Notes on Ewald summation techniques


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1 February 3, 011 Notes on Ewald summation techniques Adapted from similar presentation in PHY 71 he total electrostatic potential energy of interaction between point charges {q i } at the positions {r i } is given by W = 1 4πε 0 i<j q i q j r i r j = 1 q i q j r i r j. (1) In this expression, the first form explicitly counts all pairs, while the second form counts all interactions and divides by to compensate for double counting. For a finite system of charges, this expression can be evaluated directly, however, for a large or infinite system, the expression 1 does not converge and numerical tricks must be used to evaluate the energy. In fact, for the infinite system, one has an infinite amount of charge and the energy of interaction is undefined. If the system is neutral, it is possible to define a meaningful interaction energy by use of an Ewald transformation. he basic idea of the Ewald approach is as follows. he error function erf(x) and its complement erfc(x) are defined as: erf(x) = x e t dt and erfc(x) = 1 erf(x) = e t dt. () π π 0 x Ewald noted that 1 r = erf( 1 ηr) + erfc( 1 ηr). (3) r r In this expression, the first term goes to a constant ( η/π) as r 0, but has a long range tail as r. he second term has a singular behavior as r 0, but vanishes exponentially as r. his is illustrated in the plot below where η was chosen to be η = 4.
2 10 9 erf(x)/x (1erf(x))/x 1/x hus, Ewald s idea is to replace a single divergent summation with two convergent summations. he first summation has a convergent summation in the form of its Fourier transform and the second has a convergent direct summation. hus the calculation of the electrostatic energy would be evaluated using: W = 1 q i q j r i r j = 1 q i q j erf( 1 η ri r j ) + r i r j q i q j erfc( 1 η ri r j ). r i r j (4) For an appropriate choice of the parameter η, the second summation in Eq. 4 converges quickly and can be evaluated directly. he first term in the summation of Eq. 4 must be transformed into Fourier space. In order to described these summations explicitly, we assume that we have a periodic lattice so that every ion can be located by r i = τ α + a location τ α within a unit cell and a periodic translation vector, In this way, the summation becomes: 1 = N (5) ij αβ, where N denotes the number of unit cells in the system. Since we have a periodic system, N is infinite, but the energy per unit cell W/N is well defined. he other identity that we 1 Note that for any two lattice translations i and j, i j = k, where k is also a lattice translation.
3 must use is that a sum over lattice translations may be transformed into an equivalent sum over reciprocal lattice translations according to the identity: δ 3 (r ) = 1 e i r, (6) Ω where Ω denotes the unit cell volume. he proof of this relation is given in the appendix. he first term of Eq. 4 thus becomes q i q j erf( 1 η ri r j )) = N q α q β r i r j αβ erf( 1 η τα τ β + ) τ α τ β + α q α η, (7) π where the last term in Eq. 7 comes from subtracting out the selfinteraction (i = j) term from the complete lattice sum. Using the short hand notation, τ αβ τ α τ β, the lattice sum can be evaluated: his becomes, 1 Ω erf( 1 η ταβ + ) τ αβ + = d 3 re i r erf( 1 η ταβ + r ) τ αβ + r d 3 r = 4π Ω δ 3 (r ) erf( 1 η ταβ + r ). (8) τ αβ + r 0 e i τ αβ e /η + 1 where the last term, which is infinite, comes from the = 0 contribution. 1 η 0 du, (9) If the last term of Eq. 9 cannot be eliminated, it is clear that the electrostatic energy is infinite. he term can be eliminated if and only if the system is neutral. hus, it is only meaningful to calculate the electrostatic energy of a neutral periodic system. If the actual system has a net charge of Q α q α, we could calculate a meaningful energy if we add to the actual charge density, a compensating uniform density charge density of Q/Ω. aking all of the terms into account, we find the final Ewald expression to be: u 3 W N = αβ q α q β 4π Ω e i τ αβ e /η 0 η π δ αβ + erfc( 1 η ταβ + ) 4πQ τ αβ + Ωη, (10) where the in the summation over lattice translations indicates that all selfinteraction terms should be omitted. Appendix I comments on lattice vectors and reciprocal lattice vectors In this discussion, will assume we have a 3dimensional periodic system. It can be easily generalized to 1 or  dimensional systems. In general, a translation vector can be described a linear combination of the three primitive translation vectors 1,, and 3 : = n n + n 3 3, (11)
4 where {n 1, n, n 3 } are integers. Note that the unit cell volume Ω can be expressed in terms of the primitive translation vectors according to: Ω = 1 ( 3 ). (1) he reciprocal lattice vectors can generally be written as a linear combination of the three primitive reciprocal lattice vectors 1,, and 3 : = m m + m 3 3, (13) where {m 1, m, m 3 } are integers. he primitive reciprocal lattice vectors are determined from the primitive translation vectors according to the identities: i j = πδ ij. (14) Note that the volume of the primitive reciprocal lattice is given by 1 ( 3 ) = (π)3 Ω. (15) Some examples of this are given below. Proof of Eq. 6 Consider the geometric series +M k= M e ik( 1 r) = sin ( (M + 1 ) 1 r ) sin( 1 r/). (16) he behavior of the right hand side of Eq. (16) is that it is small in magnitude very except when the denominator vanishes. his occurs whenever 1 r/ = n 1 π, where n 1 represents any integer. If we take the limit M, we find that the function represents the behavior of a sum of delta functions: sin ( (M + 1 lim ) 1 r ) = π δ( 1 (r n 1 1 ). (17) M sin( 1 r/) n 1 he summation over all lattice translations n 1 1 is due to the fact that sin( 1 r/) = 0 whenever r = n 1 1. Carrying out the geometric summations in the right hand side of Eq. 6 for all three reciprocal lattice vectors and taking the limit as in Eq. 17, e i r = (π) 3 δ( 1 (r n 1 1 ))δ( (r n ))δ( 3 (r n 3 3 )). (18) n 1,n,n 3 Finally, the right hand side of Eq. 18 can be simplified by eliminating the reciprocal lattice vectors from the δ functions: e i r = which is consistent with Eq. 6. (π) 3 1 ( 3 ) δ 3 (r ) = Ωδ 3 (r ), (19)
5 Appendix II examples In these examples, denote the length of the unit cell by a. CsCl structure here are two kinds of sites τ Cs = 0 and τ Cl = a (ˆx + ŷ + ẑ). 1 = aˆx = aŷ 3 = aẑ. (0) 1 = π a ˆx = π a ŷ 3 = π a ẑ. (1) In these terms, Eq. 10 can be broken into two summations: W N = q ( 1 + ), () where and 1 4π Ω 0 erfc( η 0 (1 ei τ Cl )e /η η π (3) erfc( η τ Cl + ). (4) τ Cl + In this expression, the sum over α and β has a total of 4 contributions pairs of identical contributions. For CsCs or ClCl interactions, τ αβ = 0 and q α = q β resulting in repulsive contributions. For CsCl or ClCs interations, τ αβ = τ Cl and q α = q β resulting in attractive contributions. his expression can be evaluated using Maple, which gives the result W N = q ( 4.071). (5) a NaCl structure here are two kinds of sites τ Na = 0 and τ Cl = a ˆx. he primitive lattice has the translation vectors: and the reciprocal vectors: 1 = a (ˆx + ŷ) = a (ŷ + ẑ) 3 = a (ˆx + ẑ), (6) 1 = π a (ˆx + ŷ ẑ) = π a ( ˆx + ŷ + ẑ) 3 = π (ˆx ŷ + ẑ). (7) a
6 Note that there are many materials that have this structure, including MgO in which Mg has a charge of +e and O has a charge of e. Zincblende structure his structure has the same bravais lattice as NaCl, but the basis vectors are τ Zn = 0 and τ S = a (ˆx + ŷ + ẑ). In this case Zn has charge of +e and S has a charge of e. 4
7 Figure 1: Diagram of CsCl structure, indicating translation vectors and atomic site vectors τ. τ
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