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Molecular Modeling

by: Mr. Clementine Gottlieb

Molecular Modeling CHEM 6970

Mr. Clementine Gottlieb
GPA 3.68

Titus Albu

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Titus Albu
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This 6 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 6970 at Tennessee Tech University taught by Titus Albu in Fall. Since its upload, it has received 24 views. For similar materials see /class/225692/chem-6970-tennessee-tech-university in Chemistry at Tennessee Tech University.


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Date Created: 10/21/15
CHEM 6970 Fall 2004 Approximation Methods in Quantum Mechanics The hydrogen atom serves as prototype for the treatment of more complex atoms By solving the Schrodinger equation for the hydrogen atom h2 2 l V2 e lwr6 Eur6 L 2me 47230 one obtains the allowed energies and the wave functions of the electron 7 Look at the He atom constituted from the nucleus and two electrons The wave function of the system will depend on 7 the position of the helium nucleus R 7 the positions of the two electrons 1 1 and 1 2 7 Schrodinger equation for the He atom 122 2 r12 2 r12 2 V V V Rrr 2M 2MB 1 2MB 2 l 1 2 282 262 e lIR 1 1 1 2 4723980 IR I ll 4723980 IR 1 239 4723980 lrl r2 EWRI II 2 7 He atom is a threebody system and is solving Schrodinger equation is more complicated 7 Consider the nucleus to be xed at the origin 52 2 2 V V 1 r 2mg 1 2 1 2 2e2 1 1 e lF1r2 lF1r2 4723980 r1 r2 4723980ll 1 I 2l ElI1 121 2 3 This equation cannot be solved exactly The interelectronic repulsion term is responsible for this Without it one can make separation of variables 7 Schrodinger equation for the He atom represents a system that cannot be solved exactly Solving these types of equations requires approximate methods 7 Variational method 7 Perturbational theory CHEM 6970 Fall 2004 Atomic units 7 Are used in atomic and molecular calculations 7 An advantage in their use is that the values expressed in atomic units are not affected by re nements in various constants me e h etc 7 Unit of angular momentum h 0 4 h2 7 Unit of length a0 L2 named bohr and denoted as do mee 7 Unit of energy E named hartree and denoted as E h e 167239 2 83 h 2 Variational Method 7 Consider an arbitrary system for which the groundstate wave function and energy satisfy H lo E0 lIo J lloH 11 0 d7 3 E0 J 11 o 11 0 d T where dr represents the volume element 7 If one substitute 110 with any other function and calculate j H dr j Mair then according to variational principle E 2 E0 E where E E0 only if z0 7 Enunciation of Variation Principle If an arbitrary wave function is used to calculate the energy then the value calculated is never less than the true energy 3 Variational principle says that we can calculate on upper bond to E 0 by using any function 7 One can choose the function called trial function that depends of some arbitrary parameters a 7 called variational parameters The energy will also depend on these parameters 3 E a yZE0 7 Optimize a 7 to get the lowest groundstate energy therefore obtaining the trial wave function CHEM 6970 Fall 2004 7 Example ground state of hydrogen atom 2 2 Hh 51021 e 2mer2 E dr 4723980r 7 Choose the trial function r e mz a is a variational parameter 3h 12 2 2 4 3 Eaa Where aamp zme 212807232 18n3ggh4 4 4 2 E 0424 L and compare to E0 0500 L mm 1672383112 1672383112 332 3 12 8 1 7 2 2 3 r e g 9r 0 and compare to le 7239 mo 7 Variational method for the He atom 7 Rewrite the Hamiltonian as 2 HHH1HH2 e i 4723980 r12 2 2 where HH39h 3 28 i 2me 4723980 rj 2 7Neglect e i term 4723980 r12 7 One has 19H39wHrJ6j j ijHrJ6j j 3 get 1 and E 7 Use a trial function 0 1 1 1 2 11 S 1 1 zjl S 1 2 7 Calculate Ez I 0H 0dz quotZS 8222 2 22 szh 167239 80 h 8 8 where Eh is a unit of energy called hartree or atomic units 1E11 26255 kJmol 272116 eV 7 Minimizing Ez 3 2mm 3 Emin 28477E11 7 Compare with accurate calculated value 29037E11 7 Experimental value 29033E11 CHEM 6970 Fall 2004 7 The value of 2mm can be interpreted as an effective nuclear charge 3 each electron partially screens the nucleus from the other 2 lt 2 7 Look at a more complex trial function cnfn C1f1 C2f2 nl 7Examplefor N2 clf1 02f2 er 1 Mir 2 2 2 I H dTC1 H11 C1C2H12 C1C2H21C2H22 EclCZ I dTC1ZS11 26162312 63322 where Hij I HfJdr is called Coulomb integral Sij ftfde is called resonance integral Htj H if 1 is Hermitian ECICZ C121H112612H12 C2H22 61511445162312 C2522 6E 6c 0 3 61H11 ES1102H12 ES120 1 6E 30 3 61H12 312C2H22ESZZ0 2 7 Coef cients cl and 02 are nonzero if and only if Hn ES11 H12 ES12 0 called a secular determinant H12 ES12 H22 ES22 2 3 H11 ES11XH22 ES22 H12 ES12 0 2 2 2 2 2 2 H11H22 EH11S22 11525104457 S11S22 H12 131112512 E S120 E2S S S2 EH2S2 H S H S H H H20 1122 12 1212 1122 2211 1122 12 7 This is a secondorder equation and the smallervalue solution is the variational approximation for the groundstate energy CHEM 6970 Fall 2004 7 For the case of larger N than the determinant in N order Hll ESH H12 7ES12 H12 7ES1N H12 ES12 H22 Eszz 0 HlN ESIN HZN ESZN HNN ESNN 7 Once E is determined one can go back in determining coefficient ci 7 A more complex function can be used for trial functions N M chfjfje j jl r2 where function f J is function of few coefficients as well 7 Slater filled the need for general and suitable trial functions that are not necessarily same as the hydrogen wave functions by introducing a set of orbitals called Slater orbitals defined as Srtlm r 6 gt anrnileigrYlm 69 2gnl 2 where N n is a normalization constant 1Zn Y 139quot 9 are the spherical harmonics 7 Properties of Slater orbitals 7 Q zeta is arbitrary and is not necessarily equal to Zn as in hydrogenlike orbitals 7 The radial part of the Slater orbitals does not have nodes 7 S r 9 is not orthogonal to Snrlm r 9 nlm 7 n can be also considered as a variational parameter CHEM 6970 Fall 2004 Perturbation Theory 7 General idea suppose we are unable to solve the Schrodinger equation H 1 E 11 for the system of interest but we can solve it for a similar system gltogtwltogt Eltogtwltogt where H H0 H1 H 0 is called unperturbed Hamiltonian operator H 1 is called perturbation If the perturbation is small the solutions of H will be similar to those of H 0 7 Example Anharmonic oscillator 2 2 H i lh3lm3 m4 2 dx2 2 6 4 2 A 0 h d 2 lkx2 is the harmonic oscillator operator 2 dx the solutions are known z0x E0 v l 2hv v v H1oc3 x4 is the perturbation 7 Perturbation theory says that the wave function and the energy of the unperturbed system can be successively corrected wwmwmwmm E Elo E1 E2 where rpm is the wave function of the unperturbed system E 0 is the energy of the unperturbed system 110 1pm are successive corrections to z0 EU E2 are successive corrections to E0 7 A basic assumption is that those successive corrections become smaller 7 Expressions are available for these terms We will work with only E 1 which is the firstorder correction to E 0 E0 I WW mulllt0 d1 E E 0 E 1 is the energy through frrstorder perturbation theory 1 z0 110 is the wave function through frrstorder perturbation theory E E 0 E 1 E 2 is the energy through secondorder perturbation theory


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