
N=3
two possibility
of labeling
Circular lines
are hole lines
Correspondence between pictures and nth order of energy
Rules to translating pictures to algebraic expression:
a) Each dot contributes a factor < label in |V| label out> to numerator
b) Each pair of adjaced dots contributes to the denominator factor
∑E
(0)
hole
(0)
− ∑ E particule
where the sums runs over label of all holes and particle lines crossing an imaginary
horizontal line separating two adjacent dots
c) The sign of expression is (-)h+l ,where h is number of hole lines and l is number of closed
loops
Ex.
Upper two dots contribute by factor
remaining by factor
1 V 2 = V12
Line A and C contributes by factor
(0)
(0)
2
E
−
2
E
1
2
line B by
(
)
2 V 1 = V21
to the numerator.
E1( 0 ) − E2( 0 )
Since we have two holes (h = 2) and one closed
loop (l = 1) => (-1)3
Algebraic expression for following pair in diagram
We can distort given diagram as long as we
don’t change the vertical ordering of dots.
First order
Translation diagram into formulas
Ex. Write down and evaluate all fifth-order diagrams that have property that
imaginary horizontal line crosses only one hole and one particle line. Show that the
3
V
V
(
V
−
V
)
12
21
22
11
sum of such diagrams is
( E1( 0 ) − E2( 0 ) ) 4
Hint: there are eight such diagrams and they can be generated by adding
three dots to the second-order diagram
Diagrammatic perturbation theory for N states
N-state system, still only one hole state |1> but N-1 particle states |n>, n=2,3, ..,4.
Any particle line can be labeled by index n. So
or generally, m and n is in depended from 2, ..N
Subset of diagrams where m=n is called diagonal
d) Sum the expression over all particle indices
Second order energy (klick on the form)
Third order of energy
What if we want expression for some state i, which is not necessary the lowest?
Label hole line by i and particle line as k, m,n from 1, 2,..,.i-1, i+1, ..,N
Diagrammatic RS
Ex. Using diagrammatic techniques to obtain 4th order perturbation theory of a
particular state i of N-state system. That is evaluate the diagrams
Where m, n, k exclude i. Obtain also result from algebraic expression and compare it together
Summation of diagrams
Two state system ,2nd order of energy
All diagram
by adding
dots
then
or
with
Δ=
⎡ V12V21 ⎤ ⎡V12V21 (V22 − V11 ) ⎤ ⎡V12V21 (V22 − V11 ) 2 ⎤ ⎡V12V21 (V22 − V11 ) 3 ⎤
Δ = ⎢ (0)
+⎢
+⎢
+⎢
+…
(0) ⎥
(0)
(0) 2 ⎥
(0)
(0) 3 ⎥
(0)
(0) 4 ⎥
−
−
−
−
E
E
(
E
E
)
(
E
E
)
(
E
E
)
2 ⎦
2
1
2
1
2
⎣ 1
⎣ 1
⎦ ⎣
⎦ ⎣
⎦
2
3
⎤
V12V21 ⎡ ⎛ V22 − V11 ⎞ ⎛ V22 − V11 ⎞ ⎛ V22 − V11 ⎞
⎜
⎟
⎜
⎟
⎜
⎟
⎢1 +
Δ = (0)
+
+
+ …⎥
E1 − E2( 0 ) ⎢ ⎜⎝ E1( 0 ) − E2( 0 ) ⎟⎠ ⎜⎝ E1( 0 ) − E2( 0 ) ⎟⎠ ⎜⎝ E1( 0 ) − E2( 0 ) ⎟⎠
⎥⎦
⎣
(1 − x) −1 = 1 + x + x 2 + x 3 + …
V12V21
E1( 0 ) − E2( 0 )
⎡
⎢
1
⎢
⎢1 − V22 − V11
⎢⎣ E1( 0 ) − E2( 0)
V22-V11<E1-E2 !
⎤
⎥
V12V21
⎥ = (0)
(0)
⎥ ( E1 + V11 ) − ( E2 + V22 )
⎥⎦
like 2nd order of perturbation
but with shifted energy
denominator
Summing to infinite order can be obtained by finite order – different partitioning into
perturbated and unperturbated part:
⎛ E1( 0 )
H 0 = ⎜⎜
⎝ 0
0 ⎞
⎟
(0) ⎟
E2 ⎠
⎛ V V12 ⎞
⎟⎟
V = ⎜⎜ 11
⎝V21 V22 ⎠
⎛ E1( 0 ) + V11
H 0′ = ⎜⎜
0
⎝
⎞
0
⎟
(0)
E2 + V22 ⎟⎠
⎛ 0 V12 ⎞
⎟⎟
V ′ = ⎜⎜
0
V
⎝ 21
⎠
Then geometric sum of diagrams is simply 2nd order PT in new partitioning.
For N-state system, adding dots to second order diagram
V1nVn1
Δ = ∑ (0)
(0)
(
E
+
V
)
−
(
E
n
n + Vnn )
1
11
n=1 term excluded
Orbital perturbation theory: one particle perturbations
Special case where unperturbed Hamiltonian is sum of one-particle Hamiltonians. Η 0 =
is
F ∑ r (i )
V = ∑ v(i )
(not really)
0
i
HF have such form, so we are interested in improving by PT.
First consider case where
∑ h (i)
(ex. Molecule in electric field F =>perturbation
i
where r(i) is position of ith electron.
i
Set of spin orbitals and orbital energies that are eigenfunction and eigenvalues of h0
h0 χ i( 0 ) = ε i( 0 ) χ i( 0 )
Ψ0 = χ1( 0 ) … χ a( 0 ) … χ N( 0 )
Ground state wave function (|Ψ0>) :
Occupied (hole) spin orbitals a, b, c and unoccupied (particle) spin orbitals r, s, t
⎛
⎞
H 0 Ψ0 = ⎜ ∑ ε a( 0 ) ⎟ Ψ0
⎝ a
⎠
In our case:
Exact wf
V
E0 = Ψ0 H Ψ0 = Ψ0 H 0 + V Ψ0 = ∑ ε a( 0) + ∑ a v a = ∑ ε a( 0) + ∑ vaa
a
H = H 0 + V = ∑ (h0 (i ) + v(i )) = ∑ h(i )
i
Φ 0 = χ1 … χ a … χ N
i
and
1
e- terms
a
a
hχ i = (h0 + v) χ i = ε i χ i
⎛
⎞
H Ψ0 = ⎜ ∑ε a ⎟ Φ0 = E0 Φ0
⎝ a ⎠
a
Perturbation expansion of exact E0 can be generally written): E0 = E0( 0 ) + E0(1) + E0( 2 ) + …
and for orbital energy εa holds:
ε a = ε a( 0 ) + a v a + ∑
i
ε
(0)
a
i
ε a = ε a( 0 ) + vaa + ∑
X =∑
r
ab
b≠a
Second order
vab vba
=0
( 0)
(0)
εa − εb
a
b
E0 = ∑ ε a =∑ ε a( 0) + ∑ vaa + ∑
a
a
a
E0( 0 )
E0(1)
ar
+…
var vra
vab vba
+
+…
∑
(0)
( 0)
( 0)
(0)
εa − εr
b≠a ε a − ε b
Total energy
Term X
−ε
(0)
i
vai via
+…
(0)
(0)
εa − εi
ε a = ε a( 0 ) + vaa + ∑
Two parts of summation over
particle orbitals and over holes
(except a).
avi iva
var vra
+
(0)
(0)
εa − εr
E0( 2 )
Matrix element vij is non-zero is spinorbitals i and j have same spin, than for closed systems:
N /2
E
(0)
0
= 2∑ ε a
a
E
N /2
(1)
0
= 2 ∑ vaa
a
E
N /2
( 2)
0
= 2∑
ar
var vra
ε a( 0) − ε r( 0 )
Ex. Derive
E0( 2) = ∑
ar
var vra
ε a( 0) − ε r( 0)
Starting with general expression for second-order energy applied
2
to N-electron system
E0( 2 ) = ∑ '
Ψ0
n
∑ v(i) n
i
E0( 0 ) − En( 0 )
where sum runs over all stages of system except ground state
Hint: The states |n> must be single excitations of type
Ψar = χ1( 0 ) … χ a( 0−)1 χ r( 0 ) χ a( 0+)1 … χ N( 0 )
Ex. Calculate the third-order energy E0(3) using the general expression.
a) Show that
B0(3) = − E0(1) ∑ '
n
b) Show that
A0(3) = ∑ '
nm
c) Show that
Ψ0 V n
2
( E0( 0 ) − En( 0 ) ) 2
= −∑
Ψ0 V n n V m m V Ψ0
( E0( 0 ) − En( 0 ) )( E0( 0) − Em( 0) )
abr
vaa vrb vbr
(ε b( 0 ) − ε r( 0 ) ) 2
=∑
abrs
var vsb Ψar V Ψbs
(ε a( 0) − ε r( 0 ) )(ε b( 0) − ε s( 0 ) )
if a = b r ≠ s
Ψar V Ψbs = vrs
if a ≠ b r = s
= −vba
= ∑ vcc − vaa + vrr
if a = b r = s
c
and zero otherwise.
d) Finally combine two term to obtain
E0(3) = A0(3) + B0(3) = ∑
ars
e) Show that for close-shell system
E
var vrs vsa
vra vab vbr
−
∑
(ε a( 0 ) − ε r( 0 ) ) abr (ε a( 0 ) − ε r( 0 ) )(ε b( 0 ) − ε r( 0 ) )
N /2
( 3)
0
N /2
var vrs vsa
vra vab vbr
= 2∑ ( 0)
−
2
∑
(0)
( 0)
(0)
( 0)
( 0)
ars (ε a − ε s )
abr (ε a − ε r )(ε b − ε r )
Cyclic polyene with N carbon atoms (N=2n=4v+2), v=1, 2, 3, ..)
Resonance energy- difference between exact total energy and energy of N/2=n localized
ethylenic units.
Occupied (hole) and unoccupied (particle) orbitals in the ith unit are denoted by |i> or |i*>
Hamiltonian is
H = H 0 + V = ∑ h0 (i ) + ∑ v(i )
i
i
ε i( 0 ) − ε (j*0 ) = 2 β independent of i and j
Non zero matrix elements
i v (i ± 1) * = ± β / 2
i v (i ± 1) = β / 2
i * v (i ± 1)* = − β / 2
Since polyene is cyclic, 0th ethylen unit is same as nth, (n+1)th is first
In this model exact resonance energy of benzene is 2β, asymptotically exact energy is
lim ER = (4 / π − 1) Nβ = 0.2732 Nβ
N →∞
We need to known only matrix elements of perturbation v
Second-order energy (close-shell)
E
N /2
avr rva
var vra
= 2∑ (0)
=
2
∑
(0)
ε a( 0) − ε r( 0 )
ar ε a − ε r
ar
N /2
( 2)
0
a runs over all n occupied |i > , r over all n unoccupied |j* >
*
*
n
n
i
v
j
j
vi
1 n n
( 2)
E0 = 2∑∑
= ∑∑ i v j * j * v i
(0)
(0)
ε i − ε j*
β i =1 j =1
i =1 j =1
Using fact hat the difference between orbitals is always 2β
For fixed i summation over j is easy : matrix elements are non-zero onlu when j = i ± 1
thus
E
( 2)
0
=
n
i v (i + 1) (i + 1)
∑
β
1
*
v i + i v (i − 1)*
*
(i − 1)* v i
i =1
=
nβ N β
[
]
(
)
(
)
β
/
2
+
−
β
/
2
=
=
= 0.25 Nβ
∑
β
2
4
1
n
2
2
i =1
For benzene second-order resonance energy is 1.5β, exact value is 2 β, in larger systems is
agreements even better- for large N energy approach 91.5% exact results.
Summation i over fixed j: orbital i
can
interact only with orbital (i±1)* .See
picture:
Plus and minus indicate that the matrix elements between the two orbitals ±β/2
Sum over j can be evaluated-start at i on the left and go to i on the right by all possible paths.
Add all contributions. Illustration-value of path i → (i+1)* →i is (+β/2)(+β/2)=β2/4 while
value of path i → (i-1)* →i is (-β/2)(-β/2)=β2/4
So the summation over j is just β2/4 for each value of i.
Calculation of third order of energy (close shell):
E
N /2
( 3)
0
N /2
var vrs vsa
vra vab vbr
= 2∑ ( 0)
−
2
∑
( 0)
(0)
(0)
(0)
(0)
( 0)
( 0)
(
ε
−
ε
)(
ε
−
ε
)
(
ε
−
ε
)(
ε
−
ε
ars
abr
a
r
a
s
a
r
b
r )
Proceeding same way as in E0(2), the first term is
Evaluation of expression over j and k
Note since i cannot interact with
i* or (i+2)* or (i-2)* the value of
this sum must be zero. Similar in
second term, so third-order
energy is zero.
n
2
n
n
iv
∑∑∑
(2 β )
2
i =1 j =1 k =1
j*
j* v k * k * v i
But don’t be hasty. In benzene orbitals (i ± 2)* is same as (i ∓ 1)*, thus pictorial representation of
the sum over j and k with i is:
So there are two paths 1) 1 → 2* → 3* →1 with value (β/2)(-β/2)
(- β/2)= β3/8
2) 1 → 3* → 2* →1 with value (-β/2)(-β/2)
( β/2)= β3/8
(3)
Than the first term E0
is
2
(2 β ) 2
Total third energy E0(3) for benzene is E0( 3) (benzene) =
∑ [( β
3
3
]
/ 8) + ( β 3 / 8) =
i =1
3β
8
3β
4
Resonance energy up to third order is 2.25β (113% of exact value).
Nβ
( 4)
E
=
In excise the fourth order energy is 0
, thus resonance nergy up to fourth order is
64
2.34 β (117% of exact value). Perturbation expression does not converge. But works better for
larger systems.
The energy up to fourth level is 0.2656 Nβ i.e. ⎛⎜ 1 + 1 ⎞⎟ Nβ
⎝4
64 ⎠
Compare with asymptotic exact value 0.2732 Nβ (i.e. 97% of exact value).
Ex. Show that the second term in
E
N /2
( 3)
0
N /2
var vrs vsa
vra vab vbr
= 2∑ ( 0)
− 2∑ (0)
( 0)
(0)
(0)
(0)
( 0)
( 0)
ars (ε a − ε r )(ε a − ε s )
abr (ε a − ε r )(ε b − ε r )
3
β
is equal to
8
for benzene.
Ex. Consider a cyclic polyene with N = 4v+2, v=1 ,2, 3, .. Carbons. Instead that all bonds are
identical suppose that they alternate in length. In context of Hückel theory this means that
resonance integrals alternate between β1and β2. For example for benzene we have
Now it can be shown that the exact energy in this case is
v
2 jπ
)
ε 0 = Nα − 2 ∑ ( β12 + β 22 + 2 β1 β 2 cos
2v + 1
j =−v
Note that when β1= β2= β, since 2 cos 2 θ = (1 + cos 2θ )
v
β is negative we recover ε 0 = Nα + 4 β ∑ cos
j =−v
jπ
2v + 1
And if β1= β and β2= 0 than E0 = Nα+Nβ (total energy in localized ethylenic description)
Purpose of this excise is to obtain the exact energy in power of β2/ β1.
and
a) Show that for benzene (v = 1) in alternating short and long bond model is
ε 0 = 6α + 2( β1 + β 2 ) − 4(β12 + β 22 − β1 β 2 )
1/ 2
Do this first by using general expression and then by setting up the Hückel matrix,
diagonalizing it and then adding up the occupied orbital energies. Note that for β1= β2= β
we obtain old result 6α+8β.
b) Setting β1= β and β2/ β1= x show that the resonance energy of benzene can be written as
1
ER = 4 β ( x − 1 + (1 + x 2 − x)1/ 2 )
2
Note that when x = 0,ER = 0 and when x = 1, ER = 2β which is exact.
1
1 2 1 3 5 4
1/ 2
(
)
1
+
y
=
1
+
y
−
y + y −
y +…
c) Using relation
2
8
16
128
and thus show that
y <1
expand ER to fourth order in x
3
3
3
E R = β ( x 2 + x 3 + x 4 + …) identifying coefficient of xn with nth2
4
32
order perturbation results, we have
3
β
2
3
=
β
4
3
=
β
32
E 0( 2 ) =
E 0( 3 )
E 0( 4 )
Note that E0(2) and E0(3) agree with previous results. Small insight into poor convergence of
perturbation expansion of the resonance energy of benzene. When x is small, then rapid
convergence. However for our problem is x = 1.
Resonance energy calculated to Mth-order as function of M is below. Converging to exact
value of 2β. Above method for obtaining E0(n) for n= 2, 3, 4 is extremely laborious.
Results below were calculated by first showing that
E0( n ) = 4 βCn−1/ 2 ( 12 ), where Cn−1/ 2 ( x) is GegenBauer polynomial.
And using recursive properties of these polynomials to shown that
(n + 1)E0( n+1) = (n − 12 ) E0( n ) − (n − 2) E0( n−1)
Diagrammatic representation of orbital perturbation theory
Holes – a, b, …
Particles - r, s,
Fourth rule: Sum the expression over all particle and hole indices.
Using these rules:
For close shell systems:
N
∑
N /2
= 2∑
Ex. Find fourth order energy for closed-shell cyclic polyene
a)
and
Show that
so that
Thus the resonance energy calculated for cyclic polyene with N>6 up to fourth order is
(1/4+1/64) Nβ = 0.2656 Nβ, which compares with asymptotically exact value of 0.2732 Nβ
(97% of exact).
b) For benzene, show that the diagrammatic result for the fourth-order energy agrees with the
independently calculated results from previous exercise (with short and long bonds).
Perturbation expansion of the correlation energy
H = H0 +V
Improving of HF of N-electron system
HF
E0( 0 ) = ∑ ε a
H 0 Ψ0 = E0( 0 ) Ψ0
a
E0(1) = Ψ0 V Ψ0 = −
1
ab ab
∑
2 ab
(0)
(1)
HF energy is E0 = E0 + E0 = ∑ ε a − 12 ∑ ab ab
a
First correlation from
E0( 2 ) = ∑ '
n
0 = Ψ0
ab
0V n
2
E0( 0 ) − En( 0 )
Summation over but the
ground state of system
|n> these states cannot be single excitation due to
Ψ0 V Ψar = Ψ0 H − H 0 Ψar = Ψ0 H Ψar − f ar = 0
Brillouin’s theorem
Triply excited states do not mix with |Ψ0 > due
to two-particle nature of perturbation!
Spinorb. are
eigenfunction of
Fock op.
H 0 Ψabrs = ( E0( 0 ) − (ε a + ε b − ε r − ε s )) Ψabrs
Ψabrs
Double excitation
Summing over all possible double excitation
2
E0( 2 ) = ∑
a <b
r <s
Ψ0
∑r
i< j
−1
ij
Ψabrs
ε a + εb − ε r − ε s
=∑
a <b
r <s
ab rs
2
εa + εb − εr − ε s
FO
E0( 2 ) = ∑ eab
Sum of contribution from each pair of
electron in occupied orbitals
a <b
FO
eab
=∑
r <s
ab rs
2
εa + εb − εr − ε s
“Pair theory”
At level of first-order pairs, pair theory gives same correlation as 2nd PT.
a = b or r=s
E0( 2 ) =
close shell
E0( 2 ) =
0
E
N /2
= 2∑
abrs
ab rs rs ab
εa + εb − ε r − ε s
N /2
−∑
abrs
2
1
∑
4 abrs ε a + ε b − ε r − ε s
ab rs rs ab
ab rs rs ba
1
1
−
∑
∑
2 abrs ε a + ε b − ε r − ε s 2 abrs ε a + ε b − ε r − ε s
( 2)
0
ab rs
ab rs rs ba
εa + εb − ε r − ε s
Ex.Derive
E
( 2)
0
E
( 2)
0
and
ab rs rs ab
ab rs rs ba
1
1
= ∑
− ∑
2 abrs ε a + ε b − ε r − ε s 2 abrs ε a + ε b − ε r − ε s
N /2
= 2∑
abrs
from
E0( 2 ) =
ab rs rs ab
εa + εb − ε r − ε s
ab rs
2
1
∑
4 abrs ε a + ε b − ε r − ε s
N /2
−∑
abrs
ab rs rs ba
εa + εb − ε r − ε s
E =
(3)
0
third order of energy
1
8
ab rs cd ab rs cd
∑ (ε +ε −ε −ε )(ε +ε
abcdrs a
+ 18 ∑
b
r
s
c
d
−εr −εs )
ab rs rs tu tu ab
(ε +εb −εr −εs )(εa +εb −εt −εu )
abrstu a
+∑
Illustration - Second and third order of
H2 in minimal basis
Exact correlation is
ab rs cs tb rt ac
(ε +εb −εr −εs )(εa +εc −εr −εt )
abcrst a
Ecorr = Δ − (Δ2 + K122 )1/ 2
2Δ = 2(ε 2 − ε1 ) + J11 + J 22 − 4 J12 + 2 K12
where
= 2(ε 2 − ε1 ) + 11 11 + 22 22 − 4 12 12 + 2 11 22
Ecorr = E0( 2) + E0(3) + …
Expansion of correlation energy to Taylor series up to third order
where
E
( 2)
0
K122
=
2(ε1 − ε 2 )
E
Single hole orbitals a = b = 1,
similarly r = s = 2 then
becomes
E
(2)
0
=2
( 3)
0
K122 ( J11 + J 22 − 4 J12 + 2 K12 )
=
4(ε1 − ε 2 ) 2
N /2
E
( 2)
0
= 2∑
11 22 22 11
2 (ε 1 − ε 2 )
abrs
−
ab rs rs ab
εa + εb − ε r − ε s
11 22 22 11
2 (ε 1 − ε 2 )
ab rs rs ba
N /2
−∑
abrs
εa + εb − ε r − ε s
11 22
2
K 122
=
=
2 (ε 1 − ε 2 ) 2 (ε 1 − ε 2 )
Ex. Derive
E
and
from
E
( 3)
0
( 2)
0
K122
=
2(ε1 − ε 2 )
K122 ( J11 + J 22 − 4 J12 + 2 K12 )
=
4(ε1 − ε 2 ) 2
Ecorr = Δ − (Δ2 + K122 )1/ 2
The N-dependence of RS perturbation expansion
Proof that PT is proportional to number of particles:
Supermolecule N noninteracting H2 (in DCI was proportional to N1/2 in limit of large N
Label the orbitals:
All two electron integrals involving orbitals from different units are zero. HF wave function is
Ψ0 = 111112 12
Zeroth-order
E
( 0)
0
N
1N 1N
= Ψ0 H 0 Ψ0 = 2∑ 1i f 1i = 2 Nε1
i =1
First order
E
N
(1)
0
= Ψ0 V Ψ0 = −∑ 1i1i 1i1i = − NJ11
i =1
HF energy of supermolecule
E0 = Ψ0 H 0 + V Ψ0 = E0( 0 ) + E0(1) = N (2ε1 − J11 )
is indeed simply N times HF of one subunit. Expression for 2nd order energy is
E0( 2 ) = ∑ '
n
2
0V n
E0( 0 ) − En( 0 )
Clearly |0> = |Ψ0> and the state |n> must be a double excitation of type
Ψ12i 1ii2i
E0( 0 ) − En( 0 ) = 2(ε1 − ε 2 )
For these excitation
Ψ0 V Ψ12i 1ii2i = 1i 1i 2i 2i − 1i 1i 2i 2i = 11 22 = K12
Summation over n can be replaced by summation over i
E
Ψ0 V Ψ
2i 2i
1i 1i
N
( 2)
0
=∑
2
NK122
=
2(ε1 − ε 2 )
2(ε1 − ε 2 )
i =1
General expression of third order energy is
where
A0(3) = ∑ ' ∑ '
n
and
m
0V n nV m mV 0
( E0( 0 ) − En( 0 ) )( E0( 0 ) − Em( 0 ) )
B0(3) = − E0(1) ∑ '
n
0V n
2
( E0( 0 ) − En( 0 ) ) 2
N times energy of one unit
E0(3) = A0(3) + B0(3)
B0(3) doesn’t appear to have N dependence, since B0(3) is proportional to N2 :
N
K122
N 2 J11 K122
( 3)
=
B0 = −(− NJ11 )∑
2
4(ε1 − ε 2 ) 2
i =1 ( 2ε 1 − 2ε 2 )
But this term is cancelled by part of term A0(3).
22
In A0(3) both |n> and |m> must be states of type Ψ1i 1ii i
N
A0(3) = ∑
i =1
so that
Ψ12i 1ii2i V Ψ12i 1ii2i K122
4(ε1 − ε 2 ) 2
Only diagonal term because integral involving different units are zero.
Ψ12i 1ii2i V Ψ12i 1ii2i = Ψ12i 1ii2i H − H 0 Ψ12i 1ii2i
= − NJ11 + J11 + J 22 − 4 J12 + K12
so we have
or
E
A
( 3)
0
( 3)
0
=A
N 2 J11 K122
NK122 ( J11 + J 22 − 4 J12 + 2 K12 )
=−
+
4(ε1 − ε 2 ) 2
4(ε1 − ε 2 ) 2
( 3)
0
+B
( 3)
0
NK122 ( J11 + J 22 − 4 J12 + 2 K12 )
=
4(ε1 − ε 2 ) 2
Ex.Derive
Ψ
2i 2i
1i 1i
and
V Ψ
2i 2i
1i 1i
= Ψ
2i 2i
1i 1i
H − H0 Ψ
2i 2i
1i 1i
= − NJ11 + J11 + J 22 − 4 J12 + K12
E
N
(1)
0
= Ψ0 V Ψ0 = −∑ 1i1i 1i1i = − NJ11
i =1
Diagrammatic representation of the perturbation expansion of the correlation energy
Nth order of energy (zero order is HF), HF is good for start, due to Brillouin theorem.
Hugenholtz diagram
In
2
E
( 2)
0
ab rs
1
= ∑
4 abrs ε a + ε b − ε r − ε s
the numerator have four indices – result of two particle
nature of perturbation
Dot with two line in and two line out
In diagram – connect dots in all possible ways:
1) Each dot has four lines
2) Each diagram is linked
3) Each diagram is distinct
4) Diagram containing more than one dot do not have lines which start and end at
the same point.
Consequence of Brillouin’s
Fourth order
(unlabeled)
Labeling
Hugenholtz diagram
Translation into algebra:
a)
b)
c)
d)
e)
Second order
Two equivalent line (r,s) and (a,b)
Case A
ab rs rs ab
⎛1⎞
2+l A
(
)
−
⎜ ⎟
∑
⎝ 2⎠
abrs ε a + ε b − ε r − ε s
Case B
ab rs sr ab
⎛1⎞
2+lB
(
)
−
⎜ ⎟
∑
⎝ 2⎠
abrs ε a + ε b − ε r − ε s
2
2
A and B must be equal – l factor. Number of closed loops can be found only after we
written down a string of matrix elements
Case A
lA = 2
Case B
lB = 1
rs ab = − sr ab
E0( 2 ) =
A=B
ab rs rs ab
1
∑
4 abrs ε a + ε b − ε r − ε s
Third order
l=2
l=2
a→r →c→a
b→s→d →b
a→r →t →a
b→s→u →b
l=3
a→r→a
b→s→b
c→t →c
Total energy E0(3)
E
( 3)
0
ab rs cd ab rs cd
1
= ∑
8 abcdrs (ε a + ε b − ε r − ε s )(ε c + ε d − ε r − ε s )
+
+
ab rs rs tu tu ab
1
∑
8 abrstu (ε a + ε b − ε r − ε s )(ε a + ε b − ε t − ε u )
∑ (ε
abcrst
ab rs cs tb rt ac
a
+ ε b − ε s − ε r )(ε a + ε c − ε r − ε t )
Ex. Show that the fourth-order diagram
is equal to
−
1
2
∑
abcderst
rs ac at de dc tb eb rs
(ε a + ε c − ε r − ε s )(ε c + ε d + ε e − ε r − ε t − ε s )(ε b + ε e − ε r − ε s )
Goldstone diagrams
Two particle interaction is not dot but dashed line
This eliminates the antisymmetric elements.
Direct exchange
Exchange diagram
Rules:
Table: Second
and third order
of Goldstone
diagrams
Second order of energy
Direct: Reflecting in perpendicular plane – diagram remains invariant to reflection
Weight factor
1/2
Two holes + two closed loop
Exchange
1/2
Two holes, but only one loop
r →b→s→a→r
Close shell – converting from spin orbital to spatial: direct multiplied by 22
exchange
by 21
Ex. Goldstone diagrams in Table can be obtained by “pulling apart: the second and thirdorder Hugenholtz diagrams. For example if we push
Push all third-order diagram in Table in similar way find which Goldstone diagram come to
Hugenholtz. For above Hugenholtz diagram,verify that its mathematical value indeed the
sum of the values of corresponding Goldstone diagrams.
Example: H2
Two second order diagram
Third order diagram
Sum over 12 expresions
E0(3)
Summation of diagrams
Transforming to second order diagrams; adding interaction lines, (labels below and above
interaction line is same)
All have weight factor ½, same number of holes and particles – same sign
Shifted energy denominator
Sum over all diagrams
Similar as Epstein-Nesbet pair energy
If we sum over all double excitation diagrams (imaginary line cross only two holes and two
particles line )
double excited MBPT (D-MBPT) of Bartlett and co. Tis is
equivalent to linear CCA.
If D only diagonal
D-MBPT is exact third order of PT
EN
Linked cluster theorem
Algebraic term proportional to N2 can be represented by diagram with separate pieces (unlinked
diagrams). But such diagrams never appear in final result of the nth-order of energy.
So:
Nth order can be written only with linked diagrams.
Ex.Calculate E0(3) for supermolecule consist of N noninteracting minimal basis H2 molecules
evaluating the Goldstone diagrams in Table. Compare your result with
E
( 3)
0
=A
( 3)
0
+B
( 3)
0
NK122 ( J11 + J 22 − 4 J12 + 2 K12 )
=
4(ε1 − ε 2 ) 2
which was obtained algebraically explicitly cancelling term proportional to N2
Hint: Simply show that value of each Goldstone diagram for the supermolecule is N times the result for
a single molecule.
Calculations
inc
SCF/6-31G** differ 0.011 a.u.
Equilibrium bond length(a.u) for ten-electron series
SCF largest diff. to exp. is 0.03
MP2 below 0.01 a.u.
MP2 reduce SCF error
to half
Problems with multiple bonds
Error SCF 0.03 a.u.
MP2 overestimates
correction
Base is near HF limit
MP2 better than SDCI
PT poor convergence, non monotonic
H2 r
infinity
SCF dissociation into 2 H
CI
infinity