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Helium Atom, Approximate Methods

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Helium Atom, Approximate Methods Helium Atom, Approximate Methods 22nd April 2008 I. The Helium Atom and Variational Principle: Approximation Methods for Complex Atomic Systems The hydrogen atom wavefunctions and energies, we have seen, are deter- mined as a combination of the various quantum "dynamical" analogues of classical motions (translation, vibration, rotation) and a central-force inter- action (i.e, the Coulomb interaction between an electron and a nucleus). Now, we consider the Helium atom and will see that due to the attendant 3-body problem for which we cannot determine a close-for, first-principles analytic solution, we will have to find recourse in approximate methods. The Helium atom has 2 electrons with coordinates r1 and r2 as well as a single nucleus with coordinate R. The nuclues carries a Z = +2e charge. The Schrodinger equation is: 2 2 2 h¯ 2 h¯ 2 h¯ 2 1 2 (R; r1; r2) + −2M r − 2me r − 2me r ! 2e2 2e2 e2 + (R; r ; r ) = E (R; r ; r ) −4π R r − 4π R r 4π r r 1 2 1 2 o j − 1j o j − 2j o j 1 − 2j! where the symbol "nabla", when squared, is given by: @2 @2 @2 2 = + + r @x2 @y2 @z2 Keep in mind that the R, r, and r represent the Cartesian coordinates of each paticle. This is a 3-body problem and such problems are not solved exactly. Thus, the problem will be reformulated in terms of 2 variables. The first approximations: Mme , fix the nucleous at the origin (R) = 0. Thus, the Schrodinger equation in relative variables is: 1 2 2 2 h¯ 2 2 2e 1 1 e 1 2 (r1; r2) + + (r1; r2) = E (r1; r2) 2me −∇ − r −4πo r1 r2 4πo r2 r1 j − j Recall that the 2, representing the kinetic energy operator, in spherical r polar coordinates is: 1 @ @ 1 @ @ 1 @2 r2 + sinθ + 2 @r 1 @r 2 @θ 1 @θ 2 2 2 r1 1 1 r1sinθ1 1 1 r1sin θ1 @φ1 The Independent Electron Approximation to Solving the Helium Atom Schrodinger Equation If we neglect electron-electron repulsion in the Helium atom problem, we can simplify and solve the effective 2-body problem. Solve the relative motion problem (separate out the center of mass • motion as we have seen earlier) Center of mass is assumed to be the nucleus; good approximation for • heavier nuclei The Hamiltonian is now: ^ H = KEe1 + KEe2 + Vne1 + Vne2 2 2 h¯ 2 2 2e 1 1 = 1 2 + 2me −∇ − r − 4πo r1 r2 h¯2 2e2 1 h¯2 2e2 1 = − 2 + − 2 2m 1 4π r 2m 2 4π r e r − o 1 ! e r − o 2 ! Under the independent electron approximation, if we take the total He atom wavefunction as a product of the individual electron wavefunctions (here ap- proxmiated as hydrogen-like wavefunctions): Ψ(r1; r2) = (r1) (r2) H^ Ψ(r1; r2) = H^1 (r1) + H^2 (r2) 2 This yields E = E1 + E2. Recal the hydrogen-like energies and wavefunc- tions are: Ψ (r) = R (Zr=a )Y (θ; φ) (Z=1 for hydrogen, Z=2 for helium) • nlm nl o lm 2 Z Eh 1 Energies are: E = 2 • n − 2 n Thus, approximate wavefunctions and energies for helium: Ψn1; l1; m1; n2; l2; m2(r1; r2) spin = Ψn1; l1; m1(r1)Ψn2; l2; m2(r2) spin Z2E 1 1 E = h + n1;l1;m1;n2;l2;m2 2 2 2 − n1 n2 How sound is the independent electron approximation (no electron-electron repulsion model)? We can compare the predicted ionization potentials with the exerimental values. The ionization potential is the energy required to extract an electron from an atom. For the Helium atom, this can be represented as an equation such as: He He+ + e− ! The energy change associated is: ∆E = E E 2 He;1s − He;(1s) = 54:4eV ( 108:8eV ) = 54:4eV − − − The ground state energy for Helium can also be contrasted as: 2 0 2 Eh 1 1 E = E1s;1s = 2 + 2 − 2 1 1 = 4E − h = 108:8eV significantlytoonegative − 3 The experimental value for the ionization potential (from mass spectroscopic measurements) is 24.6 eV; so the independent electron approach entails sig- nifican error. Because the electrons are not allowed Coulombic repulsion, the energy required to remove a particular electron is higher than the ex- perimental value. This is also validated by the much lower (and thus more attractive/favorable) ground state energy predicted by the independent elec- tron model compared to experiment. Moving Beyond the Independent Electron Model: Perturba- tive and Variational Methods noindent Perturbation Theory The Helium atom Hamiltonian, we recall, is: 2 2 2 ^ h¯ 2 2 2e 1 1 e H = 1 2 (r1; r2) + + 2me −∇ − r − 4πo r1 r2 4πo r2 r1 j − j = H^ 0 + H^ 1 H^ 0 0 = E0 0 We see that the electron-electron repulsion term can be treated as a "per- turbation" to the independent electron Hamiltonian. in this sense, we can choose to define, develop, and include various orders of perturbative cor- rections to the energies and wavefunctions. WThough we are currently applying this to the problem of electron-electron repulsion, we will see later that modern advanced methods for electronic strucuture calculations em- ploy perturbation methods to account for important electron correlation. For now, we concern ourselves with the development of perturbation theory and application to correct for two-body Coulomb repulsion in the Helium atom. First Order Perturbation Theory First, expand the total wavefunction up to first order contributions: 0 1 n = n + n 0 1 En = En + En ^ 0 ^ 1 0 1 0 1 0 1 H + H n + n = En + En n + n Expanding the expression: ^ 0 0 ^ 1 0 ^ 0 1 ^ 1 1 0 0 1 0 0 1 1 1 H n + H n + H n + H n = En n + En n + En n + En n 4 The first terms on each side are equivalent. Last terms are approxi- mated/assumed to be neglibly small{this is a perturbation energy and wavefunction. This leaves: Expanding the expression: ^ 1 0 ^ 0 1 1 0 0 1 H n + H n = En + En n 1 1 0∗ Solve for n and En. Left multiply by n and integrate: Expanding the expression: 0∗ ^ 1 0 0∗ ^ 0 1 0∗ 1 0 0∗ 0 1 dτ n H n + dτ n H n = dτ n En n + dτ n En n Z Z Z Z ∗ ^ 0 0∗ ^ 0 1 ^ 0 0 1 Since H is Hermitian, dτ n H n = dτ H n n. R R 0∗ ^ 1 0 1 0 0∗ 0 0∗ 1 1 dτ n H n + dτ nEn n = En dτ n n + En Z Z Z 1 Solving for En , the fist order perturbative correction to the inde- pendent electron energy level for helium gives: 1 0∗ ^ 1 0 En = dτ n H n Z 0 1 Expilcitly, this means, En = En +En; we have added a "small" perturbative correction to the reference independent electron energy for a given energy level, n! Now, what about the correction to the wavefunction? Let's recall: ^ 1 0 ^ 0 1 1 0 0 1 H n + H n = En + En n 1 We will solve for n by expanding n as a linear combination of unper- turbed wavefunctions as follows: 5 0 0 n = n + anj j jX=6 n 0∗ Left multiply by k and integrate: 0∗ ^ 1 0 0∗ ^ 0 1 0∗ 1 0 0∗ 0 1 dτ k H n + dτ k H n = dτ k En n + dτ k En n Z Z Z Z 0∗ ^ 1 0 0∗ ^ 0 0 0∗ 1 0 0∗ 0 0 dτ k H n + dτ k H anj j = dτ k En n + dτ k En anj j Z Z jX=6 n Z Z jX=6 n 0∗ ^ 1 0 0 0∗ 0 1 0∗ 0 0 0∗ 0 dτ k H n + anjEj dτ k j = En dτ k n + En anj dτ k j Z jX=6 n Z Z jX=6 n Z a E0 E0 = E1 dτ 0∗ 0 dτ 0∗H^ 1 0 nk k − n n k n − k n Z Z Now, we consider two cases: k = n E1 = dτ 0∗H^ 1 0 ! n n n Z dτ 0∗H^ 1 0 H^ 1 k = n a = k n kn 6 ! nk E0 E0 ≡ E0 E0 R n − k n − k Thus, the first order correction to the wavefuntion is: 0 1 0 0 n = n + n = n + anj j jX=6 n ^ 1 0∗ ^ 1 0 0 Hjn 0 0 dτ k H n 0 = n + 0 0 j = n + 0 0 j En Ej ! R En Ek ! jX=6 n − jX=6 n − Higher Order Corrections For higher order corrections, energies and wavefunctions are expanded in 6 like manner: 0 1 2 0 1 2 n = n + n + n En = En + En + En The second-order correction to the energy is thus determined to be (see other sources for derivation): H^ 1 2 0∗ ^ 1 1 0∗ ^ 1 jn 0 En = dτ n H n = dτ n H 0 0 j En Ej Z Z jX=6 n − ^ 1 ^ 1 HnjHjn = 0 0 En Ej jX=6 n − Helium Atom: First Order Perturbation Correction to Account for Electron-Electron Repulsion From the above discussion, the first order correction to the ground state energy of Helium is: 1 0∗ ^ 1 0 En = dτ n H n Z If we assume an unperturbed wavefunction as the product of the 2 Helium electrons in 1s orbitals: 2 3 1=2 0 Z −Zr1 −Zr2 (r1; r2) = φ1s(r1)phi1s(r2) = e e 0 π ! 1 1 @ A H^ 1 = r12 Z6 1 E1 = dτ 0∗H^ 1 0 = dτe−Zr1 e−Zr2 e−Zr1 e−Zr2 n n n π2 r Z Z 12 Z6 e−2Zr1 e−2Zr2 5 = dτ = ZE π2 r 8 h Z 12 7 Thus, the ground state energy with first order correction is : 5 11 E = E0 + E1 = 4E + ZE = − E = 74:8eV 0 0 0 − h 8 h 4 h − Neglect of electron-electron repulsion: E = -108.8 eV • 0 First order correction: E = -78.8 eV • 0 13th order correction: E = -79.01 eV • 0 Experiment: E = -70.0 eV • Perturbation theory results may be greater or less than the true energy, • unlike variational results Perturbation, unlike variational theory, can be used to calculate any • energy level, not just the ground state.
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