Rev.Adv.Mater.Sci.Potentials of interatomic 20(2009) interaction 1-13 in molecular dynamics 1 POTENTIALS OF INTERATOMIC INTERACTION IN MOLECULAR DYNAMICS Alexander I. Melker Department of Metal Physics and Computer Technologies, St. Petersburg State Polytechnic University, Polytekhnicheskaya 29, 195251, St. Petersburg, Russia Received: February 05, 2008 Abstract. In this contribution we have analyzed the principles lying at the basis of interatomic potentials which are used in molecular dynamics studies. The different types of potentials were considered: ab initio, semi-empirical, and empirical. The classical, ab initio, and bond-charge molecular dynamics were discussed. 1. INTRODUCTION and then integrates numerically the equations of motion of N particles. At N >>1, the system has a Over the course of the past half-century Molecular statistical property. This allows to watch the mo- Dynamics has progressed from a small branch of tion of individual particles and observe dynamical scientific investigation to an independent science. structure of the system; and simultaneously calcu- The onset was done in 1955 when specialists in late the average properties of the system with re- computational mathematics asked Enrico Fermi to spect to time and ensemble of particles (energy, set them a task which must satisfy three condi- pressure, temperature, stress-strain diagrams, tions: etc.). The calculated averages can and must be • it must be interesting for physics, compared with experimentally observed values. • it has no analytical solution, Therefore, once the problem for study was put, • it is not beyond the capacity of existent comput- you have to choose the most appropriate method ers. of numerical integration of differential equations of A computer allows to obtain an exact numerical motion and the most appropriate law of interaction solution, which helps to understand the kernel of for the particles of a system studied. The methods the problem. After such a computer prompt, one of integration were considered thoroughly in [1]. can separate main and secondary factors, formu- Now we only mention that a real professional does late the problem in a new fashion, and construct a not rely on numerous commercial programs be- simpler nonlinear equation having an analytical cause ‘a person must know how the calculations solution. As a rule, at that, there appears a new are going on. If one does not understand how it is scientific direction with a lot of interesting compu- done, one sees only bare numbers, but their genu- tational and analytical results. Such new method ine significance is hidden in calculations’ [2]. Just of scientific investigation, the combination of analy- the same approach is valid to a greater extent for sis and computer simulations, becomes more and the choice of interaction law. more important in a scientific discovery. In this contribution we consider interatomic po- Molecular dynamics is a type of computer ex- tentials which were deduced from the first principles periments where one assigns the law of interac- as well as semi-empirical potentials and briefly tion for N particles, initial and boundary conditions, empirical ones. Corresponding author: Alexander I. Melker, e-mail: [email protected] © 2009 Advanced Study Center Co. Ltd. 2 A.I. Melker 2. THOMAS–FERMI ATOM The one-particle Schrödinger equation has the solution The atom model suggested by Thomas and Fermi (L.H. Thomas, 1926; E.R. Fermi, 1927) has be- ψarkrf = expai f , come a classic and can be found in many books on quantum mechanics, e.g. in [3]. The model is at that based on two ideas: one of them is taken from h22k quantum mechanics and another from electrostat- Eafk = . ics. The starting point of the quantum mechanical 2m theory of an atom is Schrödinger’s equation for Here k is any vector which does not depend on electrons coordinates. Let us restrict the motion of electrons HEΨΨ= . by volume V. In this case Here Ψ, E is the wave function and the total energy ψaf2 = of all electrons respectively and the Hamiltonian z rrd 1 has the form V and hence p 2 1 e 2 =−j 2 + H ∑∑∑Ze ∑ . 1 j 2m j r i < j r ψaf= a f j ij rkrexpi . V The first term gives the kinetic energy of electrons, the second one interactions of the electron with a If we select the volume V as a cube with the side L nucleus, and the third describes interaction of the and put the periodic boundary conditions electrons with each other. In order to obtain the ψψaxLyz+=+=,,f a xyLz , ,f simplest description of electron motion, let us ex- clude the interaction of electrons with each other ψψaxyz,,+= Lf a xyz ,,f , and with the nucleus. Then we obtain the model of free electrons in which Schrödinger’s equation we exclude the influence of a surface. takes the form The periodic boundary conditions mean that an electron does not reflect going to a surface, but p 2 h2 goes out the volume and simultaneously enters the j Ψ∆ΨΨ=− = ∑∑j E . volume through an opposite plane, e.g. in one-di- 22mm j j mensional case the electron moves in a ring. By Let us write down the many-particle wave func- virtue of the boundary conditions tion Ψ as the product of one-particle functions ψ expaiikra += Lff expa krf , each of them being dependent on only one-elec- tron coordinates i.e. Ψaf= ψ af a f = rr1,...,Nj∏ r . expikr 1 . j This means that Substitute this formula in the Schrödinger equa- tion. The latter disintegrates into one-electron equa- kL⋅=2πn, tions where h 2 n = 012, , ,... −=∆ψaf ψ af jjrrE , 2m or for which the sum of one-particle energies is equal F 2πI ===α to the total energy kααn, xyzn , , , α012 , , ,... H L K ∑ EE= . j are integers. j Consider how to interpret these results. The We will investigate only a one-electron equation, vector equation r . k = const defines a plane nor- so delete the index j. mal to the vector k. On this plane the function Potentials of interatomic interaction in molecular dynamics 3 exp(ikr) has a constant value, whereas along any can add any constant value, it is reasonable to ac- ε straight line, which is parallel to the vector k, the cept that max = 0. In this case function changes periodically. For this reason, the function exp(ikr) is termed a plane wave and the 2 2m k =− Uraf, vector k is referred to as a wave vector. max h 2 It follows from the boundary conditions that the wave vector can take only the values which and the problem is reduced to the question how to comppponents are integers of 2π/L. One can de- find U(r). fine these mathematically allowed values as quan- With this purpose, refer to electrostatics. Let tum numbers which characterize electron states in the electron density at the distance r from the the system. Let us introduce k–space and construct atomic nucleus is equal to n(r). Then the electro- ϕ the lattice in this space, the knots of which are the static potential (r) produced by joint action of the allowed values of the wave vector. Since the elec- nucleus and electrons satisfies Poisson’s equation tron energy is proportional to the wave vector (Siméon Denis Poisson, 1812) squared, the lattice gives all the possible values of ∆ϕ = 4πenar f, electron energy. Therefore, we can find the physi- cal state of the system with the help of the geo- at this U(r) = -eϕ(r). metrical scheme developed. Now we have two equations which incorporate Suppose that the system contains N electrons. the electron density; one equation is quantum one The electron energy has a minimum in the ground and another is Poisson’s equation. Using the quan- state. To obtain this state, it is necessary to fill N/2 tum equation, one can exclude the electron den- cells of least energies, each cell having the vol- sity from Poisson’s equation and obtain so-called ume (2π/L)3 and containing two electrons with op- Thomas–Fermi’s equation posite spins. Since the energy does not depend on 32/ the direction of a wave vector, the filled cells cre- 42emea f ∆ϕ = ϕ 32/ . ate a sphere. The sphere is spoken of as Fermi’s 3πh 3 sphere and its radius k is known as the Fermi wave F It is reasonable to use the atomic units, i.e. to put vector (Enrico Fermi, 1926). Therefore e = m = h = 1. Then N af2π 3 N 4 33 82 ==4ππk , 32/ F ∆ϕ = ϕ . 2 V V 3 3π and hence the electron density of the system is The solution of this equation must satisfy the bound- ary condition in the immediate proximity to the 3 N k nucleus, where the potential ϕ(r) at r → 0 must n ==F . V 3π 2 coincide with the nucleus potential Ze/r, i.e. in the atomic units rϕ(r) → Z. Now introduce the potential energy U(r). Tak- Let us find the derivatives of the expression rϕ(r) ing into consideration this energy, one can write for any electron the following equality dafrϕ dϕ =+ϕ r , dr dr h 22k max +=Uraf ε . d2 afrϕ d2 ϕϕd 2m max =+r 2 . dr 2 drr2 d From this, it follows Since ϕ depends only on r, the Laplace operator 2m reduces to k 2 =−afε Uraf. max2 max h d2 2 d ∆= + , ε drrr2 d Here max is the maximal value of the total energy. The total energy is constant if the system consid- so ered is in a stationary state, i.e.