Ideas of Quantum Chemistry P106 doc

If the origin is located at its centre as “most people would do”, then we have a quite simple de-scription of ρr by using the moments, namely, the only non-zero moment is the charge, i.e

A



r∈A



B



s∈B

PrsSrs

Afterwards we may choose the following partitionings:

Atomic partitioning:

N=

A

qA qA=

r ∈A

 

B



s ∈B

PrsSrs



where q are called the Mulliken charges They are often calculated in practical

ap-Mulliken

charges plications and serve to provide information on how much of the electronic

den-sity ρ is concentrated on atom A Such a quantity is of interest because it may

be directly linked to the reactivity of atom A, often identified with its ability to

be attacked by nucleophilic or electrophilic agents.2 Also, if we measure the di-pole moment, we would like to know why this moment is large in a molecule By performing Mulliken analysis, we might be able to identify those atoms that are responsible for this This might be of value when interpreting experimental data

Atomic and bond partitioning: The summation may also be performed in a slightly different way

N=

A



r s∈A

PrsSrs+

A<B

2

r∈A



s∈B

PrsSrs=

A

¯qA+ 

A<B

¯qAB

The first term represents the contributions ¯qAof the atoms, the second pertains

to the atomic pairs ¯qAB

The latter populations are large and positive for those pairs of atoms for which chemists assign chemical bonds

The bond population ¯qABmay be treated as a measure of whether in the A− B atomic interaction, bonding or antibonding character prevails.3If, for two atoms,

¯qAB< 0, we may say that they are not bound by any chemical bond, if¯qABis large, then we may treat it as an indication that these two atoms are bound by a chemical bond or bonds

2 We have to remember that, besides electrons, this atom has a nucleus This has to be taken into account when calculating the atomic net charge.

3 Prs is the sum (over the occupied orbitals) of the products of the LCAO coefficients of two atoms

in each of the occupied molecular orbitals The equal signs of these coefficients (with Srs > 0) means a

bonding interaction (recall Chapter 8 and Appendix R on p 1009) and such a contribution increases Prs.

The opposite signs of the coefficients (with Srs> 0) corresponds to the antibonding interactions and

in such a case the corresponding contribution decreases Prs If Srs < 0, then the words “bonding” and “antibonding” above have to be exchanged, but the effect remains the same This means that the product PrsSrs in all cases correctly controls the bonding (PrsSrs > 0) or antibonding (Prs Srs < 0) effects.

S POPULATION ANALYSIS 1017

Example 1 The hydrogen molecule. Let us take the simplest example First, let us

consider the electronic ground-state in the simplest molecular orbital

approxima-tion, i.e two electrons are described by the normalized orbital in the form (a b

denote the 1s atomic orbitals centred on the corresponding nuclei; note that this is

the famous bonding orbital)

ϕ1= N1(a+ b) where N1= (2 + 2S)− 1

, and S≡ (a|b) Then,

Psr=

i

2c∗ricsi= 2c∗r1cs1= (1 + S)−1 independent of the indices r and s Of course,

S=



1 S

S 1

 and therefore P S=



1 1

1 1





Thus, Tr (P S)= 2 = the number of electrons = P11S11+ P22S22+ 2P12S12= qA+

qB+qAB, with qA= qB= (1+S)−1, and q

AB= 2S 1+S > 0 Thus we immediately see

that the HH bond has an electronic population greater than zero, i.e the atom–atom

interaction is bonding.

Let us now consider H2with two electrons occupying the normalized orbital of

a different character4ϕ2= N2(a− b), with N2= (2 − 2S)− 1

, then

Psr=

i

2cri∗csi= 2cr2∗cs2= (1 − S)−1

for (r s)= (1 1) and (r s) = (2 2) while Prs= −(1 − S)−1for (r s)= (1 2) and

(r s)= (2 1)

Now, let us calculate

P S=



1 −1

−1 1



and Tr(P S)= 2 = the number of electrons = P11S11+ P22S22+ 2P12S12= qA+

qB+ qAB, but now qA= qB= (1 − S)−1and q

AB= − 2S 1−S < 0 Thus, a glance at

qABtells us that this time the atoms are interacting in an antibonding way

A similar analysis for polyatomic molecules gives more subtle and more

inter-esting results

Other population analyses

Partitioning of the electron cloud of N electrons according to Mulliken population

analysis represents only one of possible choices For a positively definite matrix5S

(and the overlap matrix is always positively definite) we may introduce the powers

4 We do not want to suggest anything, but this orbital is notorious for antibonding character.

5 I.e all the eigenvalues positive.

of the matrix6Sx, where x is an arbitrary real number (in a way shown in Appen-dix J on p 977), and we have S1−xSx= S Then we may write7

N= Tr(PS) = TrSxP S1−x

Now, we may take any x and for this value construct the corresponding partition

of N electronic charges into atoms If x= 0 or 1, then we have a Mulliken popula-tion analysis, if x=1

2 then we have what is called the Löwdin population analysis,

Löwdin

population

analysis etc

Multipole representation

Imagine a charge distribution ρ(r) Let us choose a Cartesian coordinate system

We may calculate the Cartesian moments of the distribution:

ρ(r) dV , i.e the total charge, then

xρ(r) dV , 

yρ(r) dV ,

zρ(r) dV , i.e the components of the dipole moment,

x2ρ(r) dV ,

y2ρ(r) dV ,

z2ρ(r) dV ,

xyρ(r) dV ,

xzρ(r) dV ,



yzρ(r) dV – the components of the quadrupole moment, etc The moments mean

a complete description of ρ(r) as concerns its interaction with another (distant) charge distribution The higher the powers of x y z (i.e the higher the moment) the more important distant parts of ρ(r) are If ρ(r) extends to infinity (and for atoms and molecules it does), higher order moments tend to infinity This means trouble when the consecutive interactions of the multipole moments are calcu-lated (multipole expansion, Appendix X) and indeed, the multipole expansion

“explodes”, i.e diverges.8This would not happen if the interacting charge distrib-utions could be enclosed in two spheres

There is also another problem: where to locate the origin of the coordinate

sys-tem, with respect to which the moments are calculated? The answer is: anywhere.

Wherever such an origin is located it is OK from the point of view of mathematics However, such choices may differ enormously from the practical point of view For example, let us imagine a spherically symmetric charge distribution If the origin is located at its centre (as “most people would do”), then we have a quite simple de-scription of ρ(r) by using the moments, namely, the only non-zero moment is the charge, i.e.

ρ(r) dV If, however, the origin is located off centre, all the moments would be non-zero They are all needed to calculate accurately the interaction of the charge distribution (with anything) As we can see, it is definitely better to lo-cate the origin at the centre of ρ(r)

Well, and what if the charge distribution ρ(r) were divided into segments and each segment represented by a set of multipoles? It would be all right, although, in view of the above example, it would be better to locate the corresponding origins

at the centre of the segments It is clear that, in particular, it would be OK if the segments were very small, e.g., the cloud was cut into tiny cubes and we consider

6 They are symmetric matrices as well.

7 We easily check that Tr(ABC) = Tr(CAB) Indeed, Tr(ABC) =i k lAikBklCli, while Tr(CAB) =



i k l CikAklBli Changing summation indices k → i, l → k, i → l in the last formula, we obtain Tr(ABC).

8 Although the first terms (i.e before the “explosion”) may give accurate results.

S POPULATION ANALYSIS 1019

every cube’s content as a separate cloud.9But, well , what are the multipoles for?

Indeed, it would be sufficient to take only the charges of the cubes, because they

approximate the original charge distribution In this situation higher multipoles

would certainly be irrelevant! Thus we have two extreme cases:

• a single origin and an infinite number of multipoles,

• or an infinite number of centres and monopoles (charges) only

We see that when the origins are located on atoms, we have an intermediary

situation and it might be sufficient to have a few multipoles per atom.10 This is

what the concept of what is called the cumulative multipole moments is all about cumulative

multipole moments

(CAMM11) Besides the isotropic atomic charges qa= M(000)

a calculated in an ar-bitrary population analysis, we have, in addition, higher multipoles Ma(klm)(atomic

dipoles, quadrupoles, octupoles, etc.) representing the anisotropy of the atomic

charge distribution (i.e they describe the deviations of the atomic charge

distribu-tions from the spherical)

Ma(klm)= Zaxkayalzam−

r∈a



s

Dsr

rxkylzms

k  k



l  l



m  m (k  l m ) =(k l m)

 k

k

  l

l

  m

m



× xk −k 

a yl−l 

a zm−m 

a · Mk lm

where Ma(klm) is the multipole moment of the “klm” order with respect to the

Cartesian coordinates x y z located on atom a, Ma(000) stands for the atomic

charge, e.g., from the Mulliken population analysis, Za is the nuclear charge of

the atom a, (r|xkylzm|s) stands for the one-electron integral of the corresponding

multipole moment, and Dsrχ∗

rχsrepresents the electronic density contribution re-lated to AO’s: χsand χrand calculated by any method (LCAO MO SCF, CI, MP2,

DFT, etc.) We may also use multipole moments expressed by spherical harmonic

functions as proposed by Stone.12

9 The clouds might eventually overlap.

10 If the clouds overlapped, the description of each centre by an infinite number of multipoles would

lead to a redundancy (“overcompleteness”) I do not know of any trouble of that kind, but in my opinion

trouble would come if the number of origins were large This is in full analogy with the

overcomplete-ness of the LCAO expansion These two examples differ by a secondary feature: in the LCAO, instead

of moments, we have the s, p, d, orbitals, i.e some moments multiplied by exponential functions.

11W.A Sokalski and R Poirier, Chem Phys Lett 98 (1983) 86; W.A Sokalski, A Sawaryn, J Chem.

Phys 87 (1987) 526.

12A.J Stone, Chem Phys Lett 83 (1981) 233; A.J Stone, M Alderton, Mol Phys 56 (1985) 1047.

T THE DIPOLE MOMENT OF A LONE

ELECTRON PAIR

The electronic lone pairs play an important role in intermolecular interactions

In particular, a lone pair protruding in space towards its partner has a large dipole moment,1which may interact electrostatically with its partner’s multipole moments (see Appendix X, p 1038) Let us see how the dipole moment depends on the atom

to which it belongs and on the type of hybridization

Suppose the electronic lone pair is described by the normalized hybrid

h=' 1

1+ λ2

 (2s)+ λ(2px)

with the normalized 2s and 2px atomic orbitals The coefficient λ may change from −∞ to +∞ giving a different degree of hybridization Fig T.1 shows for comparison two series of the hybrids: for carbon and fluorine atoms If λ= 0, we have the pure 2s orbital, if λ= ±∞ we obtain the pure ±2pxorbital

The dipole moment of a single electron described by h is calculated2as (N=

1

√

1+λ 2):

μx= h|−x|h = −N2

2s|x|2s + λ22px|x|2px + 2λ2s|x|2px

μy= μz= 0

where x stands for the x coordinate of the electron

The first two integrals equal zero, because the integrand represents an odd func-tion3with respect to the reflection in the plane x= 0 As a result

μx= −N22λ2s|x|2px

We will limit ourselves to λ 0, which means we are considering hybrids pro-truding to the right-hand side4as in Fig T.1, and since2s|x|2px > 0 then μx

The negative sign stresses the fact that a negative electron is displaced to the right-hand side (positive x)

1 Calculated with respect to the nucleus; a large dipole moment means here a large length of the dipole moment vector.

2 Atomic units have been used throughout, and therefore μ is expressed in a.u.

3 Please recall that the orbital 2px represents a spherically symmetric factor multiplied by x.

4 The hybrids with λ < 0 differ only by protruding to the left-hand side.

1020

T THE DIPOLE MOMENT OF A LONE ELECTRON PAIR 1021

Fig T.1.The length of the dipole moment vector μlone(in a.u.) as a function of the mixing parameter

λ for carbon (upper curve) and fluorine (lower curve) atoms The figure shows the shape of different

hybrids h = √ 1

1+λ 2 [(2s)+λ(2px)] which correspond to various mixing of the 2s and 2px carbon Slater

orbitals (with exponential factor ζ = 1625) and fluorine orbitals (ζ = 260); from the left: λ = 0, λ = 1

(sp), λ = 141 (sp 2 ), λ = 173 (sp 3 ), λ = 1000 All the hybrids are shown in square windows of 10 a.u.

The fluorine orbitals are more compact due to the larger charge of the nucleus A hybrid orbital which

corresponds to λ < 0 looks exactly like that with λ  = −λ, except it is reflected with respect to the yz

plane The maximum dipole moment corresponds to the sp hybridization.

To calculate2s|x|2px we need to specify the atomic orbitals 2s and 2p As the

2s and 2p atomic orbitals, let us take Slater type orbitals:

2s= Nr exp(−ζr) 2px= Nx exp(−ζr) where the exponential factor ζ (the same for both orbitals) is calculated using

simple rules for building the Slater orbitals, see p 355

Using the integral

 ∞

0

xnexp(−αx) dx = n!α−(n+1)

we obtain the normalization constants N= ζ2 ζ

3π and N= ζ2 ζ

π The contri-bution of two electrons (“lone electron pair”) to the dipole moment is, therefore,

equal to

μlone= 2μx= −N2|λ|(2s|xpx)= −2N2NN(2λ)

rx2exp(−2ζr) dv

= −2N2NN2λ

r3x2exp(−2ζr) sin θ dr dθ dφ

= −2N2NN2λ ∞

0

dr r5exp(−2ζr)

 π

0

sin3θ dθ

 2π

0

cos2φ dφ

= −2N2NN2λ 5!

(2ζ)6

4

3π= − 4λ

(1+ λ2)ζ

2

( ζ 3πζ

2

( ζ π

5!

(2ζ)6

4

3π

= − λ

1+ λ2

10

ζ√

3 THE DIPOLE MOMENT OF A LONE PAIR μlone= − λ

1 +λ 2

10

ζ √

3 The dipole moment at λ= 0, i.e for the pure 2s orbital, is equal to 0, for λ = ∞, i.e for the pure 2px orbital it is also equal 0 It is interesting to see for which hybridization the length of dipole moment has a maximum We easily find

∂|μlone|

∂λ = 10

ζ√ 3

(1+ λ2)− 2λ2

(1+ λ2) = 0 which gives λ= ±1, independently of ζ

Thus the maximum dipole moment is at the 1: 1 mixing of 2s and 2p, i.e for digonal hybridization (for any element), Fig T.1

From Table T.1 it is seen that the dipole moment of a lone pair strongly depends

on the chemical element,5and to a lesser extent on hybridization

Table T.1. The length of the dipole moments μlone(a.u.) corresponding to doubly occu-pied hybrid atomic orbitals The orbital exponents of 2s and 2p STO’s are identical and calculated using the rules given by Slater: ζC= 1625, ζN = 195, ζO = 2275, ζF = 260 Atom Digonal λ = 1 Trigonal λ =√2 Tetrahedral λ =√3

5 From the practical point of view, it is probably most important to compare nitrogen and oxygen lone pairs Thus, coordination of a cation by amines should correspond to a stronger interaction than that

by hydroxyl groups.

U SECOND QUANTIZATION

When we work with a basis set composed of Slater determinants we are usually confronted with a large number of matrix elements involving one- and two-electron operators The Slater–Condon rules (Appendix M) are doing the job to express these matrix elements by the one-electron and two-electron integrals However, we may introduce an even easier tool called second quantization, which is equivalent

to the Slater–Condon rules

The vacuum state

In the second quantization formalism we introduce a reference state for the system under study, which is a Slater determinant (usually the Hartree–Fock wave func-tion) composed of N orthonormal spinorbitals, where N is the number of elec-trons This function will be denoted in short by 0or in a more detailed way by

N(n1 n2    n∞) The latter notation means that we have to do with a normal-ized N electron Slater determinant, and in parenthesis we give the occupancy list (ni= 0 1) for the infinite number of orthonormal spinorbitals considered in the basis set and listed one by one in the parenthesis This simply means that some spinorbitals are present in the determinant (they have ni= 1), while others are absent1 (ni= 0) Hence, ini= N The reference state is often called the vac-uum state The subscript 0 in 0 means that we are going to consider a

single-determinant approximation to the ground state Besides the reference state, some

normalized Slater determinants of the excited states will be considered, with other

occupancies, including those corresponding to the number of electrons which differs from N.

The creation and annihilation of electrons

Let us make quite a strange move, and consider operators that change the number

of electrons in the system To this end, let us define the creation operator2 ˆk† of the electron going to occupy spinorbital k and the annihilation operator ˆk of an electron leaving spinorbital k:

1 For example, the symbol 2(001000100000   ) means a normalized Slater determinant of dimen-sion 2, containing the spinorbitals 3 and 7 The symbol 2(001000   ) is a nonsense, because the num-ber of “ones” has to be equal to 2, etc.

2 The domain of the operators represents the space spanned by the Slater determinants built of spinor-bitals.

Richard Feynman, in one of his books, says jokingly that he could not understand the very sense of the operators If we annihilate or create an electron, then what about the system’s electroneutrality? Happily enough, these operators will always act in creator–annihilator pairs.

1023

CREATION AND ANNIHILATION OPERATORS

ˆk†N(   nk  )= θk(1− nk)N+1(   1

k   ) ˆkN(   nk  )= θknkN−1(   0

k   ) where θk= (−1)$j<kn j

The symbol 1k means that the spinorbital k is present in the Slater determi-nant, while 0kmeans that this spinorbital is empty, i.e is not present in the Slater determinant The factors (1− nk) and nk ensure an important property of these operators, namely that

any attempt at creating an electron on an already occupied spinorbital gives zero, similarly any attempt at annihilating an empty spinorbital also gives

zero

It can be easily shown,3that (as the symbol suggests) ˆk†is simply the adjoint operator with respect to ˆk

The above operators have the following properties that make them equivalent

to the Slater–Condon rules:

ANTICOMMUTATION RULES

 ˆk ˆl

+= 0

 ˆk† ˆl†

+= 0

 ˆk† ˆl

+= δkl where the symbol[ ˆA ˆB]+= ˆA ˆB+ ˆB ˆA is called the anticommutator.4It is simpler

anticommutator

than the Slater–Condon rules, isn’t it? Let us check the rule [ ˆk† ˆl]+= δkl We have to check how it works for all possible occupancies of the spinorbitals k and l, (nk nl): (0 0), (0 1), (1 0) and (1 1)

Case: (nk nl)= (0 0)

 ˆk† ˆl

+N(   0k   0l  )= ˆk†ˆl+ ˆlˆk†

N(   0k   0l  )

= ˆk†ˆlN(   0k   0l  )+ ˆlˆk†N(   0k   0l  )

3Proof Let us take two Slater determinants a=  N+1(   1

k   ) and b=  N (   0k  ), in both

of them the occupancies of all other spinorbitals are identical Let us write the normalization condition for bin the following way: 1 = b|θk ˆka = θkb| ˆka = θk ˆk #  b|a, where as ˆk # has been denoted the operator adjoint to ˆ k, θkappeared in order to compensate for (θ2k= 1) the θk produced by the annihilator On the other hand, from the normalization condition of a we see that 1 = a|a =

θ k ˆk †  b|a Hence, θk ˆk #  b|a = θk ˆk †  b|a or ˆk # = ˆk † , This is what we wanted to show.

4 The above formulae are valid under the (common) assumption that the spinorbitals are orthonor-mal If this assumption is not true, only the last anticommutator changes to the form [ ˆk † ˆl]+ = Skl , where S stands for the overlap integral of spinorbitals k and l.

U SECOND QUANTIZATION 1025

= 0 + ˆlθkN+1(   1

k   0l  )

= θkˆlN +1(   1

k   0l  )

= θkδklθkN(   0k  )

= δklN(   0k  )

So far so good

Case: (nk nl)= (0 1)

 ˆk† ˆl

+N(   0k   1l  )= ˆk†ˆl+ ˆlˆk†

N(   0k   1l  )

= ˆk†ˆlN(   0k   1l  )+ ˆlˆk†N(   0k   1l  )

= θkθlN(   1k   0l  )− θkθlN(   1k   0l  )

= δklN(   0k   1l  )

This is what we expected.5

Case: (nk nl)= (1 0)

 ˆk† ˆl

+N(   1k   0l  )= ˆk†ˆl+ ˆlˆk†

N(   1k   0l  )

= ˆk†ˆlN(   1k   0l  )+ ˆlˆk†N(   1k   0l  )

= (0 + 0)N(   1k   0l  )

= δklN(   1k   0l  )

This is OK

Case: (nk nl)= (1 1)

 ˆk† ˆl

+N(   1k   1l  )= ˆk†ˆl+ ˆlˆk†

N(   1k   1l  )

= ˆk†ˆlN(   1k   1l  )+ ˆlˆk†N(   1k   1l  )

= ˆk†ˆlN(   1k   1l  )+ 0

= θ2

kδklN(   1k   1l  )

= δklN(   1k  )

The formula has been proved

Operators in the second quantization

Creation and annihilation operators may be used to represent one- and

two-electron operators.6The resulting matrix elements with Slater determinants

cor-respond exactly to the Slater–Condon rules (see Appendix M, p 986).

5 What decided is the change of sign (due to θk) when the order of the operators has changed.

6 The original operator and its representation in the language of the second quantization are not

identical in practical applications The second ones can act only on the Slater determinants or their

combinations Since we are going to work with the creation and annihilation operators in only those

methods which use Slater determinants (CI, MC SCF, etc.), the difference is irrelevant.

Fig T.1.The length of the dipole moment vector μlone(in a.u.) as a function of the... of ζ

Thus the maximum dipole moment is at the 1: mixing of 2s and 2p, i.e for digonal hybridization (for any element), Fig T.1

From Table T.1 it is seen that the dipole moment of. ..

2 The domain of the operators represents the space spanned by the Slater determinants built of spinor-bitals.

Richard Feynman, in one of his books, says jokingly