Ligand Group Orbitals

Transcription

1 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

2 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 G with following characters: D

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

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

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

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

7 n. unmoved atoms O E 6C 3C 6 C ' 8C i 6S 3 6 8S g 48 h h d x y z 1g 1u g u g u (, ) 1g x y z 1u E x y z x y E ( R, R, R ) g u G ( x, yz, ) ( xy, xz, yz) F basis h 4 unmoved atoms E F C B D 7 7

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

9 O E 6C 3C 6 C ' 8C i 6S 3 6 8S h h d 6 1u E C F B Projector : ( ) ( i) i * P R R R Projection of orbital into 1u C 4 : D operations with axis D : operations with axis EB : C, F operations with axis CF : B, E 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 RC 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) 3 D 1 D h 6 B C E F 1 B C E F d R 8S ( B C E F)

10 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

11 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. Isolated ion M M Number of d electrons: i V Cr Mn Fe Co Ni Cu Friedrich Hund, (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

12 Number of d electrons: i V Cr Mn Fe Co Ni Cu O M M M Hund s rule prompts the ground state quantum numbers: electron number ground state d d d d d d d d d D F F D S D F F D For instance, with 3 electrons ground state is 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

13 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 a ( j) z a j ( j) ( j) a z a m j Characters of representation 1 sin[ ] 5 j a sin[ a] ( ), ( ), a a mj j sin sin j ( j) im ja a e j a G d a matrix of j representation is D ( R ) jm' R jm im a j diag( e ) j ( ) rd ( R ) e. j im a O z M C a Sin[ ] Sin[ ]

14 C 5 a Sin[ ] 1 Sin[ ] 1 1 C a Sin[ ] Sin[ ] O M one finds O E 6C 3C 6 C ' 8C i 6S 3 6 8S h h d 6 G d Inversion is like identity E for d states. For the d orbitals (or even D) states, i= parity=e

15 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 rotation. 1 sin[ ] 5 j a sin[ ] Since ( a), and j, ( ) 1 a sin sin O E 6C 3C 6 C ' 8C i 6S 3 6 8S h h d 6 G d Reflections have character 1 because: Inversion i is the result of a rotation and a reflection: i= C Improper rotations: Since a reflection is a factor 1,improper rotations are like proper ones: a ( ) 5 5 sin[ ] sin[ ] ( S 3 6) ( R6 ) 1, ( S4) ( R4 ) 1. sin sin 6 4 j j m j 1 sin[ j a] im j e a a sin 15 15

16 one finds O E 6C 3C 6 C ' 8C i 6S 3 6 8S h h d 6 G d Number of times irrep i is present in basis: 1 i * ni R R N G R he 5-fold degeneracy of d orbitals is broken. O E 6C 3C 6 C ' 8C i 6S 3 6 8S g 48 h h d x y z 1g 1u g u g u (, ) 1g x y z 1u E x y z x y E ( R, R, R ) g u ( x, yz, ) ( xy, xz, yz) G d d basis d G E d g g ( x y, z x y ) ( xy, xz, yz) 16 16

17 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 D 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

18 Crystal field theory-interactions E g O M D 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

19 pplications of Group heory to vibrations Vibrations of a methylene group (-CH -) in a molecule Symmetrical stretching symmetrical stretching Scissoring (Bending) Rocking Wagging wisting Stretch= stirare Rock= scuotere Wag=agitare wist=storcere

20 pplications of Group heory to vibrations Normal modes of molecular vibration: classical motion of i-th nucleus v nuclear displacement vector v= x, y, z,... x, y, z ( v,... v ) N N N 1 3N Born-Oppenheimer approximation: Nuclei = classical point masses which move in effective potential U m i v i U v v i Groups help with any symmetric secular problem 19

21 m i v i U v U v v Harmonic approx to potential: i i m i v i 1 U v v v pq 1 U v v= x, y, z,... x, y, z ( v,... v ) N N N 1 3N ij i U U v pq p q ij j j Equation of motion: 1 m i j U ij v j v i vv ij i j U ij force matrix secular problem; eigenvalues = vibration frequencies. 1 1

22 Equation of motion: Uijv j vi mi j Most people prefer to put masses into force matrix 1 mi vi Uijvj mi j 1 m U ij i vi mj vj j mm i j Uij Introducing Qi mi v i Wij one has the secular problem mm Q W Q Det( W I) 0 i ij j j i j Qα eigenvectors of W are called normal modes a ωα eigenfrequencies of W 0: a 0: 3 translations and 3 rotations ( rotations for linear molecules)

23 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. 3 3

24 Group heory and classification of vibrations Normal modes Qα eigenvectors of W ωα eigenfrequencies of W Vibrations belonging to different irreps are orthogonal. eigenvectors 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. Same problem as finding symmetry molecular orbitals starting with atomic p orbitals 4 4

25 Program: o diagonalize W simultaneously with as many D as possible, + Dirac s characters W 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 5 5

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

27 Using the 9X9 matrices : Cv E C ( yz) ( xz) G 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) 7 7

28 One can find the characters without writing the D matrices: E C 9 unmoved 9 1 unmoved reversed 1 Oxygen arrows moved xz yz 6 unmoved 3 reversed 3 unmoved 1 reversed 1 Characters of the reducible representation of atomic arrows: Cv E C ( yz) ( xz) G Vibrations + rotations+translations 8 8

29 Cv E C ( yz) ( xz) G C I C g B B v xz yz xy, R x, R y, R z 4 y x z read from able: Gtrasl : B B Cv E C ( yz) ( xz) G trasl 1 1 Note : R x x x x G : B B rot i j j k k 1 Cv E C ( yz) ( xz) G rot 9 9

30 C I C g B B v xz yz xy, R x, R y, R z 4 y x z C I C g G G G G v xz yz trasl rot vibr By difference: i ni R R N G nalysis in irreps: R * G vibr = 1 + B 1 Breathing mode 1 in all molecules What kind of vibration is B 1? 30

31 C I C g 4 v xz yz B B z xy, R x, R y, R y x z What kind of vibration is B 1? Projection operator on arrow basis: P( B ) E C ( xy) ( yz) 1 Recall the block matrices: 0 0 b D( ( yz)) 0 b 0, b( ( yz)) b b D( C) 0 b 0, b( C) b b D( ( xz)) 0 b 0, b( ( xz)) he block matrix for the projector is: 0 0 b b xz 0 bc b( yz) ( ) 0 1 P( B ) E C ( xy) ( yz) 0 1 b xz b C b ( yz) 0 1 b C b yz b xz 31

32 P( B 1 ) b xz 0 bc b( yz) ( ) 0 1 P( B ) 0 1 b xz b C b ( yz) 0 1 b C b yz b xz PB ( 1) C I C g 4 v xz yz B B z xy, R x, R y, R 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). 3 3

33 G( H ) : E C 3 v Vibrations of NH 3 Movements of N like (x,y,z) : 1 +E G(N) has characters a 9 unmoved 9 0 unmoved 0 c b unmoved 1 reversed 1 C I C 3 g 6 3v 3 v x y E 1 0 x, y, R, R z R C E C 3 3v 3 G( N) G( H ) G( NH ) 1 0 G G G trasl rot vibr z G vibr 1 E 33 33

34 3 1 4 With 5 atoms χ(e) = 15. Vibrations of Methane (CH 4 ) E 8C 3C 6 6S N 4 d 3 d 4 G E z r x y (3, ) ( R, R, R ) ( x, y, z) 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 ) = C all H moved. For the atom of C: arrows of the C change sign and the third does not move: χ = -1 34

35 3 1 4 E 8C 3C 6 6S N 4 d 3 d 4 G E z r x y (3, ) ( R, R, R ) ( x, y, z) 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: / 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 6 6S N 4 d 3 d 4 G G tot G trasl G rot 1 G E vibr 1

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

37 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 37

38 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 38

39 C C D E C C C 3 C ' 3 C '' i S S 3 3 g 4 B B 6h h d v 1g g 1g g 1g x y g 1u u E ( R, R ) E ( x y, xy) B B E E G 1u u 1u u tot z ( x, y) G trasl G rot G 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. G B B B E 3E 4E E vibr 1g g u 1u g u 1g 1u g u 39

40 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 Born-Von Karman Periodic boundary conditions: for some N>> 1, in = 1. Halite (sodium chloride) Band theory: - a single, large crystal. One-electron approximation: the electron moves in a periodic crystal potential V (x ) H, _=0 with H (x)= (x) and i k k k Simultaneous eigenvectors of H and all : i.k t (x) = (x+t ) = e i (x), H (x) ( k) (x) i k k i k k k i.k t unitary, eigenvalue C=e i : i N Born-Von Karman b.c. C 1 i 40

41 N i.n.k i C =e 1 with p, q, r defined by requires k t = * integer; therefore, i N pg + qg + rg 1 3 k= N Z, and the reciprocal lattice basis vectors g t Bloch's functions are: t g = i Elementary aspects: i.k ti H and = e, k k k j ik x Bloch's theorem (Floquet's theorem) : (x) = e u (x ), with u (x ) lattice periodic. k i k k ij. k k j ik x ik x For each k, since (with =1) p e =e (p+k) the Schrödinger equation has one solution: (p + k ) [ + V (x)]u (x) = u m k k k (x) ˆ within the unit cell. ˆ 41

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

43 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 ( t ) e character One solution of H for each k, expected k k k k he degeneracy of 1 d chain spectrum ( k) t cos( k) ( k) is not explained in this way: the inversion symmetry is involved (see below) 43

44 ( ) ( ). Since D k ( t ) k ( t ) e ik t is the character of translation, i * j In this case : ( ) ( ) RG 1 i( k k ') t e ( k k ') N cells tbravais LO R R N G ij Second character orthogonality theorem: Summing over irreps I labelled by k ( i i i NG ( C) ( C') n ) ( ) * CC' C kbz i( t t ') k e NC t t (, ') BZ integration cell localization! 44

45 Relativistic corrections O(v /c ) e i ev c p t c Dirac's equation mc B B e c p i ev c t yields relativistic corrections via a Foldy Wouthuysen transformation: H 4 p p e ev m 8m c 4m c 3 e.( E p) dive 8mc he correction which lifts degeneracy: SO interaction e e dv e dv In atomic physics.( E p).( r p) S. L H 4 4 m c m c r dr m c r dr SO e 1 lso,.( ) ). E p ( V p 4mc 4mc 45 How does it change the symmetry in crystals?

46 dding the Spin-orbit interaction lifts trivial spin degeneracy: p H V( x) H ' m SO HSO 1 ' V. p 4mc Eigen-spinors can still be taken of the Bloch form, i. k ti H and = e, with k k k ik x k, k, k, i k (x) = e u (x ), u (x ) lattice periodic spinor. he equation for the periodic function, with p i p k, ( p k) m V( x) u ( x) u ( x) with the spin-orbit interaction becomes k k k k ( p k) 1 V( x) V.( p k) u ( x) u ( x), k k k m 4mc and the solution u ( x) is cell-periodic. k 46

47 Space Inversion operation Lactic acid enantiomers are mirror images of each other. Mirror: P (0) *C P (0) =i (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 H ( x) ( x), spinor index, H V( x) H ' describe k, k, k, SO m hen, how to describe? ( i) p ( i) It is essentially the same problem with H V ( x) H ', SO m ( i) (0) (0) V ( x) V( x) P V( x) where P parity. (0) 1 P : (,, p p) H ' V. p is the same SO 4mc Intuitively: inverted problem ( x) ( x) with same. () i k, k, k, 47

48 (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 ( x). H ( x) ( x) P H ( x) P ( x) (0) (0) k, k, k, k, k, k, (0) (0) (0) (0) P HP P ( k, k, k, is the same for enantiomers., k Ennergy eigenvalues are the same x) P ( x), k, But what happens if the enantiomers are identical, and P is a symmetry? 48

49 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 Effect of Parity: nothing happens Effect of Parity and then up translation (0) P (0) tp t (0) P t Effect of up translation Effect of up translation then Parity 49

50 Crystals with Space Inversion symmetry (0) P x x p H V( x) H ', SO m (0) with V( x) V( x) P V( x) re the Bloch states ( x) and ( x) degenerate? PHPP ( x) P ( x). (0) k, k, k, (0) (0) p (0) Now, P HP H V( x) H ' P ( x) ( x) SO k, k, m H ( x) ( x). k, k, k, Yes : k, k, Let H ( x) ( x); then, PH k, k, k, k, (0) ( x) P ( x) and since P 1, k, k, 50

51 Crystals with Space Inversion symmetry re the degenerate ( x) and ( x) orthogonal? k, k, Yes, ( x) P ( x) e u ( x) (0) ikx k, k, k, belongs to nondegenerate irrep k since t ( x) e ( x). ikt k, k, ( x) is the same as ( x). k, k, hey belong to different eigenvalues of the unitary translation operator. hen H ( x) ( x) k, k, k, k, k, (in simple tight binding chain,too, it is P which implies it). 51

52 Question: u ( x) u ( x)? k, k, Yes ( x) ( x) ( x) e u ( x) ikt k, k, k, k, ikt coincides with e u ( x) k, k, k, u ( x) u ( x) 5

53 ime Reversal operator Suppose we can solve Schrödinger equation with no magnetic field (real Hamiltonian) and that the Hamiltonian depends on time. () t i H ( t) ( t), with Real H ( t) t Question: can we use the knowledge of ( t) to solve time-reversed dynamics: '( t) i H( t) '( t)? t Yes, we can. Set t ' t. Can we solve by setting '( t ') ( t)? No, but we shall find the time-reversal operator such that '( t ') ( t). and obtain ime-reversed operators: ˆ ' ˆ 1 Introduce Kramers operator : t t t ' * K K Ki i i K K 53

54 Schroedinger equation at time ( ) i H( ) ( ) pply K, using the fact that H is real: Ki KH i H * ( t) * Now set t and get i H ( t) ( t) t * ( ) ( ) * ( ) ( ) ( ) ( ). '( t) compare to time-reversed dynamics i H( t) '( t) t '( t) * ( t) yields '( t ') ( t) with t'=-t ime-reversed operators: p' p 1 KpK p. K 54

55 ime reversal operator for Pauli equation with B e i [ H ( t). B( t)] ( t), = 0 t m c, H 0 = spin independent part of the Hamiltonian, which is complex, e e p ( t) ( t) p H ( t) c c V( t) 0 m Can we use () t to solve the time-reversed dynamics? Yes. Is complex conjugation still sufficient? No. '( t) ime-reversed dynamics i [ H '( t). B'( t)] '( 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. '( t) * time-reversed dynamics i [ H ( t). B( t)] '( t) 0 t We wish such that '( t') ( t) 55

56 e Original problem: i [ H ( t). B( t)] ( t), = 0 t mc '( t) * time-reversed dynamics i [ H ( t). B( t)] '( t) 0 t, K must obviously occur in to change sign of t. Write for t ( ) Original problem: i [ H ( ). B( )] ( ); apply K, 0 * ( ) * * i [ H ( ) *. B( )] ( ). Now set t, 0 * ( t) * * * i H ( t) * ( t). B( t) ( t). 0 t '( t) * Compare to time-reversed dynamics i [ H ( t). B( t)] '( t) : 0 t * * almost OK, but ( t) does not work because of instead of - 56

57 One should expect that the spin must be reversed by time reversal. * Next, I show that '( t) i ( t) a Note: i, flips spin =, a 1 0 ( i )( i ). 0 1 * ( t) c.c. Pauli equation: i H ( t) * ( t). B( t) ( t) 0 t * * * Multiply on the left by i, which commutes with H 0 * * * * i ( i ) ( t) H ( t) * ( i ) ( t) ( i ). B( t) ( t). 0 t and using ( i )( i ) =-1 i ( i ) ( t) * H ( t) * ( i ) ( t) * ( i ) * ( i ). B( t) ( i ) ( t) *. 0 t

58 i ( i ) ( t) * H ( t) * ( i ) ( t) * ( i ) * ( i ). B( t) ( i ) ( t) *. 0 t his is what we want. o see this, note that * ( i ) ( i ) in fact, ( i ) ( i ) * * ( i ) ( i ) * * ( i ) ( i ). * * i ( i ) ( t) H ( t) * ( i ) ( t). B( t)( i ) ( 0 t) t * * * '( t) * compare time-reversed dynamics i [ H ( t). B( t)] '( t) 0 t 58 * '( t) i ( t) (Not only t -t, not only K, but also reverse spin).

59 hus, the time reversal operator is i yk ime-reversal of matrix elements: for spinors * * ( ) 1 1 * * 1 1 K K 0 1 Square of i K K : y K K i 1 0 Who could have predicted this - sign? 59

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

61 Question: 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? nswer: is a symmetry only when H is time independent and there is no B. Kramers theorem: in Pauli theory,time-reversal symmetry (i.e. H with no magnetic field and no time dependence) implies degeneracy (even with spin-orbit interaction) p 1 H V( x) H ' H ' V. p SO SO m 4mc

62 We must show that H 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 just because is a symmetry: the spin-orbit interaction is L.S and both L and S are reversed. i K [, H ] 0 H H y H E H E H E _ 1 Proof that : * * a 0 1a a a a 0 * * 1 0 (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. ( x) ie u ikx * k y k belongs to k and has the same energy 6

63 ( x) ie u ikx * k y k belongs to k and has the same energy reverses spin Next, I show that z reverses under. Note that i K implies since [, ] 0 y z z y z and that i K i K K K y y y K K. herefore, time reversal reverses spin: k z k z k k z k k k z k anticommutation Hermiticity of z In conclusion : ( x) ( x) k, k, [, H] 0 also with spin-orbit. k, k, 63

64 Summary on symmetry and degeneracy: P x x P H 0 0 Parity k ( ) k ( ), 0 k, k, -k k ime reversal ( x) ( x), H 0 k k k k -k k 64

65 Coniugation In molti cristalli coesistono P 0 e C 0 P 0 [ P, H] 0, [, H] 0 [ C, H] 0 C x P x P x x (0) (0) k ( ) k ( ) k ( ) k ( ) k k k -k k -k k spin degeneracy at every k point even with SO 65