THE ELECTROSTATIC INTERACTION ENERGY IRyan P. A. Bettens,

Department of Chemistry,

National University of Singapore.

FORCES IN NATURE

All of the interactions in nature are accounted for by four absolutely fundamental forces: Weak nuclear force Strong nuclear force Coulomb force Gravitational force

As far as we are concerned in chemistry the interaction of electrons with nuclei is paramount.

We need only explicitly consider the coulomb force in understanding how atoms and molecules interact.

THE COULOMB FORCE

All of matter is made up of charged particles. That is, electrons and protons.

The coulomb force describes how these charged particles interact with one another.

The expression for the magnitude of the force between charged particles is:

The consequent expression for the potential energy between the charged particles is:

204

1

ij

ji

r

qqF

ij

ji

r

qqV

04

1

SI UNITS AND ATOMIC UNITS

For convenience we will switch to atomic units.

In atomic units, the charge on an electron is 1, and all distances are measured in “bohr”, a0 = 5.291 772 49(24) × 10-11 m.

The number for an energy calculated in atomic units is often called a “hartree”.

1 hartree = 4.359 748 2(26) × 10-18 J and is twice the ionization energy of the H atom.

In atomic units the formulae for the magnitude of the force and potential energy of the previous slide become:

2ij

ji

r

qqF

ij

ji

r

qqV Here q is in electron charge

and rij is in a0.

INTERMOLECULAR INTERACTION

When we consider two molecules interacting this occurs via Coulomb’s law.

Each molecule is a collection of positively charged nuclei and negatively charged electrons.

The interaction energy is just:

a bN

i

N

j ij

ji

r

qqV

1 1

INTERMOLECULAR INTERACTION

However, electrons are not stationary in a molecule, but are moving very rapidly (some near the speed of light!) around the nuclei.

Furthermore, the electrons can not be thought of as discrete particles, as they exist in molecular orbitals which are standing matter waves in and about the nuclei.

DERIVING THE INTERACTION ENERGY

We begin with the following expansion for the distance between two charges:

where the Cl,m are the renormalized spherical harmonics and are defined as:

and the Yl,m are the spherical harmonics.

0

22,11,1 ,,)1(1

l

l

lmmlml

ml

l

CCr

r 21 rr

,12

4, ,, mlml Y

lC

SPHERICAL HARMONICS

There is nothing complicated about the Yl,m

they are simply math functions. Check out this Table of Spherical Harmonics

to see just what they look like. We shall now get a feeling for how well the

expansion of 1/r12 works. We will consider two charged particles

existing only in the xz plane for simplicity, i.e., yi = 0. Setting the y-coordinates to zero means our

spherical harmonics are nicely real.

EXPANSION TO L = 1

So you can see how the expansion formula works, let’s do the expansion up to l = 1.

We need the Cl,m functions, so here they are:

14

1

1

4

1

40,00,0

YC

r

iyx

r

iyxYC

2

1

8

3

3

4

3

41,11,1

r

z

r

zYC

4

3

3

4

3

40,10,1

r

iyx

r

iyxYC

2

1

8

3

3

4

3

41,11,1

EXPANSION TO L = 1

When particle 1 is at (x1, 0, z1) and particle 2 is at (x2, 0, z2) then we can write the expansion as:

If we set x1 = 0 and z1 = −1, and x2 = 0 and z2 = 10 then r1 = 1 and r2 = 10 and r12 = 11 with 1/r12 = 0.090909.

Using our expansion we find: Compare 0.091 to 0.090.

...)1()1()1()1(1 )2(

1,1)1(

1,11

11

1)2(

0,1)1(

0,10

11

1)2(

1,1)1(

1,11

11

1)2(

0,0)1(

0,00

10

0

12

CCr

rCC

r

rCC

r

rCC

r

r

r

...2

1

2

1

2

1

2

111

2

2

1

12

2

2

1

12

2

2

1

12

12

r

x

r

x

r

r

r

z

r

z

r

r

r

x

r

x

r

r

rr

...11

2

2

1

12

2

2

1

12

12

r

z

r

z

r

r

r

x

r

x

r

r

rr

...100

9

...100

1

10

1

...10

10

1

1

100

10

10

11

12

r

EXAMPLE OF THE 1/R12 EXPANSIONr1 r2 r12 1/|r12| 0.090909 Residual % Residual

X 0 0 0 Expand0 0.100000 -0.009091 -10.0000

Z 10 -1 -11 Expand1 0.090000 0.000909 1.0000

|ri| 10.000000 1.000000 11.000000 Expand2 0.091000 -0.000091 -0.1000

C 0,0(i ) 1.000000 1.000000 Expand3 0.090900 0.000009 0.0100

C 1,-1(i ) 0.000000 0.000000 Expand4 0.090910 -0.000001 -0.0010

C 1,0(i ) 1.000000 -1.000000

C 1,1(i ) 0.000000 0.000000

C 2,-2(i ) 0.000000 0.000000

C 2,-1(i ) 0.000000 0.000000

C 2,0(i ) 1.000000 1.000000

C 2,1(i ) 0.000000 0.000000

C 2,2(i ) 0.000000 0.000000

C 3,-3(i ) 0.000000 0.000000

C 3,-2(i ) 0.000000 0.000000

C 3,-1(i ) 0.000000 0.000000

C 3,0(i ) 1.000000 -1.000000

C 3,1(i ) 0.000000 0.000000

C 3,2(i ) 0.000000 0.000000

C 3,3(i ) 0.000000 0.000000

C 4,-4(i ) 0.000000 0.000000

C 4,-3(i ) 0.000000 0.000000

C 4,-2(i ) 0.000000 0.000000

C 4,-1(i ) 0.000000 0.000000

C 4,0(i ) 1.000000 1.000000

C 4,1(i ) 0.000000 0.000000

C 4,2(i ) 0.000000 0.000000

C 4,3(i ) 0.000000 0.000000

C 4,4(i ) 0.000000 0.000000

-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

RELATING 1/R12 TO INTERACTING MOLECULES

A is the vector from the origin to the origin of water a, and B is the vector from the origin to the origin of water b.

a is the vector from the origin of water a to an electron in this water. Likewise for b.

The electron in water a is located at A + a. The electron in water b is located at B + b.

A

a

B

b

The distance between the two water molecules is given by |B − A| = R, and we can write the vector R to be the vector from molecule a to molecule b. Of course |R| = R.

The distance between the two charges is given by |(B + b) − (A + a)|

We can rewrite this distance as |B − A + b − a| and compare this with |r1 − r2| = r12.

We identify r1 = B − A and r2 = a − b.

Thus provided |r1|= R >|r2| we can validly use our expansion and if R » |r2| then the expansion will converge quickly.

REGULAR AND IRREGULAR SPHERICAL HARMONICS

Fancy words for something so simple. The regular spherical harmonics are just:

The irregular spherical harmonics are just:),(,, ml

lml CrR

),(1

,1, mllml Cr

I

0

22,11,1 ,,)1(1

l

l

lmmlml

ml

l

CCr

r 21 rr

REGULAR SPHERICAL HARMONIC ADDITION THEOREM

M

L

m

l

m

lRR

ll

LR mlml

l l

l

lm

l

lm

MLLllML

2

2

1

1,,

2/1

0 0 21,, 2211

1 2

1

11

2

22

21 !2!2

!121 baba

21 llL

ML

m

l

m

l

2

2

1

1is called a Wigner 3j symbol. It is just a simple number.

aa MLL

ML RR ,, 1

0

22,11,1 ,,)1(1

l

l

lmmlml

ml

l

CCr

r 21 rr

0

22,11,1 ,,1

)1(1

l

l

lmml

lmll

m CrCr

21 rr

022,11,1

,,1

)1(1

l

l

lmml

l

mllm CCR

babaR

0,,)1(

1

l

l

lmmlml

m RI baRbaR

ba mlR ,

We’ll need to use our addition theorem to separate the a and b away from this single regular spherical harmonic.

Will only work if |R|>|a − b|

SORTING OUT

This final expression can now be substituted back into our expansion.

ba mlR ,

m

l

m

l

m

lRR

ll

lR mlml

l l

l

lm

l

lm

mllllml

2

2

1

1,,

2/1

0 0 21,, 2211

1 2

1

11

2

22

21 !2!2

!121 baba

bb22

2

22 ,, 1 mll

ml RR

21 lll

m

ll

m

l

m

lRR

ll

llR ml

lml

l l

l

lm

l

lm

mllml

21

2

2

1

1,,

2/1

0 0 21

21, 22

2

11

1 2

1

11

2

22

21 1!2!2

!1221 baba

m

ll

m

l

m

lRR

ll

llR mlml

l l

l

lm

l

lm

mlml

21

2

2

1

1,,

2/1

0 0 21

21, 2211

1 2

1

11

2

22

1

!2!2

!1221 baba

We are interested in the full interaction potential energy, not just a single inverse distance.

0 0

2/1

21

21,

1 2

21

21

1

11

2

22

1

21 !2!2

!1221)1(

1

l l

ll

llm

l

lm

l

lm

mlmll

m

ll

llI R

baR

m

ll

m

l

m

lRR mlml

21

2

2

1

1,, 2211ba

0 0,

2/1

21

21

1 2

21

21

1

11

2

22

21

1

!2!2

!1221

1

l l

ll

llm

l

lm

l

lmmll

l Ill

llR

baR

m

ll

m

l

m

lRR mlml

21

2

2

1

1,, 2211ba

m

ll

m

l

m

lIRR

ll

ll

ll mmmmllmlml

l 21

2

2

1

1

, ,,,,,

2/1

21

21

21 21

212211

1

!2!2

!1221

1Rba

baR

Aa Bb

ba

Aa Bb ab

ba qq

r

qqH

baRˆ

m

ll

m

l

m

lIRqRq

ll

llH

ll mmmmll

Bbmlb

Aamla

l 21

2

2

1

1

, ,,,,,

2/1

21

21

21 21

212211

1

!2!2

!1221ˆ Rba

m

ll

m

l

m

lIQQ

ll

llH

ll mmmmll

Bml

Aml

l 21

2

2

1

1

, ,,,,,

2/1

21

21

21 21

212211

1 ˆˆ!2!2

!1221ˆ R

Aa

mlaAml RqQ a

1111 ,,ˆ

Bb

mlbBml RqQ b

2222 ,,ˆ RR mllllmll C

RI ,1, 212121

1

Interaction energy depends on 121 llRMolecular Multipoles

l = 0, monopole; l = 1, dipole; l = 2, quadrupole; l = 3, octapole; etc.

COMPLETING THE HAMILTONIAN

The derivation does not finish at the previous last line.

There are still two more steps to be applied that assist significantly in calculation.

Firstly, the previous multipoles are in the laboratory axis system, and for convenience need to be transformed into the molecule fixed axis system.

Secondly, the Hamiltonian is presently complex, but it is particularly convenient to work only with real quantities, so a second transformation is required.

THE FINAL HAMILTONIAN

After rotating our axes so that the multipoles refer to molecule fixed axes and making our Hamiltonian real we arrive at the following expression:

Here designates a real component of the molecule fixed multipole.

The

tensors contain all the orientation dependant terms as well as the

22112,2

21 21

1,1,,,

,

ˆˆˆ

llB

ll

A TQQHll

2211 ,,, llT

121 llR

THE INTERACTION ENERGY: AB INITIO Because the electrons in each molecule are

not at specific fixed locations the interaction energy is evaluated using the wavefunction.

This interaction energy can be directly evaluated with an ab initio calculation of the entire system.

Here represents the wavefunction of the two (or more) interacting molecules.

While this is the most accurate method for computing an interaction energy, the calculation must be repeated for any change in relative orientation.

Aa Bb

baqqHEbaR

ˆ

APPROXIMATING In order to make use of our derived

expansion we need to make the following approximation:

Here A represents the ground state wavefunction of an isolated molecule A, i.e., without any other molecule present. Similarly for B.

Making such an approximation does not allow electrons in molecule A to exchange with electrons in molecule B.

By preventing such an exchange means that short-range exchange repulsion (steric repulsion) is neglected.

BA

ELECTROSTATIC INTERACTION ENERGY

If we now operate on our approximate wavefunction with the Hamiltonian we obtain:

AB

Aa Bb

baBA qqHE

baRˆ

AB

ll

Bl

Alll

BA QQTEl

21 21

2212211,

,,,,,eleˆˆ

B

ll

Bl

BAAl

All QQT

l

21 21

2212211,

,,,,,ˆˆ

21 21

2212211,

,,,,,ll

Bl

Alll QQTl

Note the change from the multipole operators to multipoles.

CENTRAL MULTIPOLES The multipoles of the previous page all

represent multipoles for the entire molecule. When l = 0, we are talking about a monopole,

which is the overall charge on the molecule. When l = 1, we are talking about a molecular

dipole moment. A dipole moment has three components: a dipole

along the x-direction, a dipole along the y-direction and a dipole along the z-direction.

For example, water at equilibrium has a non-zero molecular dipole orientated along the z-direction only if the z-axis is the symmetry axis of water.

When l = 2, we are talking about a molecular quadrupole moment.

MOLECULAR QUADRUPOLE A quadrupole in Cartesians coordinates has

nine components like so:

However, the xy = yx and xz = zx and yz = zy as well as xx +yy +zz = 0. So in reality there are only five unique components.

We can choose to use multipoles in Cartesian coordinates, or we may choose to use spherical polar coordinates. In the former case our multipoles are Cartesian

tensors, and in the latter they are spherical tensors.

zzzyzx

yzyyyx

xzxyxx

θ

SPHERICAL VS CARTESIAN TENSORS

Spherical tensors are more convenient for many reasons, but one difference is that spherical tensors only ever have 2l+1 unique components.

We will adopt A. Stone’s implementation of spherical tensors and use the l, labels for the components.

The components are labeled as follows: l, ≡ l,0; l,1c; l,1s; l,2c; l,2s; l,3c; l3s; etc.

The link between Cartesian multipoles and Stone’s spherical multipoles up to the quadrupole is provided here: qQ 00

zQ 10

xcQ 11

ysQ 11

zzQ 20

xzcQ 3

221

yzsQ 3

221

yyxxcQ 3

122

xysQ 3

222

A FEELING FOR QUADRUPOLES

From The Theory of Intermolecular Forces, A. J. Stone (2002) Clarendon, Oxford.

(CO2 ) = −3.3au

(HF ) = +1.76au

(C2H2 ) = +5.6au

(C6H6 ) = −6.7au

xx(H2O ) = −1.86au

yy(H2O ) = +1.96au

zz(H2O ) = −0.10au

ELECTROSTATIC INTERACTION ENERGY Because we used the product of the ground

state wavefunctions of molecules A and B to approximate the interaction energy we obtain is known as the electrostatic interaction energy.

Of course the presence of molecule B will perturb the wavefunction of molecule A, and visa versa, but this perturbation is not taken into account in this interaction energy.

The perturbing effect on each wavefunction due to the presence of both molecules is independent of the errors incurred by preventing exchange of electrons between the molecules, i.e., is independent of steric repulsion.

INDUCTION AND DISPERSION We have stated that by using the ground state

wavefunctions of molecules A and B, the interaction energy so computed is the electrostatic interaction energy, Eele.

One can take account of the interaction energy due to the perturbed wavefunctions via perturbation theory.

The first order interaction energy is Eele.

The second order interaction energy is known as the induction and dispersion interaction energies.

The total interaction energy can then be written as E ≈ Eele + Eind + Edis.

We use “≈” because we are still neglecting electron exchange between A and B.

EXAMPLE ELECTROSTATIC INTERACTION ENERGY EXPANSION

Consider a water molecule in its equilibrium geometry interacting with a Na+.

The water molecule possess C2v symmetry, and thus has the following non-zero multipoles:

Na+ possesses only a monopole, i.e., The electrostatic interaction energy is

therefore:

,...,, OH2,2

OH0,2

OH0,1

222cQQQ

Na0,0Q

...Na0,0

OH2,20,0,2,2

Na0,0

OH0,20,0,0,2

Na0,0

OH0,10,0,0,1ele

222

QQTQQTQQTE cc

Varies as R−1−0−1 = R−2 Varies as R−2−0−1 = R−3

Dipole-monopole Quadrupole-monopole

CONVERGENCE Recall that our

expansion is only valid if|R| > |a – b|

This amounts to saying that the effective physical extent of the two molecules had better be smaller than the distance between the molecules.

Furthermore, even if the series does converge it may do so very slowly if the distances involved are fairly similar.

qA A qB B l1+l2+1 -89.204545

Z0 2.0 -5.000 -1.0 5.000 E0,0,0,0 / mH 1 -100.000000

Z1 -0.5 -4.000 -0.5 6.000 E0,0,1,0 / mH 2 10.000000

Z2 -0.5 -6.000 0.5 4.000 E1,0,0,0 / mH 2 0.000000

Q0,0 1.00000 -1.00000 E1,0,1,0 / mH 3 0.000000

Q1,0 0.00000 -1.00000 E0,0,2,0 / mH 3 0.000000

Q2,0 -1.00000 0.00000 E2,0,0,0 / mH 3 1.000000 -100.000000

Q3,0 0.00000 -1.00000 E0,0,3,0 / mH 4 0.100000

Q4,0 -1.00000 0.00000 E3,0,0,0 / mH 4 0.000000 -90.000000

Q5,0 0.00000 -1.00000 E1,0,2,0 / mH 4 0.000000 -90.000000

R 10 E2,0,1,0 / mH 4 -0.300000

eab 1 E0,0,4,0 / mH 5 0.000000 -88.900000

rAz 1 E4,0,0,0 / mH 5 0.010000

rBz -1 E2,0,2,0 / mH 5 0.000000 -89.190000

E1,0,3,0 / mH 5 0.000000

E3,0,1,0 / mH 5 0.000000 -89.189000

E0,0,5,0 / mH 6 0.001000

E5,0,0,0 / mH 6 0.000000

E1,0,4,0 / mH 6 0.000000

E4,0,1,0 / mH 6 -0.005000

E2,0,3,0 / mH 6 -0.010000

E3,0,2,0 / mH 6 0.000000

Eele / mH -89.204000Residual -0.000545% error 0.001

-20

-15

-10

-5

0

5

10

15

20

CONDITION FOR CONVERGENCE It can be shown mathematically that the

divergence sphere for the central multipole expansion is a sphere that encloses just the nuclei of the molecule.

Consider the example: For benzene this is a sphere of radius 2.4 Å. For hexafluorobenzene the radius is 2.8 Å. Thus if a central multipole expansion is to be used to

compute the interaction energy of these two molecules they need to be further from each other than 2.4+2.8 = 5.2 Å.

In the benzene…hexafluorobenzene complex the two planes are 3.5 Å apart.

A central multipole expansion of the electrostatic interaction can not be used.

IMPROVING CONVERGENCE

There is a solution to the previous issue. That is to use a distributed multipole approach. This means that for a given molecule instead of

using a single site about which the multipoles occur we use several sites.

Often the sites chosen are those of the nuclei. You are already familiar with this type of

approach. Chemists often use distributed monopoles in

describing the charge distribution on molecules. E.g., H2O:

−½+

½+

DISTRIBUTED MULTIPOLE INTERACTION ENERGY By including a number of multipole sites in

molecules A and B, our electrostatic interaction energy is now computed as the sum of interactions over all these sites thus,

If we only had distributed monopoles then this expression becomes:

However, as we shall see later, using only distributed monopoles to describe the interaction energy is appallingly inaccurate and suffers convergence problems like the central expansion.

Aa Bb ll

bl

al

abll QQTE

l

21 21

2212211,

,,,,,ele

Aa Bb ab

ba

Aa Bb

baab

R

qqQQTE 0,00,00,0,0,0ele