A 2D Formulation for the Helium Atom Using a Four- Spinor Dirac-Like Equation and the Discussion of an Approximate Ground State Solution

Physical Science International Journal 9(4): 1-12, 2016, Article no.PSIJ.24355 ISSN: 2348-0130

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A 2D Formulation for the Helium Atom Using a Four- Spinor Dirac-like Equation and the Discussion of an Approximate Ground State Solution

D. L. Nascimento 1* and A. L. A. Fonseca 1

1Institute of Physics and International Center of Condensed Matter, University of Brasília, 70919-970, Brasília, DF, Brazil.

Authors’ contributions

This work was carried out in collaboration between both authors. Both authors read and approved the final manuscript.

Article Information

DOI: 10.9734/PSIJ/2016/24355 Editor(s): (1) Shi-Hai Dong, Department of Physics School of Physics and Mathematics National Polytechnic Institute, Mexico. (2) Christian Brosseau, Distinguished Professor, Department of Physics, Université de Bretagne Occidentale, France. Reviewers: (1) Balwant Singh Rajput, Kumaun University, Nainital, India. (2) S. B. Ota, Institute of Physics, Bhubaneswar, India. Complete Peer review History: http://sciencedomain.org/review-history/13496

Received 16 th January 2016 th Original Research Article Accepted 14 February 2016 Published 29 th February 2016

ABSTRACT

We present a two-dimensional analysis of the two-electron problem which comes from the classical conservation theorems and from which we obtain a version of the Dirac equation for the helium atom. Approximate solutions for this equation are discussed in two different methods, although in principle it can be solved analytically. One method is variational, of the Hylleraas type, the execution of which is left for a later communication. In contrast, the other method will have a more complete treatment, in which the set of equations will be separated into its angular and radial components. Furthermore, an exact solution for the angular component will be displayed as well as an approximate solution for the radial component, valid only for the fundamental state of the atom.

Keywords: Helium atom; Dirac equation; relativistic quantum mechanics; variational methods; Hylleraas method; semi analytic solutions.

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*Corresponding author: E-mail: [email protected];

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

1. INTRODUCTION parameter. The gradual superposition of the corresponding Hamiltonians yields a system of Since the beginning of quantum chemistry in the differential equations that is dependent on the 1920s the implementation of purely parameter of penetration, which should be used computational calculations of the Hartree-Fock at the end of the calculation to obtain the type [1] has prevailed, due to their high minimum energy of the two-electron system. performance and easy implementation, over Considering now the Hylleraas-like procedure, more analytical structures such as those of the the Hamiltonian is used in the traditional way in Hylleraas type [2]. Nevertheless some authors which the system is treated as a whole, without have tried to give an analytical basis to their distinction of individual equations for each iterative calculations, some of which have electron. For both procedures we try to express become the source of inspiration to start this line the system of equations in a truly covariant form, of work [3-6]. in which we can introduce later the retardation effects without breaking this fundamental Following this analytical aim [7,8], we try to requirement of the Theory of Relativity. explore the Classical Theorems of Conservation before any quantization procedure is performed. In the last section of the paper we separate the We believe that they can reduce the dimensions angular and radial components of the Dirac-like of the coordinate systems that are necessary to system of partial differential equations for the formulate the problem in the quantum domain. helium atom [13]. We find the angular More recently [9], we demonstrated that it is eigenfunctions that allow us to separate the possible to use, in the analytical solution of the system of radial equations and an asymptotic Dirac equation [10] for the hydrogen atom, Dirac form of the wave function that is a solution of this 2x2 matrices rather than the usual 4x4 matrices, system for the ground state of the atom. From which leads to a considerable reduction in this we get a determination of the atom energy complexity of the problem. Moreover, our method eigenvalue that agrees with the experimental led us to a new approach to the relativistic data within 0.1% of accuracy and we also check Hylleraas procedure in which the Dirac equation that it tends to the exact value of the ion energy is derived from an extremum problem. This when the outer electron is displaced to infinity. procedure was used to carry out numeric calculations for hydrogen-like atoms that resulted 2. THEORY in extremely accurate energy eigenvalues with respect to the exact values, which are well In the infinity mass nucleus rest frame, the known [11]. relativistic classical Hamiltonians for the individual electrons of the Helium atom in natural = = 2 In this paper we treat the problem of the helium units h c 1 and α =e ≅ 1 / 137 are atom in a similar way to what we did in the case of the hydrogen atom. This treatment has 2α α allowed us to use Dirac 4x4 matrices instead of H=p2 + m 2 − + 1 1 r r the 16x16 matrices of the Breit theory for the 1 12 , same atom [12], although it should be mentioned 2α α that we do not consider here the time retardation H=p2 + m 2 − + 2 2 r r effects. Therefore we have developed a dual 2 12 , (1a,b) procedure: on the one hand we obtain a Dirac- like system of partial differential equations and =2 + 2 − θ on the other a Lagrangian density to carry out a where r12 r 1 r 22 rr 1 2 cos . We see that the variational calculation of the Hylleraas type, repulsion energy entries fully for each electron in whose execution is however left to an upcoming this case, on the other hand, if we consider the article. energy of the whole system, not taking into account the electrons individually, we arrive at For the Dirac-like procedure we take into account the usual classical Hamiltonian the trivial fact of the Theory of Relativity that we cannot add together the geodesics of individual α α α =2 ++ 2 2 +−−+ 2 2 2 particles. Then we consider a system formed by Hp1 m p 2 m , (1c) + a single electron plus the nucleus, i.e., the He r1 r 2 r 12 ion, as a substrate on which an outer electron is introduced gradually through a penetration in which the repulsion energy entries only once.

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

Still in the infinity mass nucleus rest frame, we on the other hand, σ =1 correspond to the limit choose to use a coordinate system in which the when the two electrons form a single system with motion occurs in the plane defined by the perfectly symmetric positions so that the system = Hamiltonian becomes two times the Hamiltonian nucleus and the two electrons, i.e., pz 0 , whose z axis may be moving at constant velocity of one of the electrons, which was chosen by with respect to the z axis of another inertial convenience to be the electron 2. In fact, it will be σ system, so that the system is invariant against seen that r12 becomes a function of , so that space translations in this direction. In this frame, the equations of the system are solved for r = the only non vanishing components of the 12 classical angular momentum of each electron constant and then, at the final of the calculation, this constant is varied through σ in order the J= r × p and J= r × p and of the total 1 1 1 2 2 2 equilibrium configuration may be obtained. = + angular momentum J J1 J 2 are their z components, namely, J= xp − yp , We now search for a Dirac equation 1z 11 y 11 x corresponding to the quantization of the classical = − = + J2z xp 22 y yp 22 x and Jz J1 z J 2 z Eqs. (1c) and (1d), in the infinity mass nucleus respectively. Now, we know from Classical rest frame. The quantization is done in a way Mechanics [14] that the Poisson Bracket for similar to that performed by Breit [12], in which each electron angular momentum with respect to each square root is “linearized” individually: the Classical Hamiltonian (1c) is not null and that 2 they are symmetric with respect to each other, 225+=γ − γ 3 + γ 0 p1m( p 1y pm 1 x ) − i.e., {HJ,} = rrr3 sinθ = − { HJ , } , in which , (2a) 1z 1212 2 z θ = θ− θ , so that the summation of them is null 2 2 1 p221+=m(γ p − γ 2 pm + γ 0 ) 2 2y 2 x . (2b) and hence the total angular momentum J z becomes a constant of the motion. This happens We need five 4× 4 anticommuting matrices, because the repulsion force between the µ electrons is a non central force and hence it which are the four usual γ Dirac matrices produces a torque in each electron that makes it 5 together with the γ matrix which always oscillating about the axis that join the nucleus to the other electron. In the Poisson Bracket it appears connected with Dirac´s theory: appears due to the implicit derivatives of 1/ r 12 1 0  0 iσ  γ 0 = = with respect to x1, y 1 or x2, y 2 , which produces   γ   0− 1 −iσ 0 the symmetric terms because   ,   , 2 2 r=()() xx − +− yy in Cartesian 0 1  12 1 2 1 2 γ5= − i γγγγ 0123 =   coordinates. Therefore, a 2D formulation of the 1 0   , (3) problem is, at least in principle, perfectly possible and we shall present two possibilities of it below. σ σ σ where x , y , z are the Pauli spin matrices. In this way, besides the usual Hamiltonian for the As is well known 1 , the γ matrices obey whole system (1c), an alternative approach for γγµ ν+ γγ ν µ = δ µν µ ν = the problem would be to define an effective 2 , for , 0,1,2,3,5 , that is, Hamiltonian function for the two electron system are unitary and anticommute in pairs, as required which would be composed of a inner Hamitonian to make equal the two sides of Eqs. (2). + H of the ion H and another Hamiltonian H 1 e 2 which would take into account an outer electron, 1 There are several possibilities of defining these matrices which is superposed to the former through a σ according to the rules of the Clifford Algebra; we have chosen penetration factor , that is µ ν the only one that makes all products γ γ to be real and

µ H=(1 −σ) H + 2 σ H positive along with γ be diagonal for µ = 0 and anti- σ 1 2 . (1d) diagonal for µ > . These conditions are necessary for the 0 We see that σ = 0 corresponds to the ion limit Eq.(6b) and Eq.(8) below reduce to the one-electron 2D Dirac equation [9] when σ → 0 , which is a fundamental when →∞ and the electron 2 is not present; r12 contour condition of our approach.

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

= − ∇ = − ∇ By using the momentum operators p1i 1 and p2i 2 , the linear Hamiltonian-like matrix operator associated with Eq.(1c) and Eq.(1d) becomes

2α 2 α α Hiˆ =∂−∂()γγ35− +∂−∂ i() γγ 21 − + 2 m γ 0 + xy11r xy 22 r r 1 2 12 . (4a)

α  α  α ˆ =−σγγ35 ∂−∂−2 + σγγ 21 ∂−∂ − 2 + + σγ 0 + Hiσ ()1()  2 i()  () 1  m  xy1 1 r xy2 2 r r 1 2  12 . (4b)

In our coordinate system the total angular momentum operator becomes the z component alone, i.e.

Jˆ = iy ∂−∂+ ix iy ∂− ix ∂ z1 x1 1 y 1 2 x 2 2 y 2 . (5)

ˆ ˆ ˆ Now, as it is well known, the operator J z does not commute with H or Hσ . However, it can be ˆ= ˆ +1α + 1 α verified immediately that total angular operator M J z21 z 2 2 z , which includes the electron ˆ ˆ ˆ ˆ  = ˆ ˆ  = spins, commutes with both H and Hσ , that is H, M  0 as well as Hσ , M  0 . Here, in the definition of the operator Mˆ were introduced the two-electron spin matrices −σ 0  −σ 0  α= −i γ5 γ 3 = z and α= −i γ2 γ 1 = z and the diagonalization problems to be 1, z σ  2, z −σ  0 z  0 z  ˆψ= ψ ˆ ψ= ψ ˆψ= ψ ψ= ( χχχχ ) solved become therefore H E or Hσ E and M j , where 1, 2 , 3 , 4 is a four-spinor.

Now, in order to get a truly covariant equation, we left-multiply the energy eigenvalue problem by γ 0 , for both operators in (4), so that we get

()φ− γ0 + γγ 05 ∂−∂+ γγ 03 γγ 01 ∂−∂+ γγ 02  ψ = 12 Ei( x y) i( x y ) 2 m 0  1 1 2 2  , (6a)

()()φ− γ0 +− σγγ 05 ∂−∂+ γγ 03 σγγ 01 ∂−∂++ γγ 02 () σψ = 12 σ Ei1( xy) 2 i( xy ) 10 m  11 22  , (6b)

α α α 2( 1−σα) σα ( 1 + σα) φ =−2 − 2 + φ =− −4 + where 12 and 12 σ are the total potential energy r1 r 2 r 12 r1 r 2 r 12 functions. It may immediately be seen that Eqs.(6) can be put in the explicit covariant form σζπµ+ σζ π µ  ψ = 111µ 22 µ 2  0 , by rewriting them in terms of the matrix operators ζ= − γγγγγ05 03 0 ζ= − γγγγγ01 02 0 1µ (1, , , ) , 2µ (1, , , ) and the effective momentum operators

µ 2α α E  πˆ =m, −∂−∂− i , i , + −  , where k = 1,2 ;σ= σ = 1, σ = 2 for (6a) and σ=1 − σ , k xk y k σ σ 1 2 3 1 rk3 r 12 2 k  σ= σ σ = 2 2 , 3 1 for (6b).

Explicitly, Eq.(6a) becomes the following system of linear partial differential equations

q+ χ−∂+∂( i) χ +−∂+∂( i ) χ = 0 1xy11 3 xy 22 4 , (7a)

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

At this point we shall need to split the paper in q+ χ+∂−∂( i) χ −∂+∂( i ) χ = 0 2xy11 4 xy 22 3 , (7b) two parts. In the first one, we shall present a variational version of Eqs.(7) that allows us to χ+−∂+∂ χ +−∂+∂ χ = q− 3( xy i) 1( xy i ) 2 0 make numerical calculations of the energy 11 22 , (7c) eigenvalues in the Hylleraas scheme, as we did with high accuracy in the case of one-electron χ+∂+∂ χ −∂+∂ χ = q− ( i) ( i ) 0 atoms [11]. But we shall only introduce the 4xy11 2 xy 22 1 , (7d) problem, which will be treated fully in a next and Eq.(6b) as well becomes paper. To do this, firstly we solve the last two χ χ equations of (7) for 3 and 4 , what yields χσ−−( ) ∂+∂ χσ + −∂+∂ χ = qσ + 11( xy i) 3 2( xy i ) 4 0 11 22 , (8a) (∂−∂i) χ +∂−∂( i ) χ χ = xy111 xy 22 2 3 q− qσ + χσ+−(1)( ∂−∂ i) χσ − 2( ∂+∂ i ) χ = 0 , 2xy11 4 xy 22 3 , −∂+∂( i) χ +∂+∂( i ) χ (8b) χ = xy112 xy 22 1 4 q− , (9) qσ − χσ+−(1)( −∂+∂ i) χσ + 2( −∂+∂ i ) χ = 0 3 xy111 xy 22 2 , (8c) then we substitute them into the first two of (7), left-multiply each one by the complex conjugated qσ − χσ+−(1)( ∂+∂ i) χσ − 2( ∂+∂ i ) χ = 0 ∗ ∗ 4xy11 2 xy 22 1 χ χ , vectors 1 and 2 respectively and sum up the (8d) resulting equations to form a real quadratic

function in (χ χ ) which defines the following in which the new potential functions 1, 2 = ±(φ − ) =+( σ) ±( φ − ) Lagrangean density q± 2 m12 E and qσ± 1 m12 σ E were introduced for shortness.

2 2 2 2 ∂χχ −∂i +∂ χχ +∂ i +∂ χχ +∂ i +∂ χχ −∂ i = xy1111 xy 11 22 xy 22 11 xy 22 22 +χ 2 + L q + 1 q− ∂χ∗ +∂ χχ ∗ ∂ +∂ χ −∂ χ∗ +∂ χχχ ∗ ∂ +∂ (i )( i )( i )( i )  2 +ℜ x11 yx 12 12 y 2 2 x 2 1 yxy 21 122 1 + χ   q+ 2  q−    . (10)

The extremum problem

δ Ldx dx dy dy = 0 ∫ 1 2 1 2 (11) is then solved, in the same way as done with the hydrogen-like atoms [11], by the requirement that (χ χ ) the integral in (11) be stable against small variations of the algebraic forms of 1, 2 about the corresponding exact solutions of (7) or some suitable approximation of them:

N µ ν λ χ= χ c rrr (12) l lap ∑ l µνλ 1 2 12 µ+ ν + λ = 0 for l = 1,2 in which χ are the approximations for the exact solutions of (7), c µνλ are the variational lap l coefficients corresponding to each function χ and N is the least integer necessary to a given order lap of precision to be reached. The variation becomes thus

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

∂ Φ=−( j1)θ +( j + 1 ) θ + g = 312 12 2 23 , ∫ Ldx1 dx 2 dy 1 dy 2 0, ∂ 1 1 c µνλ = Φ=+( j)θ +( j −) θ + g l N 0,1,2,... (13) 412 12 2 24 , (15c,d) which produces two systems of linear equations = ( ) = θ with gk g k rrr1, 2 , 12 . Since r12 r 12(, rr 1 2 , ) , in c , the determinants of which generate a lµνλ the dependence of fk and gk on r12 has polynomial function on the atom energy, the evidently no effect on the values of the angular roots of which yield the energy eigenvalues for momentum, but this dependence is necessary the atom. It should be remarked that the need to when considering a complete solution of the χ know ap in advance is in fact the great l problem in the radial variables r1, r 2 , r 12 . limitation of the Hylleraas methodology because, in practice, only asymptotic solutions are known, Formally, the substitution of χ in Eq. (8) makes so that the use of arbitrary intermediary functions k becomes the only way to perform the calculation. all complex phases and angular variables vanish It is by this reason that we are proposing below and yields a new set of linear equations the sigma variation procedure, in which we retain depending only on the radial variables r1, r 2 , r 12 . the almost exact form of the one-electron To see this we consider the first order derivatives solution and perform the variation through a appearing in (8) expressed in the polar system macro parameter that is related to the average (r, r ,θ , θ ) and considering also the implicit values of the radial variables. The Hylleraas-like 1 2 1 2 problem will be reconsidered formally in a next dependence of the Cartesian coordinates in = θ paper. r12 r 12(, rr 1 2 , ) , from this we get the following differential operators 3. APPROXIMATE GROUND STATE ±θ ± θ i1 i 2 SOLUTION FOR THE SIGMA ± θ   − i 1 i re1 re 2 ∂±∂=ier ∂±∂+θ  ∂ r HAMILTONIAN x1 y 1 1 1 12 r  r 1 12 , (16a) Now, considering the second part of the paper, we shall perform the variation through the ±θ ± θ i1− i 2 parameter sigma which allows using one- ±iθ i  re re ∂±∂=2 ∂±∂−1 2 ∂ xier y 2 θ  r 12 electron solutions to provide an analytical 2 2 r2  r approximation for the solution of the Eqs. (8). 2 12 Thus, in order to separate the angular part of (16b) Eqs. (8), we must first address the angular Substituting the solution (14) together with the momentum problem, Mˆψ= j ψ , which written operators (16) into (8) and next separating it in in the polar coordinates of the electrons 1 and 2 their real and imaginary parts would bring two sets of linear partial differential equations in the becomes (∂+∂θ θ ) χ =i( j + λ) χ , where 1 2 k k k radial variables r, r , r connecting f and g λ = λ = − λ= λ = 1 2 12 k k 1 1 , 2 1 , 3 4 0 , are the diagonal with their derivatives, whose analytical solution is −1 (α + α ) = + completely out of hand at the moment. values of and j j1 j 2 . In 2 1z 2 z accordance, the most general forms for the However, in this work, we shall limit ourselves to solutions of the angular equation are the set of search for solutions satisfying the constraint eigenfunctions ρ = r = constant, so that in effect we arrive at 12 Φ i k = ( ) = χ = f e f= f( rrr, , ) = fk f k rr1, 2 and further assume that gk 0 , k k , k k 1 2 12 , k 1,.., 4 , (14) which will simplify considerably the resulting equations and also yields the variational relation where the phase functions are ρ= ρ σ ( ) that will be used to get the equilibrium Φ=+( j1)θ +( j + 1 ) θ + g configuration of the system. In these 112 12 2 21 , circumstances, we arrive at a unique system of Φ=−( j1)θ +( j − 1 ) θ + g equations given by the real part of (8), the 212 12 2 22 , (15a,b) imaginary one vanishing identically, that is

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

1 1 1 ∂fj−  ∂ f j −  −−ββ ++ µν −−σ32 +1 − σ 42 − 2 = 1122rr ss 1 2 qfσ + ()1 f  2 f  0 fe= arr , k = 1,2,3,4 (18) 1∂rr 3 ∂ rr 4 k∑ k µν 1 2 11  22  , µ+ ν = 0 (17a) because the approximation made above restricts ∂f j+1 ∂f j + 1  severely the possibility of obtaining energy sub- +−()σ4 +1 2 − σ 3 + 2 2 = qfσ + 21 f 4  2 f 3  0 states, which depend strongly on power series ∂rr ∂ rr 11  22  of higher degrees. The substitution of Eq. (18), (17b) together with its first derivatives

∂fj+1  ∂ f j − 1  1 + µ −−()σ12 +1 − σ 22 + 2 = −βr − β r s  s+µ −1 s + ν qfσ − 3 1 f1  2 f 2  0 ∂ =1 1 2 2 1 − β 1 2 ∂ ∂ rfe∑ aµν   rr , rr11  rr 22  1 k k 1 1 2 µ+ ν = 0 r1  (17c) k = 1,2,3,4 , (19a)

∂fj−1  ∂ f j + 1  +−()σ22 −1 − σ 12 + 2 = 1 qfσ − 4 1 f2  2 f 1  0 −β − β s +ν  +µ + ν − ∂ ∂ ∂ =1r 1 2 r 2 2 − β s1 s 2 1 rr11  rr 22  rfe2 k∑ a k µν 2  rr 1 2 µ+ ν = r (17d) 0 2  , (19b) Since the limitations described above do not into the system (17) yields the new set of allow us to get general solutions, we shall limit µ ν ourselves to get the simplest solution of Eq. (17), equations (summation on , is omitted for valid only for the ground state of the atom: shortness):

 2α( 1 − σ ) 4ασ  js−−1  js −− 1  γ− − +−+ σβ112 ++ σβ 22 2 = ρ  aµν ()1  aµν 2  a µν 0 2 rr1 1 r 32 r 4 1 2  1 2  (20a)

 2α( 1 − σ ) ασ  js++1   js ++ 1  γ− −4 +− σβ222 +− σβ 11 2 −= ρ  aµν 2  aµν () 1   a µν 0 2 rr2 2 r 3 r 1 4 1 2  2  1  (20b)

 2α( 1 − σ ) 4ασ  js−−1   js ++ 1  γ+ + ++ σβ2 2 2 +−− σβ 1 1 2 = ρ  aµν 2  aµν () 1   a µν 0 1 rr32 r 2 1 r 1 1 2  2  1  (20c)

 2α( 1 − σ ) 4ασ  js−−1  js ++ 1  γ+ + −−+ σβ112 +− σβ 22 2 = ρ  aµν ()1  aµν 2  a µν 0 1 rr4 1 r 22 r 1 1 2  1 2  (20d)

(1+σ) α (1+σ) α where γ=+()1 σ m +− E and γ=+()1 σ m −+ E . 1ρ ρ 2ρ ρ

Now we start with the determination of the coefficients and parameters by observing that the system of equations formed by each negative power 1/ r1 and 1/ r2 must vanish separately in order the coefficients ak 00 do not vanish:

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

−2α a +−−( js1 ) a = 0 −++( js1 ) a +4α a = 0 100 1 12 300 , (21a) 2 22 100 400 , (22c)

−2α a +++( js1 ) a = 0 ( js−−1 ) a +4α a = 0 200 1 12 400 , (21b) 2 22 200 300 . (22d)

−++( 1 ) +α = For this condition to be fulfilled it is necessary js1 12 a 1002 a 300 0 , (21c) that the determinants of the systems (21) and (22) vanish, from which we get −−−( js1 ) a +2α a = 0 1 12 200 400 , (21d) =−+1 2 − α 2 =−+1 2 − α 2 s12 j 1 4 and s22 j 2 4 . and equally Now get back to the original system (20), assume that = , for µ≠ ν = , and −4α a +−−( js1 ) a = 0 akµν a k νµ 0,1 100 2 22 400 , (22a) equate the coefficients of the system of

equations for the same powers, from which we −4α a −++( js1 ) a = 0 200 2 22 300 , (22b) get the system of recurrence equations

γa+−(1 σβ) a + 221 σβ a −+ ασ( ) a −−( 1 σ) sasa − 20 σ = 2100ρ 1300 2400 110 1310 2410 , (23a)

γa+2 σβ a +−( 1 σβ) a −+ 21 ασ( ) asa − 2 σ +−( 1 σ ) sa = 0 2200ρ 2300 1400 210 2310 1410 , (23b)

(1−σβ) a + 2 σβ aa +++−− γ 21 ασ( ) a( 1 σ) sasa − 20 σ = 1100 2200 1300ρ 310 1110 2210 , (23c)

21σβa−−( σβγ) aa + ++ 2121 ασ( ) asa − σ +−( σ ) sa = 0 2100 1200 1400ρ 410 2110 1210 . (23d)

In order the series (18) can stop, the part of the coefficients ak 00 in the recurrence must vanish a µ≠ ν = separately of that of the coefficients kµν for 0,1 , that is

γa+−(1 σ) β a + 2 σβ a = 0 2ρ 100 1 300 2 400 , (25a)

γa+2 σβ a +−( 1 σ) β a = 0 2ρ 200 2300 1400 , (25b)

(1−σ) βa + 2 σβ a += γ a 0 1100 2200 1ρ 300 , (25c)

2σβa−−( 1 σβ) a + γ a = 0 2100 1200 1ρ 400 , (25d) and also

−+21ασ( )a −−( 1 σ) sa − 2 σ sa = 0 110 1 310 2 410 , (26a)

−+21ασ( )a − 2 σ sa +−( 1 σ ) sa = 0 210 2 310 1 410 , (26b)

21ασ( +)a −−( 1 σ) sa − 2 σ sa = 0 310 1110 2210 , (26c)

21ασ( +)a − 2 σ sa +−( 1 σ ) sa = 0 410 2110 1210 . (26d)

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

Therefore in order the system (25) have a non ψ= ψ + ψ out of which a300 1 a 400 2 is a general trivial solution its determinant must vanish, from kernel vector. These basis vectors generate by 2 2 γγρ ρ − 4 σβ its turn relations among the power series what we get β = 1 2 2 as a function 1 = −( − σ) β γ 1− σ coefficients given by a1001 1300 a / 2 ρ , of β and the other parameters. = − σβ γ = 2 a2002 2300 a / 2 ρ , a400 0 for the former

vector and a= − 2σβ a / γ , Now, the non trivial solution for the 100 2 400 2 ρ = −( − σ) β γ = homogeneous system of equations (25) in the a2001 1400 a / 2 ρ , a300 0 for the later coefficients ak 00 can be obtained from the kernel one. associated to β , whose basis is given by the two 1 The last step in order to be able to make the linearly independent column vectors evaluation of the energy eigenvalue of the system is as follows. First form a null line vector −( − σ) β γ  11 / 2 ρ  corresponding to the system (23), i.e.,   = [ ] − 2σβ / γ R23 abcd ,23 ,23 ,23 and second make a ψ = 2 2 ρ  1   contraction of it with one of the kernel vectors.  1  Since it may be seen that both kernel vectors  0    , produce the same energy eigenvalue, so that the − σβ γ solutions in a300 and a400 are degenerated, we  22 / 2 ρ    have chosen to make the contraction with the −()1 − σ β / γ ψ = 1 2 ρ  first kernel vector, that is, Rψ = 0 . This 2   1  0  operation, as expected, eliminates the   1 coefficients and produces a new relation   , (27a,b) ak 00

connecting the coefficients ak10 :

α− σ2 β  2( 1 ) 1   2α() 1 + σ β  −−++()σ3  + σ 2 −−+ 3 + 1sja1 1  110 2  sja2 2  210 γ 2  γ 2 2ρ  2 ρ  2 ()1−σ β   ++−31 +σβ2 3 ++++ ασ() = sj114 222  sj 21 a 310 0 2γ ρ  2   2  . (28)

Third, decrease the indices by one step, in order we can obtain another relation for the coefficients ψ ak 00 , and use the relation given by the kernel vector 1 to eliminate them:

2α( 1 − σ2 ) β  βα()+ σ β  β 1 1  2 1 2 1 − −−++−−()1σs j   () 1 σ1 4 σ 2  −−+s j  2 + γ2 1 1 γγ2 2 2 γ 2ρ    2ρ 2 ρ  2 ρ ()−σ2 β 11 1  1 ++−sj +4σβ2  ++++= sj 210. ασ() 211γ 222  2 2ρ (29)

Before we can follow, we should note that the determinant of the system (26) is not null, so that the = only possible solution for (26) is the trivial solution, i.e., ak10 0 , so that the solution (18) for the system of differential equations reduces to the elementary form

−β − β = ss1 2 11 r 22 r = fk a k 00 r1 r 2 e , k 1, 2, 3, 4 , (30)

= −( − σ) β γ = − σβ γ = where we have redefined the coefficients ak 00 as a1001 12 / ρ , a2002 2 / 2 ρ , a300 1 = and a400 0 .

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

β= β Now, we write by convenience 2h 1 , where h is to be determined below. As a consequence, β substituting 1 found above and after a little of algebra, we get from (29)

2 41ασ2()()+2 1 −+ σ 2 4 σ 2h 2  ( γγ − )   1ρ 2 ρ γρ γ ρ = 1 2 ()1−σ2()s ++1 4 σ 2 h() s + 3 12 2 2 , (31) which is the fundamental relation that connects the electron parameters. Next, we substitute into (31) γ γ the expressions for 1ρ and 2ρ defined above to get finally an algebraic expression for the energy eigenvalues we are searching for

α(1+ σ) ( 1 + σ )m E = + ρ 2 2 α2()()+ σ − σ + σ 2 2  41 1 4 h  1+ ()1−σ2()s ++1 4 σ 2 h() s + 3 12 2 2 . (32)

However, this is not yet the final step, since we still need to find out a value for h and the connection ρ between σ and , so that we can obtain a numerical evaluation of the atom energy. This is done by considering the values of the radii for which the probability given by the radial function (30) is a = β maximum, that is, for which the first derivatives (19) vanish. From this come the relations r10 s 1/ 1 = β and r20 s 2/ 2 among the most likely orbital radii and the points of maxima of the radial part of the wave function.

ρ = r Further we assume that 12 at the equilibrium configuration may be approximated by ρ = + =σ r10 r 20 and also a linear connection r10 r 20 between the electron equilibrium radii which → ∞ σ → =σ assures the contour condition r20 when 0 . From these relations we finally get h s2/ s 1 . At this point we have finally fulfilled all the steps toward getting an expression for the energy eigenvalues in terms of the basic electron properties along with the variation factor σ:

2σαm2 (1+ σ ) 2 (1 + σ ) m E = + C C 1 2 , (33) where use has been made of the parameters

2 =−σ2 ++1 σ 3 +3  ++ ασσσ 22 −+ 2242 C1(1 )( ssss 12 ) 1 4( 2 2 ) 2  4(1 )(1 ) ss1 4 2 , (34a)

4ασ2 (1+ ) 2 (1 − σ ) 22s + 4 σ 42 s  = + 1 2  C2 1 2 (1−σ )(2ss ++1 ) 4( σ 3 s + 3 ) s  112 2 2 2  . (34b)

We also get, together with Eq. (33), a determination of the equilibrium distance between the electrons C as a function of σ , i.e., ρ = 1 , as was aimed at the beginning of the paper. And finally 2σm α (1+ σ ) σ ρ the equilibrium radii becomes r = ρ and r = . 10 1+ σ 20 1+σ

Nascimento and Fonseca; PSIJ, 9(4): 1-12, 2016; Article no.PSIJ.24355

At this point we can make a plot of the energy consistent with the dynamics of the physical excess ∆E = E −(1 + σ ) m of the system against system. Besides, the equilibrium radii found = = ρ = the effective mass (1+σ ) m . After considering the r10 0.130 , r20 0.732 and 0.862 seem unit conversion factors, the Hartree mα 2 ≅ 27 eV also to be in a reasonable agreement with the ° known values [15]. and the Bohr radius a=1/ ( mα ) ≅ 0.53 A , we 0 get dimensionless forms for the energy excess ρ 4. CONCLUSION ∆E and for the distance as follows: In this work we have considered a 2D formulation 2σ (1+ σ )2  (1 + σ )  1 of the Helium Atom, derived a four-spinor Dirac- ∆=E + −−1 σ  C C α 2 like equation and found the suitable matrices. 1 2  , (35) The work has been developed within two methods of approach: on the one hand through a total Hamiltonian in a Hylleraas context that ends C1 ρ = in an extremum problem to be solved in a next 2σ (1+ σ ) . (36) paper. And on the other hand through a pair of Hamiltonians for an ion-atom and for an outer electron respectively. This second approach stands for a process controlled by a macro- parameter of variation which is connected with the average values of the radial variables and that contains the ion helium atom ground state as a limit case. For this case we have discussed the general structure of the equations, separated the system of equations, found the angular eigenfunctions that decouple the system and a solution for the radial equation, in the approximation of constant inter-electron distance. This made possible to calculate the ground state energy eigenvalue of the atom, whose value agrees with the experimental data within 0.1% of accuracy.

COMPETING INTERESTS Fig. 1. Excess energy variation ∆E in au against the dimensionless parameter σ Authors have declared that no competing interests exist. The plot of the energy excess (in au) as a function of σ is shown in Fig. 1 (above), from REFERENCES which we immediately see that the ion ground state limit ∆E = − 2 occurs for σ = 0 or ρ = ∞ . 1. Hartree DR. The calculation of atomic The minimum of the energy excess for structures. New York (NY): John Wiley; = = 1957. j1 j 2 1 corresponds to the inner orbital state of the parahelium atom (or the state 1s-1s of the 2. Hylleraas EA. Neue Berechnung der Spectroscopy). By solving the equation Energie des Heliums im Grundzustande, d sowie des tiefsten Terms von Ortho- ∆E = 0 , we see that the equilibrium value Helium. Z. fur Physik. 1929;54:347-366. dσ σ = 3. Eisenhart LP. Separation of the variables occurs approximately for 0.17753 , for whose of the two-particle wave equation. Proc. N. value we get an energy ground state of A. Soc. 1949;35:490-94. ∆ = − E 2.9059 , which agrees with the 4. Kato T. On the existence of solutions of ∆ = − experimental value EExp 2.9033 within 0.1% the Helium wave equation. Trans. Am. of accuracy. This means that the approximations Math. Soc. 1951;70:212–218. done to get the determination of the ground state 5. Nakatsuji H, Nakashima H. Analytically energy of the atom, although rather rough, were solving the relativistic Dirac-Coulomb

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equation for atoms and molecules. Phys. 10. Dirac PAM. The quantum theory of the Rev. Letters. 1995;95:050407/1-4. electron. Proc. Royal Society. 1928; 6. Pont M, Shakeshaft R. Decomposition of A117(778):610-24. the two-electron-atom eigenvalue problem. 11. Nascimento DL, Fonseca ALA, Monteiro Phys. Rev. A. 2005;51(1):257-265. FF, Amato MA. A variational approach for 7. Nascimento DL, Fonseca ALA, Portilho O. numerically solving the two-component Determination of the Helium atom ground radial Dirac equation for one- particle state through the Hamilton–Jacobi systems. J. Mod. Phys. 2012;3:350-354. equation. Int. J. Q. Chem. 2005;103:505– 12. Breit G. The effect of retardation on the 508. interaction of two electrons. Phys. Rev. 8. Nascimento DL, Fonseca ALA. New 1929;34:553-573. approach to researches in relativistic 13. Nascimento DL, Fonseca ALA. Derivation quantum chemistry using Hamilton-Jacobi of a four-spinor Dirac-like equation for the Equation, in Quantum Chemistry Research Helium atom and an approximate ground Trends, chap. 5; 173-204. Edited by state solution. arXiv:1304.5581; 2013. Kaisas MP, Nova Science Publishers, Inc; 14. Goldstein H. Classical mechanics, 2 nd Ed. 2007. Addison-Wesley, Reading, Mass; 1980. 9. Nascimento DL, Fonseca ALA. A 2D 15. Available:http://periodictable.com/Propertie spinless version of Dirac’s equation written s/A/AtomicRadius.an.html in a noninertial frame of reference . Int. J. Q. Chem. 2011;111;1361-69. ______© 2016 Nascimento and Fonseca; This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( http://creativecommons.org/licenses/by/4.0 ), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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