INTRODUCTION I have decided to do a project on how to ...

<strong>INTRODUCTION</strong>I <strong>have</strong> <strong>decided</strong> <strong>to</strong> <strong>do</strong> a <strong>project</strong> on how <strong>to</strong> approximate the ground state of a helium a<strong>to</strong>m.My <strong>project</strong> is aimed at students that <strong>have</strong> successfully completed the course of quantummechanics one and are interested in finding out about this area quantum mechanics.After quantum mechanics 1 students will be able <strong>to</strong> determine directly the λ functions fortwo particle systems such as the hydrogen a<strong>to</strong>m and hydrogen like ions. In mycoursework <strong>project</strong> I will show the non−relativistic Schrodinger equation for the nextsimplest a<strong>to</strong>m the helium a<strong>to</strong>m which contains three particles, cannot be solved directly.In fact the Schrodinger equation has not been solved directly for any system whichcontains more than two interacting particles. For all such systems one has <strong>to</strong> resort <strong>to</strong>approximation methods. In my <strong>project</strong> I <strong>have</strong> considered the method of ignoring electronrepulsion, and a first order perturbation method. I wanted <strong>to</strong> write about the variationmethod but chose not <strong>to</strong> as I didn’t want <strong>to</strong> make this <strong>project</strong> <strong>to</strong>o big. Since most of thea<strong>to</strong>mic and molecular systems I will deal with are relatively complicated, it is thenimportant that I develop a few basic approximation techniques.THE HELIUM ATOM1

The helium a<strong>to</strong>m, as depicted in below is a three−particle system which consist of twoelectrons and a nucleus whose mass is 4.0026 amu (6.6461*10 −24 g) and whose charge is+2e. I will assume that here, just as in the treatment of the hydrogen a<strong>to</strong>m, the <strong>to</strong>talSchrodinger equation may be separated in<strong>to</strong> two equations, one involving thetranslational energy of the a<strong>to</strong>m and the other involving the relative motion of theelectrons and the nucleus. I will ignore the translation energy associated with the motionof the centre of mass of the a<strong>to</strong>m in space and will concentrate my attention on therelative motion of particles within the a<strong>to</strong>m and on the energy of the relative motion.Furthermore, in my treatment of relative motion, I will assume that the nucleus isstationary. Although this is no exactly true, the mass of the nucleus is so much larger(about four thousand times larger) than the combined mass of the electrons that theresultant error is not significant.zElectron 2( x2 , y2, z2)1r 2r 12(1 1,1)Electron 1x , y z1r 1ynucleusx1. The potential energy of attraction between the first electron and the nucleus, −2e 2 /r 1 ,where r 1 is the distance between the first electron and the nucleus.2. The potential energy of attraction between the second electron and the nucleus,−2e 2 /r 2 , where r 2 is the distance between the second electron and the nucleus.3. The potential energy of repulsion between the two electrons +e 2 /r 12 , where r 12 is thedistance between two electrons.2

FIRST APPROXIMATION: IGNORING THE ELECTRON REPULSIONSince the presence of the electron repulsion term in equation (6) prevents me fromobtaining a direct solution, as a first approximation I will assume that the electrons <strong>do</strong>not repel one another and proceed with the calculations. By ignoring the <strong>to</strong>tal potentialenergy of the free helium a<strong>to</strong>m is given from equation (1) as2 22e2eV = − −(7)r rand the Schrodinger amplitude equation is2⎡ − h⎢ 2⎣8πme122 22 2 2e2e⎤0 0 0( ∇ + ∇ ) − − ψ = E ψ(8)12r1where the eigenvalues E 0 are the staionary state energies of reletive motion of the heliuma<strong>to</strong>m assuming no electron repulsion and the eigenfunction ψ 0 are the wave functions forthe helium a<strong>to</strong>m in which electron repulsion is ignored. The associated non repulsion0Hamil<strong>to</strong>nion opera<strong>to</strong>r is?H is thenr2⎥⎦2? 0 − hH28 mso that (8) may be written ase2 22 2 2e2e( ∇ +∇ ) − −(9)12r1r2=π)? 00 0 0H ψ = E ψ(10In order <strong>to</strong> separate the variables associated with the first electron from those associatedwith the second electron, I will assume that the <strong>to</strong>tal wave function ψ 0 may be expressedas a product of two functions, such that the first function γ 1, depends only on thecoordinates of the first electron, and the second function γ 2, depends only on thecoordinates of the second electron. That is( x y , z , x , y , z ) ( x , y , z ) γ ( x , y , z ) (11)0ψ1, 1 1 2 2 2= γ1 1 1 1 2 2 2 2or more simply0ψ = γ1γ2(12)Substitution of equation (12) in<strong>to</strong> equation (8) followed by division of both sides byγ 1γ2yields2222⎡ 1 h 2 2e⎤ ⎡ 1 h 2 2e⎤0⎢−∇2 1γ1− ⎥ + ⎢−∇2 2γ2− ⎥ = E⎣ γ18πmer1⎦ ⎣ γ28πmer2⎦(13)4

4E4EE1= and E2=nnH21so that E 0 from equation (16) becomesE⎡ 1 1⎢ +2⎣n1n2= 14EH2H22⎤⎥⎦(22)If each of the electrons in the nonrepulsive model of the helium a<strong>to</strong>m is in its ground,that is if n 1 =1 and n 2 =1, the ground state energy of the helium assuming no electronrepulsion is0E 2E= 8E+ E He= +Hewhere is the ground state energy of a helium ion.−11But EHis − 2.18×10 erg, which is equivelent <strong>to</strong> −13.6eV.Thus E 0 , the ground stateenergy for the helium a<strong>to</strong>m in my first approximation is0E = 8( − 13.6eV)(24)EHeThe best experimental value for the energy _of the helium a<strong>to</strong>m in the ground state isgiven in terms of the first ionization potential I, which is the minimum energy absorbedin the reaction in which the first electron leaves the ground state of a helium a<strong>to</strong>m <strong>to</strong>become a free electron:He (ground state) → He + + e(25)Since the minimum kinetic energy corresponds <strong>to</strong> an ejected electron having notranslational kinetic energy, I can write the balance for equation (25) asEHe+ I = E +He(26)But I am able <strong>to</strong> solve the Schrodinger amplitude equation forequation 23 that0E = / 2 = −54.4eV+ EHeHE He+(23)directly and show byand I is experimentally determined <strong>to</strong> be 24.6eV. Thus from equation (26), theexperimental value for the ground state energy of a helium a<strong>to</strong>m is given asE = −54.4eV− 24.6eV 79.0eVHe=E 0A comparison ofHe (experimental) from equation (27) with E (calculated) in equation(24) indicates that in the first approximation method, the calculated value for the groundstate energy is of a helium a<strong>to</strong>m is 38% lower (algebraically) than the experimentalvalue. Such a large error indicates that the electron repulsion cannot be ignored.(27)6

SECOND APPROXIMATION:A FIRST ORDER PERTABATION METHODI <strong>have</strong> shown that the correct Schrodinger wave equation for the helium a<strong>to</strong>m isH ? ψ = Eψ ,(2)Where222 2 2? − h 2 − h 2 2e2eeH = ∇1− ∇2− − +(28)228πme8πmer1r2r12Equation (2) cannot be solved directly, which means that we cannot directly evaluate thetrue wave function, ψ . I <strong>have</strong> shown that if electron repulsion ignore electron repulsionis ignored, then? 0 0 0 0H ψ = E ψ(29)can be directly solved where, from equation (9)222 2? 0 − h 2 h 2 2e2eH = ∇1− ∇2− −(30)228πme8πmer1r20and where the eigenfunctionsψ are the wave functions for the helium a<strong>to</strong>m in whichelectron repulsion is ignored. Substituting equation (30) in<strong>to</strong> equation (28) yield2? ? 0 eH = H +(31)r12and substitution of equation (31) in<strong>to</strong> equation (2) yields2⎛? 0 e ⎞⎜H + ψ Eψ(32)r⎟ =⎝12 ⎠an equation which cannot be directly solved. I will now compare equations (29) andequations (32). The difference in the opera<strong>to</strong>rs in the two eigenvalue equations is therepulsion terme 2 / r12. The smaller the value of thee 2 / r12the closer the two equationswill be. If I lete 2 / r12equal zero (ignore electron repulsion) then the two equations willbe the same. This means that E values are given by E o values and ψ functions are given0byψ functions.7

When using the perturbation method I will assume thate 2 / r12is small enough <strong>to</strong> beconsidered as a minor modification or perturbation of the opera<strong>to</strong>r0 ?H , an opera<strong>to</strong>r forwhich we may directly calculate the eigenfunctions0ψ . I will further hope that theperturbation of the opera<strong>to</strong>r0ψ will not be <strong>to</strong>o different from ψ (which cannot bedirectly evaluated)Thus when evaluating the ground state energy of a helium a<strong>to</strong>m in a given state, I willcombine the use of the correct hamil<strong>to</strong>nion opera<strong>to</strong>r H ? = H ? 0+ e 2 / r12with an incorrect0wave functionψ for that state which is regarded <strong>to</strong> be fairly close <strong>to</strong> what the correct0wave function should be. The nonrepulsive opera<strong>to</strong>r?H is said <strong>to</strong> be perturbed <strong>to</strong> the firs<strong>to</strong>rder by terme 2 / r12, and the present method of approximation is called a first orderperturbation method. The previous approximation method in which I ignored electron0repulsion al<strong>to</strong>gether is called a zero order perturbation method, since I used?H directlyas the opera<strong>to</strong>r. This means that I added no perturbation terms <strong>to</strong> the nonrepulsive0hamil<strong>to</strong>nion that I used. It is important <strong>to</strong> mention that for a given state,ψ is not aneigenfunction of H ? and therefore the operation of H ? 0onψ will not yield an exactenergy eigenvalue. Rather, because of the perturbation of energy due <strong>to</strong> the repulsionterm, the value of E obtained by the operation H ? 0onψ for a given state will depend onthe relative positions of the two electrons so that an approximate average value of E ~must be calculated through use of the mean value postulate as~E =∞∫ψ−∞∞∫− ∞*0ψH?ψ*0ψ00dτdτ=∞∫− ∞ψ*0( H?0 2 0ψ + e ψ / r )∞∫− ∞ψ*0ψ0dτ12dτSubstituting of equation (29) in<strong>to</strong> equation (33) yields~E =∞∫−∞but since for any given state,ψ*00E ψ0dτ+∞∫−∞ψ∞∫−∞*00E is constantψψ*002( e / r )dτ12ψ0dτ(33)(34)8

ø0*0= ø = ã1ã2⎛ ⎞⎜ ⎟= ⎝ ð ⎠31 ⎛ 2 ⎞− 2r2 / a0−2r2/ a0⎜⎝ a0⎟⎠ee(40)∞∫−∞∞∫−∞ø ø dô = 1øand∞∫−∞ø*1*2ø*01ø21dô02dô= 1∞∞= ∫ ∫− ∞ − ∞ø*1ø1.ø*2ø2dô1dô2= 1(41)Substituting equations (40) and (41) in<strong>to</strong> equation (37) yields∞ 2⎛ 8 ⎞ eE’ =⎜⎟3 ∫ −2 ⎝ ða0⎠ −∞r2 / a0− 2r2/ a012 e eor, in terms of the co−ordinates of each of the particles,22∞r 2( ) dô∞ 2⎛ 8 ⎞ eE’ =⎜⎟3 ∫ ∫ −⎝ ða0 ⎠ −∞ − ∞r12( ) 2 r 2 22 / a0− r2/ ae e0 dô ô 1d2(43)5 5= − EH =− ( −13.6eV) = + 34.0eV2 2(44)0’Substituting of the value for E form equation (24) and E from equation (44) in<strong>to</strong>equation (36) yieldsE ~ = −108 . 8 + 34.0 = −74.8eV(45)for the approximate ground state energy of the helium a<strong>to</strong>m, as calculated by a firs<strong>to</strong>rder perturbation method. A comparison of E ~ (−74.8eV) with the experimental value(−79.0eV) shows that the calculated value is about 5.3% <strong>to</strong>o high algebraically. If I usethe calculated E ~ value <strong>to</strong> compute a value for the ionisation potential, then I getI E ~calc= E + −He= −54.4 − − 74.8 = 20.4( ) eVAs compared with the experimental value of I, which is 24.6eV, the value calculated byperturbation method is 17% <strong>to</strong>o low. It appears that a comparison of calculated andexperimental first ionisation potentials is a more sensitive measure of the accuracy of thegiven approximation method than is a comparison of ground state energies, the reasonsbeing that the ionisation potential is a measure of a difference in energy levels.(42)10