CHAPTER 8

1 CHAPTER 8 THE helium ATOMThe second element in the periodic table provides our rst example of aquantum-mechanical problem whichcannotbe solved exactly. Nevertheless,as we will show, approximation methods applied to helium can give accuratesolutions in perfect agreement with experimental results. In this sense,it can be concluded that quantum mechanics is correct for atoms morecomplicated than hydrogen. By contrast, the Bohr theory failed miserablyin attemps to apply it beyond the hydrogen helium atom has two electrons bound to a nucleus with chargeZ= 2. The successive removal of the two electrons can be diagrammed asHeI1 !He++e I2 !He+++ 2e (1)The rst ionization energyI1, the minimum energy required to remove the rst electron from helium , is experimentally eV. The second ionizationenergy,I2, is eV. The last result can be calculated exactly since He+is a hydrogenlike ion. We haveI2= E1s(He+) =Z22n2= 2 hartrees = 54:42 eV(2)The energy of the three separated particles on the right side of Eq (1) is, byde nition, zero.

2 Therefore the ground-state energy of helium atom is givenbyE0= (I1+I2) = 79:02 eV = 2:90372 hartrees. We will attempt toreproduce this value, as close as possible, by theoretical odinger Equation and Variational CalculationsThe Schr odinger equation for He atom, again using atomic units and as-suming in nite nuclear mass, can be written 12r21 12r22 Zr1 Zr2+1r12 (r1;r2) =E (r1;r2)(3)The ve terms in the Hamiltonian represent, respectively, the kinetic ener-gies of electrons 1 and 2, the nuclear attractions of electrons 1 and 2, and1the repulsive interaction between the two electrons. It is this last contri-bution which prevents an exact solution of the Schr odinger equation andwhich accounts for much of the complication in the theory. In seeking anapproximation to the ground state, we might rst work out the solutionin the absence of the 1=r12-term. In the Schr odinger equation thus simpli- ed, we can separate the variablesr1andr2to reduce the equation to twoindependent hydrogenlike problems.

3 The ground state wavefunction (notnormalized) for this hypothetical helium atom would be (r1;r2) = 1s(r1) 1s(r2) =e Z(r1+r2)(4)and the energy would equal 2 ( Z2=2) = 4 hartrees, compared to the ex-perimental value of 2:90 hartrees. Neglect of electron repulsion evidentlyintroduces a very large signi cantly improved result can be obtained with the functionalform (4), but withZreplaced by a adjustable parameter , thus:~ (r1; r2) =e (r1+r2)(5)Using this function in the variational principle [cf. Eq ( )], we have~E=R (r1; r2)^H (r1; r2)d 1d 2R (r1; r2) (r1; r2)d 1d 2(6)where^His the full Hamiltonian as in Eq (3), including the 1=r12-term. Theexpectation values of the ve parts of the Hamiltonian work out to 12r21 = 12r22 = 22 Zr1 = Zr2 = Z ; 1r12 =58 (7)The sum of the integrals in (7) gives the variational energy~E( ) = 2 2Z +58 (8)2 This will be always be an upper bound for the true ground-state can optimize our result by nding the value of whichminimizestheenergy (8).

4 We ndd~Ed = 2 2Z+58= 0(9)giving the optimal value =Z 516(10)This can be given a physical interpretation, noting that the parameter inthe wavefunction (5) represents ane ectivenuclear charge. Each electronpartially shields the other electron from the positively-charged nucleus byan amount equivalent to 5/8 of an electron charge. Substituting (10) into(8), we obtain the optimized approximation to the energy~E= Z 516 2(11)For helium (Z= 2), this gives hartrees, an error of about 2%(E0= 2:90372). Note that the inequality~E > E0applies in the late 1920's, it was considered important to determine whetherthe helium computation could be improved, as a test of the validity ofquantum mechanics for many electron systems. The table below gives theresults for a selection of variational computations on Z(r1+r2)Z= 2 2:75e (r1+r2) = 1:6875 2:84765 (r1) (r2)best (r) 2:86168e (r1+r2)(1 +c r12)best , c 2:89112 Hylleraas (1929)10 parameters 2:90363 Pekeris (1959)1078 parameters 2:90372 The third entry refers to theself-consistent eldmethod, developed byHartree.

5 Even for the best possible choice of one-electron functions (r),there remains a considerable error. This is due to failure to include thevariabler12in the wavefunction. The e ect is known aselectron fourth entry, containing a simple correction for correlation, gives aconsiderable improvement. Hylleraas (1929) extended this approach with avariational function of the form (r1; r2; r12) =e (r1+r2) polynomial inr1; r2; r12and obtained the nearly exact result with 10 optimized parameters. Morerecently, using modern computers, results in essentially perfect agreementwith experiment have been and the Exclusion PrincipleThe simpler wavefunctions for helium atom, for example (5), can be inter-preted as representing two electrons in hydrogenlike 1sorbitals, designatedas a 1s2con guration. According to Pauli's exclusion principle, which statesthat no two electrons in an atom can have the same set of four quantumnumbers, the two 1selectrons must havedi erentspins, one spin-up or ,the other spin-down or.

6 A product of an orbital with a spin functionis called aspinorbital. For example, electron 1 might occupy a spinorbitalwhich we designate (1) = 1s(1) (1)or 1s(1) (1)(12)Spinorbitals can be designated by a single subscript, for example, aor b, where the subscript stands for asetof four quantum numbers. In atwo electron system the occupied spinorbitals aand bmust be di erent,meaning that at least one of their four quantum numbers must be two-electron spinorbital function of the form (1;2) =1p2 a(1) b(2) b(1) a(2) (13)automatically ful lls the Pauli principle since it vanishes ifa=b. More-over, this function associates each electron equally with each orbital, whichis consistent with theindistinguishabilityof identical particles in quantummechanics. The factor 1=p2 normalizes the two-particle wavefunction, as-suming that aand bare normalized and mutually orthogonal. The func-tion (13) isantisymmetricwith respect to interchange of electron labels,meaning that (2;1) = (1;2)(14)4 This antisymmetry property is an elegant way of expressing the Pauli note, for future reference, that the function (13) can be expressedas a 2 2 determinant: (1;2) =1p2 a(1) b(1) a(2) b(2) (15)For the 1s2con guration of helium , the two orbital functions are thesame and Eq (13) can be written (1;2) = 1s(1) 1s(2) 1p2 (1) (2) (1) (2) (16)For two-electron systems (butnotfor three or more electrons), the wave-function can be factored into an orbital function times a spin function.

7 Thetwo-electron spin function 0;0(1;2) =1p2 (1) (2) (1) (2) (17)represents the two electron spins in opposing directions (antiparallel) with atotal spin angular momentum of zero. The two subscripts are the quantumnumbersSandMSfor the total electron spin. Eq (16) is called thesingletspin state since there is only a single orientation for a total spin quantumnumber of zero. It is also possible to have both spins in thesamestate,provided the orbitals are di erent. There are three possible states for twoparallel spins: 1;1(1;2) = (1) (2) 1;0(1;2) =1p2 (1) (2) + (1) (2) 1; 1(1;2) = (1) (2)(18)These make up thetripletspin states, which have the three possible orien-tations of a total angular momentum of States of HeliumThe lowest excitated state of helium is represented by the electron con gu-ration 1s2s. The 1s2pcon guration has higher energy, even though the 2sand 2porbitals in hydrogen are degenerate, because the 2spenetrates closerto the nucleus, where the potential energy is more negative.

8 When electronsare in di erent orbitals, their spins can be either parallel or antiparallel. Inorder that the wavefunction satisfy the antisymmetry requirement (14), thetwo-electron orbital and spin functions must haveoppositebehavior underexchange of electron labels. There are four possible states from the 1s2scon guration: a singlet state +(1;2) =1p2 1s(1) 2s(2) + 2s(1) 1s(2) 0;0(1;2)(19)and three triplet states (1;2) =1p2 1s(1) 2s(2) 2s(1) 1s(2) 8<: 1;1(1;2) 1;0(1;2) 1; 1(1;2)(20)Using the Hamiltonian in Eq(3), we can compute the approximate energiesE =Z Z (1;2)^H (1;2)d 1d 2(21)After evaluating some erce-looking integrals, this reduces to the formE =I(1s) +I(2s) +J(1s;2s) K(1s;2s)(22)in terms of the one electron integralsI(a) =Z a(r) 12r2 Zr a(r)d (23)the Coulomb integralsJ(a; b) =Z Z a(r1)21r12 b(r2)2d 1d 2(24)6and the exchange integralsK(a; b) =Z Z a(r1) b(r1)1r12 a(r2) b(r2)d 1d 2(25)The Coulomb integral represents the repulsive potential energy for two in-teracting charge distributions a(r1)2and b(r2)2.

9 The exchange integral,which has no classical analog, arises because of the exchange symmetry (orantisymmetry) requirement of the wavefunction. BothJandKcan beshown to be positive quantities. Therefore the lower sign in (22) repre-sents the state of lower energy, making the triplet state of the con guration1s2slower in energy than the singlet state. This is an almost universalgeneralization and contributes to Hund's rule, to be discussed in the