3.5 Matrix elements and selection rules

The direct (outer) product of two irreducible representations A and B of a group G,

gives us the chance to find out the representation for which the product of two functions forms a basis. This representation will in general be reducible.

Theorem: The character of the direct product representation matrix is equal to the product of the characters of the separate irr. rep. matrices.

Note that we are dealing with the representations within one group, since we deal with a system with well-defined symmetry. Thus “products” of functions such as <Ψb|Ψa>, where each function forms the basis for an irreducible representation Γa and Γb of the group, form a basis for an (often reducible) representation Γa ⊗ Γb.

€

A ⊗ B = C = a11B a12Ba21B a22B⎛

⎝ ⎜

⎞

⎠ ⎟ =

a11b11 a11b12 a12b11 a12b12

a11b21 a11b22 a12b21

a21b11

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

Using the reduction formula, we can decompose the reducible representation of the products into the irreducible ones of the group.

An expression of the form <Ψf|O|Ψi>, where O is some operator which connects the initial with the final state, i.e. a transition matrix element is an integral over all space.

0 ≠ <Ψf|O|Ψi>

0 = <Ψf|O|Ψi>

<Ψf|O|Ψi>

r

If the product function does not contain a function that forms a basis for the totally symmetric irreducible representation, i.e. if that function is not of even parity under all symmetry operations of the group, then by necessity that integral will be zero.

Theorems about direct product representation

Theorem: Matrix elements of an operator H which is invariant under all operations of a group are identically zero for functions which belong to different irreducible representations

Why? “H invariant under all operations of a group” means it belongs to the totally symmetric irreducible representation. Then H ⊗ Ψa belongs to the irreducible representation to which Ψa belongs.

Thus the functions in <Ψf | H | Ψi> transform as Γf ⊗ Γtot symm ⊗ Γi = Γf ⊗ Γi

and functions which belong to different irreducible representations of a group are orthogonal.

Generalization to operators which are not invariant with respect to all operations of the group:Similar theorems apply, e.g. Mfi =<Ψf |O| Ψi> . Find out for which irreducible representation the functions and the operator form a basis, multiply and see whether the product forms a basis for the totally symmetric irreducible representation as

Γf ⊗ ΓO ⊗ Γi = Γtot symm ?

Theorem: A matrix element <Ψf |O| Ψi> must be identically zero if the direct product representation Γf ⊗ ΓO ⊗ Γi does not contain the totally symmetric irreducible representation.

Selection rules: a worked example

Consider an optical dipole transition matrix element such as used in absorption or emission spectroscopies

€

∂ω∂t

= 2πh

ψ f H ʹ′ψ i δ(E f − Ei − hω) Fermi’s golden rule

The operator for the interaction between the system and the electromagnetic field is

€

Hʹ′ = emc

(r A ⋅ v p +

r p ⋅

r A )

r p ⋅

r A

∂ω∂t

= 2πh

ψ f

v A ⋅ v p ψ i

2δ(E f − Ei − hω)

= 2πh

A2 ψ fv p ψ i

2δ(E f − Ei − hω)

often, neglected

approximately.

the momentum operator p transforms as a spatial coordinate:

€

∂ω∂t

= 2πhAx ψ f

∂∂x

ψ i

2

∂ω∂t

= 2πhAy ψ f

∂∂y

ψ i

2

∂ω∂t

= 2πhAz ψ f

∂∂zψ i

2

Now split up the transition matrix element to search for possible polarization selection rules (s-polarized light: E vector along y, p-pol light: E vector along x and z)

z

x

y

plane of incidence

example: optical dipole transitions in a Ni atom in a fourfold site on an Cu(100) surface; C4v symmetry applies

electric dipole transitions: final state operator initial state

dx2 - y2 z px,y irr.rep: B1 A1 E

dxz,yz z px,y irr.rep: E A1 E

B1⨂A1⨂E = E transition not allowed

E⨂A1⨂E = Γred = A1 + A2 + B1 + B2 contains A1 - transition is allowed !

Magnetic dipole transitions

magnetic dipole moment operator

µM = r × p = mr × (dr/dt) axial vector, i.e vector product, or antisymmetric second rank tensor* !

R = A × B, that is Rx = AyBz - AzBy etc.Such axial vectors transform as the rotations R given in the character tables. Then the same procedure can be applied as in the case of the electric dipole operator.

*Transforms as a polar vector under rotations but is invariant under inversion (two sign changes in cross product)

Basics of photoelectron spectroscopy

• photon polarization - linear, circular• photon angle of incidence• electron emission direction• electron spin• sample state (T,M)

hν

Ekin

Φ

kineticenergy, Ekin

vacuum level

Fermi levelvalenceband

bulk corelevels

122

adsorbate core level

initial state is a molecular orbital, or a Bloch wave of an electron in the solid.final state is an electron travelling in the plane of detection - must be symmetric with respect to the travelling plane (normal emission - totally symmetric!)

x

y

z

electric dipole transitions: final state operator initial state

(polarization of light)

x px

irr.rep: A1 B1 B1

x py irr.rep: A1 B1 B2

x dz2 irr.rep: A1 B1 A1

allowed

not allowed

not allowed

study effect of symmetry lowering by surface

coordinatesx -> E y -> E z -> A1

allowed electric dipole transitions: final state operator initial state

A1 E (s,p) A1(p) A1 A1g

A1 E (s,p) A1(p) E1 E1u

A1 E (s,p) A1(p) A1 A2u

A1 E (s,p) A1(p) E1 E1g

No crossing rule

Not a selection rule, but a rule governing interactions between wavefunctions of specific symmetries. Consider eigenfunctions of a Hamiltonian H0 with different eigenvalues Ψ1 and Ψ2, which transform as the same irreducible representation. Assume there is now a perturbation V that has the same symmetry as the Hamiltonian H0. The integrals are then

€

H11 = ψ1 H0 +V ψ1

H22 = ψ2 H0 +V ψ2

H12 = ψ2 H0 +V ψ1 = H21

eigenvalues

€

E1,2 = H11 + H12

2±

12

H11 −H12( ) 2

+ 4H122

cannot have the same energy if H12 not zero

With

€

H12 << H11 − H22

this leads to

€

E1 = H11 + H122

(H11 − H22 )

E1 = H11 − H122

(H11 − H22 )

interaction term

V0

E

minimum separation |H12|

So two energy levels which stem from wavefunctions that form bases of the same irreducible representations cannot cross when they interact.

4. Vibrations in molecules

Why study vibrations in molecules? Because they provide information about the molecular environment of an atom, the bonding in the molecule, and indeed the molecular structure itself. Vibrational spectroscopy is a standard technique in many fields of science: chemistry, biology, physics,...

Two techniques:

Infrared (transmission, reflection) spectroscopy

Raman spectroscopy

Which energy range are we speaking of? Most important molecular vibrations are in the range

80 meV - 500 meV

100 - 4000 cm-1 (1 eV = 8067 cm-1)

sample detectorIR beam

Visible or UV beam

sample

Monochro-mator

Monochro-mator

detector

Physical mechanisms

a) IR spectroscopy

Many molecules have a permanent dipole moment (charge transfer between atoms in the molecule µ = er)

If the molecule has a natural vibration frequency, it can interact with incident light by means of the dipole - in fact by the dynamic dipole moment, i.e. change in dipole moment as the atoms in the molecule move

-> vibrations which change the dipole moment are IR active modes

Raman spectroscopy

The Raman effect is based on an induced dipole moment in a molecule. The induced dipole moment is proportional to the applied external field

µind = α E α polarizability tensor

Classical explanation is based on a kind of frequency modulation -> change in polarizability during excitation, “side bands” occur at lower and higher frequency (ω0 ± ωi , “Stokes” and “anti-Stokes” line) of the incident radiation

The Raman effect: Incident light with frequency ω is modulated by the vibrations of atoms in the medium changing the susceptibility. Two side bands produced:

IR and Raman spectrum of benzene

4.1 Number and symmetry of normal modes

Complex internal motion of atoms in a molecule can be described by a superposition of a small number of simple vibrations -> “normal modes” with fixed frequency

Number of normal modes in a molecule:

N atoms, with 3 degrees of freedom each

Molecule as such has 3 translational and 3 rotational degrees of freedom (2 for a linear molecule)

So a molecule with N atoms has 3N - 6 vibrational degrees of freedom (3N - 5 for a linear molecule).

Example: CO32- carbonate ion

A1 A2

E

E E

E

describe the resultant motion of an atom in a vibration by a superposition of the movement of each atom along three basic vectors, i.e. a Cartesian system on each atom.

x1

y1

z1

x2

y2

z2x3

y3

z3

x4

y4

z4An important theorem governing the shape of molecular vibrations, or better vibrational wave functions:

Theorem: Each of the normal modes of a molecule transforms as (forms the basis for) an irreducible representation of the molecule.

This means that when we consider the movement of all atoms in a complex vibration, i.e. we consider a transformation of the type

The transformation matrix Γij is a reducible representation of the movement of the atom in the vibration; a superposition of several normal modes

We only deal with the characters of the matrices as usual, so we are only interested in those

coordinates that transform into + or - themselves and thus contribute to the character.

Example: H2O

The identity operation transforms all coordinates into themselves -> character χ(E) = 9

Operation C2: atoms H1 and H2 interchange - no contribution. On the oxygen atom, x-> -x, y-> -y, z-> z, hence

χ(C2) = -1

Likewise, operation σxz: atoms H1 and H2 interchange - no contribution. On the oxygen, x-> -x, y-> y, z-> z, hence χ(σxz) = 1.

Similarly, χ( σyz ) = +3

x

y

z

Molecule lies in y-z plane

C2v E C2 σxz σyz

Γtotal 9 -1 1 3

Reducing this representation gives Γtotal = 3a1 + a2 + 2b1 + 3b2

Identify vibrations

In the reduction of Γtotal, Γtrans is the set of irreducible representations for which x, y,

and z form a basis. The set or irreducible representations Γrot for the rotations are those

that transform as Rx, Ry, and Rz. Need to subtract both.

Thus Γvib = Γtotal - Γtrans - Γrot = 3a1 + a2 + 2b1 + 3b2 - (a1 + b1 + b2) -(a2 + b1 + b2)

= 2a1 + b2 Three normal modes of vibration in water of symmetries a1, a1 and b2.

a1

ν1

a1

ν2

b2

ν3

How to determine the motion of each atom in the normal mode in question? In simple cases this is possible by inspection. Better: write down matrix of reducible representation in block diagonal form, see which coordinates transform into themselves, construct the motion

4.2 Vibronic wave functions

At room temperature, most molecular vibrations will be occurring with their zero point energy.

V

r

Ev = (v + 1/2)hν

v = 0v = 1v = 2

v = 0 zero point vibration

Vibrational wave function ψi(vi) with vi vibrational quantum number

The total vibration is approximated by the product of the vibrational wave functionsψvib = ψ1(vn) ψ2(vm) ψ3(vo)

Excitation from ground state to first excited state ψ1(0) ψ2(0) ψ3(0) -> ψ1(1) ψ2(0) ψ3(0), for example.

4.3 IR and Raman selection rules

Selection rules say which transitions are allowed when using IR or Raman spectroscopies.Remember the physical basis for a vibrational transition:IR active vibration: dipole moment of molecule changes during vibrationRaman active vibration: polarizability changes during vibration

Infrared spectroscopy: matrix elements of the type < ψf|O|ψi> where O is the

electric dipole operator. Entire matrix element must be totally symmetric

Theorem: The vibrational ground state |ψi> of a molecule transforms as the totally symmetric irreducible representation.

Γf ⊗ Γo ⊗ Γi must contain the totally symmetric irreducible representation for the transition matrix element to be nonzero

Example: oriented H2O molecule with normal modes 2a1 and b2

< ψf O ψi>product contains

IR transition

d/dz a1 a1 a1 a1allowed

d/dx a1 b1 a1 b1Not allowed

d/dy a1 b2 a1 b2Not allowed

irr. representations of wave functions and operator

O is the dipole operator as in the case of electronic transitions, i.e. d/dx, d/dy, d/dx.

Raman spectroscopy

Raman spectroscopy : matrix elements of the type

Mfi ~ < ψf|αij|ψi>. αij transforms as the square of spatial coordinates e.g.

axy transforms as x·y. As before, the entire matrix element must be totally

symmetric, i.e. the product of the irreducible representations of the initial state wave function, the operator and the final state wave function must contain the totally symmetric irreducible representation.

Again, the example: oriented H2O with normal modes 2a1 and b2

< ψf O~xi.xj ψi> product

containsRaman

transition

αzx a1 a1• b1 a1 b1Not allowed

αzy a1 a1 •b2 a1 b2Not allowed

αyy a1 b2 • b2 a1 a1allowed

The exclusion principle: In a centrosymmetric molecule (one that has an inversion center) an IR-active vibrational mode cannot be Raman active and vice versa. Why? In point groups with an inversion center all irreducible representations can be classified as either gerade (suffix g) or ungerade (suffix u). Even basis functions transform as gerade irred. representations, odd ones as ungerade ones.

The electric dipole operator transforms as ungerade, but a component of the polarizability tensor as gerade. Since the initial state wave function ψi is always totally symmetric, the final state must belong to the same irr. Representation as the operator in order for the product to contain the totally symmetric irr. representation. The final state must then either be odd to allow an IR transition, or even to allow a Raman transition - but it cannot be both. Hence IR and Raman transitions are mutually exclusive in centrosymmetric molecules.

overtones

v = 0 zero point vibration

V

r

Ev = (v + 1/2)hν

v = 0v = 1v = 2

in an anharmonic potential, transitions from v = 0 to v = 2 can occur, with roughly twice the frequency of the v = 1 transition

1. An overtone vibration transforms as product of the irreducible representation of the single transition.

since totally symmetric representation has always at least one spatial coordinate as basis function

-> overtone of Raman inactive vibration will be Raman active

2. The product of any degenerate irreducible representation with itself has a totally symmetric component in it

V

r

v = 0v = 1v = 2

5. Electron bands in crystalline solids

5.1 Symmetry properties of crystals

The complete symmetry group of a crystal, the space group, contains point operations, translations, and combinations of both.

Consider only translations - e.g. with {R} = 0. Then the entire symmetry classification of wave functions is by Bloch’s theorem with quantum number k ψk(r+a) = eik.a ψk( r) with a primitive lattice translation.

Point operations also form a group {R}. The electronic states must be considered as in the case of molecules or in a crystal field splitting szenario

-> We need to unify both point and space symmetry aspects.

The Bravais lattices

Symmorphic and non-symmorphic space groups

b) Non-symmorphic space group in which a combination of space and point group elements forms new symmetry operations , such as screw axes and glide planes (157 groups).

Two cases:

a) space group in which point and space operations are all separate and form the entire group -> symmorphic (73 groups)

Example TiO2 rutile : C4 axis followed by trans-lation along c/2 along z and a/2 along x and y.-> Screw axis {C4|τ} where τ = (a/2,a/2,c/2)

here: only symmorphic space groups

Space groups in 2 dimensions - wallpaper groups

http://www.clarku.edu/̃djoyce/wallpaper/trans.html

The Crystallographic Notation according to the Int’l Union of Crystallography{p1, p2, pm|p1m, pg|p1g, pmm|p2mm, pmg|p2mg, pgg|p2gg, cm|c1m, cmm|c2mm, p4, p4m|p4mm, p4g|p4gm, p3, p3ml, p3lm, p6, p6m|p6mm}

The crystallographic notation consists of four symbols which identify the conventionally chosen “cell,” the highest order of rotation, and other fundamental symmetries.

Usually a “primitive cell” (a lattice unit) is chosen with centers of highest order of rotation at the vertices. In two cases a “centered cell” is chosen so that reflection axes will be normal to one or both sides of the cell. The “x-axis” of the cell is the left edge of the cell (the vector directed downward).

The interpretation of the full international symbol (read left to right) is as follows:

(1) letter p or c denotes primitive or centered cell;

(2) integer n denotes highest order of rotation;

(3) symbol denotes a symmetry axis normal to the x-axis: m (mirror) indicates a reflection axis, g indicates no reflection, but a glide-reflection axis;

(4) symbol denotes a symmetry axis at angle α to x-axis, with α dependent on n, the highest order of rotation: α = 180 degree for n=1 or 2, α = 45 degree for n = 4, α = 60 degree for n == 3 or 6; the symbol m, g, l are interpreted as in (3). No symbols in the third and fourth position indicate that the group contains no reflections or glide-reflections.

Excerpt from Doris Schattschneider's paper (The Plane Symmetry Groups: Their recognition and notation. American Math Monthly. vol 85, pp. 439-450.)

R.M. LAMBERT

glide line

a quick reminder of band structures - here from a “chemical” point of view

χ1 χ2 χ3 χ4 χ5 χ6

k=0

+χ1 +χ2 +χ3 +χ4 +χ5 +χ6

k=π/a

χ1 −χ2 χ3 −χ4 χ5 −χ6

a one-dimensional band structure

χn atomic orbital

ψn wave function of linear chain

from R.Hoffman, “Solids and Surfaces - a chemist’s view of bonding in extended structures” VCH Weinheim 1988

k is phase factor here, a distance

k takes on the meaning of momentum

Å Å Å

Dependence of band width on distance in chain

Band dispersion in chain of p orbitals

k=π/a

k=0

s-derived band structure in a square array of atoms

Real-life analog: oxygen planes in Cu oxide high temperature superconductors Γ Μ ΓX

k ->

E

DOS

N(E):density of states:

px,y-derived band structure in a square array of atoms

A quick reminder of electronic bands in solids from a physical point of view

massless particle

dispersion (e.g. photon)

energy

wave vector k0

massless particle

E = ℏωp = ℏk = hν/c

free electron parabola

energy

wave vector k

0

massive particle dispersion (e.g.

electron)

E = ℏ2k2/2mp = ℏkk = √(ℏω/2m)

effective electron mass m*

1/m*=1/ħ2 ∂2E/∂k2

Electron group velocity v:

v=1/ħ ∂E/∂k

conductivity σ

σ=ne2τ/m*

Electronic bands determine most physical properties of a solid through

G

periodic ion core potential

energy

0 π/aπ/a kx

2D band structure of

band insulator

ky

kx

EF

Emetal

ky

kx

E

E

kykx

EF

wave vector

energy

kx

kx

ky

Bands in repeated zone scheme

silicon

ΔL Λ Γ X

silicon

Γ1

X1

X5

X3

X1L1

L3

L1

L3

L1

Γʼ25

ΔL Λ Γ X

Γ15

indirect transition: need phonon - small energy, large momentum

Silicon is a semiconductor with an indirect band gap - not useful for light emitting devices (However, see article about Si Raman laser in recent Nature paper).

Indirect band gap not a disadvantage for solar cell applications

direct, k-conserving transition

Γʼ25

GaAs

X1

X5

X3

L3

0

6

L1

L3

L1

Γ15X1

ΔL Λ Γ X

Γ1

X1L1

-10

Why gallium arsenide is good as a light-emitting material and silicon isnʼt...

0

6

-10

k

F. Herman,”An atomistic approch to the nature and properties of materials”, Wiley 1967

band structure of Al

free electron band structure

electronic bands in 3 dimensions

How can we electronic structure ? -> photoelectron spectroscopy

hν

Ekin

Φ

kineticenergy, Ekin

vacuum level

Fermi levelvalenceband

bulk corelevels

122

adsorbate core levelsample

photons

lens

entranceslit

innerhemisphere

outerhemisphere

exit slit

channelelectronmultiplier

!

kout =2mh2

Ekin

kin =2mh2(Ekin +V0)

kout ,|| = kin,|| = k||

electron wave vector inside and outside the solid

refraction due to potential step V0 at surface

kin

k⊥in

k∣∣out

k⊥out

k∣∣in

kout

Θin

Θout

ℏω

Evac

EF

valence band

core levels

Φ

Ekin

E

ℏω

assuming a free electron final state (red)!

-> determine the band structure of a solid from the spectral function assuming direct, k-conserving transitions

Ener

gy

Count rate: color

Angle→kx

image for one angle β

βΘ

3 D data set: each voxel contains photo-emission intensity for one (E, k||x,k||y)

photoelectron spectroscopy

ky

kx

constant energy “movie”

kx

ky

EBky

Basis functions for space groups

We need basis functions for the irreducible representations of the space group. We introduce cyclic boundary conditions:

tn1,n2,n3 = tn1 + N,n2,n3 = tn1,n2+N2,n3 = tn1,n2,n3+N3

This crystal has N1 primitive unit cells along the a1 direction, etc.

Since the boundary conditions are cyclic, we use a cyclic group. The generating element is

{E|ai} = A .

Then tn = An = {E|ai}n = {E|a}n+N with AN = E the identity.

Thus the irreducible representations are the N roots of

Γ{E|ai} = Γ(A) = E1/N = [e-2πim]1/N = e-2πm/N with m integer. All representations are 1-dimensional since in a cyclic group each element is in its own class.

Reciprocal space: introduce the reciprocal lattice vectors bi withai •bi = 2π δij and b1 = 2π (a2 x a3)/[a1 • (a2 x a3)] and cyclic permutation.The reciprocal lattice vectors span the reciprocal lattice Kn = n1b1 + n2b2 + n3b3 .

We define a new set of vectors k = p1b1 + p2b2 + p3b3 where the pi are restricted to values 0 ≤ pi ≤ 1, with pi =mi/Ni .We then take k to label the irreducible representations of the translation group.Γk{E|tn} = e-ik•tn ; there are as many points in k space as there are cells.

k is plotted in the Brillouin zone

schematic diagram of the first and second Brillouin zone. The discrete number of k points for a finite lattice is indicated in the first Brillouin zone

General basis function Φk( r).

apply space group operation, both point and translation:

{E | tn} Φk( r) = Φk( r - tn)

and {E | tn} Φk( r) = eikn•tn Φk( r - tn). This equation is the definition for a basis function according to R Φk( r) = ΣΦk( r) Γ( r)ij as before.

Wave equation for Bloch waves:

Let us use the ansatz

Φk( r ) = eik•r uk( r) , because for u constant we have free electron waves. Then

Φk( r - tn) = eik•(r - tn) uk( r - tn) and

eik•tn Φk( r ) = eik•(r - tn) uk( r), thus

uk( r ) = uk( r - tn), i.e. uk( r) is “lattice periodic”.

This is a symmetry related proof of Bloch’s theorem.

Periodic potential V( r) leads to Bloch waves as basis functions for the cyclic translation group.

Now we include point operations:{R|0}V( r) = V( r). We only consider symmorphic space groups, such that we can separate {R|tn} = {E|tn} {R|0}.If we apply the operation {R|0} on both sides of the wave equation, we obtain

with V( R-1r) = V( r).

The Hamiltonian must be invariant with respect to all symmetry operations of the group. Thus the term Rk•p reduces the symmetry of the Hamiltonian, since it has only the symmetry of the group of the k vector.

Instead of considering the full (rotation, reflection etc.) symmetry of the unit cell of the lattice we must take into account the symmetry of the group of the k vector, which may be much lower.

Definition: The group of the k vector is that subgroup of the point group Pg of the lattice which leaves k invariant:

R k = k + Kn modulo a reciprocal lattice vector. This group is called P(k). Its elements depend on the magnitude and direction of the wave vector k !

Definition: The star of k is the set of inequivalent k vectors obtained by applying all {R|0} of Pg to k .

General and special points and lines in k space

The direct lattice has C4 and C2 rotation axes in center, and mirror planes σv and σd. These generate for a general point G eight k values which are not equivalent (because the waves are running in different directions).

The Bloch waves are hence different. The group of the k vector P(k) contains just the identity. The point operations do not add anything to the translation operations -> irreducible representations, basis functions etc. are completely determined by the translation symmetry. The star of k has eight arrows.

If P(k) has only the element {E|0}, k is called a general point in the Brillouin zone; else it is a special point or on a special line.

kx

ky

k1

k2k3

k4

k5

k6 k7

k8

example: Brillouin zone of square lattice

If the point in k space is moved to lie on an axis, e.g. the x axis, then there are only four inequivalent k vectors; the star of k has only four components.

k1

k2

k3

k4

ky

kx

If the point in k space lies on the Brillouin zone boundary, k1 is equivalent to k4 since they differ by one reciprocal lattice vector; the star of k has only four components.

ky

kx

k1

k2k3

k4

k5

k6 k7

k8

Kn

Kn

ky

kx

k1

k2k3k4

k5

k6 k7

k8

Kn

KnFinally, if the k vector lies at one corner of the Brillouin zone or at the center, then there is just one arrow in the star of k -> all eight symmetry operations are in the group of the k vector.

Theorem: If h(Pg) is the order of the point group of the space group, h(P(k)) is the order of the group of k, and h(s(k) is the number of waver vectors in the star of k (not a group), then

h(Pg) = s(k) h(p(k))

In other words, the point symmetry operations are split between those that are in P(k) (a subgroup) and those in the star of k.

Example: simple 2D square lattice (e.g. the Brillouin zone of electron states on

For Brillouin zones, new symmetry labels exist (why, after all, make life simple...)ky

kxΓ Δ X

Z

M

Σ

The groups of k for the special points and lines in the 2D Brillouin zone are

Δ: {E,σvy} = Cs

Σ: {E,σdx}= Cs

X: {E,C2, σvx,svy} = C2v

Γ, M: {E,C2,2C4, 2σv, 2σd} = C4v

So degeneracies etc. must be derived in going from the full C4v group of the lattice to these groups.

2D band structure of carbon monoxide on Ni(110)

C4v E 2C4 (z) C2 2σv 2σd

Γ1,M1, (A1) 1 1 1 1 1

Γ2,M2 , (A2) 1 1 1 -1 -1

Γ3,M3 ,(B1) 1 -1 1 1 -1

Γ4,M4 ,(B2) 1 -1 1 -1 1

Γ5,M5 ,(E) 2 0 2 0 0

In order to consider band degeneracies and level splittings, we look at the character tables for the Γ and M point, which have C4vsymmetry.

C2v E 2C2 (z) σvx σv

y

X1, (A1) 1 1 1 1

X2 , (A2) 1 1 -1 -1

X3 ,(B1) 1 -1 1 -1

X4 ,(B2) 1 -1 -1 1

For the X point, which has C2vsymmetry:

For the X point, which has C2vsymmetry:

Cs (Δ,Σ,Z) E σ

Δ1,Σ1,Z1 1 1

Δ2,Σ2,Z2 1 -1

Compatibility relations Compatibility relations for the electron states in a 2D square lattice, and possible band degeneracies in a hypothetical band structure.

We can see from the table that a Γ5 state must split into two bands Σ1 and Σ2 when the k vector is increased from the center of the Brillouin zone along the Σ direction. Moreover, we see that the Σ2 band can cross the Σ1 but two Σ1 bands cannot cross because of the no-crossing rule

Important reminder:

The symmetry of the Brillouin zone reflects just the symmetry of the bare lattice without a basis.

-> if the lattice has a symmetry-lowering basis this may not be correct, or may be just approximate and need to be considered

pictorial representation of wave functions

(from M. Tinkham, Group Theory and Quantum Mechanics, McGraw-Hill 1964

Character tables for band symmetries

Σ4

Σ3

Σ2

Σ1

Λ1

Λ2

Λ3

Δ2’

Δ5

Δ1

Δ1’

Δ2

L3’

L2’

L1’

L3

L2

L1

Γ25

Γ15

Γ12’Γ2’Γ1’Γ25’Γ15’Γ12

Γ2

Γ1

X1’

X2’

X3’

X4’

X5’

X5

X4

X3

X2

X1

W4

W5

W1

W2

W3

band compatibilities in a 3D solid

compatibility relations for a 3 D simple cubic lattice

Similar reasonings apply to 3D bands; here again we need to learn new symmetry labels, as outlined in the enlarged character table.

bands with spin-orbit splitting

Theorem: A rotation around an axis results in a transformation of the spherical harmonics such that they can be expressed as a linear combination

Let us choose a simple way to evaluate the representations of odd dimensions (1,3,5,7...). Consider a rotation around an axis by an angle α:

without loss of generality since all rotations by θ are in the same class. So

which is just a diagonal matrix (for each m)

And the character is just the sum over m

Reminder from section 3: irreducible representations of the full rotation group

Crystal Double Groups

The irreducible representations for the atomic levels s,p,d,f etc., by using the spherical harmonics as basis functions for the full rotation group:

Theorem: The irreducible representations with are the only odd-dimensional irreducible representations of the full rotation group for integer l.

However, this was only valid for integer angular momentum J = |L +S|; then (2J+1) is odd. For an odd number of electrons J is half-integral (spin moment), (2J + 1) is even, and we have do deal with a different kind of representations of the full rotation group.

It can be shown that the above formula holds for integer and half-integer J.

Now consider a rotation by 2π. We find that for half-integer J we get a double valued character.

A rotation by 2π should leave everything unchanged, but here we obtain the negative of the character!

For a rotation around 4π, however, the original value is recovered- hence it appears that this rotation is the identity.

We solve this unusual situation by introducing a new symmetry element R which represents a rotation by 2π, and which has the property R2 = E.

The new group has twice as many elements as the original one -> “crystal double group”. For the octahedral group this is

Oʼ E R 8C3 8RC33C4+3RC4

6C2+6RC2

6C4 6RC4

Γ1 1 1 1 1 1 1 1 1Γ2 1 1 1 1 1 -1 -1 -1Γ3 2 2 -1 -1 2 0 0 0Γ4 3 3 0 0 -1 -1 1 1Γ5 3 3 0 0 -1 1 -1 -1Γ6 2 -2 1 -1 0 0 √2 -√2Γ7 2 -2 1 -1 0 0 -√2 √2Γ8 4 -4 -1 1 0 0 0 0

from: Dresselhaus, Dresselhaus and Jorio, Group Theory - Application to the Physics of Condensed Matter Springer 2008 (figure given without references)

application: band structure in solids, including spin-orbit coupling

So for situations in which spin-orbit splitting is important, for example in the band structures of the heavy elements, or in a case of crystal field splitting of a heavy element, the electronic states need to be classified according to the double group representations.

bands near the zone center in Germanium affected by spin-orbit coupling

from: Dresselhaus, Dresselhaus and Jorio, Group Theory - Application to the Physics of Condensed Matter Springer 2008 (figure given without references)

hh

lh

so