Ligand Group Orbitals - PDF Free Download

Ligand Group Orbitals he O h Group is the point Group of many interesting solids, including complexes like CuSO 4 5H O and FeCl 3 where a transition metal ion at the center of an Octahedron LCO model of their properties is often called ligand field theory. Il solfato di rame anidro (bianco) ridiventa pentaidrato (blu) aggiungendo acqua. 1 1

Exact or approximate symmetry of many complexes E C F B Binding occurs between central atom orbitals and ligand orbitals of like symmetry. For instance, if...f represent s orbitals they produce a representation Γ with following characters: D

O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 C4,C unmoved atoms: character= O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 F basis 3 3

n. unmoved atoms O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 F basis S4 0 unmoved atoms E F D C B O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 0 4 F 4basis

n. unmoved atoms O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u C 0 unmoved atoms 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E g u 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 0 F basis O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 0 0 F basis 5 5

n. unmoved atoms O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 0 0 F basis C3,S6 0 unmoved atoms O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, + ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 0 0 0 0 0 F basis 6

n. unmoved atoms O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u Γ 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 6 0 0 0 0 4 0 F basis σ d unmoved atoms Number of times irrep i is present in basis: 1 i ni = χ R χ R N G R ( ) ( ) Γ= E * 1g g 1u Let us see the Projection of orbital into 1u 8 8

O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S h 4 3 4 h d 6 1u 3 1 1 1 0 3 1 1 1 0 E C F B Projector : = ( ) () i i * P χ R R R Projection of orbital into 1u C 4 : D operations with axis D : operations with axis EB : CF, operations with axis CF : BE, C : 1 operations with axis D : 1 operations with axis EB : D 1 operations with axis CF : D C ' : with axis B : B, with axis C : C with axis E : E, with axis F : F with axis CB: D, with axis EC: D 1 operation with axis CF : D Class 4 3 4 6 R C ( ) 1 u contribution E 3 3 6C + B+ C+ E+ F 1 + B+ C+ E+ F 3C + D 1 D 6 C' D+ B+ C+ E+ F 1 ( D+ B+ C+ E+ F) 8C ( B+ C+ E+ F) 0 0 I D 3 3D 6S D+ B+ C+ E+ F 1 ( D+ B+ C+ E+ F) 3σ + D 1 + D h 6σ + B+ C+ E+ F 1 + B+ C+ E+ F d R 8S ( B+ C+ E+ F) χ 0 0 9 9

he normalized 1u projection is ψ 1 =( D)/. Operating in the same way on D we again get ψ 1. Operating on the other functions, we obtain ψ =(B E )/ and ψ 3 = (C F )/. In this way one easily builds the ligand group orbitals. Ψ 1 can make bonds with p x orbitals of central atom Ψ can make bonds with p y orbitals of central atom Ψ 3 can make bonds with p z orbitals of central atom E C F B D 10

Crystal field theory O M transition metal ion M in an Oxygen cage with O h symmetry. In many cases the main effect is the splitting of the ion levels by the crystal field. In some cases the on-site interactions are important, in other cases they can be neglected in a first approximation. M M ++ +++ Number of d electrons: i V Cr Mn Fe Co Ni Cu 3 4 5 6 7 8 9 1 3 4 5 6 7 8 Friedrich Hund, 1896-1997 (aged 101) Hund s Rule 1: ground state has maximum S, highest L compatible with S Hund s Rule : ground state has maximum J for > half filling, lowest J for < half filling 11 11

Number of d electrons: i V Cr Mn Fe Co Ni Cu O M M M ++ +++ 3 4 5 6 7 8 9 1 3 4 5 6 7 8 Hund s rule prompts the ground state quantum numbers: electron number ground state d d d d d d d d d 1 3 4 5 6 7 8 9 D F F D S D F F D 3 4 5 6 5 4 3 For instance, with 3 electrons ground state is ψ = 1 0, M = 3 g L 4 F he same ion can behave in very different ways in different compounds: Ferrous Fe Fe + Fe( H O) green,paramagnetic 6 (3 d ) 4 Fe( CN) 6 yellow,diamagnetic + 6 his is explained by crystal field theory

Crystal field theory-independent electron approximation For the d orbitals (or even many-body D) states, in octahedal symmetry, using the character of the reducible representation j for the class of rotations by an angle α 1 sin[ ] 5 j j + α sin[ α] ( j) im jα Γ ( ), ( ) d = χ α = e = j = χα =, α α mj = j sin sin C C C 4 π 5π 1 π 1 α = Sin[ ] = Sin[ ] = χ = 1 4 4 5π π α = π Sin[ ] = 1 Sin[ ] = 1 χ = 1 π 5π 3 π 3 α = Sin[ ] = Sin[ ] = χ = 1 3 3 3 O M one finds O E 6C 3C 6 C ' 8C i 6S 3σ 6σ 8S h 4 3 4 h d 6 Γ 5 1 1 1 1 d 13 Inversion is like identity E for d states. For the d orbitals (or even D) states, i= parity=e. 13

Hence we can deduce the character of reflections,as follows. For the 5 d orbitals (or even D) states, parity=+, so a reflection is like a π rotation. 1 sin[ ] 5 j + α sin[ π ] Since χα ( ) =, and j =, χπ ( ) = = 1 α π sin sin O E 6C 3C 6 C ' 8C i 6S 3σ 6σ 8S h 4 3 4 h d 6 Γ 5 1 1 1 1 5 1 1 d Reflections have character 1 because: Inversion i is the result of a π rotation and a reflection: i= C σ Improper rotations: 1 j sin[ j + α] im j Since a reflection is a factor 1,improper rotations are like proper ones: χα ( ) = e α = α mj = j sin 5 π 5 π sin[ ] sin[ ] χ( S 3 6) = χ( R6) = 1, χ( S4) χ( R4) 1. π = = = π = sin sin 6 4 14 14

one finds O E 6C 3C 6 C ' 8C i 6S 3σ 6σ 8S h 4 3 4 h d 6 Γ 5 1 1 1 1 5 1 1 1 1 d Number of times irrep i is present in basis: 1 i * ( ) χ( ) ni = χ R R N G R he 5-fold degeneracy of d orbitals is broken. O E 6C 3C 6 C' 8C i 6S 3σ 6σ 8S g = 48 h 4 3 4 h d 6 1 1 1 1 1 1 1 1 1 1 x + y + z 1g 1u g u g u 0 0 1 0 0 1 (, ) 1g x y z 1u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E x y z x + y E 0 0 1 0 0 1 3 1 1 1 0 3 1 1 1 0 ( R, R, R ) g u 3 1 1 1 0 3 1 1 1 0 ( x, yz, ) 3 1 1 1 0 3 1 1 1 0 ( xy, xz, yz) 3 1 1 1 0 3 1 1 1 0 Γ 5 1 1 1 1 5 1 1 1 1 d d basis d Γ = E d g g ( x y, z x + y ) ( xy, xz, yz) 15 15

In complexes usually the energy splitting is Δ = E(E g ) E( g ) > 0, since the g orbitals stay far from the negative ligands. E g O M g hus in the absence of Coulomb interactions in the ion one would fill the available levels according to the aufbau principle, starting with g. In such cases the color of crystals is easily explained. 16 16

Crystal field theory-interactions E g O M g In crystal field theory one tries to predict the magnetic properties by diagonalizing a many-electron Hamiltonian which is the sum of the isolated ion Hamiltonian and the one of the crystal field. Many papers have been published on the electron spectroscopies of transition metal compoounds using Group theory methods. If Δ << U, one treats Δ as a perturbation of the isolated ion multiplet: Hund rule and paramagnetism for partial occupancy If Δ >> U, Hund s rule holds (high spin is preferred) within the degenerate g and E g levels, but E g starts being filled only after g is full, and 6 electrons yield a diamagnetic complex. 17 17

pplications of Group heory to vibrations Groups help with any symmetric secular problem Normal modes of molecular vibration: classical motion of i-th nucleus U ( v) (Born-Oppenheimer) m v v nuclear displacement vector v= δx, δy, δz,... δx, δy, δ z = ( v,... v ) ( ) i i = 1 1 1 N N N 1 3N Harmonic approx to potential: U ij U v 1 = force matrix i ω m vv 1 1 1 i ij i j v i ( v) U = v Equation of motion: ω vi = Upqv pvq = Uijv j mi vi pq mi j 18 ij U i 18

1 ω = Equation of motion: vi U ij v j mi j People prefer to put masses into force matrix ω mi vi = Uijvj mi j 1 i ij j j ω m U ij ivi mjvj j mm i j = Uij Introducing Qi = mi v i Wij = one has the secular problem mm ω Q = W Q Det( W ω I) = 0 Qα eigenvectors of W =normal modes ωα eigenfrequencies of W i j ω α > 0: ω α = 0: 3 translations and 3 rotations ( rotations for linear molecules) 19

Group heory and classification of vibrations Each nucleus has 3 displacements Cartesian reference which may be rotated/reflected R: unitary matrix that rotates / reflects the whole molecule leading to an identical geometry. rigid molecule would be sent to itself. vibrating molecule is sent to a vibrating molecule, with a transformed vibration. R sends each component of atomic displacement to a linear combination of components. his associates to R a matrix D(R). R is a symmetry if [R,W]=0. he set of matrices is a representation of the symmetry Group. 0 0

Group heory and classification of vibrations Normal modes Qα = eigenvectors of W ωα eigenfrequencies of W Vibrations belonging to different irreps are orthogonal. W commutes with all the R and cannot mix vibrations belonging to different irreps Reducing D to irreps, secular determinant is put in block form. 1 1

Program: o diagonalize W simultaneously with as many D as possible, + Dirac s characters Ω Practical use: reduce W to block diagonal form by linear combinations of the Q components: = U U + z Example:Water Molecule on xz plane y x

C : 1 7, 8, 3 9, 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 DC ( ) = 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 DC ( 0 0 b 1 0 0 ) has a block structure 0 b 0, with b= 0 1 0 b 0 0 0 0 1 0 0 b 1 0 0 D( σ ( yz)) also has a block structure 0 b 0, with b = 0 1 0 b 0 0 0 0 1 b 0 0 1 0 0 D( σ ( xz) ) has a block structure 0 b 0, with b = 0 1 0 0 0 b 0 0 1 3 3

Using the 9X9 matrices : Cv E C σ( yz) σ( xz) Γ 9 1 1 3 We could arrive to this result without writing the D matrices, taking into account that for each operation: the atoms that change position contribute 0 to the character; each arrow (cartesian movement) that it remains invariant contributes +1, and every arrow that changes sign contributes -1, more generally, the cartesian shifts of an atom that does not change position behave like (x, y, z), so if the arrow is rotated by θ the contribution is cos(θ). However, D matrices are needed to find eigenvectors (see below) 4 4

One can find the characters without writing the D matrices: E C 9 unmoved χ = 9 1 unmoved reversed χ = 1 Oxygen arrows moved σ σ ( xz) ( yz) 6 unmoved 3 reversed χ = 3 unmoved 1 reversed χ = 1 Cv E C σ( yz) σ( xz) Γ 9 1 1 3 5 5

Cv E C σ( yz) σ( xz) Γ 9 1 1 3 C I C σ σ g = B B v xz yz 1 1 1 1 1 1 1 1 1 1 xy, R 1 1 1 1 xr, 1 1 1 1 yr, z 4 y x z read from able: Γ trasl : + B + B Cv E C σ( yz) σ( xz) Γ 3 1 1 1 trasl 1 1 Note : R x x x x Γ : + B + B rot i j j k k 1 Cv E C σ( yz) σ( xz) Γ 3 1 1 1 rot 6 6

C I C σ σ g = B B v xz yz 1 1 1 1 1 1 1 1 1 1 xy, R 1 1 1 1 xr, 1 1 1 1 yr, z 4 y x z C I C σ σ g = Γ Γ 9 1 3 1 3 1 1 1 Γ 3 1 1 1 Γ v xz yz trasl rot vibr By difference: 3 1 3 1 4 1 i ni = χ R χ R N G nalysis in irreps: R * ( ) ( ) Γ vibr = 1 + B 1 Breathing mode 1 in all molecules What kind of vibration is B 1? Remark. similar analysis applies in the application to molecular orbitals in the LCO method when p prbitals are involved. 7

C I C σ σ g = 4 v xz yz B B 1 1 1 1 1 1 z 1 1 1 1 xy, R 1 1 1 1 xr, 1 1 1 1 yr, y x z What kind of vibration is B 1? Projection operator on arrow basis: P( B ) = E C + σ( xy) σ( yz) 1 Recall the block matrices: 0 0 b 1 0 0 D( σ( yz)) = 0 b 0, b( σ( yz)) = 0 1 0 b 0 0 0 0 1 0 0 b 1 0 0 DC ( ) = 0 b 0, bc ( ) = 0 1 0 b 0 0 0 0 1 b 0 0 1 0 0 D( σ( xz)) = 0 b 0, b( σ( xz)) = 0 1 0. he block matrix for the projector is: 0 0 b 0 0 1 1+ b( σ ( xz) ) 0 b( C ) b( σ( yz) ) ( ( )) ( ) ( ) ( ) ( σ( )) 0 1 σ ( ) P( B ) = E C + σ( xy) σ( yz) = 0 1 + b σ xz b C b σ( yz) 0 1 b C b yz + b xz ( ) 8

0 0 1 0 P( B 1 ) 0 = 0 0 0 0 1+ b( σ ( xz) ) 0 b( C ) b( σ( yz) ) ( σ ( )) ( ) ( σ ) ( ) ( σ( )) 0 1 σ ( ) P( B ) = 0 1 + b xz b C b ( yz) 0 1 b C b yz + b xz 0 1 0 1 0 0 0 0 0 ( ) 0 1 0 0 PB ( 1) 0 = 0. 0 0 0 0 C I C σ σ g = 4 v xz yz B B 1 1 1 1 1 1 z 1 1 1 1 xy, R 1 1 1 1 xr, 1 1 1 1 yr, y x z he only arrows are 3 and 9, and are opposite. One H shifts up along the molecular axis and the other goes down; such a vibration indeed changes sign under C and σ(yz). 9 9

Γ( H ): E C σ 3 v Vibrations of NH 3 Movements of N like (x,y,z) : 1 +E Γ(N) has characters 3 0 1 3 a 9 unmoved χ = 9 0 unmoved χ = 0 c b unmoved 1 reversed χ = 1 C3v I C3 3σ v g = 6 1 1 1 1 z 1 1 1 R ( ) ( x y) E 1 0 xy,, R, R C E C 3σ 3v 3 Γ( N) 3 0 1 Γ( H ) 9 0 1 3 Γ( NH ) 1 0 Γ Γ Γ trasl rot vibr 3 3 0 6 0 z 1 3 0 1 30 Γ = 30 vibr 1 E

3 1 4 With 5 atoms χ(e) = 15. Vibrations of Methane (CH 4 ) E 8C 3C 6σ 6S N = 4 d 3 d 4 G 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 (3, ) E z r x y 3 0 1 1 1 ( R, R, R ) 3 0 1 1 1 ( xyz,, ) r x y z 8C 3 : all the atoms move except one H and the C: for each, take an arrow along the rotation axis (χ= +1),while the other two, on the perpendicular plane, are transformed as the coordinates (x,y) of this plane and contribute rd(r) = cos(π/3 ) = 1. herefore χ(8c 3 ) = 0. 1 4 3C all H moved. For the atom of C: arrows of the C change sign and the third does not move: χ = -1 31

3 1 4 E 8C 3C 6σ 6S N = 4 d 3 d 4 G 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 (3, ) E z r x y 3 0 1 1 1 ( R, R, R ) 3 0 1 1 1 ( xyz,, ) r x y z 6σ d CH remains in place; each atom has arrows in plane and one reflected and χ = 3. S 4 all H moved. For the atom of C: π/ rotation around z, (x, y, z) (y, x, z); then reflection (y, x, z). So, χ = 1. hus the characters of the representation of vibrations are: E 8C 3C 6σ 6S N = 4 d 3 d 4 G Γ 15 0 1 3 1 tot Γ 3 0 1 1 1 = trasl Γ 3 0 1 1 1 = rot 1 Γ 9 0 1 3 1 E vibr 1

Benzene D 6h Group D E C C C 3 C' 3 C'' i S S σ 3σ 3σ g= 4 B B 6h 6 3 3 6 h d v 1g g 1g g 1g x y g 1u u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E 1 1 0 0 1 1 0 0 ( R, R ) E 1 1 0 0 1 1 0 0 ( x y, xy) B B E E 1u u 1u u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 ( xy, ) 1 1 0 0 1 1 0 0 R z C C

Benzene C 6 H 6 here are 1 atoms and 36 coordinates, therefore χ(e) = 36. he rotations C, C 3 and C 6 around the vertical axis move all the atoms and have character 0. Rotation C around to a diagonal of the hexagon leaves 4 atoms in place: for each one arrow is invariant and the others two change sign. herefore, χ(c ) = 4. Rotation C around an axis to opposite sides has character 0. S 3 and S 6 move all the atoms and have character 0. C C 34

he reflections σ h in the plane of the hexagon leaves two arrows invariant for every atom and changes sign to the third, therefore χ(σh) = 1. he reflection for a plane containing the C axis has character 0. he reflection χ(σ v ) for a plane containing C leaves 4 atoms in place, with two arrows invariant and one changed of sign for every atom. herefore χ(σ v ) = 4. he characters of Γ trasl are the sums of those of u and E 1u ; those of Γ rot are the sums of those of g and E 1g. C C 35

C C D E C C C 3 C' 3 C'' i S S σ 3σ 3σ g = 4 B B 6h 6 3 3 6 h d v 1g g 1g g 1g x y g 1u u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 E 1 1 0 0 1 1 0 0 ( R, R ) E 1 1 0 0 1 1 0 0 ( x y, xy) B B E E Γ 1u u 1u u tot 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 z 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 ( xy, ) 1 1 0 0 1 1 0 0 36 0 0 0 4 0 0 0 0 1 4 0 Γ 3 0 1 1 1 3 0 1 1 1 trasl Γ 3 0 1 1 1 3 0 1 1 1 rot Γ 30 4 0 0 0 0 1 4 0 vibr R z he characters of Γ transl are the sums of those of u and E 1u ; those of Γ rot are the sums of those of g and E 1g. Γ = B B B E 3E 4E E vibr 1g g u 1u g u 1g 1u g u 36

Space-ime Symmetries of Bloch States in solids t 1, t, t 3 primitive translation vectors ranslation Group { i } set of lattice translation operators = combinations of primitive translations with integer coefficients Usually, Periodic boundary conditions: for some N>> 1, in = 1. Halite (sodium chloride) - a single, large crystal. Band theory: One-electron approximation: the electron moves in a periodic crystal potential V (x ) i.k t Bloch wave function ψ (x) = ψ (x+t ) = e i ψ (x), i k k i k H, _=0 with H ψ (x)= εψ (x) and i k k k i N Born-Von Karman b.c. C = 1 i.k t i unitary, eigenvalue C=e : 37

N i.n.k t π i C =e = 1 requires k t = * integer; therefore, i N pg + qg + rg 1 3 k= N with p, q, r Z, and the reciprocal lattice basis vectors g defined by Bloch's functions are: t g = πδ i Elementary aspects: i.k ti Hψ = ε ψ and ψ = e ψ, k k k j ik x Bloch's theorem (Floquet's theorem) : ψ (x) = e u (x ), with u (x ) lattice periodic. k i k k ik x ik x For each k, since (with =1) p ˆ e =e (p+k) ˆ the Schrödinger equation has one solution: (p + k ) [ + V (x)]u (x) = ε u m k k k (x) ij. k within the unit cell. 38 k j

chille Marie Gaston Floquet (15 December 1847, Épinal 7 October 190, Nancy)

Group theory aspects: No degeneracy is predicted, since belian Groups have only onedimensional representations. belian group Each translation i by a lattice vector is a class. k is the label of an irrep, neither H nor translations can mix different k ( k ) ik. t ( k ) ik. t D ( t) = e 1-dim representation χ ( t) = e = character One solution for each k, ε k he degeneracy of 1 d chain spectrum ε() k = c t os( k) = ε( k) is not explained in this way: the inversion symmetry is involved (see below) 40

1 LO R R N e k k D t = χ t = e ( k ) ( k ) ik. t Since ( ) ( ) is the character, i * j i( k k') t : χ ( ) χ ( ) = Gδij = δ( ') R G Ncells t Bravais Second character orthogonality theorem it ( t') k e = NCδ t t k BZ i () i () i * G χ ( C) χ ( C') = δcc' nc (, ') N BZ integration cell localization! 41

Relativistic corrections for low speeds e i ev cσ p t c ψ ψ Dirac's equation mc B = B e ψ ψ cσ p i ev c t yields relativistic corrections via a Foldy Wouthuysen transformation: 4 p p e e H = + ev σ.( E p) dive 3 m 8mc 4mc 8mc he correction which lifts degeneracy: SO interaction e e dv e dv In atomic physics σ.( E p) = σ.( r p) = SL. = H 4mc 4mcr dr mcr dr SO 1 lso, e.( E p) ( V). p 4mc σ = 4mc σ How does it change the symmetry in crystals? 4

dding the Spin-orbit interaction lifts trivial spin degeneracy: p H = + V( x) + H' m SO HSO 1 = σ ' Vp. 4mc Eigen-spinors can still be taken of the Bloch form, ψ i. k ti Hψ = εψ and ψ = e ψ, with k k k i k ik x k, λ k, λ k, λ (x) = e u (x ), u (x ) lattice periodic spinor. he equation for the periodic function, with p= i p+ k, ( p+ k) + V( x) u ( x) = ε u ( x) with the spin-orbit interaction becomes k k k m ( p+ k) 1 + V( x) + σ V.( p + k) u ( x) = ε u ( x), kλ kλ kλ m 4mc and the solution u ( x) is cell-periodic. kλ k 43

he kp. Method (b y Evan O. Kane) ( p+ k) 1 H = + V( x) + σ V.( p+ k) m 4mc can be written, separating the terms linear in k, p k 1 H= + + Vx ( ) + σ Vp. + π. k, with m m 4mc p 1. π = + σ V. m 4mc If the problem is solved at k=0 one can expand for small k. (his is also used in the nonrelativistic limit). he perturbation mixes different bands.

hen one treats π in second order and k /m in first-order obtaining for the band γ ε ε γ 1 γ, k = 0 π. k δ, k = 0 δ, k = 0 π. k γ, k = 0 k = + + m ε 0 0 m ( k ) εγ ( 0) k.p theory- NonDegenerate bands p 1 Since π = + σ V is a vector, the diagonal elements γ, k = 0 π. k δ, k = 0 m 4mc vanish by symmetry. ( ) ε ( ) δ γ γ δ 1 1 ( k) ε ( ) kk * * = 0 + = inverse effective mass tensor m m γ γ µ ν µν µν. Sometimes, small gap strong interband interaction large inverse mass small effective mass. Example: 0.136 0.006 4*10 * 4 CdxHg1 xe x = gap < ev m m

k.p theory-degenerate bands in uniaxial crystals- (No Spin-orbit case) (see Kittel, Quantum heory of Solids Chapter 9) Consider the interesting case of a uniaxial crystal with a band of symmetry (x,y) at k=0 interacting with a higher band of symmetry 1 separated by a gap E G. he degenerate band is split by the interaction. Let us recall the results of degenerate perturbation theory in second order (Landau-Lifschitz paragraph 39). If V is the perturbation kp. V =. m and n,n belong to the degenerate level, while m denotes the other states, one must solve the secular equation V V Det V + Eδ = 0 nn' nm mn' (0) (0) m En Em nn'

E E = E (0) (0) n m G V = kp.. m Inversion is a symmetry. Since V is proportional to p which is odd, the firstorder energy vanishes by parity. Det VnmVmn' m E Eδ nn' = 0 G

Det VnmVmn' m E Eδ nn' = 0 Eigenvaues of G V VV xs sy xs EG E G kp., where V =. VV ys sx Vys m EG E G E (x,y). By symmetry, s p y = s p x = 0, while s p x = s p y. hen, u x y x y skp x s k p y y k p s s k p x set,,, me me me x x y y y y y y = k x = k y = k xk y G G G k k k k k has eigenvalues 0, k k x x y x y y light electrons heavy electrons he band is split in two isotropic bands with widely different effective masses.

Space Inversion operation Lactic acid enantiomers are mirror images of each other (0) P x= x Stereoisomers that are mirror images are callend enantiomers derived from 'ἐνάντιος', opposite, and 'μέρος', part or portion. wo enantiomers of a generic aminoacid p Let Hψ ( x) = ε ψ ( x), λ = spinor index, H = + V( x) + H' describe k, λ k, λ k, λ SO m hen, how to describe? () i p () i It is the same problem with H = + V ( x) + H', SO m ( i) (0) (0) V ( x) = V( x) = P V( x) where P = parity. σ σ = σ (0) 1 P : (,, p p) H ' Vp. is the same SO 4mc Intuitively: inverted problem ψ ( x) = ψ ( x) with same. ( i) k, λ k, λ ε k,λ 49

(0) P x= x (0) (0) (0) (0) ( i) p ( i) Indeed formally: P P = 1, P HP = H = + V ( x) + H' SO m (0) is the Hamiltonian for the inverted enantiomer, ready to act on P ψ ( x). k, λ (0) (0) Hψ ( x) = ε ψ ( x) P Hψ ( x) = ε P ψ ( x) k, λ k, λ k, λ k, λ k, λ k, λ (0) (0) (0) (0) P HP P ψ ( x) = ε P ψ ( x), k, λ k, λ k, λ ε is the same for enantiomers. k, λ But what happens if the enantiomers are identical, and P is a symmetry? 50

Space Inversion symmetry If the crystal is enantiomer of itself,that is, [P (0),H] = 0, adding this element to the translations produces a non-belian Group which implies degeneracy. Site at origin (0) P Effect of Parity: nothing happens (0) P t Effect of Parity and then up translation t (0) tp Effect of up translation Effect of up translation then Parity 51

Crystals with Space Inversion symmetry re ψ ( x ) and ψ ( x ) degenerate? (0) (0) p (0) Yes, as above : P HP = H = + V( x) + H ' P ψ ( x) = ψ ( x) SO k, λ k, λ m P HP P ψ ( x) = ε P ψ ( x) Hψ ( x) = ε ψ ( x). (0) (0) (0) (0) k, λ k, λ k, λ k, λ k, λ k, λ re ψ ( x ) and ψ ( x ) orthogonal? (0) ikx Yes, ψ ( x) = P ψ ( x) = e u ( x) belongs to irrep k k, λ k, λ k, λ since tψ ( x) = e ψ ( x). k, λ k, λ k, λ k, λ k, λ ikt k, λ k, λ hey belong to different eigenvalues of the unitary translation operator. hen Hψ ( x) = ε ψ ( x) k, λ k, λ Degeneracy! wo orthogonal eigenfunctions degeneracy implies it). (in simple tight binding chain,too, it is P which ε = ε k, λ k, λ 5

Question: ψ ( ) = ( x)? ( k ) ik. t Yes. Since D ( t) = e is a 1 dim representation ikx ψ k, λ ( x) = e uk( x) is the only solution for k, ε and k ikx ψ k, λ ( x) = e uk( x) is the only solution for k, ε k ψ ( x) = ψ ( x) k, λ k, λ k, λ x ψ k, λ Question: u ( x) = u ( x)? k, λ k, λ Yes ikt ψk, λ( x) = ψ k, λ( x) ψk, λ( x) = e uk, λ( x) ikt coincides with e u k, λ ( x) u ( x) = u ( x) k, λ k, λ 53

ime Reversal operator Suppose we can solve Schrödinger equation magnetic field (real Hamiltonian) with no φ() t i = Ht () φ(), t with Real Ht () t Can we use the knowledge of φ( t) to solve time-reversed dynamics: φ '( t) i = H( t) φ '( t)? t Yes, we can. Set t' = t. Can we solve by setting φ'( t') = φ( t)? No, but we shall find the time-reversal operator such that φ'( t') = φ( t). and obtain ime-reversed operators: ˆ ' = ˆ 1 Introduce Kramers operator : KφKψ = ψφ φ = φ = = t t t' * K K Ki i i 54

Schroedinger equation at time τ pply K, using the fact that H is real: φ( τ ) i = H( τ) φ( τ) τ φ τ φ τ Ki KH i H τ τ * ( ) ( ) * = ( τ) φ( τ) = ( τ) φ ( τ). φ ( t) t * * Now set τ = t and get i = H( t) φ ( t) φ '( t) compare to time-reversed dynamics i = H( t) φ '( t) t φ * '( t) = φ ( t) = K yields φ'( t') = φ( t) with t'=-t 1 ime-reversed operators: p' = p = KpK = p. 55

ime reversal operator for Pauli equation with B ψ e i = [ H() t + λσ. Bt ()] ψ (), t λ = 0 t m c, H 0 = spin independent part of the Hamiltonian, e e p + () t tp () H() t= c c + Vt () 0 m Can we use ψ () t to solve the time-reversed dynamics? Yes. Is complex conjugation still sufficient? No. ψ '( t) ime-reversed dynamics i = [ H '( t) + λσ. B'( t)] ψ '( t) 0 t Primes H 0 and B are needed: indeed the currents change sign under time reversal, hence the vector potential and the magnetic field also change sign. hus, B =-B, =-: herefore the imaginary part of H 0 (term in p) changes sign. ψ '( t) * time-reversed dynamics i = [ H ( t) λσ. B( t)] ψ '( t) 0 t 56

ψ e Original problem: i = [ H( t) + λσ. Bt ( )] ψ ( t), λ =, 0 t mc ψ '( t) * time-reversed dynamics i = [ H ( t) λσ. B( t)] ψ '( t) 0 t K must obviously occur to change sign of t. Write τ for t ψ ( τ ) i = [ H ( τ) + λσ. B( τ)] ψ ( τ ), 0 τ apply K, * ψ ( τ ) * * i = [ H ( τ)* + λσ. B( τ)] ψ ( τ). 0 τ Now set τ = t, * ψ ( t) * * * c.c of Pauli equation: i = H ( t) * ψ ( t) + λσ. B( t) ψ ( t). 0 t ψ '( t) * Compare to time-reversed dynamics i = [ H ( t) λσ. B( t)] ψ '( t) : 0 t * * ψ ( t) does not work because of + λσ instead o f -λσ One should expect that the spin must be reversed by time reversal. * Next, I show that ψ'( t) = iσψ ( t) 0 1 1 0 Note: iσ =, flips spin, ( iσ )( iσ ) =. 1 0 0 1 57

* ψ ( t) c.c. Pauli equation: i = H ( t) * ψ ( t) + λσ. B( t) ψ ( t) 0 t Multiply on the left by iσ * * * σ ψ * = σ ψ * + λ * σ σ ψ * i ( i ) ( t) H ( t)*( i ) ( t) ( i ). B( t) ( t). 0 t and using ( iσ )( iσ ) =-1 σ ψ * = σ ψ * λ σ σ * σ σ ψ * i ( i ) ( t) H ( t)*( i ) ( t) ( i ) ( i ). B( t) ( i ) ( t). 0 t * Next, note that ( iσ ) σ ( iσ ) = σ in fact, ( iσ ) σ ( iσ ) = σσσ = σσσ = σσσ = σ * * 1 1 1 1 1 ( iσ ) σ ( iσ ) = σσσ =+ σσσ = σ * * ( iσ ) σ ( iσ ) = σσσ = σσσ = σσσ = σ. * * 3 3 3 3 3 i ( iσ ) ψ( t) = H ( t)*( iσ ) ψ( t) λσ. B( t)( iσ ) ψ( 0 t) t * * * ψ '( t) * compare time-reversed dynamics i = [ H ( t) λσ. B( t)] ψ '( t) 0 t ψ = * '( t) iσψ ( t) (Not only t -t, not only K, but also reverse spin). 58

KφKψ = ψφ hus, the time reversal operator is = iσ yk ime-reversal of matrix elements: also for spinors 0 1 Inverse of = iσ K = K: y 1 0 0 1 iσ = 1 0 0 1 0 1 = K K 1 0 1 0 0 1 0 1 1 1 0 1 0 1 = = = 59

ime-reversal of dynamical variables: 0 1 0 1 1 p' = p = Kp( ) K = KpK = p 1 0 1 0 1 0 1 0 1 L' = L = KL( ) K = KLK = L 1 0 1 0 * 1 ( iσ ) σ ( iσ ) = σ S = S, 1 LS. = LS. LS. invariant Spin-orbit interaction is time-reversal invariant: Dirac s heory is. 60

In the case of P, we found that if it is a symmetry, it bears degeneracy. What about?, when is it a symmetry? If it is, does it imply degeneracy? is a symmetry only when H is time independent and there is no B. Kramers theorem: in Pauli theory,timereversal symmetry (i.e. H with no magnetic field and no time dependence) implies degeneracy. p H = + V( x) + H' m SO

Now I show that has twofold degeneracy (even with the spin-orbit interaction) since an H eigenspinor and its time-inverted spinor have the same energy and are orthogonal. he time-reversed spinor has the same energy because is a symmetry: = iσ K [, H ] = 0 H = H y Hφ = Eφ Hφ = Eφ Hφ = Eφ _ 1 Moreover, φ φ. Proof that φ φ : * * α 0 1 α β α φ= φ= = φφ = ( β α) = β β α 0 * * 1 0 β (note: flips spin, indeed). his implies degeneracy. Note: nothing to do with different irreps. his is spin degeneracy. However in solids time-reversed Bloch states also belong to opposite k. 6

ψ ( x) = ie σ u belongs to k and has the same energy ikx * kλ y kλ reverses spin Indeed, I show that σ z reverses under. Note that = iσ K implies σ = σ since [ σ, σ ] + = 0 y z z y z and that φψ = iσ Kφ iσ Kψ = Kφσ Kψ y y y = KφKψ = ψφ. herefore, time reversal reverses spin: ψ σ ψ = σ ψ ψ = σ ψ ψ = ψ σ ψ k z k z k k z k k k z k anticommutation Hermiticity of σ z In conclusion : ψ ( x) = ψ ( x) k, λ k, λ [, H] = 0 ε = ε also with spin-orbit. k, λ k, λ 63

Summary on symmetry and degeneracy: 0 0 Parity P ψk( x) = ψ k( x) P, H = 0 ε k, λ = εk, λ -k k [ ] ime reversal ψ ( x) = ψ ( x), H = 0 ε = ε kλ k λ kλ k λ -k k 64

Coniugation + C= 0 P 0 [ P, H] = 0, [ H, ] = 0 [ CH, ] = 0 Cψ x P ψ x P ψ x ψ x (0) (0) kλ( ) = kλ( ) = k λ( ) = k λ( ) ε = ε = ε kλ k λ k λ -k k -k k spin degeneracy at every k point even with SO 65