Slide 1 Chapter 7 Multielectron Atoms Part B: Electron Spin and the Pauli Principle.

Chapter 7

Multielectron Atoms

Part B: Electron Spin and the Pauli Principle

• Electron Spin and the Pauli Principle

• Inclusion of Spin in Helium Atom Wavefunctions

• Spin Angular Momentum of Ground State Helium

• The Wavefunctions of Excited State Helium

• Excited State Helium Energies: He(1s12s1)

• The Energy of Ground State Helium


KE(1) KE(2) PE(2)PE(1) PE(12)

2 21 2

1 2 12

1 2 1 2 1

2 2H

r r r

1 212

1H H H

r H1 and H2 are the one electron Hamiltonians for He+

The Helium Hamiltonian and Wavefunctions

2 21 2

1 2 1 2

1 1 2 2 1

2 2H

r r r

The Helium Hamiltonian (Chapter 7) is:

In the ground state, both electrons are in 1s orbitals and thewavefunction can be written as:

1 1 1 2( ) ( ) 1 (1)1 (2)s sr r s s

We will assume that each 1s orbital is already normalized.

1 1 2 212

1( )H H r H r

r

1 1 1 2( ) ( ) 1 (1)1 (2)s sr r s s

The Helium Ground State Energy

1 2 1 2, ( , ) 1 (1 )1 ( 2 ) 1 (1 )1 ( 2 )G S

E r r H r r s s H s s

1 21 2

11 (1)1 ( 2 ) 1 (1)1 ( 2 )s s H H s s

r

1 2

12

1 (1)1 (2) 1 (1)1 (2) 1 (1)1 (2) 1 (1)1 (2)

11 (1)1 (2) 1 (1)1 (2)

s s H s s s s H s s

s s s sr

1 2

12

1 (2) 1 (2) 1 (1) 1 (1) 1 (1) 1 (1) 1 (2) 1 (2)

11 (1)1 (2) 1 (1)1 (2)

s s s H s s s s H s

s s s sr

1 21 2

11 (1) 1 (1) 1 ( 2 ) 1 ( 2 ) 1 (1)1 ( 2 ) 1 (1)1 ( 2 )

G SE s H s s H s s s s s

r

1 21 2

11 (1) 1 (1) 1 ( 2 ) 1 ( 2 ) 1 (1)1 ( 2 ) 1 (1)1 ( 2 )

G SE s H s s H s s s s s

r

1 1 1 1s s s sGSE J

11

1 1 21 (1 ) 1 (1 ) 1 ( 2 ) 1 ( 2 )s s H s s H s 1

This is the energy of an electron in the 1s orbital of a He+ ion.

2

1 11 2

11 (1)1 ( 2 ) 1 (1)1 ( 2 )s sJ s s s s

r2

J1s1s is called the Coulomb Integral.This is the total repulsion energy between the two 1s electrons.

We will compare this energy of ground state Helium withthe energy of excited state Helium in a later section.

Further comments on the Coulomb Integral

1 11 2

11 (1)1 ( 2 ) 1 (1)1 ( 2 )s sJ s s s s

r

Coulomb Integral: Electron-Electron Repulsion.

To better understand this integral, it is convenient to rewrite itin SI units with the traditional integral format.

2

*

1 1 1 20 12

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

eJ s s s s d r d r

r

2 22 2 21 2

1 20 12 0 12

[1 (1)] [1 (2 )][1 (1)] [1 (2 )]

4 4

e s d r e s d re s sd r d r

r r

1 2

0 124

dq dq

r

From this last equation, we see that the Coulomb Integral is reallyjust adding up the product of the two charges divided by thedistance between them over all possible volume elements.

Electron Spin

We’ve known since Freshman Chemistry or before that electrons havespins and there’s a spin quantum number (there actually are two).

Yet, we never mentioned electron spin, or the Pauli Exclusion Principle(actually the Pauli Antisymmetry Principle), in our treatment of ground state Helium in Chapter 7.

This is because Helium is a closed shell system.That is, its electrons fill the n=1 shell.

As we shall see, in open shell systems, such as the Lithium atom(1s22s1) or excited state Helium (e.g. 1s12s1), the electron’s spinand the Pauli Principle play an important role in determiningthe electronic energy.

A Brief Review of Orbital Angular Momentum in Hydrogen

, ( ) ( , )n l m n l l mψ ( r , θ φ ) R r Y The wavefunction for theelectron in a hydrogen atom is:

An electron moving about the nucleus in a hydrogen atom hasorbital angular momentum.

2 2( 1 )n l m n l mL ψ ψ ^

In addition to being eigenfunctions of the Hamiltonian (witheigenvalues En), the wavefunctions are eigenfunctions of the

angular momentum operators, L2 and Lz:^ ^

ˆz n lm n lmL ψ m ψ

2 2( 1 )L n l m n l m ^

ˆzL n lm m n lm

Shorthand

Do Electrons Spin??

I don’t know. I’ve never seen an electron up close and personal.

What can be said is that their magnetic properties are consistent withthe hypothesis that they behave “as though” they are spinning.

When a beam of electrons is directedthrough a magnetic field, they behavelike little magnets, with half of theirNorth poles parallel and half antiparallelto the magnetic field’s North pole.

Because a rotating charge is known tobehave like a magnet, the electronsare behaving as though they are spinningin one of two directions about theiraxes.

Spin Angular Momentum and Quantum Numbers

A rotating (or spinning) charge possesses angular momentum.

To characterize the spin angular momentum of an electron, two newquantum numbers are introduced, s and ms (analogous to l and ml), with s = ½ and ms = ½.

In direct analogy to orbital angular momentum, spin angular momentumoperators are introduced with the properties that:

ss m

The state of the electron is characterized by s and ms and is written as:

1 1 1 1

2 2 2 2or

2 2( 1 )s sS s m s s s m ^

ˆz s s sS s m m s m

and1 1 1 1 1 1 1 1 1 1ˆ ˆ2 2 2 2 2 2 2 2 2 2z zS S

2 2 2 21 1 3 1 1 1 1 3 1 1

2 2 4 2 2 2 2 4 2 2S S ^ ^

Because one always has s = ½, the standard shorthand is:

1 1 1 1

2 2 2 2and

1 1ˆ ˆ2 2z zS S

2 2 2 23 3

4 4S S ^ ^

Orthonormality of the Spin Wavefunctions

One can define integrals of the spin functions in analogy to integralsof spatial wavefunctions, keeping in mind that one is not really usingcalculus to evaluate integrals. Their values are defined below:

Therefore, by definition, the spin wavefunctions are orthonormal.

* 1d By definition

* 1d By definition

By definition* 0d By definition* 0d

The Pauli Principle

The Permutation Operator

By definition, this operator permutes (i.e. exchanges) two particles(usually electrons) in a wavefunction.

ˆ ( , ) ( , )i j i j j iP r r r r

For a 2 electron system: 1 2 1 2 2 1 1 2 1 2ˆ ( , ) ( , ) ( , )P r r r r p r r

This is an eigenvalue equation, with eigenvalue pij.

Permuting two identical particles will not change the probability

density:2 2ˆ ( , ) ( , )i j i j i j i jP r r p r r

Therefore: 1i jp

( , )ij i jp r r

The Pauli Principle

Postulate 6: All elementary particles have an intrinsic angular momentum called spin. There are two types of particles, with different permutation properties:

Electrons (s = ½) are fermions.

Therefore, electronic wavefunctions are antisymmetric with respectto electron exchange (permutation).

Fermions include electrons, protons, 3He nuclei, etc.

Bosons include 4He nuclei (s=0), 2H nuclei (s=1), etc.

Bosons: Integral spin (0, 1, 2,…) Pij() = +

Fermions: Half integral spin (1/2, 3/2,…) Pij() = -

Note that the permutation operator exchanges both the spatial andspin coordinates of the electrons.

Inclusion of Spin in Helium Atom WavefunctionsThe Hamiltonian for Helium does not contain any spin operators.Therefore, one can take the total wavefunction to be the productof spatial and spin parts.

If we use the approximation that the spatial part can be representedby 1s orbitals for each electron, then 4 possibilities for thetotal wavefunction are:

Shorthand Notation: 1 1 1 2( ) 1 (1 ) ( ) 1 ( 2 )s sr s r s

1 1 1 1 1 2 2( ) ( )s sr r

Electron 1 has spin. Electron 2 has spin.

1 21 (1 ) 1 ( 2 )s s

2 1 1 1 1 2 2( ) ( )s sr r

Electron 1 has spin.Electron 2 has spin.

1 21 (1 ) 1 ( 2 )s s

3 1 1 1 1 2 2( ) ( )s sr r


1 21 (1 ) 1 ( 2 )s s

4 1 1 1 1 2 2( ) ( )s sr r


1 21 (1 ) 1 ( 2 )s s

1 1 21 (1 ) 1 ( 2 )s s

2 1 21 (1 ) 1 ( 2 )s s

3 1 21 (1 ) 1 ( 2 )s s

4 1 21 (1 ) 1 ( 2 )s s

None of these 4 functions satisfies thePauli Antisymmetry Principle.

1 2 1 1 2 1 21 (1 ) 1 ( 2 )P P s s

1 2 3 1 2 1 21 (1 ) 1 ( 2 )P P s s

22 11 ( 2 ) 1 (1 )s s

1 2 2 1P Similarly:

2 11 ( 2 ) 1 (1 )s s 3

1 2 4 4P Similarly:

We can construct a linear combination of 1 and 2 that does satisfythe Pauli Principle.

A wavefunction that satisfies the Pauli Principle

1 2N

1 2 1 2 1 2P P N 2 1N Thus, is antisymmetric with respect to electron exchange, as requiredby the Pauli Principle.

1 2 1 21 (1 ) 1 ( 2 ) 1 (1 ) 1 ( 2 )N s s s s

Two electrons in an atom cannot have the same set of 4 quantumnumbers, n, l, ml and ms. That is, if two electrons have the samespatial part of the wavefunction (100 for both electrons in theHelium ground state), then they cannot have the same spin.

1 2 1 2 1 2 1 1 2 2 2 1P N N P P N

The wavefunction, , can be written as the product of aspatial and spin part:

Note: The sum of 1 and 2 would not be a satisfactory wavefunction.

1 2 4 4P Because 1 2 3 3P andneither of these functions can be used in theconstruction of an antisymmetric wavefunction

This is the basis for the more famous, but less general, formof the Pauli Principle, known as the Exclusion Principle:

1 2 1 21 (1 ) 1 ( 2 ) 1 (1 ) 1 ( 2 )N s s s s 1 2 1 21 (1)1 (2)N s s

1 1 21 (1 ) 1 ( 2 )s s

2 1 21 (1 ) 1 ( 2 )s s

3 1 21 (1 ) 1 ( 2 )s s

4 1 21 (1 ) 1 ( 2 )s s

Normalization of the Antisymmetric Wavefunction

1 1 1 2 1 2 1 2 1 2 1 2( ) ( ) 1 (1 )1 ( 2 )s sN r r N s s

1 * d We must integrate over both the spin andspatial parts of the wavefunction.

21 2 1 2 1 1 2 1 1 21 1 (1)*1 (1) 1 (2) *1 (2) *

N s s dr s s dr d d

=

1=

1

We assume that the individual spatial wavefunctions have alreadybeen normalized.

or

21 2 1 2 1 2 1 21 1 (1 ) 1 (1 ) 1 ( 2 ) 1 ( 2 )N s s s s

=

1

=

1

=

1

=

1

1 1 1 2 1 2 1 2

1( ) ( )

2s sr r 1 2 1 2

11 (1)1 (2)

2s s

21 2 1 2 1 2 1 21 1 (1 ) 1 (1 ) 1 ( 2 ) 1 ( 2 )N s s s s

21 2 1 2 1 2 1 2N

21 2 1 2 1 2 1 2 1 2 1 2 1 2 1 21 N

21 1 2 2 1 1 2 2 1 1 2 2 1 1 2 21 N

=0

=

0

=

0

=

0

=

1

=

1

21 2 N 1

2N

Spin and the Energy of Ground State Helium

Would its inclusion have affected the results?

1 1 1 2 1 2 1 2

1( ) ( )

2s sr r 1 2 1 2

11 (1)1 (2)

2s s

Earlier in this chapter, prior to reducing electron spin, we showedthat the energy of ground state helium is given by:

1 1 1 1s s s sG SE J

We will examine this question below.

E H 1 2 1 2 1 2 1 2

1 11 (1)1 (2) 1 (1)1 (2)2 2

s s H s s

The expression for the expectation value of the energy is given by:

1 2 1 2 1 2 1 2

1 11 (1)1 (2) 1 (1)1 (2)2 2

s s H s sE H

1 2 1 2 1 2 1 2

11 (1)1 (2) 1 (1)1 (2

2 E s s H s s

We can factor out the spin part of the wave functionbecause H is independent of spin

11 (1)1 (2) 1 (1)1 (2 2

2 E s s H s s 1 (1)1 (2) 1 (1)1 (2 s s H s s

Thus, inclusion of the spin portion of the wavefunction has noeffect on the computed energy in a closed shell system suchas ground state Helium.

Note: It can be shown that one arrives at the same conclusion ifa more sophisticated spatial function is used to characterize thetwo electrons.

Spin Angular Momentum of Ground State Helium

z-Component of Spin Angular Momentum

For a two electron system, the operator for Sz is

1 2ˆ ˆ ˆz z zS S S

Therefore 1 2 1 2 1 2

1ˆ ˆ ˆ2

z z zS S S

1 1 2 2 1 2 1 1 2 2 1 2

1ˆ ˆ ˆ ˆ ˆ2

z z z z zS S S S S

1 2 1 2 1 2 1 2

1 1 1 1 1ˆ2 2 2 22

zS

0

Therefore, the eigenvalue of Sz is 0. The z-component ofangular momentum is MS = 0.

^

Total Spin Angular Momentum

Therefore, we will just present the results.

The S2 operator for a two electron system and the calculation ofthe eigenvalue of this operator is significantly more complicatedthan the calculation of the z-component.

^

The result is 2 0S ^

Thus, for ground state Helium: S=0 and MS=0

We say that GS helium is a “singlet” because there is onlyone possible combination of S and MS (0 and 0).

This calculation requires application of spin raising and loweringoperators (introduced in various texts**), and is a digression from our prime focus.

**See for example, “Quantum Chemistry”, by I. N. Levine (5th. Ed.) Sect. 10.10

Generalization

Some possible combinations of S and MS that can be encounteredare given in the table below

In general, the spin wavefunctions of multielectron atoms are

eigenfunctions of S2 and Sz, with eigenvalues S(S+1)ħ2 and MSħ .^^

z S S SS SM M SM

2 2( 1)S SS SM S S SM ^

S MS Designation

0 0 Singlet

1/2 1/2, -1/2 Doublet

1 1, 0, -1 Triplet

3/2 3/2, 1/2, -1/2, -3/2 Quartet

The Wavefunctions of Excited State Helium

In ground state Helium, we were able to write the wavefunctionas the product of spatial and spin parts.

1 1 1 2 1 2 1 2

1( ) ( )

2s sr r

I have included the normalization constant with the spin function,which is what it is normalizing (it is assumed that the spatial partincludes its own normalization constant)

In ground state Helium, the spatial wavefunction is symmetricwith respect to electron exchange. Therefore, it is necessary forthe spin function to be antisymmetric with respect to exchangein order to satisfy the Pauli Principle.

1 2 1 2

11 (1)1 (2)

2s s

If one of the electrons is excited to the 2s orbital to give He(1s12s1),the spatial wavefunction can be either symmetric or antisymmetricwith respect to electron exchange, broadening the possibilities forvalid spin functions.

1 2( , ) 1 (1)2 (2) r r s sNeither nor 1 2( , ) 2 (1)1 (2)

r r s s

are valid spatial wavefunctions because they are neither symmnetricnor antisymmetric with respect to the exchange of the two electrons.

Symmetric and Antisymmetric Spatial Wavefunctions

However, one can “build” combinations of these wavefunctions that are either symmetric or antisymmetric with respect to electron exchange.

1 2

1( , ) 1 (1)2 (2) 2 (1)1 (2)

2

sym r r s s s s

Symmetric

1 2 1 2 1 2( , ) ( , )s y m s y mP r r r r

We have denoted this as a symmetric function, because it is easy

to show that:

Antisymmetric

For this function 12 1 2 1 2( , ) ( , )antisym antisymP r r r r

1 2

1( , ) 1 (1)2 (2) 2 (1)1 (2)

2

antisym r r s s s s

Symmetric and Antisymmetric Spin Wavefunctions

Two symmetric spin wavefunctions are: 12 and 12 because

1 2 1 2 1 2P 1 2 1 2 1 2P and

We could not use either of these symmetric spin functions forground state Helium because the symmetric spatial function requiredthat we must have an antisymmetric spin function to satisfythe Pauli Principle.

A third symmetric spin wavefunction is: 1 2 1 2

1

2

It is straightforward to apply the permutation operator, P12, to thisfunction to prove that it is symmetric with respect to exchange.

As shown when discussing ground state Helium, a spin wavefunctionthat is antisymmetric with respect to electron exchange is:

1 2 1 2

1

2

S and MS of the Spin Wavefunctions

1 2 1 2 1 2ˆ ˆ ˆz z zS S S

1 1 2 2 1 2ˆ ˆ

z zS S 1 2 1 2

1 1

2 2 1 21

Therefore, MS=+1 for 1 2

1 2 1 2

1

2 Similarly, MS=0 for

MS=-1 for 1 2

Therefore, S=1 for the 3 symmetric spin wavefunctions.

Together, these functions are a triplet with S=1 and MS=+1,0,-1.

Using advanced methods,** (you are not responsible for it), one can show that when the S2 operator is applied to any of the 3 symmetric spin functions, the eigenvalue is 2ħ2 [ = S(S+1) ħ2 ].

**e.g. Introduction to Quantum Mechanics in Chemistry, by M. A. Ratnerand G. C. Schatz, Sect. 8.3

When Sz operates on the antisymmetric spin function 1 2 1 2

1

2

one finds that MS=0.

It can be shown that when S2 operates on this function, theeigenvalue is 0. Therefore, S=0 for the antisymmetric spin function.

Therefore the antisymmetric spin wavefunction is a singlet, with S=0 and MS=0.

The Total Wavefunction for Excited State Helium

Spatial Wavefunctions Spin Wavefunctions

1 2

1( , ) 1 (1)2 (2) 2 (1)1 (2)

2sym r r s s s s

1 2

1( , ) 1 (1)2 (2) 2 (1)1 (2)

2antisym r r s s s s

001 2 1 2

1

2spin Singlet

101 2 1 2

1

2spin

111 2sp in

1 11 2sp in

Triplet

One can write the total wavefunction as the product of spin andspatial parts.

1 2 1 2( , ) ( , )SSMspa t sp inr r

Singlet Wavefunction

001 2( , )sym sp inr r

1 2 1 2

1 11 (1)2 (2) 2 (1)1 (2)

2 2Sing s s s s

Triplet Wavefunctions

11 2( , ) SM

an tisym sp inr r

1 1 2

11 (1)2 (2) 2 (1)1 (2)

2T s s s s

2 1 2 1 2

1 11 (1)2 (2) 2 (1)1 (2)

2 2T s s s s

3 1 2

11 (1)2 (2) 2 (1)1 (2)

2T s s s s

Excited State Helium Energies: He(1s12s1)

The expectation value for energy is given by:

1 2 1 2

1 2 1 2

*

*

H dr dr d dHE

dr dr d d

The Helium Hamiltonian is:

2 21 2

1 2 12

1 1 2 2 1

2 2H

r r r

KE(1) KE(2) PE(2)PE(1) PE(12)

2 21 2

1 2 12

1 2 1 2 1

2 2H

r r r

1 212

1H H H

r

H1 and H2 are the one electron Hamiltonians for He+

Triplet State Energy1 1

1 2 1 2 1 2 1 2

1 11 2 1 2 1 2 1 2

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , )

S S

S S

M Mantisym spin antisym spin

Trip M Mantisym spin antisym spin

r r H r rE

r r r r

Because the Hamiltonian does not contain any spin operators,the above expression can be simplified.

1 11 2 1 2 1 2 1 2

1 11 2 1 2 1 2 1 2

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , )

S S

S S

M Mspin spin antisym antisym

Trip M Mspin spin antisym antisym

r r H r rE

r r r r

1 2 1 2

1 2 1 2

( , ) ( , )

( , ) ( , )

antisym antisym

Tripantisym antisym

r r H r rE

r r r r

Note that the energy does not depend directly on the spinwavefunction.

It is the fact that the triplet state symmetric spin wavefunctionrequires us to use the antisymmetric spatial wavefunction thataffects the calculated energy.

1 2 1 2

1 2 1 2

( , ) ( , )

( , ) ( , )

antisym antisym

Tripantisym antisym

r r H r rE

r r r r

1 2 1 2( , ) ( , )an tisym an tisymr r H r r

We have assumed that the spatial wavefunction is normalized,in which case the denominator is 1.

1 2

1, 1 (1)2 (2) 2 (1)1 (2)

2

antisym r r s s s s

The energy can then be calculated from:

1 11 (1)2 (2) 2 (1)1 (2) 1 (1)2 (2) 2 (1)1 (2)

2 2tripE s s s s H s s s s

1 11 (1)2 (2) 1 (1)2 (2) 2 (1)1 (2) 2 (1)1 (2)

2 21 11 (1)2 (2) 2 (1)1 (2) 2 (1)1 (2) 1 (1)2 (2)

2 2

tripE s s H s s s s H s s

s s H s s s s H s s

1 2 3 4tr ipE I I I I

1 2 3 4tripE I I I I

1

11 (1)2 (2) 1 (1)2 (2)

2I s s H s s

1 1 212

1 1 1 11 (1) 1 (1) 2 (2) 2 (2) 1 (1)2 (2) 1 (1)2 (2)

2 2 2I s H s s H s s s s s

r

1 1 2

12

1 11 (1)2 (2) 1 (1)2 (2) 1 (1)2 (2) 1 (1)2 (2)

2 21 11 (1)2 (2) 1 (1)2 (2)

2

I s s H s s s s H s s

s s s sr

1 1 2

12

1 12 (2) 2 (2) 1 (1) 1 (1) 1 (1) 1 (1) 2 (2) 2 (2)

2 21 11 (1)2 (2) 1 (1)2 (2)

2

I s s s H s s s s H s

s s s sr

1 212

1 11 (1)2 (2) 1 (1)2 (2)

2s s H H s s

r

1 1 2 1 2

1 1 1

2 2 2s s s sI J where 1 212

11 (1)2 (2) 1 (1)2 (2)s sJ s s s s

r

Similarly,

2 1 212

1 12 (1)1 (2) 2 (1)1 (2)

2I s s H H s s

r

2 2 1 1 2

1 1 1

2 2 2s s s sI J

1 1 212

1 1 1 11 (1) 1 (1) 2 (2) 2 (2) 1 (1)2 (2) 1 (1)2 (2)

2 2 2I s H s s H s s s s s

r

1 1 212

1 11 (1)2 (2) 1 (1)2 (2)

2I s s H H s s

r

1 2 3 4tr ipE I I I I

3

11 (1)2 (2) 2 (1)1 (2)

2I s s H s s

3 1 2

12

1 12 (2) 1 (2) 1 (1) 2 (1) 1 (1) 2 (1) 2 (2) 1 (2)

2 21 11 (1)2 (2) 2 (1)1 (2)

2

I s s s H s s s s H s

s s s sr

3 1 212

1 11 (1)2 (2) 2 (1)1 (2)

2I s s H H s s

r

3 1 2

12

1 11 (1)2 (2) 2 (1)1 (2) 1 (1)2 (2) 2 (1)1 (2)

2 21 11 (1)2 (2) 2 (1)1 (2)

2

I s s H s s s s H s s

s s s sr

0||

0||

312

1 11 (1)2 (2) 2 (1)1 (2)

2I s s s s

r

3 1 212

1 11 (1)2 (2) 2 (1)1 (2)

2I s s H H s s

r


312

1 11 (1)2 (2) 2 (1)1 (2)

2I s s s s

r

3 1 2

1

2 s sI K where1 2

12

11 (1)2 (2) 2 (1)1 (2)s sK s s s s

r

Similarly,

4 1 212

1 12 (1)1 (2) 1 (1)2 (2)

2I s s H H s s

r

4 1 2

1

2 s sI K


1 2 1 2 2 1 1 2 1 2 1 2

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2s s s s s s s s s s s stripE J J K K

1 2 1 2 1 2s s s s s stripE J K

where 1 212

11 (1)2 (2) 1 (1)2 (2)s sJ s s s s

r

2 2

1 2 1 212

1 (1) 2 (2)s s

s sJ dr dr

r

1 212

11 (1)2 (2) 2 (1)1 (2)s sK s s s s

r

1 2 1 212

[1 (1)2 (1)] [1 (2)2 (2)]s s

s s s sK dr dr

r

1

1. Energy of electron in 1s He+ orbital

2

2. Energy of electron in 2s He+ orbital

3

3. Coulomb (repulsion) Integral

4

4. Exchange Integral

2 2

1 2 1 212

1 (1) 2 (2)s s

s sJ dr dr

r

1 2 1 212

[1 (1)2 (1)] [1 (2)2 (2)]s s

s s s sK dr dr

r

1 2 1 2 1 2s s s s s stripE J K

1 2 3 4

4. Exchange Integral

3. Coulomb (repulsion) Integral

The integrand of the Coulomb integral represents the repulsion oftwo infinitesimal electron densities, (1)=1s(1)2 and (2)=2s(2)2,separated by a distance, r12. The repulsion is summed over allinfinitesimal electron densities.

Arises purely from the antisymmetry of the spatial function with respectto electron exchange. It has no classical analog.

If the above calculation had been performed with a simple productwavefunction, spat = 1s(1)2s(2), there would be no exchange integral

Always positive

Usually positive

Singlet State Energy

Triplet: One of 3 components of the Triplet

2 1 2 1 2

1 11 (1)2 (2) 2 (1)1 (2)

2 2T s s s s

1 2 1 2( , ) ( , )antisym antisymtripE r r H r r

1 2 1 2 1 2s s s s s sJ K

Singlet: 1 2 1 2

1 11 (1)2 (2) 2 (1)1 (2)

2 2Sing s s s s

1 2 1 2 1 2s s s s s sSingE J K

1 2 1 2 1 2s s s s s sSingE J K

1 2 1 2

1 11 (1)2 (2) 2 (1)1 (2)

2 2Sing s s s s

2 1 2 1 2

1 11 (1)2 (2) 2 (1)1 (2)

2 2T s s s s

1 2 1 2 1 2s s s s s sTripE J K

1 212

11 (1)2 (2) 2 (1)1 (2)s sK s s s s

r

1 212

11 (1)2 (2) 1 (1)2 (2)s sJ s s s s

r

Because the exchange integral is almost always positive, theenergy of excited triplet state Helium is lower than that of theexcited state singlet.

The physical basis for the lower energy of the triplet is thatthe wavefunction (and therefore the probability) is small whenthe coordinates of the two electrons are close to each other.

Therefore, the electron-electron repulsion energy is minimized

Slide 1 Chapter 7 Multielectron Atoms Part B: Electron Spin and the Pauli Principle.

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Transcript of Slide 1 Chapter 7 Multielectron Atoms Part B: Electron Spin and the Pauli Principle.