18 Multi-electron Atom

So far we have dealt with the properties of one particle moving in severaldi!erent potentials. This has allowed us to solve everything exactly andobtain analytic expressions for the wavefunction and energy of the system.However, there are very few problems for which the Schrodinger equation canbe solved exactly. In fact, as soon as the potential is more complicated thanwhat we already have seen or that we have more that two interacting particleswe cannot solve the equations exactly. Before we start discussing how wecan obtain approximate solution let start by considering the Hamiltonian ofa system with many electrons with a fixed nucleus.

18.1 The many electron Hamiltonian

Remember the form for the Hamiltonian for the hydrogen-like atom

H = T + V = !1

2"2 !

Z

R(793)

in atomic units (me = ! = 1, e = !1), where R is the distance between thenucleus and the electron, and Z is the atomic number of the nucleus. Similarfor a system of n electrons the Hamiltonian is

H = !1

2

n!

i=1

"2i !

n!

i=1

Z

Ri+

1

2

n!

i,j=1

!1

rij(794)

where the first term is the kinetic energy operator for each electron, thesecond term is due to the attraction between the electron and the nucleus, andthe last term accounts for the repulsion due to electron-electron interactions.The factor 1

2 in front of the double sum prevents counting the electron-electron interactions twice, and the prime excludes the i = j terms. If wecompare with the Hamiltonian for the hydrogen-like atom we realize that wecan express the many-electron Hamiltonian as

H =n

!

i=1

"

!1

2"2

i !Z

Ri

#

+1

2

n!

i,j=1

!1

rij(795)

=n

!

i=1

h(i) +1

2

n!

i,j=1

!1

rij(796)

96

where h(i) is the hydrogen-like Hamiltonian for the i’th electron. Lets con-sider the simplest system having more than one electron which is the heliumatom.

For He the Hamiltonian reads (see figure for definitions of distances),

H = !1

2"2

1 !1

2"2

2 !2

R1!

2

R2+

1

r12(797)

= h(1) + h(2) +1

r12(798)

We already know how to solve for h(1) and h(2), but what about the lastterm ? Since the term depends on both the position of electron 1 and 2we cannot separate their motion. This means that we cannot use the trickof separating variables which we have used so far and we cannot obtain ananalytic solution to the Schrordinger equation.

18.2 The independent electron approximation

The simplest approximation is simply to ignore the electron-electron repul-sion. Obviously this is a very bad approximation but it allows us to separatevariables and thereby illustrate some important physics of a many-electronsystem. Since we neglect the interactions between the electron s we treatthe electrons as independent of each other, hence the name ”independentelectron approximation”. The approximate Hamiltonian is then

Happrox. = h(1) + h(2) (799)

where each of the individual hydrogen-like Hamiltonians obey a one-electronSchrodinger equation

h(1)!i(1) = "i!i(1) (800)

97

where !i is the atomic orbital and "i is the orbital energy. This mean thanan orbital is simply a one-electron wavefunction. We already know the so-lution to these one-electron Schrodinger equation from our treatment of thehydrogen atom, i.e.

!i # {1s, 2s, 2p, · · ·} , "i = !1

2

Z2

n2(801)

Previously, we have shown that if we can separate the Hamiltonian intoindependent terms we can write the total wavefunction as product of eigen-functions of the individual terms. Therefore, the wavefunction for He in theindependent electron approximation becomes

#(He) = !i(1)!j(2) (802)

Now, lets show that this is wavefunction is an eigenfunction of the approxi-mate Hamiltonian,

Happrox#(He) = (h(1) + h(2))!i(1)!j(2) (803)

= h(1)!i(1)!j(2) + h(2)!i(1)!j(2) (804)

where h(1) and h(2) only works on variables of electron 1 and electron 2,respectively. This gives us

Happrox#(He) = h(1)!i(1)!j(2) + !i(1)h(2)!j(2) (805)

= "i!i(1)!j(2) + !i(1)"j!j(2) (806)

= ("i + "j)!i(1)!j(2) (807)

= ("i + "j)#(He) (808)

and we see that the total energy is the sum of the individual orbital energies.

18.3 Electron exchange symmetry

We can now write the ground state of the He atom in the independent electronapproximation as 1s(1)1s(2), for which the electron configuration is 1s2. Nowlets instead consider the first excited state of He. What would we expect ?There are only two di!erent ways of arranging the electrons in a simpleproduct

#(1, 2) = 1s(1)2s(2), or,#(2, 1) = 2s(1)1s(2) (809)

98

which both have the same energy but describes di!erent physics, e.g. electrondistribution. This can be illustrated by considering the electron distribution

|#(1, 2)|2 =8

$exp(!4r1)

1

$(1 ! 2r2 + r2

2)exp(!2r2) (810)

and

|#(2, 1)|2 =8

$exp(!4r2)

1

$(1 ! 2r1 + r2

1)exp(!2r1) (811)

Therefore, we see that|#(1, 2)|2 $= |#(2, 1)|2 (812)

The two wavefunctions therefore di!er by exchange of electron indices, some-thing that is wrong. While in classical mechanics it is always possible to dis-tinguish between identical particles, however, this is not the case in quantummechanics due to the uncertainty principle. Therefore, in quantum mecha-nism identical particles are indistinguishable and the probability has to beinvariant under exchange of indices. This implies that the wavefunction itself most by either symmetric or antisymmetric under the exchange. Let Pij

be an operator that when acting on a function interchange the indices i andj such that

Pij#(ij) = ±#(ij) (813)

For He we can construct the following symmetric and antisymmetric wave-function from linear combinations of the original functions

#s =1%2[1s(1)2s(2) + 2s(1)1s(2)] (814)

and

#a =1%2[1s(1)2s(2) ! 2s(1)1s(2)] (815)

If we now operate on these wave functions with P12 we see that

P12#s =1%2[1s(2)2s(1) + 2s(2)1s(1)] = +#s (816)

P12#a =1%2[1s(2)2s(1) ! 2s(2)1s(1)] = !#a (817)

If we now plot ( see figure) the propability density of these new functionswe will see that the antisymmetric function has a depletion of density as the

99

distance goes to zero, which is called a Fermi hole. Similar the symmetryfunction has a increase in density near the region called a Fermi heap. Doesthis look like something familiar ?

18.4 Stern-Gerlach experiment

Up to this point, we have treated the particles as structureless with a massand a charge which can be described by a wavefunction specified by thespatial coordinates, x, y, z, as variables. However, our empirical evidencepoints to the need for attributing an intrinsic angular momentum to theparticles as well. The most direct evidence comes from the work of Sternand Gerlach in their now famous experiment from 1922, although it was notrealized at that time. In the experiment they passes an unexcited beam ofsilver atoms though an inhomogeneous magnetic field, see figure.

If a particle have a magnetic moment µ (we have seen that the angularmomentum of a charged species gives rise to a magnetic moment µ = !%eL)they will be deflected due to a force, F & µb"B. Thus, particles withdi!erent moments will be deflected di!ering amounts by the magnetic field.If the particles are classical, ”spinning” particles, then the distribution oftheir spin angular momentum vectors is taken to be truly random and eachparticle would be deflected up or down by a di!erent amount, producingan even distribution on the screen of a detector. Consider the electronicconfiguration of silver, Ag = [Kr]4d105s1, what would you expect classicalfor the moment and what would you expect to see in the experiment?

100

Thus, the magnitude magnetic moment of the electron is fixed, and thedirection it points is quantized and can take on one of two values. Since anycomponent of L has 2l + 1 eigenvalues we would expect that the magneticmoment to have 2l + 1 eigenvalues. However, experimentally only two dis-tinct traces where seen (although, nothing was seen due to very thin layerdeposited until Stern’s breath full of sulfur from cheap cigars developed AgSwhich is ”jet” black ). The solution to this comes from the postulate byGoudsmit and Uhlenbeck in 1925 that the electron have an intrinsic angularmoment now called spin with an angular magnitude of 1/2!. They postu-lated the existents of the spin in order to explain the fine structure of atomicspectra. However, it was not until 1928 when Dirac combined Einsteins rel-ativity with quantum mechanics that the treatment of electron spin arisednaturally. This is well beyond the scope of this class.

18.5 Spin and the Pauli exclusion principle

The spin aungular momentum is denoted by S and similar to the orbitalangular momentum they obey the following eigenvalue equations

S2 |sms' = !2s(s + 1) |sms' , Sz |sms' = !ms |sms' (818)

where the allowed values of ms = !s,!s + 1, · · · , s ! 1, s. Electrons havea spin of s = 1

2 which gives two values for ms = ±12!. The eigenstate

corresponding to$

$

12 ,

12

%

is traditional called & or spin-up and$

$

12 ,!

12

%

is called% or spin-down. We will not express then spin function in explicit form butrather keep the symbolic representation & and %. Therefore, the wavefunction# = !(1)&(1) refer to an electron in space orbital !(1) with spin &, and iscalled a spin-orbital. Since the electrons are indistinguishable we requirethat the wavefunction is either symmetric or antisymmetric with respect tointerchange of electron space and spin indices. Lets consider the possiblecombination of space and spin wavefunctions for the ground state of He

1s(1)1s(2)

&

'

'

(

'

'

)

&(1)&(2) !symmetric&(1)%(2) !unsymmetric%(1)&(2) !unsymmetric%(1)%(2) !symmetric

(819)

101

Similar to what we did for the unsymmetric space wavefunctions we need tomake linear combinations in to form correct wavefunctions

1s(1)1s(2)

&

'

'

(

'

'

)

&(1)&(2) !symmetric1"2[&(1)%(2) + %(1)&(2)] !symmetric

1"2[&(1)%(2) ! %(1)&(2)] !antisymmetric

%(1)%(2) !symmetric

(820)

Instead of one wavefunction we now have four di!erent wavefunctions that allobey the correct exchange properties, where three of these are symmetric andonly one that is antisymmetric. Experimentally we know that the groundstate of He is a singlet. This indicates that the wavefunction should beantisymmetric under electron exchange. In general experimental evidenceshowed that for electrons only the antisymmetric wavefunction occur. Thislead Pauli to postulate his now famous principle, here given in its mostgeneral form.

The total wavefunction must by antisymmetric under the in-terchange of any pair of identical fermions and symmetrical

under the interchanges of any two pair of identical bosons.

where fermions are spin half-integer particles (electrons, protons) and bosonsare spin integer particles (photons, &-particles). Lets consider the conse-quences of the Pauli principle by considering the ground state of Li and fora minute forget what we already know . Similar to He we assume that wecan write the wavefunction as a product of hydrogen-like functions as 1s3

#(Li) =

&

'

'

'

'

'

'

'

'

'

'

(

'

'

'

'

'

'

'

'

'

'

)

&(1)&(2)&(3) !symmetric&(1)&(2)%(3) !unsymmetric&(1)%(2)&(3) !unsymmetric%(1)&(2)&(3) !unsymmetric&(1)%(2)%(3) !unsymmetric%(1)&(2)%(3) !unsymmetric%(2)%(2)&(3) !unsymmetric%(1)%(2)%(3) !symmetric

(821)

Is is possible to find any linear combinations that are totally antisymmetric? No, which is in good agreement with our experimental observation that nostates have 1s3 electronic configuration. If this was possible we would nothave a periodic table! Therefore, from the Pauli exclusion principle we seethat no two electrons and occupy the same spin-orbital. For Li the lowestfunction most therefore have a space wavefunction of the form 1s(1)1s(2)2s(3)

102

18.6 Slater-determinants

This procedure for generate wavefunctions that exhibit the correct symmetryeasily becomes very tedious. Slater realized that writing the wavefunctionas a determinant was an easy way of generating wavefunctions that are an-tisymmetric. A 2 x 2 determinant is given by

$

$

$

$

a11 a12

a21 a22

$

$

$

$

= a11a22 ! a12a21 (822)

For a general A(n ( n) determinant we need to use the cofactor (or Laplaceexpansion of the determinant as

det(A) =n

!

k=1

(!1)i+kaikdet(Aik) (823)

where Ajk is the determinant that remain after removing the ith row and thekth column of A These determinant are now known as Slater determinants.The properties of determinant ensures the Pauli principle since 1) changesign when interchange two rows (or columns), 2) if two rows are identical thedeterminant is zero, i.e. no two electrons can be in the same spin-orbital and3) if a row is zero then the determinant is zero. The general form for a Nelectron wavefuntion is

#Na =

1%N !

$

$

$

$

$

$

$

$

$

!1(x1) !2(x1) · · · !N(x1)!1(x2) !2(x2) · · · !N(x2)

......

. . ....

!1(xN ) !2(xN) · · · !N(xN)

$

$

$

$

$

$

$

$

$

(824)

where !i is a general spin-orbital. Therefore, for Li the slater determinantwould be

#(Li) =1%6

$

$

$

$

$

$

1s(1)&(1) 1s(1)%(1) 2s(1)&(1)1s(2)&(2) 1s(2)%(2) 2s(2)&(2)1s(3)&(3) 1s(3)%(3) 2s(3)&(3)

$

$

$

$

$

$

(825)

=1%6

$

$

$

$

$

$

1s(1) 1s(1) 2s(1)1s(2) 1s(2) 2s(2)1s(3) 1s(3) 2s(3)

$

$

$

$

$

$

(826)

Since we can always construct the full Slater determinant if we know a listof all the spin-orbitals a simplified notation is

# = |!1!2!3 · · ·!n' . (827)

103

Example: Write the normalized Slater determinant for the beryllium in the1s22s2 configurations. Explain how you would expand it.

#(Be) =1%4!

$

$

$

$

$

$

$

$

1s(1) 1s(1) 2s(1) 2s(1)1s(2) 1s(2) 2s(2) 2s(2)1s(3) 1s(3) 2s(3) 2s(3)1s(4) 1s(4) 2s(4) 2s(4)

$

$

$

$

$

$

$

$

(828)

18.7 The Hamiltonian and spin

Since the hamiltonian is independent of spin, at the non-relativitis level oftheory, we know that it commutes with the spin operators

[H, Sz] = [H, S2] = [S2, Sz] = 0 (829)

thus, we can specify simultaneous eigenfunctions of these operators. Whatabout the energies:

H"(x, y, z, ') = [H#(x, y, z)]g(') = [E#(x, y, z)]g(') = E"(x, y, z, ')(830)

and our energies are the same as we previously found.

18.8 Singlet and Triplet excited states of Helium

For the 1s2s state of He we saw that there is two di!erent space functionsthat have the correct symmetry, one symmetric and one symmetric. Wehave also learned that for electron it is the total wavefunction that needsto be antisymmetric with respect the interchange of both space and spinlabels. Therefore, we need to combined the symmetric space part with anantisymmetric spin part and a symmetric spin part with the antisymmetricspace functions. The give us for the 1s2s

#s,a(He) =1%2

[1s2s + 2s1s]1%2

[&% ! %&] (831)

called a singlet state, and,

#a,s(He) =1%2

[1s2s ! 2s1s]

&

(

)

&&1"2[&% + %&]

%%(832)

104

call a triplet state. Since all four state satisfy the correct symmetry and arelinear independent it indicates that there are 4 distinct physical states. Is itpossible to write a single Slater determinant for each of these states ? Theanswer is no and is very important to realize. It is not always possible withonly one Slater determinant to write a wave function that has the correctsymmetry of the full wavefunctions,

#s,a(He) =1%2

*

1%2

$

$

$

$

1s 2s1s 2s

$

$

$

$

!1%2

$

$

$

$

1s 2s1s 2s

$

$

$

$

+

(833)

but instead we need to make linear combinations of Slater determinants. Allfour state are degenerate with the independent electron model, would weexpect this to be the case for the full Hamiltonian ?

Let evaluate the energy of these state using the full Hamiltonian. What isthe expression for the full Hamiltonian ? and how do we evaluate the energy?

)E' =

,

##H#d( (834)

+

,

##"

h1 + h2 +1

r12

#

#d( (835)

=

,

##"

!1

2"2

1 !1

2"2

2 !2

R1!

2

R2+

1

r12

#

#d( (836)

Since the Hamiltonian do not dependent on spin we can integrate the spinout first. This means that the energy is completely determined by the spacefunctions. We can now substitute the space function for the triplet andsinglet case in the energy expression

)E1,3' =

, ,

[1s#2s# ± 2s#1s#]

"

h1 + h2 +1

r12

#

[1s2s ± 2s1s] d1d2 (837)

using that the orbitals are ortho-normal we can simplify the integrals

)E1,3' =

,

1s#h11sd1 +

,

2s#h12sd1 (838)

+

, ,

1s#2s#1

r121s2sd1d2 (839)

±, ,

1s#2s#1

r122s1sd1d2 (840)

= E1s + E2s + J ± K (841)

105

where J is the Coulomb integral and K is the exchange integral. J representthe Coulomb repulsion between two electrons where K is similar but withone indices exchanged. Therefore, the triples state is lower in energy than thesinglet since K (and J) is positive. This is in agreement with our experimentalobservation of the the 1s2s configurations for which two states are found andthe lowest split into 3 state in a magnetic field.

18.9 Angular momentum in many-electron atoms

When dealing with many-electron systems we need to account for the spinand orbital angular momentum of the electrons. We therefore need to under-stand the rules for adding angular momentum. For many electron atom theindividual angular momentum operators do not compute with the Hamil-tonian, however, their sum does. The total orbital angular momentum isdefined as the vector sum of the orbital angular momenta of the individualelectrons

L =!

i

Li (842)

with eigenvaluesL2 |LML' = !

2L(L + 1) |LML' (843)

and for the projection

Lz |LML' = !ML |LML' (844)

n general the addition of two angular momenta with quantum number l1 andl2 will results in a total quantum number whose number J has the followingpossible values

L = |l1 + l2|, |l1 + l2 ! 1|, · · · , |l1 ! l2| (845)

andM = m1 + m2 (846)

The total orbital angular momentum quantum number L of an atom is de-noted by S for L=0, P for L=1, D for L=2, F for L=3, and so forth.

106

18.10 Ladder operators

Similar for S we have

S2 |SMS' = !2S(S + 1) |SMS' (847)

and for the projection

Sz |SMS' = !MS |SMS' (848)

So far we have seen how to work with S2 and Sz operators. Similar towhat we did for the orbital angular momentum we can define ladder operatorsas

S+ = Sx + iSy (849)

for the raising operator and

S$ = Sx ! iSy (850)

for the lowering operator. Operating with these operators on our spin func-tions gives

S+% = !& (851)

S+& = 0 (852)

S$% = 0 (853)

S$& = !% (854)

(855)

Using these we can now operate with the other two components on the func-tions and get

Sx% = 1/2(S+ + S$)% = 1/2!& (856)

Sy% = 1/(i2)(S+ ! S$)% = !i/2!& (857)

Sx& = 1/2!% (858)

Sy& = i/2!% (859)

(860)

107

18.11 Eigenvalues of a two-electron spin function

The excited state configuration of He (1s2s) have the following possible spineigenfunctions

&(1)&(2) (861)1%2[&(1)%(2) + %(1)&(2)] (862)

%(1)%(2)1%2[&(1)%(2) ! %(1)&(2)] (863)

(864)

Where the possible total spin is S = (1/2 + 1/2), (1/2 ! 1/2 = (1, 0) withdegeneracy d = (2L+1)(2S +1) which gives 3 and 1, respectively. Thus, theterm symbols are 3S1 and 1S0, respectively. Now lets use the spin operatorsand verify this. First lets start with the projection

Sz&(1)&(2) = S1z&(1)&(2) + S2z&(1)&(2) (865)

=1

2!&(1)&(2) +

1

2!&(1)&(2) (866)

= !&(1)&(2) (867)

Similar for the other functions we find

Sz%(1)%(2) = !!%(1)%(2) (868)

Sz[&(1)%(2) ± %(1)&(2)] = 0 (869)

For S2 we get:

S2 = (S1 + S2) · (S1 + S2) = S21 + S2

2 + 2(S1xS2x + S1yS2y + S1zS2z) (870)

Operating with this on the eigenfunctions give

S2&(1)&(2) = &(2)S21&(1) + &(1)S2

2&(2) + 2S1x&(1)S2x&(2) (871)

+S1yS2y + S1zS2z (872)

= (3/4!2 + 3/4!

2 + 1/2!2)&(1)&(2) = 2!

2&(1)&(2) (873)

Similarly for the other eigenfunctions we get

S2%(1)%(2) = 2!2%(1)%(2)(874)

S2[&(1)%(2) + %(1)&(2)] = 2![&(1)%(2) + %(1)&(2)]S2[&(1)%(2) ! %(1)&(2)] = 0(875)

Therefore, we see that for the singlet we indeed get S = 0, ms = 0 and forall of the triplet states we S = 1, ms = !1, 0, 1 as expected.

108

18.12 Term symbols

Example: Find the possible quantum number L for states of the Carbonatom with the following electron configuration 1s22s22p3p ? Since the selectrons have zero angular momentum they do not contribute. The 2p hasl = 1 and the 3p l = 1 gives L from 1+1=2 to |1!1| = 0, therefore L = 0, 1, 2

Example : Find the possible values for S for the states arising from1s22s22p3p. S electrons will contribute nothing due to Pauli principle, i.e.1/2 - 1/2 = 0. For the last two electrons we can get S = 0 and S = 1.

For a given S we have 2S + 1 values for MS where 2S + 1 is called themultiplicity. For 2S + 1=1,2,3,4 is called singlet, doublet, triplet, quartet.The total angular momentum is then

J = L + S (876)

The allow us to characterize the di!erent electronic states in a multi-electronatoms using what is know as a term symbol

2S+1LJ (877)

where

• S is the total spin quantum number. 2S+1 is the spin multiplicity:the maximum number of di!erent possible states of J for a given (L,S)combination.

• L is the total orbital quantum number in spectroscopic notation. Thesymbols for L = 0,1,2,3,4,5 are S,P,D,F,G,H respectively.

• J is the total angular momentum quantum number.

for which a total number of states are given by (2S +1)(2L+1). An alterna-tive statement of Hund’s rule is that the term with the highest multiplicityis lowest in energy. Remember that Hund’s rule works well for the groundstate, however, can fail for excited states.

Example: Write down the possible term symbols for Carbon atom in the1s22s22p3p configuration.

109

L = 2, S = 1 gives 3D3,3 D2,3 D1,L = 2, S = 0 gives 1D2,L = 1, S = 1 gives 3P2,3 P1,3 P0,L = 1, S = 0 gives 1P1,L = 0, S = 1 gives 3S1, andL = 0, S = 0 gives 1S0.

18.13 The atomic hamiltonian

Although our non-relativistic hamiltonian do not include spin, there is asmall term in the true Hamiltonian which comes from the interactions be-tween the spin and orbital angulr momentum and is referred to as spin-orbitinteractions. The atomic hamiltonian is then given by a sum of three terms

H = H0 + Hrep + HS.O (878)

where the first term is a sum of hydrogen-like Hamiltonians

H0 =n

!

i=1

-

!1

2"2 !

Z

ri

.

(879)

the second term is the electron-electron repulsion term

Hrep =1

2

n!

i,j

!1

rij(880)

and the third term is the spin-orbit interactions

HS.O =n

!

i=1

)Li · Si (881)

The repulsion operator splits antisymmetric and symmetric space wavefunc-tions into terms and the spin-orbitat interactions splits the terms into theindividual levels, resulting in the fine-structure of electronic spectra. Finallywe can use a magnetic field perturbation to split the individual levels instates. The magnetic field perturbation is given

Hmag = %e(J + S) · B = %eB(Jz + Sz) (882)

and is known as the Zeeman e!ect.

110

111