Evaluation of Coulomb and exchange integrals for higher excited states of helium atom by using spherical harmonics series

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1 J Math Chem () 5:86 DOI.7/s ORIGINAL PAPER Evaluation of Coulomb and exchange integrals for higher excited states of helium atom by using spherical harmonics series Artit Hutem Sutee Boonchui Received: 6 September / Accepted: March / Published online: 5 April Springer Science+Business Media, LLC Abstract In this work we study the higher excited states of Helium Atom. The purpose is to evaluate Coulomb and exchange integral via spherical harmonics series. The Coulomb and exchange integrals energy shift is evaluated up to sixth order. This is the energy when the atom is perturbed by Coulomb potential between electrons. The energy levels obtained from both integrals are in agreement with the experimental data. For highly-excited states, the calculated energy approaches ev, in agreement with the graphical results from the book by Powell and Crasemann []. Keywords Coulomb integral Exchange integrals Perturbation-theory Addition theorem of spherical harmonics Introduction Modern physical theory rests upon the basic fact that matter is composed of relatively few types of elementary particles (electrons, positrons, protons, neutrons, etc.). Each type is characterized by a few properties, such as mass, charge, and intrinsic angular momentum or spin, which enter into the equations of the theory as invariable parameters. All electrons, for example, are intrinsically identical in every aspect. Helium atom is a typical three-body system with strong correlated motion of two electrons influence by Coulomb potential. The numerical calculation of the energy A. Hutem S. Boonchui (B) Forum for Theoretical and Computational Physics Department of Physics, Faculty of Science, Kasetsart University, Bangkok 9, Thailand fscistb@ku.ac.th A. Hutem S. Boonchui Center of Excellence in Forum for Theoretical Science, Chulalongkorn University, Bangkok, Thailand

2 J Math Chem () 5:86 87 levels and the wave functions of a helium atom is one of the most interesting and fundamental problems in the atomic physics. The perturbation method can be used to predict with a very high precision for energy values [ 4]. Scherr and Knight used -term trial functions to get extremely accurate approximations to the wave function correction up to sixth order and thus to the energy corrections up to thirteenth order [5,6]. The perturbation-theory series expansion for the He-atom energy can be proved to converge [7,8].Baker et al. [9] calculated the energy corrections up to order of 4. Drake and Yan used linear variational functions containing (Sect. ) to calculate the ground-state energy and many excited-state energy of helium that are thought to be accurate to part in 4 or better [,]. These workers similarly calculated (Li) variational energy for the ground state and two excited states with part in 9 accuracy or better [ 5] In, Duan et al. [6 9] showed a new method to evaluate the energy levels of helium atom for a quantum three-body system, where the motion of the center of mass and the global rotation of the system are completely separated from the internal motion. The plan of this paper is as follows. The helium atom has three-body system. We shall consider the nucleus to be at rest and place the origin of the coordinate system at the nucleus. We will explain the method in Sect. and calculate the Coulomb integral and exchange integral. In Sect., using the addition theorem of spherical harmonic [] into the Coulomb integral and exchange integral to calculate total energy of each state. The discussion and conclusion are given in Sect. 4. Basic theory The helium atom has two electrons and a nucleus of charge +e. We shall consider the nucleus to be at rest and place the origin of the coordinate system at the nucleus. The coordinate of two electrons and are (x, y, z ) and (x, y, z ); see Fig.. If we take the nuclear charge to be +Ze instead of +e, we can treat heliumlike ions such as H, Li +, Be + The basic Hamiltonian is given by Ĥ ˆp m + ˆp m Ze Ze e + r r, () Hˆ ˆp m + ˆp m Ze Ze, Ĥ r e r, () Fig. Interparticle distances in the helium atom

3 88 J Math Chem () 5:86 where m is the mass of the electron, and r are the distances of electrons and from the nucleus, and is the distance from electrons to. The first two terms are the operators for the electrons kinetic energy. The third and fourth terms are the potential energy of attraction between the electrons and the nucleus. The final term is the potential energy of interelectronic repulsion. Note that the potential energy of a system of interacting particles cannot be written as the sum of potential energy of the individual particles; the potential energy is a property of the system as a whole. The Schrödinger equation involves six independent variables, three coordinates for each electron. In spherical coordinates, we obtain ψ ψ(,θ,φ, r,θ,φ ). The operator ˆp h is given by the Laplacian spherical coordinates with,θ,φ replacing r,θ,φ. The variable is [(x x ) + (y y ) + (z z ) ] /, and by using the relations between Cartesian and spherical coordinates, we can express in terms of,θ,φ, r,θ,φ. Suppose the e / r e / were absent. Because of the e / term, the Schrödinger equation for helium cannot be separated in any coordinate system, and we must use approximation methods. The perturbation method separates the Hamiltonian Eq. () into two parts, Ĥ and Ĥ, where Ĥ is the Hamiltonian of an exactly solvable problem and Ĥ is the electron-electron interaction. Then, with the identity question ignored, the wave function would be just the product of two hydrogen atom wave functions with Z change into Z The total spin is constant, so the state is either singlet or triplet. The spatial part of the wave function for the important case where one of the electrons is in the ground state and the other in a higher excited states characterized by (nlm)is ψ(, r ) [ψ ( )ψ nlm (r ) ± ψ (r )ψ nlm ( )]. () We have, from the expectation value f ψ (x, t) ˆ f ψ(x, t)dx (4) The next step is to evaluate the first-order perturbation correction to the energy. E () H e (5) Let us define expectation values of Ĥ as e r e ψ (, r ) r ψ(, r )d d r. (6)

4 J Math Chem () 5:86 89 By substituting Eq. ()into(5), we have e (ψ ()ψnlm (r ) ± ψ (r )ψnlm (r e )) r r (ψ ( )ψ nlm (r ) ± ψ (r )ψ nlm ( )d d r ) e [ψ ()ψnlm (r )ψ ( )ψ nlm (r ) r ±ψ ()ψnlm (r )ψ (r )ψ nlm ( ) ±ψ (r )ψnlm ()ψ ( )ψ nlm (r ) +ψ (r )ψnlm ()ψ (r )ψ nlm ( )] d d r (7) r e e [ ψ ( ) ψ nlm (r ) ± ψ ()ψnlm (r )ψ (r )ψ nlm ( ) r ±ψ (r )ψ nlm ()ψ ( )ψ nlm (r ) + ψ (r ) ψ nlm ( ) ] d d r r (8) Consider the symmetry between and r,theneq.(8) can be rewritten as e ψ ( ) r ψ (r )ψnlm () r e r ψ nlm(r ) d d r ± Re e r ψ ( )ψ nlm (r )d d r. (9) The first term on the right-hand sides of Eq. (9)areCoulomb integral [ 4]written as J ψ ( ) e r ψ nlm(r ) d d r. () r The second term on the right-hand sides of Eq. (9) are Exchange integral [ 4] written as K r ψ (r )ψ nlm () e r ψ ( )ψ nlm (r )d d r. ()

5 9 J Math Chem () 5:86 We consider just (s)(nl). We write the energy of this state as E E + E nlm + ΔE. () In first-order perturbation theory, ΔE is obtained by evaluating the expectation value of e /. We can write e (J ± K ). () By substituting Eq. () back into Eq. (), we obtain total energy of helium atom The lower energy level is, therefore, E E + E nlm + J ± K. (4) E t E + E nlm + (J K ), (5) and is triply degenerate(-) it is called a triplet [4]. The higher energy level, E t E + E nlm + (J + K ), (6) is non-degenerate, and is therefore a singlet [4], as is the ground state (Table ). The physical interpretation for this is as follows: In the singlet (Eq. (6)) case the spatial function is symmetric and the electrons have a tendency to come close to each other. Therefore, the effect of the electrostatic repulsion is more serious; hence, a higher energy results. In the triplet (Eq. (5)) case, the spatial function is antisymmetric and the electrons tend to avoid each other. Helium in spin-singlet states is known as parahelium, while helium in spin-triplet states is known as orthohelium. See Fig.. for a schematic energy level diagram of the helium atom. Numerical results of the Coulomb integral and exchange integral The Coulomb integral and exchange integral for higher excited states of Helium Atom is calculated by using Spherical Harmonics Series. We use mathematica program in the calculation of energy.. Calculation of the Coulomb integral The Coulomb integral can be written as J r ψ ( ) e r ψ nlm(r ) d d r, (7)

6 J Math Chem () 5:86 9 Table The total energy of the helium atom in higher excited state Spectral term Calculated Experimental [9] % Difference ss S ss S s4s S s5s S s6s S s7s S s8s S ss S ss S s4s S s5s S s6s S s7s S s8s S sp P sp P s4p P s5p P s6p P s7p P s8p P sp P sp P s4p P s5p P s6p P s7p P s8p P sd D s4d D s5d D s6d D s7d D s8d D sd D s4d D s5d D s6d D

7 9 J Math Chem () 5:86 Table continued Spectral term Calculated Experimental [9] % Difference s7d D s8d D s4f F s5f F s6f F s7f F s8f F s4f F s5f F s6f F s7f F s8f F s5g G s6g G s7g G s8g G s5g G s6g G s7g G s8g G s6h H s7h H s8h H s6h H s7h H s8h H s7i I s8i I s7i I s8i I and using the addition theorem of spherical harmonics []: r m ( 4π ) + r< l Y r l + m (Ω )Yl m (Ω ), (8) where r (r < ) is the larger (smaller) of and r respectively. Thus a different expansion is presented for each region see Fig.. r is the vector indicating the position of

8 J Math Chem () 5:86 9 Fig. Schematic energy level diagram for higher excited state configurations of helium atom Fig. The graph shows the relation between individual electrons and atomic nuclei the electron far away from the nucleus. r < is the vector indicating the position of the electron near the nucleus. Substituting Eq. (8) back into Eq. (7) yields J e ( 4π ) + d dr dω dω r R ()Y (Ω ) m Ω Ω r ( r ) < Y (Ω ) Y r l + m (Ω )Yl m (Ω )r R nl (r )Ylm (Ω )Y lm (Ω ) (9) Substituting Y (Ω )Y (Ω ) (4π) / into Eq. (9) gives J e r m ) + ( Ω d dr r R ()r R nl (r ) We define the new variable C i so that C r Ω dω dω Y l m (Ω )Y m (Ω )Y lm (Ω )Y lm (Ω ) ( r < r l + d dr r R ()r R nl (r ) ) ( r < r l + () ). ()

9 94 J Math Chem () 5:86 Fig. 4 Showing the area of double integrals Substituting the definition [] ( r ) < r l + r l r l + r l r l + for < r for r into Eq. (), we have C r r rl + Rnl (r )R ()d dr + r l + r r The first term on the right-hand sides of Eq. () arec written as r l + R r l + nl (r )R ()d dr () C r r rl + R r l + nl (r )R ()d dr () The second term on the right-hand sides of Eq. () arec written as C r r r l + R r l + nl (r )R ()d dr (4) Let us consider Fig. 4. The area integrals of the region R I, R II are f (x, y) dxdy R I f (x, y) dxdy R II a y a a x f (x, y) dxdy, (5) f (x, y) dydx. (6)

10 J Math Chem () 5:86 95 when R I : x y, y a R II : x a, x y a. Following the Double Integral method in Ref. [], Eq. (5) can change the order of the integrated hazard. There is value in the Eq. (6), we have a y f (x, y) dxdy a a x f (x, y) dydx. (7) Equation () is hence transformed to C r r rl + R r l + nl (r )R ()d dr r r l + R r l + nl (r )R ()d dr (8) Equation () is compared with that of the Eq. (8). It is found that the value of and r relation is r. Equation () has to be transformed into Eq. (8) in order to get the Coulomb integral done. Inserting Eq. ()into() the Coulomb integral is written as J e m ( C + ) Ω Y m (Ω l )Yl m (Ω )Ylm (Ω )Y lm (Ω ) (9) Ω We consider the first excited state(sp) of helium n, l, m ±, into Eq. (9), yields J e m Ω ( C + ) Y Ω Ω m (Ω )Y m (Ω )Y m (Ω ) Y (Ω ) Y (Ω ) Y m (Ω )dω dω () Using properties of the orthonormality relations of spherical harmonics into Eq. (), we have J e C ()

11 96 J Math Chem () 5:86 By substituting Eq. () into(), we have J e + r R (r ) r r R ()d dr r R (r ) R ()d dr r () Substituting Eq. () into Mathematica program, will the value of the Coulomb integral so that J 59Ze 4a. ev. (). Calculation of the exchange integral The exchange integral can be written as K e ψ (r )ψnlm ( ) r ψ ( )ψ nlm (r )d d r (4) r Substituting the addition theorem of spherical harmonics series into Eq. (4), we finally obtain the exchange integral as K e ( 4π ) + ( r < d d r ψ (r )ψnlm () m r l + r Y m (Ω )Yl m (Ω )ψ ( )ψnlm (r ). (5) We consider the first excited state(sp) of helium n, l, m ±, into Eq. (5), yields K e ( 4π ) + ( r < d d r ψ (r )ψm () m r l + r Y m (Ω )Yl m (Ω )ψ ( )ψm (r ). (6) We consider the wave function to be ψ,, (r) R, (r)y, (Ω) and using properties of the orthonormality relations of spherical harmonics into Eq. (6), we get K e { r R (r )R (r ) r r R ( )R ( ) ( r < r l + ) ) )d } dr. (7)

12 J Math Chem () 5:86 97 Substituting the definition [] into Eq. (7), we have K e + e r ( r ) < r l + r l r l + r l r l + for < r for r r r r R (r )R (r )R ( )R ( )d dr r r r r R (r )R (r )R ( )R ( )d dr (8) The first term on the right-hand sides of Eq. (8) arek a written as K a e r r r r R (r )R (r )R ( )R ( )d dr. (9) The second term on the right-hand sides of Eq. (8) arek a written as k a e r r r r R (r )R (r )R ( )R ( )d dr. (4) From Eq. (9) used change the order of the integrated, we obtain K a rewritten as K a e r r r r R (r )R (r )R ( )R ( )dr d. (4) It is easy to see that K a K a. The exchange integral Eq. (8) can therefore be rewritten as K e r R (r )R (r ) R ( )R ( )d dr. (4) r Substituting m, into Eq. (5), we get the exchange integral like Eq. (4). Substituting Eq. (4) into Mathematica program (see Sect..), will the value of the exchange integral so that K Ze 656a.94 ev. (4)

13 98 J Math Chem () 5:86. The calculation of mathematica program Calculation of the energy of the Coulomb integral and exchange integral( e a unit) Case. Evaluation of the Coulomb integral (ss) J e + r R (r ) r r R ()d dr r R (r ) R ()d dr, (44) r where, l, n. The calculation is as follows. In : R[r_] :( a z ) E zr a In : R[r_] :( a z ) ( zr In : R[r_] :( a z ) E zr a In : R[r_] :( a z ) ( zr In : JA[r_] : r r(r[r]) dr In : JA[r] Out:ConditionExpression[( a + z r 7a a + z r 7a rz e a rz e a z(a+rz) ) a )E zr a )E zr a z(a+rz) ), Re[ z a a ] ] In : JA[r_] :( In : JA r (R[r]) (JA[r])dr 69z Out : ConditionExpression[ 768a, Re[ a z ] ] In : JA ( 768a 69z ) Out : 768a 69z In : JB[r_] : r r (R[r]) dr In : JB[r] Out : ( rz exp a (a +arz+r z ) ) a rz exp a (a +arz+r z ) In : JB[r] :( ) a In : JB r(r[r]) (JB[r])dr Out : ConditionExpression[ 768a 99z, Re[ a z ] ] In : JB ( 768a 99z, Re[ a z ) Out : 768a 99z In : J JA+ JB Out : 85z 89a Case. Evaluation of the exchange integral (ss) K e r R (r )R (r ) dr (45) r

14 J Math Chem () 5:86 99 where, l, n. The calculation is as follows. In : R[r_] :( a z ) E zr a In : R[r_] :( a z ) ( zr a + z r 7a In : R[r_] :( a z ) E zr a In : R[r_] :( a z ) ( zr a + z r 7a In : K [r_] : r rr[r]r[r]dr In : K [r] [ Out : ConditionalExpression In: K [r_] : )E zr a )E zr a ] exp ( 4rz a )z(8a +8a rz 6ar z +r z ) 4 a 4 ( ) exp ( 4rz a )z(8a +8a rz 6ar z +r z ) 4 a 4 In : K r R[r]R[r]K [r]dr Out : ConditionalExpression[((89z)/(6556a)) In : K ( 6556a 89z ) Out : ( 89z 6556a ) 4 Conclusion For the state ss, we obtain the Coulomb integral and exchange integral as follows: r J e r R (r ) r R ()d dr + r R (r ) R ()d dr, (46) r K e r R (r )R (r ) R ( )R ( )d dr. (47) r For the state sp, we get r J e r R (r ) r R ()d dr + r R (r ) R ()d dr, (48) r K e r R (r )R (r ) R ( )R ( )d dr. (49) r

15 J Math Chem () 5:86 For the state sd, we obtain the Coulomb integral and exchange integral as follows: r J e r R (r ) r R ()d dr + r R (r ) R ()d dr, (5) r K 5 e r 4 R (r )R (r ) R ( )R ( )d dr. (5) For the state s4f, we have r J e r R 4 (r ) r R ()d dr + r R 4 (r ) R ()d dr, (5) r K 7 e r 5 R (r )R 4 (r ) R ( )R 4 ( )d dr. (5) r r For the state s5g, we obtain the Coulomb integral and exchange integral as follows: r J e r R54 (r ) r R ()d dr + r R 54 (r ) R ()d dr, (54) r K 9 e r 6 R (r )R 54 (r ) R ( )R 54 ( )d dr. (55) r For the state s6h, we obtain r J e r R65 (r ) r R ()d dr + r R 65 (r ) R ()d dr, (56) r r r

16 J Math Chem () 5:86 K e r 7 R (r )R 65 (r ) R ( )R 65 ( )d dr. (57) For the state s7i, we obtain the Coulomb integral and exchange integral as follows: J e r R76 (r ) r r r 4 r R ()d dr + r R 76 (r ) R ()d dr, (58) r K e r 8 R (r )R 76 (r ) R ( )R 76 ( )d dr. (59) For highly-excited states, the calculated energy approach ev, in agreement with the graphical results from the book by Powell and Crasemann [] (ss, sp, sd, s4s, s4p, s4d, s4f). Acknowledgments I would like to thank Dr. Chanun Sricheewin and Dr. Nattaporn Chattham for their useful discussions. This work is supported by The Graduate School Kasetsart University and Department of Physics, Faculty of Science, Kasetsart University, Bangkok, Thailand 9. r r 5 References. J.L. Powell, B. Crasemann, Quantum Mechanics (Addison-Wesley, Reading, Mss, 96), pp H. Eyring, J. Walter, G.E. Kimball, Quantum Chemistry (Wiley, United State of America, 944), p. 69. T.-Y. Wu, Quantum Mechanics (World Scientific Publishing Co Pte Ltd, Singapore, 986), pp D. Park, Introduction to the Quantum Theory (Dover, New York, 5), pp C.W. Scherr, R.E. Knight, Two-electron atoms II. A perturbation study of some excited states. Rev. Mod. Phys. 5, 4 (96) 6. C.W. Scherr, R.E. Knight, Two-electron atoms III. A sixth-order perturbation study of the S ground state. Rev. Mod. Phys. 5, 46 (96) 7. F.C. Sanders, C.W. Scherr, Perturbation study of some excited states of two-electron atoms. Phys. Rev. 8, 84 (969) 8. R. Ahlrichs, Convergence of the /Z expansion. Phys. Rev. A 5, 65 (97) 9. J.D. Baker, D.E. Freund et al., Radius of convergence and analytic behavior of the /Z expansion. Phys. Rev. A 4, 47 (99). G.W.F. Drake, Z.-C. Yan, Energies and relativistic corrections for the Rydberg states of helium: variational results and asymptotic analysis. Phys. Rev. A 46, 78 (99). B. Zhou, C.D. Lin, A hyperspherical close-coupling calculation of photoionization from the He atom, Li+ and C4+ ions. II. Between the N andn thresholds. J. Phys. B 6, 575 (99). G.W.F. Drake, Z.-C. Yan, Variational eigenvalues for the S states of helium. Chem. Phys. Lett. 9, 486 (994). G.W.F. Drake, Z.-C. Yan, Eigenvalues and expectation values for the ss S, sp P, and sd D states of lithium. Phys. Rev. A 5, 7 (995) 4. M. Viviani, Transformation coefficients of hyperspherical harmonic functions of an A-body system. Few-body systems 5, 77 (998)

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