Multipole moments - CRM2

Chapter 7
Multipole expansion – Cartesian
formalism
The complete determination of the complete charge density and various response
functions (and the associated deformation densities) is computationally heavy and
conceptually not very rewarding. As any distribution function, the essential features
of the charge distribution can be be characterized by its moments.
The mathematical analogy between the Longuet-Higgins interaction operator and
the electrostatic interaction energy allows us to generalize the results obtained for
the latter to the interaction operator too.
7.1 Potential of a charge distribution
Potential V (R) of the charge distribution ϱ(r)
∫
V (R) = drT (R − r)ϱ(r) (7.1)
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can be expanded in Taylor series around r = 0 (|r| < |R|.
Taylor expansion of the Coulomb kernel
)
)
T (R − r) = 1 ( 1
R − r α∇ α + 1 ( 1
R 2! r αr β ∇ α ∇ β
R
− 1 ( ) 1
3! r αr β r γ ∇ α ∇ β ∇ γ + . . .
R
(7.2)
Define the (Cartesian) Coulomb interaction tensors
T (R) = 1 R
)
T α (R) = ∇ α
( 1
R
T αβ (R) = ∇ α ∇ β
( 1
R
)
)
(7.3)
T αβγ (R) = ∇ α ∇ β ∇ γ
( 1
R
)
T αβ...ν (R) = ∇ α ∇ β . . . ∇ ν
( 1
R
and the electrostatic potential
∫
V (R) = T (R)
∫
drϱ(r) − T α (R) drr α ϱ(r)
+ T αβ (R) 1 ∫
drr α r β ϱ(r) − T αβγ (R) 1 ∫
2!
3!
drr α r β r γ ϱ(r) + . . .
(7.4)
Define the multipole moments of the charge distribution
∫
q = drϱ(r)
∫
m α = drr α ϱ(r)
∫
Q αβ = drr α r β ϱ(r)
(7.5)
ξ αβ...ν = drr (n)
αβ...ν ϱ(r)
∫
O αβγ = drr α r β r γ ϱ(r)
∫
Cartesian multipole expansion of the potential
V (R) = qT (R) − m α T α (R) + 1 2! Q αβT αβ (R) − . . . + (−)n ξ αβ...ν T αβ...ν (R) (7.6)
n!
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7.2 Multipole moments
There are several conventions in the literature. The above-defined multipole moments
are symmetric tensors Q αβ = Q βα and their trace is nonzero
. ∑
= Q αα ≠ 0 (7.7)
Q αα
They can be called “unnormalized traced Cartesian” multipoles.
Unabridged multipoles (Applequist)
Traceless cartesian multipoles (Buckingham)
Potential of a quadrupole
α
µ αβ...ν = 1 n! ξ αβ...ν (7.8)
V (2) (R) = 1 2 Q αβT αβ (R) (7.9)
Since the T (R) = 1 function satisfies the Laplace equation
R
( ) 1
∇ 2 = 0
R
∑
( ) 1
∇ α ∇ α = ∑ T αα (R) = 0
R
α
α
we can add to the quadrupole potential an arbitrary quantity
(7.10)
λδ αβ T αβ (R) (7.11)
without changing its value
V (2) (R) = 1 2 (Q αβ + λδ αβ ) T αβ (R) (7.12)
Let us choose λ as the trace of Q αβ
λ = − 1 3 Q αα = − 1 ∑
Q αα (7.13)
3
The new expression of the quadrupole potential
V (2) (R) = 1 2 · 1 ∫
dr ( )
3r α r β − r 2 δ αβ ϱ(r)Tαβ (R) = 1 3
3 Θ αβT αβ (R) (7.14)
where have introduced the traceless cartesian quadrupole moment
Θ αβ = 1 ∫
dr ( )
3r α r β − r 2 δ αβ ϱ(r) (7.15)
2
α
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7.2.1 Traceless (Buckingham) multipole moments
General definition
Electrostatic potential
M (n)
αβ...ν = (−)n
n!
V (R) = ∑ n
where the notation (2n − 1)!! means
∫
( )
drϱ(r)r 2n+1 ∂ n 1
∂r α ∂ . . . r ν r
(7.16)
(−) n
(2n − 1)!! M (n)
αβ...ν T (n)
αβ...ν
(R) (7.17)
(2n − 1)!! = 1 · 3 · 5 · . . . (2n − 1) (7.18)
The traceless multipoles do not contain the full information about the ϱ(r) charge
distribution, since they are undetermined up to an arbitrary spherically symmetric
component.
On the contrary, one can reconstruct the full charge density from the traced multipoles.
Consider the Fourier transform of ϱ(r)
∫
ϱ(k) = dre ikr ϱ(r) (7.19)
and expand the exponential
∫
∫
ϱ(k) = drϱ(r) +ik α dr r α ϱ(r) − 1 ∫
2! k αk β dr r α r β ϱ(r) + . . . (7.20)
} {{ } } {{ } } {{ }
q
m α
Q αβ
7.2.2 Explicit form of higher multipole moments
◦ Quadrupole moment Θ αβ
Θ αβ = ∑ a
q a [ 3 2 a αa β − 1 2 a2 δ αβ ] (7.21)
Six distinct components, five independent components.
◦ Octopole moment Ω αβγ
Ω αβγ = ∑ a
q a [ 5 3 a αa β a γ − a 2 (a α δ βγ + a β δ γα + a γ δ αβ )] (7.22)
10 distinct components, seven independent components.
◦ Hexadecapole moment Φ αβγγ
◦ 2 n moments...
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7.2.3 Translation of multipole moments
◦ Dipole moment of a point charge distribution
µ α = ∑ i
q i r i α (7.23)
In a new frame obtained after a translation by a vector a
µ ′ α = ∑ q i (rα i − a α ) = ∑ ∑
q i rα i − a α q i = µ α − a α q tot (7.24)
i
i
i
Dipole of a neutral system (q tot = 0) is invariant to translation of origin.
◦ Quadrupole moment
Θ αβ = ∑ q ( )
i 3rαr i β i − (r i ) 2 δ αβ (7.25)
i
After translation by a
Θ ′ αβ = Θ αβ + 1 2 qtot (3a α a β − a 2 δ αβ )
− 3 2 (µtot α a β + µ tot
β a α ) − δ αβ µ tot
α a α
(7.26)
7.3 Electric field, field gradient
The electric field, field gradient
F α (R) = −∇ α V (R) F αβ (R) = −∇ α ∇ β V (R) (7.27)
and higher derivatives of the potential of a multipolar charge distribution can be
expressed with the same interaction tensors
F l 1
α 1 ...ν 1
= −∇ α . . . ∇ ν1 V (R) =
∞∑ (−) l 2
−
(2l 2 − 1)!! T l 1+l 2
α 1 ...ν 1 α 2 ...ν 2
M l (7.28)
2
α 2 ...ν 2
l 2
7.4 Energy of a charge distribution in non-uniform
field
We choose the origin inside the charge distribution, ϱ(r) and expand the external
potential in Taylor series around the origin
V (r) = V (0) + r α V α (0) + 1 2! r αr β V αβ (0) + 1 3! r αr β r γ V αβγ (0) + . . . (7.29)
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Using the expression of the interaction energy
∫
∫
∫
U = drϱ(r)V (r) = drϱ(r)V (0) + drϱ(r)r α V α (0)
+ 1 ∫
drϱ(r)r α r β V αβ (0) + 1 ∫
drϱ(r)r α r β r γ V αβγ (0) + . . .
2!
3!
(7.30)
where we recognize the definition of the unnormalized traced multipoles,
U = MV + M α V α + 1 2! M αβV αβ + 1 3! M αβγV αβγ + . . . (7.31)
By the virtue of the Laplace equation, V αα = 0, we can subtract the trace of M αβ ,
i.e. 1 3 M αα from the quadrupolar energy
1
2 M αβV αβ = 1 2 V 1
αβ(M αβ − δ αβ
3 M αα) = 1 3 V αβΘ αβ (7.32)
By similar manipulations at higher orders
U = qV + µ α V α + 1 3 Θ αβV αβ + 1
3 · 5 Ω αβγV αβγ
1
+
3 · 5 · 7 Φ 1
(7.33)
αβγδV αβγδ +
(2n − 1)!! ξ(n) α...νV α...ν
The energy is the scalar product of the field with dipole, filed gradient with quadrupole
and higher field gradients with the corresponding multipole tensors.
Attention: the result may depend on the choice of origin!
7.5 Multipole interaction energy
Take two sets of multipoles. Taylor series of the Coulomb interaction
1
|τ − τ ′ | = 1
|B + s − A − r| = 1
|R − (r − s)| =
= 1 ( ) 1
R + (r α − s α )∇ α + 1 ( ) 1
R 2! (r α − s α )(r β − s β )∇ α ∇ β + . . .
R
(7.34)
which leads directly to an expression of the interaction energy of two multipole
charge distributions in terms of their traced Cartesian moments
U(R) = q A q B T (R) + (m A αq B − q A m α )T α (R)
+ 1 2! (QA αβq B + q A Q B αβ − m A αm B β − m B α m A β )T αβ (R) + . . .
Using the traceless (Buckingham) moments
(7.35)
U(R) = ∑ l 1 l 2
(−) l 1
(2l 1 − 1)!!(2l 2 − 1)!! M (l 1)[A]
α 1 ...ν 1
T (l 1+l 2 )
α 1 ...ν 1 α 2 ...ν 2
(R)M (l 2)[B]
α 2 ...ν 2
(7.36)
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7.6 Calculation of cartesian interaction tensors
They are defined as
T = 1 R = R−1 =
1
√ R
2
x + R 2 y + R 2 z
(7.37)
Using the following derivation “rules”
∇ α R β = δ αβ ∇ α R n = nR α R (n−2) ∇ α R −n = −nR α R −(n+2) (7.38)
◦ Charge-dipole
T α (R) = ∇ α T (R) = −R α R −3 (7.39)
◦ Dipole-dipole/charge-quadrupole
T αβ (R) = ∇ α ∇ β T (R)
= −∇ α R β R −3 = −δ αβ R −3 + 3R α R β R −5
= (3R α R β − R 2 δ αβ )R −5 (7.40)
◦ Dipole-quadrupole/charge-octopole
T αβγ (R) = ∇ α ∇ β ∇ γ T (R) = δ α (3R β R γ − R 2 δ βγ )R −5
= 3δ αβ R γ R −5 + 3δ αγ R β R −5 + 3δ βγ R α R −5 + 3 · 5 · R α R β R γ R −7
= 3 ( 5R α R β R γ − R 2 (R α δ βγ + R β δ γα + R γ δ αβ ) ) R −7 (7.41)
Interaction tensors
◦ invariant to interchange of indices
◦ traceless, since ∇ 2 (1/R) = 0
◦ 2n+1 independent components
Cartesian tensor formulation
◦ simple in low orders
◦ easily extended algorithms
◦ global frame
◦ heavy formalism
◦ too much components (reducible
and redundant)
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7.7 Example: Structure of the HX dimers
Competition of three multipolar interaction. Angular dependence for linear molecules
(symmetric tops) is the following
◦ dipole-dipole
◦ dipole-quadrupole
U µµ = − µA µ B
R 3 (2 cos θ A cos θ B − sin θ A sin θ B cos ϕ) (7.42)
U µΘ = 3µA Θ B
4R 4 [cos θ A (3 cos 2 θ B − 1) − sin θ A sin 2θ B cos ϕ] (7.43)
◦ quadrupole-quadrupole
U ΘΘ = ΘA Θ B
R 5 3
4 [1 − 5 cos2 θ A − 5 cos 2 θ B − 15 cos 2 θ A cos 2 θ B
+ 2(4 cos θ A cos θ B − sin θ A sin θ B cos ϕ) 2 ]
(7.44)
The structure is characterized by θ A ≈ 0 and ϕ = 0, µ A = µ B = µ and Θ A = Θ B =
Θ. The sum of the 4 interaction terms (two for U µΘ )
U = − µ2
R 3 · 2 cos θ
+ µΘ
R 4 · 3
2 [(3 cos2 θ − 1) − 2 cos θ]
+ Θ2
R 5 · 3(3 cos2 θ − 1)
(7.45)
Introduce the parameter λ = Θ µR
U = µ2
R 3 {
−2 cos θ + λ
3
2 [(3 cos2 θ − 1) − 2 cos θ] + λ 2 3(3 cos 2 θ − 1) }
= µ2
2R 3 {
3λ(1 + 2λ)(3 cos 2 θ − 1) − 2(2 + 3λ) cos θ } (7.46)
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Electrostatic interaction energy
is plotted for the
◦ HF dimer
µ = 0.72ea 0 ,
Θ = 1.875ea 2 0
R = 5.1a 0 ;
minimum at 68 degrees
(exp. 60)
◦ HCl dimer
µ = 0.433ea 0 ,
Θ = 2.8ea 2 0
R = 6.8a 0 ;
minimum at 79 degrees
63