The Electrostatic Interaction Energy I

The Electrostatic Interaction Energy I Ryan 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:

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: Here q is in electron charge and rijis 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:

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.

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,mfunctions, so here they are:

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.

Example of the 1/r12 Expansion

Relating 1/r12 to Interacting Molecules a • 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 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:

Regular Spherical Harmonic Addition Theorem is called a Wigner 3j symbol. It is just a simple number.

Will only work if |R|>|a − b| We’ll need to use our addition theorem to separate the a and b away from this single regular spherical harmonic.

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

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

Molecular Multipoles Interaction energy depends on 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 k designates a real component of the molecule fixed multipole. • The tensors contain all the orientation dependant terms as well as the

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 Y 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.

Approximating Y • In order to make use of our derived expansion we need to make the following approximation: • Here yA represents the ground state wavefunction of an isolated molecule A, i.e., without any other molecule present. Similarly for yB. • 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.

Electrostatic Interaction Energy • If we now operate on our approximate wavefunction with the Hamiltonian we obtain: 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 qxy = qyx and qxz = qzx and qyz = qzy as well as qxx +qyy +qzz = 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.

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,k labels for the components. • The components are labeled as follows: • l,k≡ 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:

A feeling for quadrupoles q(CO2 ) = −3.3au q(HF) = +1.76au q(C2H2 ) = +5.6au q(C6H6 ) = −6.7au qxx(H2O) = −1.86au qyy(H2O) = +1.96au qzz(H2O) = −0.10au From The Theory of Intermolecular Forces, A. J. Stone (2002) Clarendon, Oxford.

Electrostatic Interaction Energy • Because we used the product of the ground state wavefunctions of molecules A and B to approximate Y, 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 C2vsymmetry, and thus has the following non-zero multipoles: • Na+ possesses only a monopole, i.e., • The electrostatic interaction energy is therefore: Dipole-monopole Quadrupole-monopole Varies as R−1−0−1 = R−2 Varies as R−2−0−1 = R−3

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.

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: d− ½d+ ½d+

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.

The Electrostatic Interaction Energy I