Crystal field effect on atomic statesMehdi Amara, Université Joseph-Fourier et Institut Néel, C.N.R.S. BP 166X, F-38042 Grenoble, France

References :Articles

- H. Bethe, Annalen der Physik, 1929, 3, p. 133 (Selected Works of Hans A. Bethe, World Scientific, 1997)- B. Bleaney and K.H.W. Stevens, 1953, Rep. Prog. Phys. 16, p 108- Books- M. Tinkham, Group Theory and Quantum Mechanics, McGraw-Hill, NY, 1964

Question :

Energy levels and wave functionsof atoms

spherical symmetry

permutation properties of an electrons set

What happens if the atom belongs to a crystal, at a site of welldefined, lower than spherical, symmetry ?

"Answer" : Properties of a quantum system in an environment of defined symmetry

Introduction

Direct group theory application : Hans Bethe 1929

Summary

-Back to the free atom

- The atom inserted in a crystalExpansion of the crystal field potentialThe various degrees of crystal field correction:

* Strong crystal field* Intermediate crystal field* Weak Crystal field

- Crystal-field splitting of a multiplet (for a cubic symmetry)Decomposition of the terms of odd multiplicityDecomposition of the terms of even multiplicityRelated theorems : Kramers, Jahn-TellerCrystal Field Operator Equivalent: Stevens method

- Conclusion

Free atom

ψ (r ) = Rn(r).Yl

m(θ ,ϕ )

Radial part(Energy in V(r))

Angular part:Spherical Harmonic

Electron in a spherical potential

nucleus

other electrons

Quantum numbers : n, l, m

V(r )

Hydrogen type Wave functions

Symmetries of the hamiltonian : Full rotational = any angle about any axis

Inversion center = nucleus

Time reversal

Schrödinger equation solutions :

Spherical Harmonics

set of functions with common l = Irreducible representation of the sphere rotation group

Rotation about the z axis :

Ylm(θ ,ϕ ) = (2l +1) ⋅ (l - m)!

4π (l - m)!⋅Pl

m (cosθ) ⋅ eimϕ

transforms into itselfRz(α )Ylm(θ ,ϕ ) = Yl

m(θ ,ϕ −α ) = e−imα Ylm(θ ,ϕ )

Rotation about the x axis : Rx(α )Ylm(θ ,ϕ ) = Ry(−

π2 )Rz (α )Ry(

π2 )Yl

m(θ ,ϕ )

Example l = 1 : set of functions

For a given l, set of 2l+1 functions with m= -l, -l+1,..., l

Y1−1(θ ,ϕ ) = 1

232π

sinθ e−iϕ

Y11(θ ,ϕ ) = −1

232π

sinθ eiϕ

Y10 (θ ,ϕ ) = 1

23πcosθ

Ry(π2 )Y1

1(θ ,ϕ ) = 12Y10 (θ ,ϕ ) + 1

2Y11(θ ,ϕ ) + 1

2Y1−1(θ ,ϕ )

Ry(π2 )Y1

0 (θ ,ϕ ) =) 12Y1−1(θ ,ϕ ) − 1

2Y11(θ ,ϕ )

Rµ (α )Ylm(θ ,ϕ ) = Cµ (α )

′m =−l

l

∑ Yl′m (θ ,ϕ )

Beyond the Hydrogen

L =

li

i∑Total orbital momentum total spin momentum:

S = si

i∑

Unfilled shell with x electrons for 2(2l+1) electronic states

First approximation : the hamiltonian commutes with total orbital and spin momentum

single electron Hartree-Fock generalization :

�

ψ ( r ) = Rnl (r).Ylm(θ ,ϕ)

Antisymmetrized x electrons wave-functions

Energy classification according to quantum numbers L and S : atomic terms 2S+1L

Tideous approach !

Hund's Rules

1) Among all S values consistent with Pauli's principle, the largest is of the lowest energy

2) Among all L values consistent with the first rule, the largest has thelowest energy.

Intra-shell electrostatic interactions

number

letter: S, P, D, F,...

Multiplet Wave Functions

Hund's ground term Atomic state : ψ = ψ L ⊗ ψ S

Orbital wave-function Spin wave-function

crystalline anisotropic environment

Partial lifting of the (2L+1) orbital degeneracy

Spin-orbit coupling :

electrons positions

(2L+1) X (2S+1) degeneracy

�

J = L + S associated J good quantum number

ψ i = L,S,J ,JzGround state Multiplet Wave Functions

Hs.o. = A ⋅L ⋅S =

A2(J(J +1) − S(S +1) − L(L +1))

ψ L

Third rule (Russel-Saunders) :For L and S resulting from Hund's rules, the J value with minimal energy is:J = |L-S|, for a less than half filled subshell (x<2l+1)J = L+S, for a more than half filled subshell (x>2l+1)

(A > 0)(A < 0)

The atom inserted in a crystal

ionic solid covalent solid metal

Unfilled shell

Electrostatic Interactions

= aspherical potential invariant under the symmetries of the crystallographic site

Crystal field hamiltonian : H CEF = −e V(ri )

i=1

x

∑

distant ions : ionic, covalent, metal

electronic exchange: metal, covalent

V(r )

Expansion of the crystal field potential

Spherical harmonics expansion :

V(r ) = Apq r p Yp

q(θ ,ϕ )q=− p

p

∑p=0

∞

∑satisfies Laplace's equation

Where to stop the expansion ?

Matrix elements of the perturbing hamiltonian : ψm(r ) = Rnl (r).Yl

m(θ ,ϕ )Single electron wave-function :

ψm2 H CEF ψm1 = −e Ylm2*(θ ,ϕ ) ⋅Yl

m1(θ ,ϕ ). Rnl (r)2.V(r )dV∫

sum of spherical harmonics of order =< 2l

Terms of order > 2l yield zero contributionTerms with odd parity are forbidden for an inversion centerThe term of order zero doesn't contribute to the splitting

+ symmetry relations between the remaining Apq

V(r ) = V0 + r2 A2

q Y2q(θ ,ϕ )

q=−2

2

∑ + r4 A4q Y4

q(θ ,ϕ )q=−4

4

∑ + r6 A6q Y6

q(θ ,ϕ )q=−6

6

∑ + ...l = 0

l = 1l = 2 l = 3

No crystal field splitting to be expected for l =1

Expansion for the cubic case

x = r sinθ cosϕ, y = r sinθ sinϕ , z = r cosθ

V2 x,y,z( ) = A2xx x2 + A2

yyy2 + A2zzz2 + A2

xyxy + A2yzyz + A2

zxzx = A20(x2 + y2 + z2 ) =A2

0 r2Order 2

Order 4

Order 6

V4 x,y,z( ) = A4xxxxx4 + A4yyyyy4 + A4zzzzz4 + A4xxyyx2y2 + A4yyzzy2z2 + A4zzxxx2z2

= A40 r4 + A4

1 (x2y2 + y2z2 + z2x2 )

V6 x,y,z( ) = A6xxxxxx (x6 + y6 + z6 )+A6xxyyyy(x4 (y2 + z2 ) + y4 (x2 + z2 ) + z4 (x2 + y2 ))+A6xxyyzz (x2y2z2 )=A6

0 r6 + A61 r2(x2y2 + y2z2 + z2x2 )+A6

2 x2y2z2

polar plot

V(r ) = V0 + r2 A2

q Y2q(θ ,ϕ )

q=−2

2

∑ + r4 A4q Y4

q(θ ,ϕ )q=−4

4

∑ + r6 A6q Y6

q(θ ,ϕ )q=−6

6

∑ + ...

Number of cubic crystal field parameters reduced to :0 for l = 11 for l = 2 (iron group ions)2 for l = 3 (lanthanides)

Cartesian coordinates

First crystal field splitting term

V(x,y,z) = V0 +V2 (x,y,z) +V4 (x,y,z) +V6 (x,y,z) + ...

V(x,y,z) = V0 (r) + (A41 +A6

2 r2 ) (x2y2 + y2z2 + z2x2 ) + A62 x2y2z2 + ...

The various degrees of crystal field correction

Intermediate crystal field Weak crystal field

Intra-shell electrostatic interactions: Hee

Spin-Orbit coupling: HSO Crystal Electric Field : HCEF

Sequence of applied perturbations

rmax

atomic shell:x electrons for 2(2l+1) states

Corrective hamiltonian : H = Hee + HCEF + HSO

("Hund's" third rule)

(Hund's rules)

Usually, different magnitudes for Hee, HCEF and HSO

Strong crystal field

HCEF

HSO

Hee

Hee

HCEF

HSO

Hee

HCEF

HSO

inner shell, heavy atomsHCEF < HSO <Hee

Weak crystal field: 4f

In real systems

Lanthanides series

exposed, close to half-filled 3d shell, lighter atoms

Hee >HCEF > HSO

Intermediate crystal field : 3d

Iron group

exposed, nearly empty of filled 3d shell, light atoms

HCEF > Hee > HSO

Strong crystal field : 3d

Iron group

L, S

L, S

L, S, J

L, S, Γ

L, S, J, Γ

L, S, Γ, ΓSO

Crystal-field splitting of a multipletG.T. approach : inside a given energy level, eigen functions of the hamiltonian transform

according to an irreducible representation of the hamiltonian space group.

Number of levels and respective degeneracies of the perturbed hamiltonian

Intermediate crystal field : Crystal Field Hamiltonian acting on themulti electronic orbital wave-functions

free atom quantum state basis

irreducible representations of the hamiltonian point group

{ L, Lz = L , L, Lz = L −1 ,..., L, Lz = −L }

Quantization of orbital momentum + starting hamiltonian of spherical symmetry

Orbital multiplet L for x electrons set of 2L+1 spherical harmonic

ψ L

YLL (θ ,ϕ ),YL

L−1(θ ,ϕ ),...,YL−L (θ ,ϕ ){ }Transform

identically

Share the same representation

Strong crystal field : Starting representation Dl Yll (θ ,ϕ ),Yl

l−1(θ ,ϕ ),...,Yl−l (θ ,ϕ ){ }

Weak crystal field : starting manifold associated with L,S andtotal angular momentum J

J = integer : { J , Jz = J , J , Jz = J −1 ,..., J , Jz = −J } YJ

J (θ ,ϕ ),YJJ−1(θ ,ϕ ),...,YJ

−J (θ ,ϕ ){ }Transforms as

J = half integer no associated set of spherical harmonics

Crystal field acting on a manifold of odd degeneracy

Generalization ?

Starting manifold :l, L or integer J Spherical harmonics representation

{ L, Lz }

{ J , Jz }

Ylm(θ ,ϕ ){ }

Dl Irreducible spherical representation

Point group irreducible representationsCrystal Field Splitting

Γι

Example : Cubic point group

Oh Point group of the octaedra C3, C3

-1

C4, C4-1, C2 = (C4)2

C2

Rotation Classes:EC3C4C2 = (C4)2

C2

Inversion/reflection Classes:II C3 = C6/reflectionI C4 =C4/reflectionI (C4)2 = σhI C2 = σv

one can restrict to the 24 elements of the O rotations group

Oh = O × I

Octaedra rotation group

Parity (-1)l for atomic states

Only half of the rows/columns is useful for the decomposition

Oh has twice as many classes and I.R. as O

l even : decomposition over the positive IRs of Oh

l odd : decomposition over the negative IRs of Oh

24 elements 24 elements

Rotations only : O characters table

-11-103Γ5

1-1-103Γ4

002-12Γ3

-1-1111Γ2

11111Γ1

6 C46 C23 (C4)28 C3E

Construction of the O group characters table

Dimensionality theorem: ni2

i∑ = g

I.R. dimensionGroup orderNumber of I.R. = number of classes

Bethe's notations

5 irreducible representations

Γ1 , Γ2 , Γ3, Γ4 , Γ5 12+12+22+32+ 32 =24

Weighted orthogonality between rowsOrthogonality between columns

Γ1 Γ2 Γ3 Γ4 Γ5

Class multiplication : χΓ (Rn )χΓ (Rm ) =nΓ

gn gmCnmp gp χΓ (Rp )

p∑

mult. constant for Rp in Rn X Rm

Characters for rotation classes

Dl (Rz(α )) =

eilα 0 ... 0

0 ei(l−1)α ... 0... ... ... ...

0 0 ... e−ilα

⎛

⎝

⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟

Rz(α )Ylm(θ ,ϕ ) = e−imα Yl

m(θ ,ϕ )

Spherical harmonics representation

Character of a rotation of angle α for Dl :

Dl

χl (Rz (α )) = Tr(Dl (Rz (α ))) = e−ilαm=−l

l

∑ = e−ilα (eiα )nn=0

2l

∑

χl (R(α )) = e−ilα ei(2l+1)α −1

eiα −1=e−ilαei(2l+1)α /2

eiα /2ei(2l+1)α /2 − e−i(2l+1)α /2

eiα /2 − e−iα /2

χl (R(α )) =sin ((l +1 / 2)α )sin (α / 2)

For a rotation of angle α about any axis

character for a rotation of angle α class

Decomposition of the terms of odd multiplicity Characters of the O group classes for a spherical harmonics representation of dimension 2l+1

χl (R(α )) =sin ((l +1 / 2)α )sin (α / 2)

E: χl(E) = 2l+1

C3 :

C2 or (C4)2:

C4:

Rotations :

χl (R(2π3)) =

sin ( 2π3(l +1 / 2))

sin ( 2π6)

=

Identity :

1 for l = 0, 3, 6,...0 for l = 1, 4,...

-1 for l = 2, 5,...

χl (R(π )) =sin (π (l +1 / 2))

sin (π2)

=1 for l = 0, 2, 4,...

-1 for l = 1, 3, 5,...

χl (R(π4)) =

sin (π4(l +1 / 2))

sin (π8)

=1 for l = 0, 1, 4, 5...

-1 for l = 2, 3, 6,...

aΓ =1g

χl (R).R∑ χ

Γ(R)* =

1g

gi χl (R).i∑ χ

Γ(R)*

Decomposition formula :

group order

class order

Application for l = 0, 1, 2, 3, 4

- 1/41/4-1/801/8Γ5

1/4-1/4-1/801/8Γ4

001/4- 1/31/12Γ3

- 1/4-1/41/81/31/24Γ2

1/41/41/81/31/24Γ1

χ(C4)x gi/g

χ(C2)x gi/g

χ((C4)2)x gi/g

χ( C3)x gi/g

χ( E)x gi/g C41-1-111

C21-11-11

(C4)21-11-11

C301-101

E97531l = 4l = 3l = 2l = 1l = 0

Γ511100Γ411010Γ311100Γ200000Γ110001

Composition table

"Normalized" O group characters table

Spherical harmonics characters

43211

In cubic field, no splitting for l ≤ 1

33211

nb levels

degeneracy

Strong crystal field 3d approach : l = 2 Γ5 (t2g)

Γ3 (eg)

(double group approach)Decomposition of the terms of even multiplicity

Generalization : χJ (R(α )) =sin ((J +1 / 2)α )

sin (α / 2)

Double valued characters :

Bethe's approach : R(2π) and the double group

for J half integral

χJ (R(α + 2π )) = sin ((J +1 / 2)(α + 2π ))sin ((α + 2π ) / 2)

=sin ((J +1 / 2)α )sin (α / 2 + π )

= −χJ (α )

χJ (R(α + 2π )2 ) = χJ (α )

Introduction of a new group element R : R = R(2π ) ≠ E and R(4π )=R2 = E

Double group (octaedral case): ′O = O × R 48 elements in 8 classes

3 additional classes: R, RC3, RC4 3 additional double valued I.R.: Γ6, Γ7, Γ8

dimensions 2 2 4

Double group : O' characters table

0

-110-11

6 C4

0001-10-11

6 C2+ 6 RC2

000-1-1211

3 (C4)2+ 3 R(C4)2

-11100-111

8 C3

42233211E

01-4Γ8

-1-2Γ7

-1-2Γ6

-103Γ5

103Γ4

0-12Γ3

-111Γ2

111Γ1

6 RC48 RC3R

22 − 2

− 2

χJ (R(α )) =sin ((J +1 / 2)α )

sin (α / 2)χJ(E) = 2J+1 Rotations :Identity :Characters of the J manifold representation2π rotation :χJ(R) = -(2J+1)

Composition table

2201

J = 9/2

43110Γ8

3

21

J = 7/2

2211nb levels

2100Γ7

2001Γ6

degen.J = 5/2J = 3/2J = 1/2

Related theoremsKramers: (anti-unitary Time reversal for half integer spins)

Jahn-Teller:

In presence of only electrostatic fields, the energy levels of a system containingan odd number of electrons have an even degeneracy.

half integral S and J double-valued representations

Any non-linear molecular system in a degenerate electronic state will be unstable andwill undergo distortion to form a system of lower symmetry and lower energy thereby

removing the degeneracy

crystal field ground-state degeneracy > Kramers minimum

Kramers ionsodd number of electrons

minimal degeneracy = 2

integral S and J single-valued representations

Non-Kramers ions

even number of electronsminimal degeneracy = 1

Symmetry lowering viacrystal distortion