A new weak field coupling scheme for complexes and application of complete d³ energy level calculations to some tetragon...

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Title:
A new weak field coupling scheme for complexes and application of complete d³ energy level calculations to some tetragonal chromium (III) complexes
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vi, 302 leaves : ill. ; 28 cm.
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English
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Collins, John Anthony, 1950-
Publication Date:

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Subjects / Keywords:
Ligand field theory   ( lcsh )
Crystal field theory   ( lcsh )
Chromium compounds   ( lcsh )
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bibliography   ( marcgt )
theses   ( marcgt )
non-fiction   ( marcgt )

Notes

Thesis:
Thesis (Ph. D.)--University of Florida, 1982.
Bibliography:
Includes bibliographical references (leaves 300-301).
Statement of Responsibility:
by John Anthony Collins.
General Note:
Typescript.
General Note:
Vita.

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Full Text






A NEW WEAK FIELD COUPLING SCHEME FOR COMPLEXES
AND APPLICATION OF COMPLETE d3 ENERGY
LEVEL CALCULATIONS TO SOME TETRAGONAL
CHROMIUM (III) COMPLEXES






BY

JOHN ANTHONY COLLINS


A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL
OF THE UNIVERSITY OF FLORIDA IN
PARTIAL FULFILLMENT OF THE REQUIREMENTS
FOR THE DEGREE OF DOCTOR OF PHILOSOPHY



UNIVERSITY OF FLORIDA


1982





























I dedicate this work to Jesus of Nazareth.












ACKNOWLEDGMENTS


With sincere gratitude I acknowledge my advisor, Professor

Jayarama R. Perumareddi, whose guidance, patience and encouragement

made this work possible. I owe a special debt of gratitude to

those who typed this manuscript: Mrs. Rita Levy, Mrs. Bobbi Rice

and, especially, Mrs. Joyce Phillips. I would like to thank

Dr. Timothy Cotton for his help with the computer programming.

Most importantly, I would like to thank my Mother and Father for

their encouragement, prayers, love and support which sustained me

throughout this endeavor.












TABLE OF CONTENTS


ACKNOWLEDGMENTS . . .

ABSTRACT . . .

CHAPTER

I. INTRODUCTION . . .

II. THE WEAK FIELD COUPLING SCHEMES .

The Interelectronic Repulsions-Free Ion Terms
The {L,S,XC,xXQ,r} Coupline Scheme .
The {L,S,XCTC,IQ} Coupling Scheme .
Checking of the Weak Field Perturbation
Matrices . . .

III. THE STRONG FIELD COUPLING SCHEME .

The Cubic Ligand Field .. .. ...
The Interelectronic Repulsions .
The Spin-Orbit Interaction . .
The Axial Ligand Field ...........
Checking of the Strong Field Perturbation
Matrices . . .

IV. THE PERCENT PURITIES OF THE EIGENFUNCTIONS IN
THE VARIOUS COUPLING SCHEMES . .

The Percent Purities of the Cubic Levels .
The Percent Purities of the Tetragonal Levels

V. APPLICATIONS TO SOME TETRAGONAL CHROMIUM (III)
COMPLEXES . . .

Trans-[Cr(PDC) . .
fCr(NH3)5(NCO)jI2
ICr(NH3)5(NCO)]+2 . .
Summary and Discussion . .

BIBLIOGRAPHY . . .

BIOGRAPHICAL SKETCH . . .


. iii

V



S 1

. 17

17
. 47
S. 146

S. 188

204

. 204
. 207
. 223
. 247

. 247


256

257
269


. 282

S 292
S 295
297

300

302


. .












Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy



A NEW WEAK FIELD COUPLING SCHEME FOR COMPLEXES
AND APPLICATION OF COMPLETE d3 ENERGY
LEVEL CALCULATIONS TO SOME TETRAGONAL
CHROMIUM (III) COMPLEXES


By

John A. Collins

December, 1982

Chairman: Jayarama R. Perumareddi
Major Department: Chemistry

In the well-known weak field sequence of perturbations,

which is called {L,S,J} coupling scheme, one considers the

spin-orbit interaction immediately after the interelectronic

repulsions but prior to the ligand field potential. For 3dn

systems the spin-orbit interaction is not large relative to

the ligand field potential. Therefore, it is of value to

obtain energy levels in a coupling scheme, {L,S,X}, which

considers the ligand field potential prior to the spin-orbit

interaction. We have devised this new {L,S,X} sequence,

for both cubic and tetragonal symmetry, for the case of d3

(and d7) electronic configuration. We have found that the

{L,S,X} sequence yields purer eigenfunctions than does the

{L,S,J} sequence. We have also compared the {L,S,X} weak








field sequence with the well-known strong field sequence. To

expand this comparison we have devised an additional tetragonal

strong field coupling scheme. We find in the cubic limit, for

both d3 and d7, that the strong field sequence yields purer

eigenfunctions than the comparable weak field sequence. We

find in the tetragonal limit that the strong field sequence

yields purer eigenfunctions for d3 but for d7 in a strong

tetragonal field it is the {L,S,X} sequence which yields the

purest eigenfunctions. In general, we find that considering

the spin-orbit interaction as the last perturbation in the

sequence yields purer eigenfunctions than would be obtained if

the spin-orbit interaction were considered earlier in the

sequence. Lastly, we have, using complete energy level calcula-

tions, fitted the spectra of some tetragonal chromium (III)

complexes in which the five lowest energy intraconfigurational

doublets have been identified.












CHAPTER I


INTRODUCTION


Crystal field theory attempts to rationalize the spectral and

magnetic properties of transition metal compounds with a simple

electrostatic model. In this model the ligands, i.e., the entities

directly bound to the central ion, are considered to be point charges

or point dipoles fixed in space about the central transition metal

ion. The electric field generated by the ligands (henceforth "the

ligand field") gives rise to a splitting of the d-orbitals of the

metal ion. This splitting of the electronic energy levels of the

free ion upon complex formation can be determined qualitatively by

symmetry arguments alone, as has been shown by Bethe.1

An important application of crystal field theory is in the

interpretation of the near infrared, visible and near ultraviolet

absorption spectra of coordination compounds. These low intensity

bands are now generally recognized as arising due to electronic trans-

itions between the no longer degenerate d-orbitals.

When quantitative calculations of the electronic energy levels

are undertaken with the ligands approximated as point charges or

point dipoles very poor agreement with experimental data is obtained.2


1H. Bethe, Ann. Physik., [5], 3, 807(1935)

2S. Sugano, Y. Tanabe and H. Kamimura, Multiplets of Transition-
Metal Ions In Crystals, Academic Press, New York, 1970.

1








This is not surprising since the simple electrostatic theory neglects

all covalent bonding and attempts to describe the many electron and,

often, the many atom ligands as just point charges or dipoles.

Molecular orbital theory, which takes account of ligand-metal

orbital overlap, gives probably the best description of bonding in

transition metal compounds. Unfortunately, quantitative molecular

orbital calculations have to be carried out for virtually every

single system and it is difficult to generalize the results for a

series of systems.

The hybrid of the best points of crystal field theory and molecu-
2
lar orbital theory is found in modern ligand field theory. Qualita-

tive aspects of the crystal field theory and the molecular orbital

theory can be obtained on the basis of symmetry arguments alone.

Ligand field theory incorporates the symmetry aspects of both the

theories but rather than attempting to calculate the energy differences

rigorously, they are taken as adjustable parameters. These ligand

field parameters are then obtained by fitting the absorption spectra

of each complex with the calculated parametric energy levels. Ligand

field theory is thus the parametric form of crystal field theory or

of molecular orbital theory.

The Hamiltonian for a free ion, many-electron system is given

below


J. P. Dahl and C. J. Ballhausen, Advan. Quantum Chem., 4, 170
(1968) and references cited therein.

C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw-
Hill Book Co., New York, 1962.










-h2 v2 h 2 ze2
8T 2M 8T2m i r.
(1)
2
+ I T + 7 (ri- s
+ e + (ri)i'*si,
i
where h is Planck's constant,

M is the mass of the nucleus,

2 2 2
2 d 3 9
2 is the Laplacian, i.e., 2 + 2
3x y y 3z

for the nucleus,

m is the mass of an electron,

V. is the Laplacian for the i electron,


ze is the effective nuclear charge,

-e is the charge of a single electron,

ri is the distance between the nucleus and the i- electron,


rij is the distance between the ih- and
th
j electrons and

+ ->
((ri).is.i is the term in the Hamiltonian due to the inter-

action of the magnetic fields generated by the spin and

orbital angular moment of the electron, this is called

the "spin-orbit interaction."








The Hamiltonian for a free ion, many-electron system can be

rewritten as

2
H = Ho + e + E(ri).*s- (2)
i

where Ho is a sum of one electron and free ion Hamiltonians and

the terms which arise due to the interelectronic repulsions and the

spin-orbit interaction are considered to be perturbations.

When a free ion forms a complex, a fourth term is added to the

Hamiltonian: the ligand field potential (VLF). This potential is

due to the coulombic repulsions between the central ion's d electron

and the ligands. The ligand field potential has the form

qke
VLF(Xi) r (3)
k ik




th
where xi is the coordinate of the i-t electron,


-qk is the magnitude of the charge on the kth ligand, and


r.ik is the distance between the ith- electron and the

kth ligand.

There are now three perturbations to be added on to the single

electron Hamiltonian: the interelectronic repulsions, the spin-orbit

interaction, and the ligand field potential.

The order in which the perturbations are considered will follow

the decreasing order of the magnitude credited to each effect. This

is reflected in the secular determinant in the extent to which each

perturbation is diagonalized.









In the strong field approach we consider the ligand field

potential first. In (j,j) coupling we carry out the spin-orbit

interaction as the first perturbation. Lastly, in the weak field

procedure we consider the interelectronic repulsions to be the first

perturbation.

For transition metal complexes, especially for first row tran-

sition metal complexes, the spin-orbit interaction is always a minor

perturbation relative to the interelectronic repulsions and the

ligand field potential.1 Therefore, the (j,j) coupling approach,

which implies that the spin-orbit interaction is the major perturba-

tion, is inappropriate for transition metal complexes (especially

first row transition metal complexes). For this reason, we will be

concerned only with the weak field and the strong field approaches.

After specifying which perturbation will be taken first,we must

still decide the order of the remaining perturbations. Clearly,

there are two possibilities for each of the three approaches we have

just described. Each particular sequence of perturbations is known

as a coupling scheme or a representation.

When complete configuration interaction is included all coupling

schemes for a given electronic configuration and symmetry will yield

the same set of eigenvalues. However, how apt the description of the

eigenfunctions is, in terms of the basis eigenvectors of the coupling

scheme, will be dependent on the sequence of perturbations chosen.

Thus, the choice of an appropriate coupling scheme is important


C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw-
Hill Book Co., New York, 962.








because only then can reasonable parentages be ascribed to the various

levels and a reasonable approximation to the true eigenfunction be

gained from the basis eigenvectors. Also, if configuration interac-

tion is neglected, the various coupling schemes for a given electronic

configuration and symmetry will no longer yield identical sets of

eigenvalues. With neglect of configuration interaction only an

appropriate coupling scheme (i.e., a coupling scheme in which the
sequence of perturbations reflects the relative magnitudes of the

perturbations) will yield a reasonable approximation to the true set

of eigenvalues.

Thus, for the octahedral (Oh) case there are two possible weak

field and two possible strong field coupling schemes. These coupling

schemes can be represented as shown in Table 1. Henceforth, we will

refer to the octahedral case as the cubic case.

To date only the {L,S,J} sequence of perturbations has been

used in complete weak field calculations.1 The {L,S,J} sequence




1A. D. Liehr, J. Phys. Chem., 64, 43(1960), (dl and d9); A D.
Liehr and C. J. Ballhausen, Ann. Phys., (N.Y.), 6, 134(1959), (d and
d8); H. A. Weakliem, J. Chem. Phys., 36, 2117(1962), (d3 and d/);
J. P. Jesson, J. Chem. Phys., 48, 161(1968), (d3 and d7); J. Ferguson,
Aust. J. Chem., 23, 635(1970), 7d3 and d7); T. M. Dunn and Wai-Kee Li,
J. Chem. Phys., 47, 3783(1967); Erratum, Ibid., 53, 2132(1970), (d4
and db); E. Konig and S. Kremer, Z. Naturforschung, 29a 31(1974),
(d4 and d6); Wai-Kee Li, Spectrochim. Acta, 24A, 1573(1968); Erratum,
Ibid., 2289(1971), (d5); E. Konig, R. Schnakig and S. Kremer,
Z. Naturforschung, 29a, 419(1974, (d5); W. Low and G. Rosengasten,
J. Mol. Spec., 12, 319(1964), (dO).














TABLE 1

Weak and Strong Field Coupling Schemes
in Cubic Symmetry


Title of the Coupling
Sequence of Perturbations Scheme


2
e 2/r, then ((r)as, then VLF {L,S,J}



e /rij, then VLF, then ((r)-.s {L,S,X}


+ + 2 C C C
VLF, then ((r)-'s, then e2/rij {x ,y ,



VLF, then e2/rij, then s(r)t-s {xC ,XC,FC



has been found convenient because it follows the free ion sequence

of perturbations. That is, when carrying out an {L,S,J} coupling

scheme one need not repeat the calculations for the first two

perturbations (the interelectronic repulsions and the spin-orbit

interaction). This duplication of the free ion case for the first

two perturbations greatly reduces the computational effort required

in an {L,S,JI coupling scheme relative to the other cubic coupling

schemes.








As we have mentioned, for complexes of the 3d transition

metals, in particular for those members with low atomic numbers, the

spin-orbit interaction's effect is small and can often be neglected.

Therefore, obtaining energy levels in the {L,S,X} weak field

coupling scheme, in which the spin-orbit interaction can be con-

sidered to be a minor perturbation relative to the ligand field

potential, could be of value.

The first and the main purpose of this investigation is to

devise the new {L,S,X) coupling scheme for the various d" elec-

tronic configurations and to demonstrate the superiority of the

{L,S,X} coupling scheme over the {L,S,J} coupling scheme for the

first row transition metal ions.

We have previously carried out the {L,S,X} coupling scheme

for the case of d2 (and d8) electronic configuration.1 Here we

extend the {L,S,X} coupling scheme to the case of d3 (and d7)

electronic configuration.

The {x ,y ,r } cubic strong field coupling scheme considers

the spin-orbit interaction as the second perturbation. Hence, the

{x ,y ,C} coupling scheme is the strong field analog to the

{L,S,J} weak field coupling scheme. The {x ,X ,C } cubic strong

field coupling scheme considers the spin-orbit interaction as the

last perturbation. It is likely that the {xC,X C,r} scheme, in which

the spin-orbit interaction is the last perturbation considered, is

more appropriate for first row transition metal complexes than is the

J. A. Collins and J. R. Perumareddi, Theoret. Chim. Acta, 54,
325 (1980).








{x ,yC,r } scheme, in which the spin-orbit interaction is the

second perturbation considered. The {x ,X ,FC} cubic strong field

coupling scheme for the case of d3 (and d7) electronic configura-

tion has been carried out by Eisenstein. We have repeated the
SC C 3 7
calculation of the {x ,X ,C } scheme for the case of d (and d7)

electronic configuration in order to have available to us the requi-

site wave functions so as to be able to extend this strong field

coupling scheme to a lower symmetry.

When the symmetry of the complex is lowered from Oh to D4h,

either by substitution or by distortion, then the ligand field

changes from cubic to tetragonal. We denote the ligand field poten-

tial of tetragonal (or quadrate) symmetry by VQ. It is also possible

to express the tetragonal potential as a sum of cubic (VC) and

axial (Va) potentials,2 as is illustrated in Fig. 1. In this

latter case, tetragonal energy levels are obtained by going through

the cubic levels and finding how they are perturbed by the axial

field potential.

We have seen that two weak field schemes are feasible for cubic

symmetry: the {L,S,J} and the {L,S,X} coupling schemes. The

{L,S,J} sequence of perturbations gives rise to two coupling schemes

in tetragonal symmetry. In the first coupling scheme the cubic and

the axial fields can be considered separately, always taking the cubic

field first since the added axial field acts as a small perturbation

1J. C. Eisenstein, J. Chem. Phys., 34, 1528(1961).
2j. R. Perumareddi, Phys. Stat. Sol. (b), 59, K127(1973).





C C-






























U -----
-" Cr"


--



LLn

r-

r*


c

-o
-o




0

E










*r-

C
(a
4a)




I-v
a




*r-


..J

e-
.,






I-


a--
r-








to the cubic field, both after spin-orbit interaction. In the second

coupling scheme the cubic and the axial fields may be taken together,

after the spin-orbit interaction.

The {L,S,X} sequence of perturbations gives rise to three

coupling schemes in tetragonal symmetry. When the cubic and the

axial fields are taken separately,the axial field may be considered

either before or after the spin-orbit interaction. The third scheme

results when the tetragonal potential is applied as a whole before

the spin-orbit interaction.

The two cubic and the five tetragonal weak field coupling

schemes are listed in Table 2 with special emphasis being placed on

the cubic parentage of the tetragonal coupling schemes.

We have carried out the {L,S,XC,C,F } and the {LSXC,XQ,FQ

tetragonal weak field coupling schemes for d3 (and d7) electronic

configuration. The {L,S,XC,rC, Q} tetragonal scheme is an extension

of the {L,S,X} cubic scheme. The {L,S,X ,r } tetragonal weak

field coupling scheme differs only slightly from the {L,S,XC,XQ, }

tetragonal weak field scheme. Therefore, we have not carried out

the {L,S,X ,r } scheme for the case of d (and d7) electronic

configuration.

The d3 wave functions and the Hamiltonian (energy) matrices for

the {L,S,XC,C, Q} and the {L,S,XC,XQ,FQ} coupling schemes will be

presented along with the details of their derivation.

J. A. Collins, Master's Thesis, Florida Atlantic University,
1976.















(I
E



an
u


a) cn


4--'

0 3
0
u...-
4- a-


o




O
S--











r-)






cu
f-





i-
"3








cr
0






cn
(U









*r





















3



*r


-0



3


1^1 r-_1
oC cr

0 0 0
S Cy cY
Cr X D

O U CY
X X X
a f> a


-1 -1 -1


O CY
o c
m A





> *












ca





u c
> >

!

-C .C


+-3 +







c m c

-= -
L n

-4- -I-


*r-S *1-3
C'- C'~



Nc N
me


0}

*r-"






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I-



4-
rc
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0







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re-








ar-








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4-

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*r**













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a,
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c,












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01



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-r

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tt)


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ti

L




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s-,


rl
cu


*


i-
luvP



v
r: +inu
cu *
.c
s


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>


c,
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U >
>= +-
cu >
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cc




CU -=
.2 +J
4+>

*0-
*r3 *p-

N N
Co' M


u
U
C

c-
4<



.



4-2
v
LLP
ft-



+*? 0
*r
e-

c.

n
*'r1



C\J








The strong field approach yields six coupling schemes when the

symmetry is lowered from cubic to tetragonal. The two cubic and the

six tetragonal strong field coupling schemes are listed in Table 3.

A special emphasis has been placed, in Table 3, on the cubic

parentage of the tetragonal coupling schemes.

The {xC,yC C ,FQ} and {x ,C ,Y ,rQ} strong field coupling

schemes consider the spin-orbit interaction as the second perturba-

tion, prior to the interelectronic repulsions. The {xQ,y Q,Q}

strong field coupling scheme also considers the spin-orbit interaction

prior to the interelectronic repulsions. It is unlikely, especially

for complexes of first row transition metal ions, that the spin-orbit

interaction would be more important than the interelectronic repul-

sions. Consequently, the {xCyC, ,Q}, {xCy Q} and

{xQ,yQ,rQ} strong field schemes are not expected to be important for

complexes of the transition metal ions (especially those of the first

row).

The {x ,XQ,r } strong field coupling scheme considers the

interelectronic repulsions prior to the spin-orbit interaction but

after the axial ligand field perturbation. This sequence of perturba-

tions is almost certainly an improvement over those of the three

previously mentioned strong field coupling schemes in which the

spin-orbit interaction is taken prior to the interelectronic repulsions.

Nonetheless, the axial ligand field is often due to a relatively small

distortion in the cubic environment and, as such, is not expected to

be of greater importance than the interelectronic repulsions. The

{xQ,XQ,r } scheme might well find application in the study of
































0)




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s-



C)
ul
















*r-





Ec:









0
0


























4-)
-o













S..
C-









V-e
LJ.





a0





0.2


(I)

E






r--


0

C- (




S-


CD






i--
r--


CI



























a)


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O


LC




t)


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C
*r--


a,


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UT

















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CA
O




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CD C7


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X X ><















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X X X












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a (v
t (



S.- S-


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4C a, c
ci, -c












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4-) .4- 4-)
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C C











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square planar complexes. A square planar complex is, of course, an

extreme case of tetragonal distortion in which two transligands are

completely removed.

The {x xC,XC, FQ} and {xC,XC, C,FQ} strong field coupling

schemes both consider the interelectronic repulsions as the second

perturbation. It is likely that of the six possible strong field

schemes the {xCXC,XQ,FrQ and {xC,XrCrCQ} schemes are the most

appropriate for trans-disubstituted and moderately distorted octa-
hedral complexes.

The {xC,XC,XQ,FQ} coupling scheme has been carried out recently

by Perumareddi for d3 (and d7) electronic configuration.1

As the second purpose of this investigation we have carried out

the {xC,C,rC,rQ} strong field coupling scheme for d3 (and d7)

electronic configuration, in order to compare this tetragonal coupling
scheme with the {xCxC,XQ,X Q} strong field coupling scheme and with

the two tetragonal weak field coupling schemes which we have carried

out. The {xC,XC, C ,Q} tetragonal scheme is an extension of the

{x ,X C} cubic scheme.

The d3 wave functions and the Hamiltonian (energy) matrices

for the {xC,XC,rC ,Q} coupling scheme will be presented along with

the details of their derivation.
The detailed comparison of the various coupling schemes for d3

(and d7) electronic configuration and both cubic and tetragonal



J. R. Perumareddi, to be published.








symmetries will be presented. This comparison will be based on the

percent purities of the eigenfunctions yielded by each scheme.

The spin-orbit interaction usually need not be included in the

interpretation of the spin-allowed d-d spectra of transition metal

complexes. This is so because the spin-orbital components of the

spin-allowed bands are seldom resolved. The interpretation of the

spin-forbidden transitions, however, requires the inclusion of the

spin-orbit interaction. As the third purpose of this investigation

we will attempt the fitting of the spectra of some tetragonal

chromium (III) complexes, taken from the literature. The complexes

that we have chosen are those in the spectra of which all the ex-

pected low energy intraconfigurational spin-forbidden doublets have

been identified.











CHAPTER II


THE WEAK FIELD COUPLING SCHEMES

The Interelectronic Repulsions--Free
Ion Termsi

In the weak field coupling schemes the interelectronic repulsions

constitute the first perturbation. Hence, we need calculate only once

the interelectronic repulsion energies for all of the weak field

schemes. Furthermore, the wave functions derived in order to

calculate the free ion interelectronic repulsion energies will serve

as the basis functions for the other perturbations in the weak field

coupling schemes.

The three electron wave functions are normalized Slater determi-
2
nants; that is for y1',2',3 wave functions for three isolated

electrons, with total generalized coordinates X1,X2,X3 respectively,

we define the three-electron wave function (Y) by


1(X1) 1(X2) ~1(X3)

'(X1,X2,X3) = '2(X1) 2(X2) -2(X3) / v3


3(X1) 3(X2) 3(X3) (4)

1E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra,
University Press, Cambridge, 1951.
C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw-
Hill Book Co., New York, 1962.








= (1(X1)P2(X23(X3) (X1)2(X33(2)

1(X2)2(X1)3(X3) + 1(X2)2(X3)3(X1)

+ (X3)2(1)(X2) 1(X3)2(X2)3(X1))//6.


These wave functions are anti-symmetric in accordance with the
Pauli exclusion principle. Throughout, we will use a shorthand nota-
tion in which the fact that the wave functions are Slater determinants
will be understood. Thus, we will write the above expansion of
Y(X1,X2,X3) as simply ('1(1)Y2(2)i3(3)) or even more briefly as


For a 3d electron the four principal quantum numbers are

n = 3 (5a)
S= 2 (5b)
m = 2, 1, 0 (5c)
and
ms = 1/2 (5d)

As usual, we have assumed the single electron hydrogen-like orbitals
for the many electron case.
For the many electron case we have the quantum numbers L,S,ML
and MS, where

ML = m(m.) and (6a)
1

MS = (msi) (6b)
*1








with the summation being over all the electrons. The L and S are

analogous to the single electron quantum numbers Z and s. Thus,

we have


ML = L, L-l,..., -L and (7a)

MS = S, S-1,..., -S (7b)


The Pauli principle states that no two electrons can have the

same wave function. Therefore, 3d electrons in the same system must

differ in either mR or ms. Furthermore, due to interelectronic

repulsions, not all the L and S combinations allowed by the Pauli

principle will have the same energy. For each electronic configura-

tion, we need to pick out the allowed combinations of L and S,

known as "terms," and, in order to calculate the energies, determine

the wave functions associated with each term.

We shall use the Russell-Saunders (L-S coupling) procedures for

finding the energies corresponding to the terms. This is a parametric

method the details of which we will now describe for the case of d3

outer electronic configuration.

For a 3d3 ion we can construct the microstate table as shown in

Table 4, obeying at all times the Pauli principle.

In writing the microstates the fact that n=3 and k=2 is

understood. Thus, the microstate (221) means that electron one has

1C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw-
Hill Book Co., New York, 1962.




20
TABLE 4
The Weak Field Microstate Table for d


+3/2 +1/2 -1/2 -3/2
ML


+5 (211) (2T1)


(270) (270)
+4
(21T) (71T)


(210) (210) (2TO0) (210) (210) (210)
+3
(710) (27-1) (2T10) (27-T)



(21-T) (2T-1) (2T-1) (21-T)
+2 (21-1) (71-1) (27-2) (2T-T) (22-7) (2-1-T)
(20U) (ITO) (200) (110)

(21-7) (2T-2) (2T-2) (71-7)
(21-2) (71-2) (20-T) (21-2) (70-1) (71-7)
+1
(20-1) (2U-1) (70-1) (0-T) (2-T1) (7--T)
(10) (11-) () 10) (1T-T)

(20-7) (20-2) (20-2) (20-2)
(20-2) (70-2) (T0-1)-2) (2) (TO-1) (20-2)
0
(10-1) (1O-2) (10-T) (TO-T) (10-T) (10-1)
(IT-2) (2-1-T) (1T-7) (7-1-T)


The rest are obvious








m equal to +2

equal to +2 and

equal to +1 and

or (221) is not

We can now

which we denote


and ms equal to +1/2, that electron two has mk

ms equal to -1/2, and that electron three has m.

ms equal to +1/2. Obviously, the microstate (221)

possible.

derive the Russell-Saunders terms (L-S coupling terms)

by


where symbols are used for L as shown in Table 5.

TABLE 5

Term Symbols


L Symbol


The spin multiplicity is given by (2S+1) and the

tiplicity by (2L+1). Thus, the total multiplicity of

by the product (2S+1)(2L+1).

The microstate (221) yields ML = 5 which, bein

ML possible, belongs to a term of L=5. Since for tl


orbital mul-

a term is given


g the maximum

his term MS








can only be +1/2 or -1/2, then S is 1/2. Therefore, we have a

2H term whose total multiplicty is 22. The 2H(22) term requires a

microstate in each box from ML = +5 to ML = -5 with M = +1/2

and -1/2, 22 in all.

Turning to the microstate (220) we have ML = 4. Now ML = 4

is maximum since the microstate (221) has already been claimed by 2H.

Thus, (220) belongs to a term with L = 4. Since for this term MS

can be only +1/2 or -1/2 then, S = 1/2. Therefore, we have a 2G

term whose total multiplicity is 18. The 2G(18) term requires a

microstate in each box from ML = 4 to ML = -4 with MS = +1/2 and

-1/2, 18 in all.

The microstate (210) has ML = 3, now the maximum possible, and

MS = 3/2, also the maximum. Therefore, we have a 4F term whose

total multiplicty is 28. The 4F(28) term requires a microstate in

each box from ML = 3 to ML = -3 with MS = 3/2, 1/2, -1/2, and

-3/2, 28 in all.

Continuing we find a 2F(14) term, two 2D(10) terms, a 4p(12)

term and a 2P(6) term.

In summary, for d3 we have the following terms: 2H(22), 2G(18),

4F(28), 2F(14), 2D-a(10), 2D-b(10), 4p(12) and 2P(6).
Now we must decide which microstates or linear combinations of
A A
microstates are eigenfunctions of the L and S terms (the circumflex

indicates an operator). The appropriate combinations of the microstates

will then serve as the wave functions for the terms and will be known

as the IL,S;ML,Ms> functions.








When only one microstate has a particular ML and MS

combination, this is straightforward. Thus, since only the microstate

(221) has ML = 5 and MS = 1/2, the function 15,1/2;5,1/2> is
L
(221) and is one of the 22 eigenfunctions of 2H. There are two
microstates with ML = 4 and MS = 1/2; these are (220) and (211).

Finding the linear combination of these two microstates that is the

eigenfunction 15,1/2;4,1/2> of 2H is our problem. Application of
1 ^ ^
the angular momentum raising and lowering operators, L. and S.,

will allow us to form the proper IL,S;ML,MS> functions for 2H. The

use of these raising and lowering operators is illustrated below.


L+IL,ML> =.f[(L+ML+1)(L-ML)]1/2IL,ML+1> (9a)

L_IL,ML> =,K (L-ML+1)(L+ML)]1/2 L,ML-1> (9b)

S+IS,Ms> = [(S+Ms+1)(S-MS)] 1/21S,Ms+1> (9c)

S_IS,MS> = [(S-MS+1)(S+Ms)] 1/2S,MS-1> (9d)

The operators are to be applied to each electron separately;

that is,

L = Z(1) + Z(2) + ...(+(n) (o1a)


and


S+ = s(1) + s.(2) + ...s(n) (10b)

when there are n electrons.

C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw-
Hill Book Co., New York, 1962.








Thus,

L_5,1/2;5,1/2> = 4[(5-5+1)(5+5)]1/2 5,1/2;4,1/2>

= [_(1) + Z (2) + Z (3)](221)

=4[(2-2+1)(2+2)]1/2(121) +,6[(2-2+1)(2+2)]1/2(211)

+~[(2-1+1)(2+1)]1/2(220)

so

15,1/2;4,1/2> = [/3(220) /2(211)]/J5.

By repeated application of these operators we can derive the
complete set of IL,S;ML,MS> functions for the 2H term.
For the 2G term we do not have a unique microstate for any ML
and MS combination to begin the analysis with. For the box cor-
responding to ML = 4 and MS = 1/2 we have the linear combination
of (220) and (211) which is the function 15,1/2;4,1/2> of 2H. We
know that the function 14,1/2;4,1/2> of 2G must be orthogonal to

15,1/2;4,1/2> of 2H. Also, we have the normalization condition.
Using these two conditions we can write down 14,1/2;4,1/2> of 2G.
This function is

14,1/2;4,1/2> = [/2(220) + /3(211)]//5

Note that the choice of phase for |4,1/2;4,1/2> of 2G is initially
arbitrary but once made must be adhered to throughout. Then by








application of the appropriate raising and lowering operators we can

derive the complete set of IL,S;ML,MS> functions for the 2G term.

For 4F we have a unique microstate with ML = 3 and MS = 3/2.

Thus, the function 13,3/2;3,3/2> is (210). Applying S_ we obtain

S_13,3/2;3,3/2> = [(3/2-3/2+1)(3/2+3/2)]1/2 3,3/2;3,1/2>

S[S_(1) + S_(2) + S_(3)](210)

=-,[(1/2-1/2+1)(1/2+1/2)]1/2(210)

+ ,[(1/2-1/2+1)(1/2+1/2)]1/2(210) +-K[(1/2-1/2+1)(/1/2+1/2)] (210)

and


3,3/2;3,1/2) = [(210) + (210) + (210)]//3

Continued application of the appropriate raising and lowering operators

allows us to derive the complete set of IL,S;ML,MS> functions for the

F term.

The use of orthonormality and the raising and lowering operators

allow us to determine the IL,S;ML,MS> functions for the other terms

(2F, 0-a, 2-b, 4P and 2P) as well. The case of the two D terms
2 2
(2D-a and D-b) is particularly difficult since two independent

functions must be found simultaneously for the same ML and MS

combinations.1

E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra,
University Press, Cambridge, 1951.








The complete set of the L,S;ML,MS> functions for the d3

terms is given in Table 6.

In the perturbation theory of degenerate systems one constructs

a secular determinant which is then diagonalized to obtain the

perturbation energies. In the case of the interelectronic repulsion

perturbation the microstates are the unperturbed basis functions and

what Russell-Saunders coupling does is to obtain the IL,S;ML,MS>

eigenfunctions as the linear combinations of these basis functions

that diagonalize the perturbation matrix. Also, for a term of

definite L and S the set of (2S+1)(2L+1) states will have the

same energy since the energy is independent of MS and ML*

The energy of a term then is calculated by the expression


E = /L H TdT (11)


where Y is the IL,S;ML,MS> function of that term and H is the

interelectronic repulsion perturbation ( I e2 /r). For 2D a 2 by 2
i secular determinant is formed.

We can expand 1/ri. in terms of the well-known spherical
1 mR
harmonics, YZ (86,):

i *
CO +i) r m m
1 41n r< m .
S--(2+1 .1+1 Y (i' i)'y (oj',ij) (12)
ij Z=0 mz=-z r

where r< is the lesser and r> the greater of the two distances

ri and rj and Y (96,) are the spherical harmonics.


1H. Eyring, J. Walter and G. E. Kimball, Quantum Chemistry,
John Wiley and Sons, Inc., New York, 1944.







TABLE 6


The IL,S;ML,Ms> Functions for


1,1/2;1,1/2> = (-4/3(21-2)+2/3(2I-2)+2/3(21-2)+4/2(20-I)+/2(20-1)
-5/2(20-1)-3/3(11-1)-3/3(100))//210

1,1/2;0,1/2> = (-4(20-2)-3/6(1i-2)+8(20-2)-4(20-2)-(10-1)+3/6(2-1- )
+2(10-1)-(10-1))//210

1,1/2;-1,1/2>= (2/3(2-1-2)-5/2(10-2)+4/2(15-2)+/2(I0-2)+2/3(2-1-2)
-4/3(2-1-2)+3/3(1-1-1)+3/3(00-1))//210

1,1/2;1,-1/2>= (-2/3(21-2)-2/3(21-2)+4/3(21-2)+5/2(20-i)-4/2(20-1)
-2(20-)-33-33 11-1 -/3 (100))/V210

1,1/2;0,-1/2>= (-8(20-2)-3/6(li-2)+4(20-2)+4(26-2)-2(10-1)
+3/6(2-1-i)+(0-I)+(10-1) )//210

I1,1/2;-1-1/2>= (5/2(i-2)-/2(10-2)-4/2(10-2)+4/3(2-1-2)-2v3(2-1-2)
-2/3(2-1-2)+3/3(i-1-i)+3/3(00-1))//210

D-a


12,1/2;2,1/2> = ((21-1)+(200)-(22-2)-(21-1))/2

12,1/2;1,1/2> = ((11-1)+(100)+(21-2)-(21-2))/2

12,1/2;0,1/2> = ((10-i)+(20-2)-(i0-1)-(20-2))/2

2,1/2;-1,1/2>= ((1-1-1)+(2-1-2)+(00-1)-(2-1-2))/2








TABLE 6--Continued


12,1/2;-2,1/2>

12,1/2;2,-1/2>

12,1/2;1,-1/2>

2,1/2;0,-1/2>


= ((i-1-2)+(oO-2)-(2-2-2)-(1-I-2))/2




= ((1i-I)+(I0O)+(zI-2)-(2I-2))/2

=~l-)(b 2-I- 1-2-)/


12,1/2;-1,-1/2> = ((I-1-I)+(2-i-2)+(00-f)-(2-I-2))/2

12,1/2;-2,-1/2> = ((-1-2)+(00-2)-(2-2)-(I-i-2))/2


2-b


12,1/2;2,1/2>


= (3(21-1)-5(22-2)+(21-1)-4(21-1)-3(200)


-2/6(110))//84


= (3(21-2)+(21-2)-4(21-2)-(11-1)+3(100)


2,1/2;0,1/2>


= (3(20-2)-2/6(11-2)-3(20-2)-3(10-1)


+3(10-I)-2/6(2-1-I))//84


2,1/2;-1,1/2>


12,1/2;-2,1/2>


= (2/6(10-2)-(2-1-2)-2/6(10-2)-3(2-1-2)
+3(00-1)-(1-1-I)+4(2-i-2))//84

= (4(1-1-2)-5(2-2-2)-3(00-2)-(1-1-2)


-3(I-1-2)-2/6(0-1-I))//84


12,12;2-11> =(4(2T-T)-5(22-2)-3(21-1)-(21- T)


12,1/2;1,1/2>


-2/6(20-1)+2/6(2D-1))/84


2,1/2;2,-1/2>


-3(200)-2/6(110))//84








TABLE 6--Continued


= (4(21-2)-3(21-2)-(21-2)-(11i-)+3(iOO)


12,1/2;0,-1/2>


= (3(20-2)-2/6(11-2)-3(20-2)-3(10-1)


+3(10-1)-2/6(2-1-1))//84

12,1/2;-1,-1/2> = (2v6(10-2)-2/6(10-2)-4(2-1-2)+3(00-1)
-(I-1-1)+(2-1-2)+3(2-1-2))//84


12,1/2;-2-1/2>


13,1/2;3,1/2>

13,1/2;2,1/2>

13,1/2;1,1/2>


13,1/2;0,1/2>


=((I-1-2)+3(1-I-2)-5(2-2-2)-3(o6-2)


= (/6(22-1)+2(210)-(210)-(210))/2/3

= ((21-1)+2(22-2)-V6(110)-(21-1))/2/3

= (-2/2(11-1)+/3(20-1)+3/2(21-2)-2v2(21-2)
+3/2(100)-/3(20-1)-V2(21-2))//60

= (2(iO-1)-4(10-1)+V6(1-2)-/6(2-1-I)+4(20-2)


-2(20-2)+2(10-I)-2(20-2))//60


3,1/2;-1,1/2>


13,1/2;-2,1/2>

13,1/2;-3,1/2>


= (2/2(1-1-I)-3/2(00-1)-/3(I0-2)+/3(10-2)
+3/2(2-I-2)-2V2(2-1-2)-/2(2-1-2))//60

= (/6(0-1-1)+(1-I-2)-2(2-2-2)-(I-1-2))/2/3

= (2(0-1-2)-(0-1-2)-76(1-2-2)-(0-1-2))/2/3


2,1/2;1,-1/2>


-2/6(20-1)+2/6(20-i))/184


-4(1-1-2)-2/6(0-1-1))//84










13,1/2;3,-1/2>

13,1/2;2,-1/2>

13,1/2;1,-1/2>



13,1/2;0,-1/2>



13,1/2;-1,-1/2>



3,1/2;-2,-1/2>

3,1/2;-3,-1/2>


TABLE 6--Continued

= (/6(22-1)+(210)+(210)-2(2ii))/2/3

= ((2i-1)+2(22-2)-/6(1i0)-(21-i))/2/3

= (-2/2(11-i)+V3(20-1)+/2(21-2)+2/2(21-2)
-3/2(21-2)+3/2(100)-V3(20-1))//60

= (4(i)-2)-2(-1)-2(10-)+/6(1I-2)-/6(2-1-i)
(20-+2(2-2)(20-2)-4(20-2))//60

= (2/2(i-1-i)-3/2(00-1)-V3(i0-2)+/3(10-2)
+2/2(2-i-2)+2(2-i-2)-3/2(2-1-2) )//60

= (/6(0-1-I)+(1-I-2)-2(2-2-2)-(1-1-2))/2/3

= ((0-1-2)+(0-i-2)-2(0-i-2)-/6(I-2-2))/2/3


2G

14,1/2;4,1/2> = (/2(220)+/3(211))//5

14,1/2;3,1/2> = (3(210)-2(210)-(210)+v6(22-1))/2/5

14,1/2;2,1/2> = (4/3(200)+3/3(21-I)+/2(110)+4/3(21-1)
+/3(21-1)+2/3(22-2))/2/35

14,1/2;1,1/2> = (/6(100)-3(20-1)+7(20-I)+V66(1-2)+2/6(11-1)
-4(20-1)-2V6(21-2)+v6(21-2))/2/35

14,1/2;0,1/2> = (J2(10-I)+/3(2-1-I)+/2(20-2)-V2(10-1)
+/3(1I-2)-V2(20-2))//14








TABLE 6--Continued


14,1/2;-2,1/2>


14,1/2;-3,1/2>

14,1/2;-4,1/2>

14,1/2;4,-1/2>

14,1/2;3,-1/2>

14,1/2;2,-1/2>


14,1/2;0,-1/2>


= (2/6(1-1-1)+4(10-2)-/6(2-I-2)+2V6(2-1-2)
+/6(00-1)-7(10-2)(-2)-+3(1/-2)6( -2))/2/35

= (V2(0-1-1)-/3(1-I-2)+4/3(1-1-2)+2/3(2-2-2)


= ((0-1-2)+2(0-1-2)+/6(1-2-2)-3(0-1-2))/2/5

= (/3(-1-1-2)+/2(0-2-2))//5

= (/2(220)+/3(211))//5

= ((210)+2(210)-3(210)+/6(22-I))/2/5

= (4/3(200)+4/3(21-I)+/2(110)-3V3(2I-1)
-/3(21-I)+2/3(22-2))/2/35

= (/6(00)-7(20-1)+3(20-1)+4(20-1)+2/6(21-2)
+2/6(11-I)-/6(21-2)-/6(21-2))/2/35

= (/2(10-I)+/3(2-1-I)+/2(20-2)-2(10-1)


+/3(11-2)-/2(20-2))//14

4,1/2;-1,-1/2> = (2/6(1-1-1)+7(10-2)-2/6(2-1-2)+/6(2-1-2)
+/6(2-1-2)+/6(00-I)-4(10-2)-3(IO-2))/2/35


14,1/2;-2-1/2>


= (/2(0-1-T)+v'3(I-1-2)+3/13(1--2)+2/3(2-2-2)


+4V3(00-2)-4/3(1-1-2))/2/35


4,1/2;-1,1/2>


+4/3(00-2)-3/3(1-1-2))/2/35


4,1/2;1,-1/2>







TABLE 6--Continued


14,1/2;-3,-1/2> = (3(0-1-2)+/6(1-2-2)-2(0-1-2)-(0-1-2))/2/5

14,1/2;-4,-1/2> = (3(-1-1-2)+/2(0-2-2))//5

2H


15,1/2;5,1/2>

15,1/2;4,1/2>

15,1/2;3,1/2>

15,1/2;2,1/2>

15,1/2;1,1/2>


15,1/2;0,1/2>


15, 1/2 ;-14/2>


15,1/2;-2,1/2>


= (221)


= (v3(220)-/2(211))//5

= (v3(22-1)-/2(210)+/8(210)-/2(210))//15

= (/6(110)-3(200)-(21-I)+3(21-(21-1)-2( )+(22-2))//30

= (8(1i-1)-6(100)+3/6(20-1)-2/6(20-I)-(21-2)
+4(21-2)-/6(20-1)-3(1-)21-2)210

= (4(10-1)-2(I0-1)+/6(1i-2)-/6(2-1-i)-(20-2)


= (6(00-1)-8(1-1-I)+3V6(10-2)-2/6(10-2)
+4(2-1-2)-3(2-1-2)-/6(10-2)-(2-1-2))//210

= (-/6(0-1-1)+3(00-2)-2(1-1-2)+3(1-1-2)


-(I-1-2)-(2-2-2))//30


15,1/2;-3,1/2>

15,1/2;-4,1/2>

15,1/2;-5,1/2>


= (v8(0-1-2)-/2(0-1-2)-2(01-2)-3(1-2-2))/15

= (-V3(0-2-2)+/2(-1-1-2))//15


= -(-1-2-2)


+2(20-2)-(20-2)-2(10-1))//42







TABLE 6--Continued


15,1/2;5,-1/2>

15,1/2;4,-1/2>


= (221)


= (/3(220)-/2(211))//5


15,1/2;,3,-1/2> = (/3(22-I)-/8(210)+/2(210)+/2(210))//15


5,1/2;2,-1/2>


= (/6(1i0)-3(200)-3(21-I)+(21-1)


= (8(1I-I)-6(Io1)+2,/6(2O-1)+/6(20-I)-3/6(20-)


-4(21-2)+(21-2)+3(21-2))//210


15,1/2;0,-1/2>


=(2(Ib-1)+2(16-I)-4(rIO-I)+/6(1T2)-/6( 2-1-f


-2(20-2)+(20-2)+(20-2))//42

5,1/2;-1,-1/2> = (6(00-i)-8(1-1-I)+/6(i0-2)+2/6(10-2)-3/6(i0-2)
+3(2-I-2)+(2-I-2)-4(2-1-2))//210


i5,1/2;-2,-1/2> =


+(1-I-2)-(2-2-2))//30

5,1/2;-3,-1/2> = (/2(0-i-2)+/2(0-I-2)-/8(0-1-2)-/3(I-2-2))//15

|5,1/2;-4,-1/2> = (-/3(0-2-2)+/2(-1--2))//5

|5,1/2;-5,-1/2> = -(-1-2-2)


1,3/2;1,3/2>

11,3/2;0,3/2>


= (/2(20-1)-/3(21-2))//5

= (2(10-1)-(20-2))//5


+2(2I-1)+(22-2))//30


15,1/2;1,-1/2>


(-/6(0-1-1)+3(00-2)-3(1-1-2)+2(1-1-2)










11,3/2;-1,3/2>

11,3/2;1,1/2>



11,3/2;0,1/2>



11,3/2;-1,1/2>



11,3/2;1,-1/2>



1,3/2;0,-1/2>



1,3/2;-1,-1/2>



1,3/2;1,-3/2>

I1,3/2;0,-3/2>

11,3/2;-1,-3/2>


13,3/2;3,3/2>

13,3/2;2,3/2>


TABLE 6--Continued

= (/2(10-2)-/3(2-1-2))//5

= (/2(20-1)+/2(20-1)+/2(20-I)-/3(21-2)
-/3(21-2)-V3(21-2))//15

= (2(10-1)+2(10-1)+2(10-I)-(20-2)
-(20-2)-(20-2))//15

= (/2(IO-2)+/210-2(1-2)+2(10-2)-3(2-1-2)
-J3(2-1-2)-/3(2-1-2))//15

= (/2(20-1)+/2(0-I)+/2(20-I)-/3(2i-2)
-/3(21-2)-/3(2I-2))//15

= (2(10-1)+2(10-1)+2(10-1)-(20-2)
-(20--(-2)-(20-2))//15

= (V2(10-2)+/2(10-2)+/2(10-2)-/3(2-i-2)
-/3(2-1-2)-/3(2-I-2))//15

= (/2(20-I)-/3(21-2))//5

=(2 (10-I)-(20-2))//5

=(2(10-2)-/3(2-1-2))//5

4F


= (210)

= (21-1)


----------










13,3/2;1,3/2>

13,3/2;0,3/2>

13,3/2;-1,3/2>

13,3/2;-2,3/2>

13,3/2;-3,3/2>

13,3/2;3,1/2>

13,3/2;2,1/2>

13,3/2;1,1/2>



13,3/2;0,1/2>



13,3/2;-1,1/2>



13,3/2;-2,1/2>

13,3/2;-3,1/2>

3,3/2;3,-1/2>

3,3/2;2,-1/2>


TABLE 6--Continued

= (/3(20-1)+/2(21-2))//5

= ((10-1)+2(20-2))//5

= (/3(10-2)+/2(2-1-2))//5

= (1-1-2)

= (0-1-2)

= ((210)+(210)+(210))//3

= ((21-1)+(21-1)+(21-1))//3

= (/3(20-1)+/3(20-1)+/3(20-i)+/2(21-2)
+/2(21-2)+v2(21-2))//15

= ((i0-1)+(16-1+((10-1)+2(20-2)
+2(20-1)+2(20-2))//15

= (/3(10-2)+/3(10-2)+/3(10-2)+/2(2-1-2)
+/2(2-I-2)+/2(2-1-2))//15

= ((1-1-2)+(1-T-2)+(1-1-2))//3

= ((0-1-2)+(0-1-2)+(0-1-2))//3

= ((210)+(210)+(210))//3

= ((i-1)+(21-)+(21-I))//3


~I~







TABLE 6--Continued

13,3/2;1,-1/2> = (/3(2-1)+/3(20-1)+/3(2(-1)+/2(2i-2)
+/2(21-2)+/2(21-2))//15

13,3/2;0,-1/2> = ((1I-1)+(10-I)+(10-i)+2(20-2)+(20-2)
+(20-2))//15

13,3/2;-1,-1/2> = (/3(16-2)+/3(10-2)+/3(10-2)+/2(2-1-2)
+/2(2-1-2)+/-1- 21-2))//15

13,3/2;-2,-1/2> = ((1-1-2)+(1-1-2)+(1-1-2))//3

13,3/2;-3,-1/2> = ((0-1-2)+(0-1-2)+(0-1-2))//3

13,3/2;3,-3/2> = (210)

13,3/2;2,-3/2> = (21-1)

13,3/2;1,-3/2> = (/3(20-1)+/2(21-2))//5

13,3/2;0,-3/2> = ((10-1)+2(26-2))//5

3,3/2;-1,-3/2> = (/3(i0-2)+/2(2-l-2))//5

13,3/2;-2,-3/2> = (1-1-2)

13,3/2;-3,-3/2> = (0-1-2)








The wave functions associated with the microstates are the single

electron wave functions that are the solutions of Schrodinger's

equation for the hydrogen atom case. These have the form

m,
o = R,(r), y (,) (13)


with Rn, (r) being the radially dependent factor. For a 3d ion we

will write R ,(r) as R3d(r).

As an example let us consider the calculation of the energy of
the 2H term. We could use any of the 22 degenerate IL,S;ML,MS>

functions which belong to 2H. For the sake of simplicity we choose

5,1/2;5,1/2> = (221). We must use the expanded Slater determinant
for the calculation. Thus,

15,1/2;5,1/2> = [(221)-(212)-(221)

+ (122)+(212)-(122)]//6

The energy of 2H is given by

E'(2H) = f/ / [(221)-(212)-(221)+(122)
1 2 3
123

+ (212)-(122)1*(e2/r12 + e2/r13 + e2/r23)[(221)


(212)-(221)+(122)+(212)-(122)]dT1 dT2 di3/6 (14)


To illustrate the method of calculation let us take a fragment

of the above expression for the energy of 2H. Let us consider only









Sf f (221) (e /r12 + e2/r13 + e2/r23)(221)dTr dT2 dT3
123


Expanding we have


l a(1)*a(1)ds1 f2 8(2)*B(2)ds2 f3 a(3)~(3)ds3


r
f f f R3d(1)R3d(2)R 3d (3)e 2 y <+lJR3d(1)R3d(2)
1 2 3 =0 r

S2 2*
R3d(3)r drIr2 dr2r3 dr3 f f f Y2 (1)Y2 (2)
123


j=3
Y21 (3)[
i

=00
ZO +z

Q=0 m =-t


4TT
(2Z+1)


*
m m.
Y (i, i)Y' (ej, j)


Y22(1)Y22(2)Y21(3)sine1deldy1sine2de2dp2sine3de3d 3


The spin functions are orthornormal, that is


a ads ds =
i a ads = / 8 ds = 1


I 8 ads = / a Bds


= 0


Continuing, we have


o R3d (1)R3d(1) R3d (2) (2)R(3)R3d(3)
0 0 0


2 < 2 2 2
e [ <+ I]r 2drlr2 dr2r3 dr3
I=0 1 1 2
),=O r-


and


(15a)


(15b)








2* 2 1"
Sf f [Y22 (1)Y22(1)Y2 (2)Y 22(2)Y 2 (3)Y21(3)] *
123


S -2
I I
R=0 m =-R


4 m7 m ,
*(2 2- [V (1' r (e2,2) +


m m m m
Y (61,'l)Y9 (13,3) + Y ( 2'2)Y (e3,23)]


sineldEd1d 1sin92d62d 2sin93de3d3 .


Further,

00 m 3)
f f f [R3d (1)R3d(1)R3d (2)R3d(2)R3d (3)R3d(3)].
123


2 0o
e [Z
Z=0o


r< 2 2 2 +
+1]rl dr r2 r dr dr3 dr
r> =0 mt=-O


47T
(2Z+1)


f[ / / Y22 (1)Y22(1)Y2 (2)Y22(2)Y21(3)Y21(3)
1 2 3
123

Yz (al1)Yz (2,0 2)sine9de ld1sin2 d2 d92sin93dO3d3 +


/ y Y22 (1)2(Y2Y2 (2)Y22(2)Y21 (3)Y21(3)Y z (e1'1)Yz (63'3)
1 2 3
r%


sine91d61d sine2de2d 2sin93de3dq3


2* 2 2* 2 1*
S Y22 (1)y22(1)Y22 (2)Y22(2)Y21*(3)Y21(3)Yn d (02',2)
123
Y am .(03,3)sinldold~lsinld2dsin83Gd3] 3








Which further reduces to


f f R3d (1)R3d(1)R3d (2)R3d(2)R3d(3)R3d(3)
123


e co r Ir r 2 dr 00
e [ r+ 1 drlr2 dr2r3 dr3
Z=0 r> t=O mR=-z
^> )v


47T
(2k+1)


m m
[ f Y2 (1)Y22(1)Y22 (2)Y22(2)Y9 (el')Y9 Z(e2'2)
12


sinOedeold 1sine2de2d02 + / Y22 ()Y22(1)





sin3d3d-3 + f f Y2 (2)Y22(2)Y2 (3)Y21(3)
2 3


*
m r
Y (e2',2)Y


due to the normality of

The product of two

of spherical harmonics.

Y22*Y2 =
'2 2


1* 1
Y Y 1 =
*2 2


((03',3)sin92dO2d42sin93dO3d#3]


the spherical harmonics.

spherical harmonics can be expressed as a sum

Thus, we find that


Y40/14/T /5 Y20/7/7 + Y00/2/


2Y 0/7/7 + /5 Y20/14/T + Y 0/2/ .
4 2 0


A list of other relevant products written as sums is given in

Table 7.





41



TABLE 7

Products of Spherical Harmonics Expressed
as Linear Combinations of
Spherical Harmonics


*
= 0 /14/,T -
4=


= 5Y+1 /14/Tr
4


*
/v5Y /7/Tv
2


*
+ Y0 /2/u
0


*1
+ /30Y+1 /14/VT
2


= /5Y1 /14/v /30YT1 /14,/T
4 2


Y2 y2
2 Y2


Y2 Y+1
2 2



2 2


y2 y0
2 2


Y2* YTl
Y2 Y2


Y01y2
2 2

*
Y- Y
2 2
Y2 Y2


*




Y-1 Y
2 2


Y+1 +1
2 2



*
1 0

Y Y2
2 2


*
= -/5Y3 /2/7$
4



= /5Y /2+/7T
4

*
= /5Y4 //147
4





*
= -2Y4- /17/T +



= v/30Y1 /14/Tv
4


*
/15Y2 /14/Tv
4


5
/5Y /14/T
2


*
- /5Y-2 /7/v
2


*
+ Y /2/T
0


*
+ /5Y /14/Tr
2


1* *
= -/30Y+1 /14/Tr /5Y+1 /14/5
4 2


S 0 2 -
2 2




42



TABLE 7--Continued


*
= -/10Yo2 /7V /15Y2 /7
4 2


0 +
3Y /7/ + J/Y /7/T +
4 2


v/2T


Y /20 r
0


y+1 Y+1
2 2


S0*
Y Y2
2 2







2* 2 1* 1
Since the products Y2 2 and Y2 2 are functions of
0Y0 0
Y 4,Y20 and Y0 we need only consider in the summation those terms
with A = 0,2,4 and m, = 0. Thus, the integrals we have been

considering reduce to


e2 ( ) R3d 3d(l)R3d (2)R3d(2)R3d (3)R3d(3)
4
r<2 2 2
( 5)r dr r2 dr2r3 dr3 +
r> 5 1 1 2 2 3 3


e2( 2 R3d(1)R3d(1)R3d (2)R3d(2)3d(3)R3d(3)


2
r< 2 2 2
(3 )r 2drlr2 dr2r3 dr3

23 d *
e (1) I / / R3d (1)R3d(1)R3d (2)R3d(2)R3d (3)R3d(3)
^1 2 2

(r-)r 2drlr2 dr2r3 dr3


In ligand field calculations the radial integrals are not solved

directly but rather defined as parameters. For the interelectronic

repulsions perturbation the radial integrals become the well-known

Condon and Shortly parameters. These Condon-Shortley parameters are
defined as

2 mco *o
Fk = (-) ) / R3d (1)R3d (2)R3d (3)R3d(1)R3d(2)R3d(3)
r
r>k1 1 12 23 3k








when k = 0,2 or 4 then Dk = 1,49 or 441, respecitvely. There-

fore, the fragment of 2H which we have been considering reduces to

3FO 7F4. To obtain the complete energy of 2H we must return to
equation (14) and proceed as we have for the (221) fragment. Equa-

tion (14) reduces to

E'(2H) = [6J(22) + 12J(21) -6K(21)]/6

In the above expression for E'(2H) we have used the following

simplified notation


f / (ab)(e2/r2)(cd)dT1dT2 = <(ab)|(cd)> (17a)
12

<(ab) (cd)> = <(cd)l(ab)> (17b)

<(ab)I(ab)> = J(ab) (17c)

<(ab)l(ba)> = K(ab) (17d)

These integrals (17a-d) are the electron correlation integrals
for d electrons. The J(ab) and K(ab) are known, respectively, as

the Coulomb and Exchange integrals. The non-zero electron correlation
integrals are listed in Table 8.
The energy of 2H is thus

E'(2H) = (Fo+4F2+F4) + 2(FO-2F2-4F4) (6F2+5F4)

E'(2H) = 3F0 6F2 12F4







TABLE 8
The Non-Zero Electron Correlation
Integrals of the d Electrons

a) J(ab)

J(+2+2)=J(+2;2)=Fo+4F2+F4=A+4B+2C

J(+2+1)=J(+2t1)=F0-2F2-4F4=A-2B+C

J(+20)=FO-4F2+6F4=A-4B+C

J(t)=(+1+l)=+11)=F +F2+16F4=A+B+2C

J(10)=FO+2F2-24F4=A+2B+C

J(00)=Fo+4F2+36F4=A+4B+3C

b) K(ab)

K(+2+1)=6F2+5F4=6B+C

K(+20)=4F2+15F4=4B+C

K(+211)=35F4=C

K(+272)=70F4=2C

K(+10)=F2+30F4=B+C

K(+11)=6F2+40F4=6B+2C

c) <(ab)l(cd)>

<(+2;2)I (t~l)>=/6(F2-5F4)=/6B

<(+2T1) (+10)>=-V6(F2-5F4)=-V6B

<(+2;1) (0+1)>=2J6(F2-5F4)=2/6B







TABLE 8--Continued

<(+2;2) (t+1i )>=-6F2-5F4=-6B-C

<(+2;2)1 (1+1)>=-35F4=-C

<(+2;2) (00)>=4F2+15F4=4B+C

<(tl1)i (00)>=-F2-30F4=-B-C

<(+1+2) (10)>=2/6(F2-5F4)=2/6B

<(+1;2)| (01)>=-V6(F2-5F4)=-/6B

<(0O2) I (;1l1)>=,6(F2-5F4)=,/6B









We can also give the energy of 2H with the Racah parameters

A, B and C by making use of the following relationships.


A = F0 49F4 (18a)


B = F2 5F4 and (18b)


C = 35F4 (18c)


So that we have, alternatively,


E'(2H) = 3A 6B + 3C


The energies of the other terms are calculated in the same

manner. In the case of 2D a two by two secular determinant must be
2 2
solved since the cross term between D-a and 2D-b is non-zero.

These interelectronic repulsions energies for the d3 terms are

given in Table 9.

In Fig. 2 the Russell-Saunders splitting is illustrated. We have

omitted in this figure the A values since they are the same for each

level and we are interested in energy differences. Note that 2H and
P are accidentally degenerate.


The {L,S,XC,X Q} Coupling Scheme

This new {L,S,X} coupling scheme for d3 considers the cubic

and axial potentials individually, both before the spin-orbit inter-

action. We must now determine how the terms are affected when we

lower the symmetry of the central ion by introducing the ligand field

potential.





48

TABLE 9

The Electronic Repulsion Energies for the d3 Terms


= 3A 15B


= 3A

= 3A 6B + 3C

= 3A 11B + 3C

= 3A + 9B + 3C


D-a


2D-b


D-a


3A+7B+7C-E1


D-b


3/21B


3A+3B+3C-E2


= 0


2 3A 6B + 3C
P = 3A 6B + 3C
























3/21B


FIGURE 2. Free Ion Energy Levels for d3








The Cubic Ligand Field

The cubic and quadrate wave functions

First we take the cubic case, that is, the octahedral arrangement

of six equivalent ligands around the central ion (cf. Fig. 1). The

point group for this symmetry is Oh. For Oh we have possible the

following symmetry operations, grouped in classes: 3C4, 3C2(=C42),

6C2 8C3, E, i, and i combined with each of the previous operations.

Since the d orbitals are even to inversion only the rotation opera-

tions will bring us new information.

We will assume here that the spin, is, and space, %0, parts

of the wave function can be separated, thus


S= 0 s (19)

and that the spin function has no angular dependence. Therefore, we

need consider only the spatial part of the wave function.

For the single electron case we have the orbitals, functions of

Z and mR, while for a many electron system we have the terms,

functions of L and ML. The angular form of the orbitals is given

by

m __ im g
Y = 0 (O)1/27 e (20)


where Z,m (e) are normalized, Legendre polynomials and e and }

are the polar angles. Analogously, the angular dependence of the

terms is

ML iML
YL L,ML 1/21 e (21)








The relationship of the polar coordinates r, 8 and Q with the

cartesian axis x, y and z is shown in Fig. 3. Obviously, no physical

quantity can depend on how we choose to locate our coordinate axis.

Thus, for each of the rotation operations of Oh we can take as the
axis of rotation the z axis. Clearly, 6 and r are invariant to

rotations about the z axis (cf. Fig. 3). So we need find only how
iML
the 0 dependent part of the wave function, e is affected by

the various rotations.

Let us take the general case of rotation by a degrees about the
z axis, R(a), and assume that this rotation returns the molecule to

an equivalent configuration. Hence, we have

iMLP iML(4+a)
R(a)e = e (22)

For a P term, L equals one and ML can have the values +1, 0

and -1. Therefore, rotation by a degrees on a P term produces the
following change.

ei I ei[+]

R(a) e0 = e (23)

Se-i\ e-i[4+a]/


The same result can be obtained by multiplying the P functions by
the matrix

(eia 0 0
0 e0 0 (24)
0 0 e-im






































Figure 3. The Polar Coordinates


r, e and 6 .


This matrix is the transformation matrix of the symmetry operation in

question, that is, rotation by a degrees about the z axis. The

set of these matrices for different rotations constitute the represen-

tation of L with the dimension (2L+1).

The above result is not dependent on the value of ML. That is,

since ML can taken all integral values between L and -L, we can

state that the general transformation matrix which corresponds to the

rotation operations of Oh is











i Lc
e


0


000


0


0


0


ei(L-1)a


000


0


0


0


0


000
ooo


ei(1-L)a


0


0


0


000
ooo


0


-i Lc
e


(25)


The sum of the diagonal elements, the character of the matrix, is in

the form of a geometric progression and is given by


X(c) = sin[(L+1/2)a]
sin(a/2)


(26)


By use of (26) we can easily derive the characters corresponding

to the rotation operations for different L values. It is well-

known that the character of each class of a reducible representation

is equal to the sum of the characters of the irreducible representa-

tions which are contained in that particular representation.

For d3 we see that the P term yields an irreducible represen-

tation in Oh, namely T1g. The D term yields a reducible

representation in Oh composed of E and T2g. The F term








yields a reducible representation in Oh composed of A2g, T1g
and T2g. The G term yields a reducible representation in Oh com-

posed of Alg, Eg, T1g and T2g. Lastly, the H term yields a

reducible representation in Oh composed of Eg, two T1g's and

T 2g These results for d3 are summarized in Tables 10 and 11.

If we use directly the free-ion basis functions, i.e., the

IL,S;ML,MS> functions, to calculate the perturbation energy due to

the cubic ligand field potential, we will need to solve two secular

determinants, a 40x40 (for spin quartets) and an 80x80 (for spin

doublets).

If the IL,S;ML,MS> functions are linearly combined such that

they are the basis functions for the irreducible representations of

the cubic group then the secular determinants can be block diagonal-

ized because of the result that the matrix elements between two

different irreducible representations are zero.

The proof of the theorem on which this latter result is based

follows.

Consider the following integral


I = I /A B dT (27)

where WA and tB are functions which serve as basis functions for

two irreducible representations of a group.

When we take the product of two representations of a group and,

thus, obtain a third representation,the characters of the new

representation will be equal to the product of the characters of the








TABLE 10


a) The Character Table of the Terms of
for the Rotation Operations of O0


L (Term) x(E) X(C4) X(C2) x(C3) X(C2)

1 (P) 3 1 -1 0 -1
2 (D) 5 -1 1 -1 1

3 (F) 7 -1 -1 1 -1
4 (G) 9 1 1 0 1

5 (H) 11 1 -1 -1 -1


b) The Character Table of the Irreducible
Representations for the Rotation
Operations of Oh

Irreducible
Representation X(E) x(C4) X(C2) X(C3) (C2)

Alg 1 1 1 1 1


A2g 1 -1 -1 1 1


E 2 0 0 -1 2


T1g 3 1 -1 0 -1


T2g 3 -1 1 0 -1












TABLE 11


The Splitting of
of d3 in


Free Ion Terms
Symmetry


Term State(s) in Oh


2D(a and b)


2E +2T2
g 2g


2A2g +'


A1g + E
ig g


T1g 2T2g


+ 2 Tg + 2T2g


2E + 22T1g + 2T2


4T1g


4A2g 41g 4T2g









two representations that were multiplied together to give the new

representation. That is, if the product of representations A and B

is C, then


X -X X
C A f


(28)


for each class. This is known as the direct product of the representa-

tions A and B.

We have assumed that iA and rB are bases for irreducible

representations of a group. The product of two irreducible representa-

tions may or may not be irreducible. If A transforms as rA and

iB as FB where FA and FB are irreducible representations of our
group, then


A B = r = C.ir
A 'B C i i


(29)


If C is reducible,then EC.P. is the sum of irreducible representa-
i 1
tions which compose FC.

Let us operate on I with one of the operations, R, of the

group to which FA and FB belong. The value of any integral will

be just a number which will not be changed by a symmetry operation.

That is


further


RI = IR = I



I = f/RAyBdT = RFABdT ,


(30)


I = f/RC FidT
i i i


(31)








Since the value of the integral does not change when we apply a sym-

metry operation to it, i.e., RI = I, then the form of the integral,

fCC.r dT, must also be invariant under all the symmetry operations of
ill
the group. This will be so only when one of the ri is the totally

symmetric representation A1. This is the case because A, is the

sole representation that is invariant under all the symmetry opera-

tions of the group.

For the product FA iB to contain the totally symmetric represen-

tation, FA and r must be identical. That is AFB .

Therefore, we can conclude that integrals of the form


/f A B dT

can be non-zero only when the representations rA and FB, for which

A and B serve as basis functions, are identical.
The matrix elements have the familiar form of


Hij = f i H' i. dT = (32)


The Hamiltonian, of which H' is part, is an expression, in quantum

mechanical operator form, of the energy of a system. Application of

a symmetry operation of a molecule cannot change the energy of a

molecule and all symmetry operations will leave the Hamiltonian in-

variant. The Hamiltonian operator thus belongs to the totally

symmetric representation A1. The integral f i H' j dT can

therefore be non-zero only if the direct product rirj contains Al

and, as we have observed, this only occurs when ri = rj. Hence, the
1








matrix elements can be non-zero only if the wavefunctions 'i and

4j belong to the same irreducible representation.

For the d3 cubic case the secular determinant is thus block

diagonalized as shown in Fig. 4.

Our aim then is to find the linear combinations of the free-ion

functions which serve as the basis functions of the various irre-

ducible representations. The IL,S;ML,MS> functions have the same

angular dependence as the spherical harmonics and these [L,ML>

functions that serve as the bases for the irreducible representations

of the cubic group are the cubic oriented spherical harmonics.

To illustrate the formation of the cubic oriented spherical

harmonics we shall take the P term as an example. From Fig. 3 it

is easy to see that the cartesian coordinates relate to the polar

coordinates as follows:


z = r cos6, (33a)

y = r sine sinp and (33b)

x = r sine cos~ (33c)


Thus, we can write the spherical harmonics for a P term as a func-

tion of x, y and z. We obtain

-1
Y1 1 = / (x-iy)/r (34a)


Y10 = v3/4 z/r and (34b)


Y11 = -v/8 (x+iy)/r (34c)


















2C
g

4x4


2TC
1g

5x5


2TC
2g


5x5


4A C
2g
1x1


The cubic secular determinant for d3
coupling scheme.


-~ .


2AC
1g

1x1


2AC
2g
1x1


4T C
1g
2x2


Figure 4.
{L,s,xC,X ,x }


4TC
2g
1x1


for the








Our phases will agree with those of Condon and Shortley, i.e., the

spherical harmonics with mR odd and positive are given a minus

sign.

The real forms of these spherical harmonics correspond to the

following combinations:

z Y1 (35a)


y i(Y1 +Y )/I 2 and (35b)


x ~ (Y1-Y1 1)/12 (35c)

These are known as the "p-orbitals" in the case of a single electron

and have the same symmetry properties as z,y and x.

Let us examine how these cartesian functions z,y and x will

be affected by the rotation operations of Oh symmetry. We will

take the z axis as the principal rotation axis, i.e., C4(z). The

other rotations, each representing a class, are C2 = C42(z), a C2

colinear with the y axis which we shall call C2'(y), the threefold

body diagonal rotation C3(z') and, of course, the identity operation

E. Using counter clockwise rotations and "replaced by" operations, the

transformation properties of z,y and x are given below.


z z

y C4(z) x (36a)
x -y


E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra,
University Press, Cambridge, 1951.










C 2(
C2 = C42(z)





C2' (y)





C3(z')






E


iz

-y
-x


-z


-X


(36b)





(36c)





(36d)





(36e)


The matrices which will accomplish the above transformations

when applied to y are listed in Table 12 underneath the appro-

priate symmetry operations. These matrices constitute a representation

of the basis functions z,y and x. The characters of each matrix are

also given in Table 12. By referring back to Table 10(b) we see that

z,y and x form a basis for the irreducible representation T1g of

Oh symmetry.
Since z,y and x form a basis for the irreducible representation

T1g, we can use the real forms of the spherical harmonics which cor-

respond to z,y and x, i.e., (28), as the wave functions for the

triply degenerate T1g representation that arises from P. These are

the cubic oriented spherical haimonics for L = 1.








TABLE 12

Matrices and Characters of the Representation
Spanned by (z,y,x) in Oh Symmetry

E C4(z) C2EC42(z) C21(y) C3(z')


1 0 0\ 1 0 0 1 0 0 -1 0 0 0/ 1 0

(z,y,x) 0 1 0 0 1 0 -1 0 0 1 0 0 0 1
0 1 0 -1 0 0 0 0 0 -1 1 0 0


X(R) 3 1 -1 -1 0


Immediately, the question arises as to which is the a, b and

c component of T C(P). Though the choice is completely arbitrary

it is important that the convention chosen is adhered to strictly in

all the following calculations.

First, let us see how the T1g representation decomposes when

the symmetry is lowered to tetragonal. It is necessary to consider

this now because it will play a role in our choosing of the a, b

and c components of Tlg,

Here, we need to consider only those rotations that are common

to Oh and D4h. These are the C4(z), C2 C42(z), C2 (y) and

E. It is obvious that the transformation properties of z,y and x

will not differ in going from Oh to D4h for these operations. But

the representation spanned by them may not be irreducible in D4h.

The characters for the basis (z,y,x) and for the irreducible represen-

tations of D4h for the operations we are dealing with are compared

in Table 13.









TABLE 13

Character Table for the Representation for
Which (z,y,x) is a Basis in D4h and
for the Irreducible Representations
of D4h


X(E) X(C4) X(C2) X(C21)

(z,y,x) 3 1 -1 -1

Alg 1 1 1 1

A2g 1 1 1 -1

B1g 1 -1 1 1

B2g 1 -1 1 -1

E 2 0 -2 0



The cartesian axes, (z,y,x), do not form a basis for an ir-

reducible representation in D4h. This is obvious since D4h has

no representation of dimension greater than two. Decomposing we find

(z,y,x) is composed of A2g and E .
Reviewing the transformation properties of z in D4h (cf.

equns. 36a-c), which are the same as those in Oh for the rotations

common to the two point groups, we see that z always transforms as

plus or minus of itself while y and x mix. This indicates that

z alone is the basis for the singly degenerate A2g representation

and (y,x) together form the basis for the doubly degenerate E
representation. This is easily confirmed by comparing the characters
representation. This is easily confirmed by comparing the characters








of z and (y,x) with those of A2g and Eg under the rotations

of D4h.

Now we can state our convention for choosing the a, b and c

components of T1g. First we choose z as the basis for the a com-

ponent of T1g. Thus, when the symmetry is lowered to D4h we have

TC -
Tga 2g

Next we decide that under C3(z') of Oh a will be replaced

by b, b by c, and c by a. From (29d) we see that C3(z')

replaces z by y, y by x and x by z. Hence, y and x

become the b and c components, respectively, of T1g. Further,

we specify that the b component of T1g becomes the a component

of E and the c component of T1g becomes the b component of

E when the symmetry is lowered to tetragonal.

That is

TC E and
Igb ga


C Q
T E
Igc b "


Finally, in Table 14 below we have summarized the wave functions,

the real forms of the spherical harmonics, for T1g from P and the

decomposition properties of T1g when the symmetry is lowered to

tetragonal.








TABLE 14

Wave Functions and Decomposition
Properties of TC (P)

Cartesian Oh D4h
Wave Function Representation Representation Representation

Y z TC A
1 Iga 2g

(YV" 1+Y1 )//2 y TC E
1 1gb ga

(Y 1"-Y 1)/V2 x TC EQ
1 1gc gb



The wave functions and decomposition of the other terms are

found in exactly the same manner. A complete list of the cubic

oriented spherical harmonics corresponding to the cubic irreducible

representations of all the d3 terms is given in Table 15. We have

used the shorthand notation IL,ML> for the spherical harmonics in

Table 15.

In Table 16 the decomposition of the cubic levels ingoing to

D4h is given.

To obtain the cubic oriented spherical harmonics in microstate

form one inserts the IL,S;ML,MS> functions (cf. Table 6) for each

corresponding IL,ML> spherical harmonic. We choose the maximum

MS for each IL,S;ML,MS> function which is to be substituted for

IL,ML>'s in writing down the cubic oriented spherical harmonics in
microstate form. The cubic oriented spherical harmonics in microstate

form are listed in Table 17.


































-it

C
















0
*-







































L-
a)

S.-
a)







C.
1-p
S.




Q1
c,
C:
*p


S A




+
A I
AA

I A
O r- *I


S-

t-

Cn







r4-





o

LO E
I- -

L.0

.J -
CL









0
U
-- 0r-









a3)

I-
< U








-0




u






-C.


ro

r-"
1-

ar


(31 N
C> 0







S -o

E 5-
L- 0
a)
Cci


A C A A-

I NCM C\j
A A



(,J I -- +
-- A I A
+ N A r-
A I r- I
N I C
O N "
- -- --
C*M *


CY)
(N




(N
S.


L L

0 0

N (N


T~i

cM
C\D


SN
X X


.0) 0') 0'


:j N C\j


N


N
















u
0')
(NJ

(NI


























A
SA
i A A
a e CY
CD n A
n oo n
+ -
C: A I + I
A +
0 A A A
*r~ C C) CV





(v +N-
A Y)
UJ -- a CD
S- D C V





A CA)
Sj A + --- +
SI A N.
A n a A
S- A n La





















CD
a +( n o CV+






















-rA
A n C" -- rQ









CJ + C










SA L ) L -


a N i C) C) N 10
C, C) CN X N



nC N \ ->-- ---
e -- m- N







S.-

C-)


S S- S -

S S. S.. S.. M.. S. M.. M .
> 0 0
C) w1)
-j CJl Cn C" C CC C i
< N- I- t-


U-


S.-
S- 0
l- CM


1 1c

A A
A A



11





A A


O O
a .
^- -

















I
N 1









N N



O N
CD C













CT) N










Lii
Cj M





























A A
c-. n ^-
A IA










+
,a- + 3 ? -
--- A -- +
A a I A



A A +
+ A A
- A a -
Id Cd +~ C















- I C a A N A
A A1

+ A A -- A
A I Cd I -

-- N N -


















r r-^ -" -h



X N I I I


v N N
SN N


A I
A A




A I
A
CM
A t
In Ln
















CM

C I
1d I

N CM


S7 N


i _r .0 .U

S> N C CM C
(U L -- I-- I- I-- LU UJ
-_ N CM r\ CM NJ CM CM CN CM









aS
CM







70












A
A

Ln I A A
A
+ an "
A L *
m I LO
S A +
A I
LO I
n I
A LAO
C M o LO 'I -. 0 LA .0
O 0 rI -
*r- + 0- I
C A +
C LO A A A C A A A
SU- LA O- A -
LU- LO *c a LA a
L Ln n L .0 LA L L
a + -- A L A
> A- I+ + + I C- j
0 Ln A A I A A I A I
I LO r- LA A -
a jI LA LA I LA
LO LA LO + LA LO LO +



S. -- O A
V) >. C A >LL









I




-j A M .J

LA LALA LO LO LO
0 C. j N N
5 &- N N N I
C V C C N N N1
A N N N N~ N N

S N dxX X x
0 N N CI C) CV) C

i I I I


S--*r- -- LA X X- N
l- N --x
S- Cco C L LA L





m CM ON CM --(M -M -cm- -U)





N N N
S1- I- 1- 1- -+ X-











-) N NC CN CN CN N
-1 N^ C1 CM CM1 CM C\J CM







71













A A
(? (A

t A n 0




LO
SI -
A C 0)



Sco co 4- ( S.

I 'h + 4 a 4 )
U A Or
C 10 A A A c-
=3 n -
LL L..I. -0 *





















U-a
) + -- LA-- S -
> A + -- 0
t3 LOn A I A 4


Sn n LO LO

-n Ca i 4-)


- .r- I -- +
3 cO.) c)
C ( S-






*r .(-





F- 0 -

S0 a0 )-





4-) I Ir0-
S- N '- o 4-
SM 0 1^ C:


4 =n
p S- -. -0 ci)


53 I I 1 -


ft -

c i) (4
0.0 -a s..



(A Y --
(- a )
O) 0 f)
(2 4-' Q
o -o





I *

I- .'-











TABLE 16
Decomposition of the Cubic Representations
in Going to Tetragonal Symmetry

Cubic Representation Tetragonal Representation

Alg Alg

A2g Blg

Ega Alg

Egb B1g

Tlga A2g

Tlgb Ega

Tlgc Egb

T2ga B2g

T2gb Ega

T2gc Egb








TABLE 17

The Cubic Oriented Spherical Harmonics
in Microstate Form for


the d3


Terms


= (-4(20-2)-3/6(11-2)+8(20-2)-4(20-2)


- (10-1)+3/6(2-1-1)+2(1-1)-(10-1))/v'210

= i(-4/3(21-2)+2/3(2I-2)+2/3(21-2)

+ 4/2(20-1)+/2(20-1)-5/2(20-1)-3/3(11-1)-3/3(100)

+ 2 /3(2-1-2)-5/2(10-2)+4v2(10-2)+/2(10-2)+2/3(2-1-2)

- 4/3(2-1-2)+3/3(1-1-1+33/3(00-1))//420


2TC = 2E
Igc gb


= (4v'3(21--2)-2/3(2i-2)-2V3(21-2)


- 4/2(20-1)-/2(20-1)+5/2(20-1)+3/3(11-1)+3/3(100)

+ 2V3(2-1-2)-5/2(10-2)+4/2(10-2)+/2(10-2)+2/3(2-1-2)

- 4/3(2-1-2)+3v3(1-1-1)+3/3(O0-1))//420


2 C
ga

2 C
gb


2
T2ga


= 2Ag
ig

S2 Q
S lg


2 Q
S 2g


D-a


= ((10-I)+(20-2)-(10-1)-(20-2))/2

= ((21-1)+(200)-(22-2)-(21-1)+(T-1-2)

+ (00-2)-(2-2-2)-(1-1-2))//8

= i((I-1-2)+(00-2)-(2-2-2)-(1-1-2)

- (21-1)-(200)+(22-2)+(21-1))//8


2 C
iga


2TC
1gb


= A2g
2g


2Q
ga








TABLE 17--Continued


2g = ((1--)+(2-1-2)+(00-)-(2-1-2)-(1 -1)
2gb = Ega
(100)-(21-2)+(21-2))//8

2 TC 2 Q

+ (00)+(21-2)-(21-2))/ -8

2D-b


2 AQ
=
ig


2 BQ
= 28g
g


= (3(20-2)-2V6(11-2)-3(20-1)-3(10-1)

+ 3(10-i)-2/6(2-1-i))//84

= (4(1-1-2)-5(2-2-2)-3(00-2)-(1-1-2)-3(i-1-2)

- 2v6(0--1-)+3(21-1)-5(22-2)+(21-1)-4(21-1)-3(200)

- 2J6(110))/2/42


2TC 2 Q
2ga 2g


2T C
2gb


2 C
Tgc
2gc


2Q
ga


2Q
= Egb
gb


= i(4(1-1-2)-5(2-2-2)-3(00-2)-(1-1-2)


- 3(I-1-2)-2/6(0-1-I)-3(21-i)+5(22-2)-(2l-1)+4(21-1)

+ 3(200)+2/6(110))/2/42

= (2V6(10-2)-(2-1-2)-2/6(10-2)-3(2-1-2)

+ 3(00-1)-(1-1-)+4(2-i-2)-3(21-2)-(21-2)+4(21-2)

+ (11-1)-3(100)+2/6(20-1)-2v/6(2-1))/2/42


= i(2V6(10-2)-(2-1-2)-2/6(10-2)


- 3(2-1-2)+3(00-1)-(1-1-1)+4(2-1-2)+3(21-2)+(21-2)

- 4(2i-2)-(1I-1)+3(100)-2/6(20-1)+2/6(20-1))/2/42


2C
Ega
ga


2 C
gb








TABLE 17--Continued

2F


2AC = 2 Q
2g = Ig


2TC
Iga


2TC
lgb


= 2A
2g


- 2 Q
ga


2TC = 2EQ
Igc gb


= i(/6(0-1-i)+(1-1-2)-2(2-2-2)-(1-1-2)
- (2i-1)-2(22-2)+/6(1i0)+(21-i))/2/6

= (2(10-1)-4(10-1)+/6(1l-2)-/6(2-1-I)

+ 4(20-2)-2(20-2)+2(10-i)-2(20-2))//60


= -i(-2/6(11-1)+3(20-1)+3/6(2I-2)-2/6(21-2)
+ 3/6(100)-3(20-1)-V6(21-2)+2/6(1-1-i)-3/6(00-1)-3(10-2)

+ 3(10-2)+3(2--)-6(2-1-2)-2/6(2-1--2)+10(-1-2)
- 5(0-1-2)-5/6(--1-2 -5-1-2)+5/6(22-1)+10(210)
- 5(210)-5(210))/8V15

= (-2V6(11-1)+3(20-1)+3/6(21-2)-2/6(21-2)

+ 3/6(100)-3(20-1)-/6(21-2)-2/6(1-1-1)+3/6(00-1)
+ 3(10-2)-3(1l-2)-3/6(2-2)6(2-1-)+6(2-1-2)+6(-1-2)
+ 10(0-1-2)-5(0-I-2)-5/6(1-2-2---2)-5-1-2)-5/6(22-1)
- 10(210)+5(210)+5(210))/8/15


2TC = 28Q
2ga 2g


2TC 2E
2gb ga


((21-1)+2(22-2)-I6(1I0)-(21-i)+/6(0-1-i)
(1-1-2)-2(2-2-2)-(I-1-2))/2/6


= i(-2/2(11-1)+/3(20-1)+3/2(2i-2)-2/2(21-2)
+ 3/2(100)-/3(20-1)-J2(21-2)+2/2(1-1-I)-3/2(00-1)
- 13(10-2)+/3(10-2)+ 3/2(2-(-2)+-2(2-1-2)
- V2(2-1-2)-2-23(-1- 3(--2)++3/2(1-2-2)+v/3(-1-2)
- 3/2(22-1)-2/3(210)+/3(210)+/3(210))/4/12








TABLE 17--Continued


2Tg = 2EQ = (-2v2(li-1)+/3(20-1)+3v2(21-2)
2gc gb
2/2(21-2)+3/2(100)-v3(20-I)-/2(21-2)-2/2(1-1-1)
+ 3/2(00-1)+v'3(10-2)- 3(10-2)-3/2(2-1-2)
+ 2/2(2-1-2)+/2(2-1-2)+2/3(0-1-2)-/3(0-1-2)-3/2(1-2-2)
13(O-1-2)-32(2-1)-23(210)+3(210)+/3(210))/412

2G


2A
lg


2 AQ
= g
"ig


2EC 2 Q
ga lg


2EC
gb




2TC
T
Iga


2 C
Igb


- 2A
2g


2Q
ga


= (v2(10-i)+/3(2-1-i)+/2(20-2)-v2(10-1)
+ 33(11-2)-V2(20-1)+V2(2I2)+/3(211)+V3(-1-1-2)
+ /2(0-2-2))/2/6

= (10(10-1)+5/6(2-1-I)+10(20-2)-10(10-1)
+ 5/6(11-2)-10(20-2)-14(220)-7/6(211)-7v6(-1-I-2)
- 14(0-2-2))/4/105

= (-4/3(200)-3/3(21-I)-72(1i0)
+ 4/3(21-1)-323(2-1)-23(22-2)-2(0-1-)+3(1-I-2)
- 4V3(1-1-2)-2/3(2-2-2)-4/3(00-2)+ 3/3(1-1-2))/2/70

= i(/3(-1-!-2)+/2(0-2-2)-/2(220)
- v3(211))//10

= (2/6(1-1-1)+4(10-2)-/6(2-i-2)
+ 2/6(2-1-2)+/6(00-1)-7(10-2)+3(10-2)-V6(2-1-2)
- /6(100)+3(20-1)-7(20-1)-v6(21-2)-2/6(11-1)
+ 4(20-1)+2/6(21-2)-/6(21-2)+3(210)-2(210)








TABLE 17--Continued


- (210)+/6(22-1)-(0-1-2)-2(0-1-2)-,6(1-2-2)
+ 3(0-1-2))/8/5


2TC
T gc
1gc


2=
gb


2TC = 2Q
2ga 2g


2TC
2gb


2 Q
ga


2TC = 2Q
2gc gb


= i(2/6(1-1-1)+4(10-2)-/6(2-I-2)
+ 2/6(2-1-2)+/6(00-1)-7(10-2)+3(10-2)-/6(2-1-2)

+ 6(100)-3(20-1)+7(20-1)+/6(21-2)+2/6(li-1)
- 4(20-1)-2/6(21-2)+/6(2I-2)+3(210)-2(210)-(2I0)

+ J6(22-1)+(0-I-2)+2(0-1-2)+/61-2-2-)-3(0-1-2))/8/5

= i(/2(0-1-1)-/3(1-1-2)+4/3(1-1-2)
+ 2/3(2-2-2)+4J3(00-2)-3/3(1-1-2)-4/3(200)
- 3/3(21-1)-/2(1i0)+4/3(21-1)-/3(21-1)-2/3(22-2))/2v70

= (/6(100)-3(20-1)+7(20-i)+/6(21-2)
+ 2/6(11-1)-4(20-1)-2/6(21-2-2)+6(22)-2/6(1-1-)
- 4(10-2)+/6(2-1)-)-2/6(2-1-2)-6(00-1)+7(0-2)
- 3(10-2)+/6(2-1-2)+21(210)-14(210)-7(210)+7/6(22-1)
- 7(0-1-2)-14(0-1-2)-7/6(1-2-2)+21(0-1-2))/8/35

= i(/6(100)-3(20-1)+7(20-1)+/6(21-2)
+ 2/6(11-1)-4(20-1)-2/6(21-2)+/6(21-2)+2v6(1-1-I)+4(10-2)
- /6(2-I-2)+2/6(2-1-2)+/6(00-1)-7(10-2)+3(10-2)
- /6(2-1-2)-21(210)+14(210)+7(210)-7/6(22-1)
- 7(0-i-2)-14(0-1-2)-7/6(1-2-2)+21(0-1-2))/8/35








TABLE 17--Continued


2EC
ga


2C
Egb


= 2Q
ig


2 C 1=2A
2Tga- 2Ag


2TC i2.-0
1gb- J2ca











2TC -1=2EQ
Igc gb


=-i (3(220)-J2(211)+/3(0-2-2)

- /2(-1-1-2))//10

= i(/6(110)-3(200)-(21-1)+3(2i-1)
- 2(21-1)+(22-2)+/6(0-1-1)-3(00-2)+2(1-1-2)
- 3(1-1-2)+(1-1-2)-(2-2-2))//60

= (4(10-1)-2(10-1)+/6(1-2)
-/6(2-1-)-(20-2)+2(20-2)-(20-2)-2(10-i))//42

= i(-21/3(-1-2-2)+21/3(221)

- 7/2(0-1-2)-72(--1-2)-7v3(1-2-2)+7/3(22-1)+14v2(0-1-2)
- 7/2(210)+14/2(210)-7/2(210)+6/3(00-1)-8/3(1-1-1)
+ 9/2(10-2)-6J2(10-2)+4/3(2-i-2)-3/3(2-1-2)

- 3J2(10-2)-J3(2-1-2)+8/3(11 6V1-)-63(10)+9/2(20-1)
- 6/2(20-1)-/3(21-2)+4J3(21-2)-3/2(20-1)
- 3/3(21-2))/16J21

= (-21J3(-1-2-2)-21J3(221)-14V2(0-I-2)

+ 7/2(0-1-2)+7/2(0-1-2)+7/3(1-2-2)+7/3(22-1)
- 7J2(210)+14/2(210)-7/2(210)+6/3(00-1)-8/3(1-1-1)
+ 9/2(10-2)-6i/2(10-2)+4/3(2-1-2)-3/3(2-1-2)-3/2(10-2)
- /3(2-1-2)-8/3(1T-1)+6/3(100)-9/2(20-1)+6/2(20-1)
+ /3(21-2)-4/3(21-2)+3/2(20-1)+3/3(21-2))/16/21


2-
H








TABLE 17--Continued


2TC -2=-2A
iga 2g


2TC -2=2EQ
Igb ga









2TC -2=2E
Igc gb









2TC 2Bg
2ga 2g




2TC 2Q
2gb ga


= (/3(220)-/2(211)-/3(0-2-2)
+ /2(-1-I-2))//10

= i(-5/3(-1-2-2)+5(221)-18/2(0-1-2)+9/2(0-1-2)
+ 9/2(0-1-2)+9/3(1-2-2)-9/3(22-1)+9/2(210)-18/2(210)
+ 9/2(210)+6/3(00-1)-8/3(1-1-1)+9/2(10-2)-6/2(i0-2)
+ 4/3(2-1-2)-3/3(2-1-2)-3/2(10-2)-v3(2-1-2)+8/3(1i-1)
- 6/3(100)+9V2(20-1)-6/2(20-1)-/3(21-2)+4/3(21-2)
- 3/2(20-1)-3/3(21-2))/16/15

= (-5/3(-1-2-2)-5/3(221)+18/2(0-I-2)
- 9/2(0-1-2)-9/2(0-1-2)-9/3(1-2-2)-9/3(22-1)+9/2(210)
- 18/2(210)+9/2(210)+6/3(00-1)-8/3(1-1-1)+9/2(10-2)
- 6/2(10-2)+4/3(2-1-2)-3/3(2-1-2)-3/2(10-2)-/3(2-1-2)
- 8/3(11-1)+6/3(100)-9/2(20-1)+6/2(20-1)+/3(21-2)
- 4/3(21-2)+3/2(20-1)+3/3(21-2))/16/15

= (/6(1i0)-3(200)-(21-i)+3(21-1)-2(21-1)
+ (22-2)-/6(0-1-I)+3(00-2)-2(1-1-2)+3(1-1-2)-(I-1-2)
- (2-2-2))//60

= -i(30(-1-2-2)-30(221)+4/6(0-I-2)
- 2/6(0-1-2)-2/6(0-1-2)-6(1-2-2)+6(221)-2/6(210)
+ 4/6(210)-2/6(210)+12(00-1)-16(1-1-i)+6/6(10-2)
- 4/6(10-2)+8(2-1-2)-6(2-1-2)-2V6(10-2)-2(2-1-2)
+ 16(11-1)-12(100)+6/6(20-1)-4/6(20-I)-2(21-2)+8(21-2)
- 2/6(20-1)-6(21-2))/16/15








TABLE 17--Continued


= (-30(-1-2-2)-30(221)+4/6(0-i-2)-2/6(0-1-2)


- 2/6(O-1-2)-6(1-2-2)-6(22-1)+2V6(210)-4v6(210)+2/6(210)

- 12(00-1)+16(1-1-1)-6V6(10-2)+4/6(10-2)-8(2-1-2)+6(2-1-2)

+ 2/6(10-2)+2(2-1-2)+16(11-1)-12(100)+6v6(20-1)-4V6(20-I)

- 2(21-2)+8(21-2)-2/6(20-1)-6(21-2))/16/15


= (2(10-1)-(20-1))//5

= i(v2(10-2)-/3(2-1-2)+/2(20-1)


- /3(21-2))//10

= (v2(10-2)-/3(2-1-2)-/2(20-1)+/3(21-2))//10


4Ag
2g


4TCa 4AQ
igb 2g

Igb ga


4TC = 4EQb
Igc gb


= i((1-1-2)-(21-1)//2

= ((10-1)+2(20-2))//5

= -i(3(20-1)+v6(21-2)+3(10-2)+/6(2-1-2)

+ 5(0-1-2)+5(210))/4/5

= (3(20-1)+/v6(21-2)-3(10-2)-/6(2-1-2)


+ 5(0-1-2)-5(210))/4/5


2TC
2gc


=2E
gb


4TCg
4 ga
4TC
Igb


4TC
Igc


=4A
42g

= 4Q
ga


S4E
gb


4F


4B








The cubic ligand field potential

The cubic potential, Vc E VOh, and the added axial potential,

Va, that constitute the quadrate field arise due to the coulombic

interactions between the central ion's d electrons and the charged

ligands. Thus, the general form of the ligand field potential is

given by (3).

In this coupling scheme, {L,S,XC,XQ,rQ}, we treat V and V
c a
as individual perturbations. Let us first consider the form of the

ligand field potential for an octahedral arrangement of the ligands.

As always, the ligands are assumed to be fixed in space at equal

distances, a, along the three cartesian axis. The charge of the

octahedral ligands is taken as -qk. Utilizing the expansion of

1/rik we have

= +R 4 r
Vc = qe[ (- ) 4
c k Z =0 m =- ( 1 r


Yt ( i)Y (ek' k)] (37)


If we consider the electron to be somewhere between the central

ion and the ligand (cf. Fig. 5), then r< is equal to ri, the

distance between the central ion and the electron, and r> is equal

to a, the central ion-ligand separation.

In Fig. 6 we have numbered the ligands and show their locations

along the cartesian axis.

Because the ligands are taken as fixed in space,the angles ek

and hk are constant for each ligand. Therefore, we can calculate










d electron
r.

Sa-
central ligand
ion


Position of a


d Electron in a Complex


'3


1

v6
I


Figure 6. The Numbering of the Ligands for an Octahedral Complex


Figure 5.


X4
1

x







m
the spherical harmonics for the ligands, i.e., Y, (ek, k). The

octahedral potential is thus given by

o + m
V0 = qke[ X DkY (6i',i)] (38)
h k Z=0 m =-z


where
*
4 r m (39)
D = __ 1 -1( a+- ) Y (9kk ) (39)
"k (2Z+1) a+1 ;
a


with

mz (2k+1) (Z- m Zl). 1/2 m im O
Y (k'k) I 2 (t +m P (cose),/1/2T e (40)



The Legendre polynomials have the form


m m 2 I- dImZI 0
P (cose) Pz (z) = (1-z2) 2 dm P. (z) and (41a)
dz


p0(z) 1 d 2_
P (z)- d (z2- 1)~ (41b)
2 ( 2l dz(


For both cubic and tetragonal symmetries the potential must be

the same at the points p() and p(,+T/2). We know that the (

dependent part of the spherical harmonics is given by


im .


~1/Z-r~e


{42)








Therefore,

im 0 im ( +T/2) im imZ(r/2)
e =e =e *e

im (Tr/2)
S= e and

m ZT m 7r
1 = cos( 2 ) + i sin( 2 ) .(43)


To satisfy (43) m, is restricted to values of 0, 4, 8, etc.

The product of the spherical harmonics for two d electrons,

i.e., the spherical harmonics with Z = 2, when written as a sum of

spherical harmonics can contain only terms with Z = 0, 2 and 4.

Thus, in the expansion of the ligand field potential, (38), we

need only consider those spherical harmonics with a equal to 0, 2

and 4 and mI equal to 0 and 4.

Since, as we have noted, the ligands are fixed in space we can

calculate Dk using the constant values of the polar angles 6k and

hk and the sine and cosine of the two polar angles for the six cubic
ligands as given in Table 18. Hence, the final form of the potential

will contain only those spherical harmonics of the electron for which

Dk is not zero. Evaluating (38) we obtain for the octahedral po-

tential


0 + (0 5 -4] (44
VOh q ke/[(12/a)YO + (7r/3a )Y4 + ( r/6a )(Y4+Y4 )] (44)
h i








TABLE 18

The 6 And p Values For The Ligands
Arranged in an Octahedron


Ligand a 4sine cosO sinp cost

1 w/2 0 1 0 0 1

2 T/2 r/2 1 0 1 0

3 T/2 T 1 0 0 -1

4 i/2 3T/2 1 0 -1 0

5 0 0 0 1 0 1

6 T 0 0 -1 0 1


0 0 0
The Y term is a constant, i.e., Y = /4-,. This Y term in
00 0
the potential, though of great importance in calculating thermodynamic

quantities,l can be neglected in deriving the spectral transition

energies since it makes a constant contribution to all the levels.

Thus, the final form of our cubic potential is


V qke(r /a)[(7/3)Y +(4 7/6)(YY4)] (45)
h i


The wave functions are linear combinations of the microstates.

The microstates are product functions of the single electron orbitals

that obey the Pauli principle and are Slater determinants written in

shorthand notation as i1(1).2(2)iP3(3) for d3 electronic


C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw-
Hill Book Co., New York, 1962.








configuration. The cubic potential is summed over electrons one, two
and three. That is


VOh = V0h(1) + VOh(2) + VOh(3)
Voh + Vh Vh


(46)


Therefore, the general form of the integrals with which we need
concern ourselves here is


/ / / $1(1)2(2)i3(3)iV0 (1) + V0 (2) + V0 (3)1
1 2 3 h h h


p4(1)p5(2)'6(3)dT2dT2dT3


(47a)


This becomes


11 '1(1)VOh(1) 94(1)dT1 2 02(2)05(2)dT2 "


f3 13(3)Y6(3)dT3 + f2 i2(2)VOh (2) 15(2)dT2


fl 41(1)4(1)dT1 f3 13(3) 6(3)dT3 +


/3 -3(3) IVOh(3) 16(3)dT3 f1 0(l)4(l)dTI f2 02(2)5(2)dr2 .


(47b)


Due to the orthonormality
(47) will be zero unless

i3 = 06 and/or '1 4 = 4


of the single electron functions we see that

2 = i5 and '3 = i6 and/or p = '4 and
and 2 = r5'








Thus, the matrix elements reduce to single electron integral.

Since the radial part, which we call R3d, and Z, which is equal to
2 for the d-electrons, are constant for the microstates these integrals
will be distinguishable only by the different mZ values. Therefore,
the single d electron integrals we must calculate are <21Vhl 2>,
<2 Vhl 1>, <21Vh |0>, <21VOh 1>, <2 VOh 12>, <1 VOh 1>,

<1 V0 0>, <11 V0 11> and <0 V0 0>.
h h h
2 2 1 +1
Only the spherical harmonics products Y2 Y2 Y Y2
0* 0 2* v2 0 4
Y2 Y 2 and Y2 Y2 contain Y4 or Y 4 (cf. Table 7), which
are the spherical harmonics present in the cubic potential (45).
Therefore, only the integrals containing these products can be non-
zero. These integrals are <21V0h 2>, <1V0h 1>, <0|VO h0> and
<21VOhl 2>.

As an illustration of the method of 'solving' the four remaining
integrals let us take as an example <21Vh 12>. Thus,


<21VOh2> = / R3d 2 V R3dY2d (48a)


From Table 7 we obtain, neglecting the constant Y 0 term,
*

Y Y2 = Y4 /14/ ~ Y2 /7/v .
2 2 4 2

Expanding (48a) we obtain

4
<* r.
<21V0 2> = / R3d R3d(qke)( -)
h o a








2x O* *
r2dr f / [Y /14/T /5 Y2 /7/] *
oo 4 -


[7A/ Y40/3 + V/70(Y44+Y 44)/6sineded) (48b)


Because of the orthonormality of the spherical harmonics (48b) reduces

to


<2|VOh 2> = (qke/6a5) R3d Rd(ri4)r2dr (48c)


<21V Oh2> = (qke/6a5) (48d)


As is the usual procedure in ligand field calculations, we will
4
make no effort to solve for the average value of r4. Rather, we
define the single electron matrix element to be the well-known cubic
parameter Dq, that is,


Dq <2 VOh 2> = (qke/6a5) (49)


Because the octahedral potential contains only fourth order

harmonics, i.e., Z = 4, we need only one parameter, Dq, for the
cubic case.
The other non-zero integrals are calculated in the same fashion

and they are given below.

<21VOh 2> = Dq (50a)








<1 V hl1> = -4Dq (50b)


<01VOh 0> = 6Dq (50c)


<21VOhl 2> = 5Dq (50d)


Using the single electron integrals tabulated above we can easily

find the values of the d3 matrix elements in terms of the cubic

field parameter Dq. The energy matrices thus obtained from the block

diagonalized secular determinant, (cf. Fig. 4) are listed in Table 19.

The Axial Ligand Field

In order to describe the tetragonal potential as the sum of a

cubic and an axial field we must define the charge on the axial

ligands, which we will call q', to be


q' = qa e (51)

where qa is the charge on the two axial ligands and qe is the

charge on the four equatorial ligands of the tetragonal complex, (cf.

Fig. 1).

The form of the axial potential is analogous to that of the cubic

potential and is given by

m +Z mt
Va = q'e[ I Y k Y (i,i)] (52)
k l=0 m =-S

For d electrons, as we have noted, only those terms in the summation








TABLE 19

The Elements of the Cubic Energy Matrices (the Hik's)
for d Electronic Configuration in the
{L,S,XC,XQ,rQl Coupling Scheme


2AC
1g

j k Hjk


2G 2G -11B+3C-2Dq

2AC




2EC
g

j k Hjk j k Hjk

2D-a 2D-a 7B+7C+3Dq 2D-b 2G 24/5Dq/7

20-a 2D-b 3/21B-15Dq//21 2D-b 2H ilODq//70

2D-a 2G -60Dq//105 2G 2G -11B+3C-2Dq/7

2D-a 2H -i/30Dq G 2H i36Dq//14

2D-b 2D-b 3B+3C-19Dq/7 2H 2H -6B+3C+2Dq

2TC
1g


-6B+3C


il8Dq//42


2p 2G
2P -1(2H)


22Dq//14


2/5Dq








TABLE 19--Continued


S k Hjk j k Hjk

2P -2(2H) 6Dq//7 2G -1(2H) i/210Dq/7

2F 2F 9B+3C+Dq 2G -2(2H) i3/6Dq

2F 2G -i/3q -1(2H) -1(2H) -6B+3C-2Dq

2F -1(2H) 10Dq//70 -1(2H) -2(2H) -2v35Dq/7

2F -2(2H) -/2Dq -2(2H) -2(2H) -6B+3C+2Dq

2G 2G -11B+3C-Dq

2T-C
2g

2D-a 2D-a 7B+7C-2Dq 2D-b 2H i40Dq/3/70

2D-a 2D-b 3/21B+10Dq//21 2F 2F 9B+3C-Dq/3

2D-a 2F -ilODq//3 2F 2G -il5Dq//35

2D-a 2G 30Dq//105 2F 2H -VlODq/3

2D-a 2H -i400q//30 2G 2G -11B+3C+13Dq/7

2D-b 2D-b 3B+3C+38Dq/21 2G 2H -i6Dq//14

2D-b 2F -i20Dq/3/7 2H 2H -6B+3C-4Dq/3

2D-b 2G -12/5Dq/7

4AC
2g

j k Hjk

4F 4F -15B-12Dq








TABLE 19--Continued


4TC
1g

j k Hjk


4p 4p 0

4p 4F -4Dq

4F 4F -15B+6Dq

4T C
S-1B-2

4F 4F -15B-2Dq








with Z = 0, 2 and 4 need be considered. Also, since for axial sym-

metry the principal rotation axis is C. the potential must be the

same for all values of p. That is, p(p) = p(p+a) with a taking all

real values. Thus,

imz) im.,((+a)
e = e (53)

which can hold true for all values of a only if m = 0. Hence, the

axial potential can be written as

Va = I q'e[ I Dk'.Y (i,i) (54a)
k Z<4

with
-z
D I4 r ) Y (kRk) (54b)
k (2k+1) a+1 Z k'Y k
a

The axial field potential for the two ligands turns out to be



Va = q'e7[(4/a)Y00 + (42/5a3)Y20 + (4ri4/3a5)Y40] (55a)


or, leaving out the constant Y,0 term,


Va = q'ev/7[4ri2/a3 )20 + (4ri4/3a5)Y40] (55b)


Y2*2 1* 1
Only the spherical harmonics products Y22 Y22, Y 21 Y2
0* 0 c o
and Y20*Y20 contain Y20 or Y4, (cf. Table 7), which are the

spherical harmonics present in the axial potential (55b). Therefore,








only the integrals containing these products can be non-zero. These

integrals are <21Val2>, and .

As an illustration of the method of 'solving' these three remain-

ing integrals let us take as an example <21Va12>. Thus,


2* 2
<21Vaj2> = a R3d 2 VaIR3dY22 d. (56a)


Expanding (56a) we obtain

2
<21Va 2> = I R3d*R3d(q'e)( 3)r2dr
0a

2-r7r 07 4 0-
f f (-y/ Y2 /7vr)(4/r Y20//5)sinededp +
0 0

4
r 2x 2 *
f R3d R(q'e)( )r2dr f (Y4 /14V/)


(r,/ Y40/3)sinededp (56b)


Because of the orthonormality of the spherical harmonics (56b) reduces

to


<21Va12> = (-4q'e/7a3) + (2a'3/21a5) (56c)


As in the cubic case we will make no effort to solve for the
2 4
average values of ri and r. Rather, we define two new parameters,

Dt and Ds, as follows