3.5 Matrix elements and selection rules
The direct (outer) product of two irreducible representations A and B of a group G,
⎛ a11b11 a11b12 a12b11 a12b12⎞ ⎜ ⎟ ⎛ a11B a12B⎞ a11b21 a11b22 a12b21 A ⊗ B = C = ⎜ ⎟ = ⎜ ⎟ ⎝ a21B a22B⎠ ⎜ a21b11 ⎟ ⎜ ⎟ ⎝ ⎠ 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. 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.
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.
<Ψf|O|Ψi>
0 ≠ <Ψf|O|Ψi>
r
0 = <Ψf|O|Ψi> 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 tot symm Γf ⊗ Γ ⊗ 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
tot symm Γf ⊗ ΓO ⊗ Γi = Γ ?
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 2π Fermi’s golden rule = ψ f Hʹψi δ(E f − Ei − hω) ∂t h The operator for the interaction between the system and the electromagnetic field is e r r r r r Hʹ = (A ⋅ pv€ + p ⋅ A ) often, p ⋅ A neglected mc
∂ω 2π v v 2 = ψ f A ⋅ p ψi δ(E f − Ei − hω) ∂t h
2 2π 2 v = A ψ f p ψi δ(E f − Ei − hω) h approximately.
€ the momentum operator p transforms as a spatial coordinate:
2 ∂ω 2π ∂ = Ax ψ f ψi ∂t h ∂x 2 ∂ω 2π ∂ = Ay ψ f ψi ∂t h ∂y 2 ∂ω 2π ∂ = A ψ ψ t z f z i ∂ h ∂
Now split up the transition matrix element to search for possible polarization€ selection rules (s-polarized light: E vector along y, z p-pol light: E vector along x and z) plane of incidence
x
y 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
2 2 dx - y z px,y irr.rep: B1 A1 E B1⨂A1⨂E = E transition not allowed
dxz,yz z px,y red irr.rep: E A1 E E⨂A1⨂E = Γ = 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
kinetic
energy, Ekin
Ekin
vacuum level Φ hν Fermi level valence band
adsorbate core level • photon polarization - linear, circular • photon angle of incidence • electron emission direction bulk core • electron spin levels • sample state (T,M)
122 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!)
electric dipole transitions: final state operator initial state (polarization of light) z x px irr.rep: A1 B1 B1 allowed
x py not allowed x irr.rep: A1 B1 B2
x dz2 not allowed irr.rep: A1 B1 A1 y study effect of symmetry lowering by surface
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 coordinates A1 E (s,p) A1(p) A1 A2u x -> E y -> E z -> A1 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
eigenvalues H11 = ψ1 H0 + V ψ1
2 H22 = ψ2 H0 + V ψ2 H11 + H12 1 (H11 − H12 ) 2 E = ± + 4H 1,2 2 2 12 H12 = ψ2 H0 + V ψ1 = H21 interaction term cannot have the same energy if H12 not zero € H H H E With 12 << 11 − 22 € H 2 this leads to E = H + 12 1 11 (H − H ) 11 22 minimum separation |H12| 2 H12 € E1 = H11 − (H11 − H22 ) V0 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,...
Monochro- Two techniques: mator sample detector Infrared (transmission, reflection) spectroscopy IR beam
Raman spectroscopy Monochro- Visible mator or UV detector beam sample
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) 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: Incident light with frequency ω is modulated by the vibrations of atoms in the medium changing the susceptibility. Two side bands produced:
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 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 A1 A2
Number of normal modes in a molecule:
N atoms, with 3 degrees of freedom each Molecule as such has 3 translational and 3 E E rotational degrees of freedom (2 for a linear molecule)
So a molecule with N atoms has 3N - 6 vibrational E E degrees of freedom (3N - 5 for a linear molecule).
2- Example: CO3 carbonate ion 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.
z1
An important theorem governing the z4 z3 y 1 shape of molecular vibrations, or
x1 better vibrational wave functions:
y4 y3 z2 Theorem: Each of the normal modes
x4 x3 of a molecule transforms as (forms the basis for) an irreducible y2 representation of the molecule.
x2
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 z themselves -> character χ(E) = 9
Operation C2: atoms H1 and H2 interchange - no contribution. y On the oxygen atom, x-> -x, y-> -y, z-> z, hence
χ(C2) = -1 x
Likewise, operation σxz: atoms H1 and H2 interchange - no Molecule lies in y-z plane contribution. On the oxygen, x-> -x, y-> y, z-> z, hence χ(σxz) = 1.
Similarly, χ( σyz ) = +3 C2v E C2 σxz σyz Γtotal 9 -1 1 3
total Reducing this representation gives Γ = 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.
vib total trans rot Thus Γ = Γ - Γ - Γ = 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.
ν1 ν2 ν3
a1 a1 b2
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
Ev = (v + 1/2)hν v = 2 v = 1 v = 0 zero point vibration v = 0
r
Vibrational wave function ψi(vi) with vi vibrational quantum number The total vibration is approximated by the product of the vibrational wave functions
ψvib = 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 vibration Raman 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
product IR transition < ψf O ψi> contains allowed d/dz a1 a1 a1 a1
Not allowed d/dx a1 b1 a1 b1
Not allowed d/dy a1 b2 a1 b2
O is the dipole operator as in the case of electronic transitions, i.e. irr. representations d/dx, d/dy, d/dx. of wave functions and operator 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
O~x .x ψ > product Raman < ψf i j i contains transition Not allowed αzx a1 a1• b1 a1 b1
Not allowed αzy a1 a1 •b2 a1 b2
allowed αyy a1 b2 • b2 a1 a1 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 in an anharmonic potential, transitions from v = 0 to v = 2 can occur, with roughly twice the frequency of the V E = (v + 1/2)hν V v v = 1 transition
v = 2 v = 2 v = 1 v = 1 v = 0 v = 0
r r v = 0 zero point vibration
1. An overtone vibration transforms as product of the irreducible representation of the single transition.
2. The product of any degenerate irreducible representation with itself has a totally symmetric component in it since totally symmetric representation has always at least one spatial coordinate as basis function
-> overtone of Raman inactive vibration will be Raman active 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.
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
Consider only translations - e.g. with {R} = 0. Then the entire symmetry classification ik.a of wave functions is by Bloch’s theorem with quantum number k ψk(r+a) = e ψk( r) with a primitive lattice translation.
-> We need to unify both point and space symmetry aspects. The Bravais lattices Symmorphic and non-symmorphic space groups
Two cases:
a) space group in which point and space operations are all separate and form the entire group -> symmorphic (73 groups)
b) Non-symmorphic space group in which a combination of space and point group elements forms new symmetry operations , Example TiO rutile : C4 axis followed by trans-lation such as screw axes and glide planes (157 2 along c/2 along z and a/2 along x and y. groups). -> 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
ψn wave function of linear chain
k is phase factor here, a distance
χ1 χ2 χ3 χ4 χ5 χ6
k=0
+χ1 +χ2 +χ3 +χ4 +χ5 +χ6 a one-dimensional band structure k=π/a k takes on the meaning of momentum
χ1 −χ2 χ3 −χ4 χ5 −χ6 from R.Hoffman, “Solids and Surfaces - a chemist’s view of bonding in extended structures” VCH Weinheim 1988 Dependence of band width on distance in chain
Band dispersion in chain of p orbitals
Å Å Å
k= 0
k=π/ a s-derived band structure in a square array of atoms
N(E): density of states:
E
Real-life analog: oxygen planes in Cu oxide high temperature superconductors Γ X Μ Γ DOS k -> 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 free electron periodic ion core Bands in repeated zone particle parabola potential scheme energy energy energy energy
massless particle dispersion (e.g. photon)
massive particle dispersion (e.g. electron)
0 0 wave π/a 0 π/a kx wave vector kx wave vector k vector k G ky E = ℏ2k2/2m E = ℏω p = ℏk ℏ p = k = hν/c kx k = √(ℏω/2m)
2D band structure of E metal band insulator E E EF
ky Electronic bands determine most kx EF physical properties of a solid through
ky ky Electron group effective electron conductivity velocity v: mass m* σ kx kx 2 v=1/ħ ∂E/∂k 1/m*=1/ħ2 ∂2E/∂k2 σ=ne τ/m* 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
Why gallium arsenide is good as a light-emitting material and silicon isnʼt...
indirect transition: need phonon - small energy, large momentum direct, k-conserving transition
6 6 L1 L1
X1 L3 L3 Γ15 X1 X1 Γ15 Γʼ25 Γʼ25 0 0 L1 L1
X5 X5
siliconsilicon GaAs X3 X3 L3 L3
X1 -10 -10 X1 L1 L1 Γ1 Γ1
LL ΛΛ ΓΓ Δ XX L Λ Γ Δ X k electronic bands in 3 dimensions
free electron F. Herman,”An atomistic approch to the nature band structure and properties of materials”, Wiley 1967
band structure of Al How can we electronic structure ? -> photoelectron spectroscopy
kinetic energy, E kin outer hemisphere
inner hemisphere
exit slit entrance Ekin channel slit electron multiplier lens vacuum level Φ hν Fermi level valence photons band
sample adsorbate core level
bulk core levels
122 E electron wave vector inside and outside the solid
kout
Θout k⊥out
kin k∣∣out
Ekin
Θin k⊥in
k∣∣in
Evac refraction due to potential step Φ V0 at surface EF
ℏω valence band 2m k E ℏ out = 2 kin ω h 2m kin = 2 (E kin + V0 ) h k = k = k core levels out,|| in,|| ||
! assuming a free electron final state (red)!
photoelectron spectroscopy
β Θ
Angle→kx
image for one angle β Energy
Count rate: color
3 D data set: each voxel contains photo-emission intensity for one (E, k||x,k||y)
-> determine the band structure of a solid from the spectral function assuming direct, k-conserving transitions EB ky
ky
ky kx
kx
constant energy “movie” 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 .
n n n+N N Then tn = A = {E|ai} = {E|a} with A = 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 with ai •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. -ik•tn Γk{E|tn} = e ; 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)
ikn•tn and {E | tn} Φk( r) = e Φk( r - tn). This equation is the definition for
a basis function according to R Φk( r) = ΣΦk( r) Γ( r)ij as before.
Let us use the ansatz
ik•r Φk( r ) = e uk( r) , because for u constant we have free electron waves. Then
ik•(r - t ) Φk( r - tn) = e n uk( r - tn) and
ik•tn ik•(r - t ) e Φk( r ) = e n 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.
Wave equation for Bloch waves: 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
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. example: Brillouin zone of square lattice k The direct lattice has C4 and C2 rotation axes in y center, and mirror planes σv and σd. These k3 generate for a general point G eight k values k2 which are not equivalent (because the waves are k4 k1 running in different directions).
kx k k The Bloch waves are hence different. The group 5 8 of the k vector P(k) contains just the identity. The k6 k7 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 the point in k space is moved to lie on an ky axis, e.g. the x axis, then there are only four k inequivalent k vectors; the star of k has only 2 four components.
k3 k1
kx If the point in k space lies on the Brillouin zone boundary, k1 is equivalent to k4 since k4 they differ by one reciprocal lattice vector; the star of k has only four components. ky
k3 k2 Kn
k4 k1 k 5 K n kx k8
k6 k7 k Kn y Finally, if the k vector lies at one corner of k the Brillouin zone or at the center, then 3 k2 k4 there is just one arrow in the star of k -> all k1 eight symmetry operations are in the group of the k vector. K n kx k5 k8 k6 k7
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 The groups of k for the special M points and lines in the 2D Brillouin zone are Σ Z
Δ: {E,σvy} = Cs Γ k Σ: {E,σdx}= Cs Δ X x
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) C E 2C (z) C 2 2 In order to consider band degeneracies and 4v 4 2 σv σd
level splittings, we look at the character Γ1,M1, (A1) 1 1 1 1 1 tables for the Γ and M point, which have C 4v Γ2,M2 , (A2) 1 1 1 -1 -1 symmetry. Γ3,M3 ,(B1) 1 -1 1 1 -1
Γ4,M4 ,(B2) 1 -1 1 -1 1
Γ5,M5 ,(E) 2 0 2 0 0
x y C2v E 2C2 (z) σv σv
For the X point, which has C2v X1, (A1) 1 1 1 1 symmetry: X2 , (A2) 1 1 -1 -1
X3 ,(B1) 1 -1 1 -1
X4 ,(B2) 1 -1 -1 1
For the X point, which has C2v σ Cs (Δ,Σ,Z) E symmetry:
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
Σ1
Σ2
Σ3
Σ4
Λ1
Λ2
Λ3
Δ1
Δ1’
Δ2
Δ2’
Δ5 L1
L2
L3
L1’
L2’
L3’
Γ1
Γ2
Γ12
Γ15’
Γ25’
Γ1’
Γ2’
Γ12’
Γ15
Γ25 X1
X2
X3
X4
X5
X1’
X2’
X3’
X4’
X5’
W1
W2
W3
W4
W5 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 Reminder from section 3: irreducible representations of the full rotation group
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 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
3C4+ 6C2+ Oʼ E R 8C3 8RC3 6C4 6RC4 3RC4 6RC2
Γ1 1 1 1 1 1 1 1 1
Γ2 1 1 1 1 1 -1 -1 -1
Γ3 2 2 -1 -1 2 0 0 0
Γ4 3 3 0 0 -1 -1 1 1
Γ5 3 3 0 0 -1 1 -1 -1
Γ6 2 -2 1 -1 0 0 √2 -√2
Γ7 2 -2 1 -1 0 0 -√2 √2
Γ8 4 -4 -1 1 0 0 0 0 application: band structure in solids, including spin-orbit coupling
from: Dresselhaus, Dresselhaus and Jorio, Group Theory - Application to the Physics of Condensed Matter Springer 2008 (figure given without references)
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. hh lh so
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)