Group. Benzene D 6h z B B. E ( x y, xy) ( x, y) A B B C 2

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

2 Benzene C 6 H 6 There are 1 atoms and 36 coordinates, therefore χ(e) = 36. The 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. Therefore, 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

3 The 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. The reflection for a plane containing the C axis has character 0. The 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. Therefore χ(σ v ) = 4. The characters of Γ trasl are the sums of those of A u and E 1u ; those of Γ rot are the sums of those of A g and E 1g. C C 3

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

5 Space-Time Symmetries of Bloch States in solids t 1, t, t 3 primitive translation vectors Translation Group {T i } set of lattice translation operators = combinations of primitive translations with integer coefficients Born-Von Karman Periodic boundary conditions: for some N>> 1, T 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,T _=0 with H (x)= (x) and i k k k Simultaneous eigenvectors of H and all T : i.k t T (x) = (x+t ) = e i (x), H (x) ( k) (x) i k k i k k k i.k t T unitary, eigenvalue C=e i : i N Born-Von Karman b.c. C 1 i 5

6 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 T = 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 6

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

8 Group theory aspects: No degeneracy is predicted, since Abelian Groups have only onedimensional representations. Abelian group Each translation T 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 1X1 matrix of representation ( t ) e character One solution of H for each k, expected k k k k The degeneracy of 1 d chain spectrum ( k) t cos( k) ( k) is not explained in this way: the inversion symmetry is involved (see below) 8

9 ( ) ( ). Since D k ( t ) k ( t ) e ik t is the character of translation T, i j In this case : ( ) ( ) LOT R R N 1 N RG i( k k ') t generalizes to e ( k k ') cells tbravais 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! Kroneker 9

10 Relativistic corrections O(v /c ) ea i ev c p t c A A Dirac's equation mc B B ea 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 The correction which lifts degeneracy: Spin-Orbit interaction 1 dv In atomic physics due to central symmetry E r r dr e e dv e dv.( E p).( r p) S. L H 4 4 m c m c r dr m c r dr e 1 Also,.( ) ). E p ( V p 4mc 4mc 10 How does it change the symmetry in crystals? SO

11 Adding the Spin-orbit interaction lifts trivial spin degeneracy: 4 p e Neglecting and 8m c 8m c p e H ev.( E p) m 4 3 mc dive p H V( x) H ' m SO HSO 1 ' V. p 4mc Eigen-spinors can still be taken of the Bloch form, since the translation symmetry is not changed even with Dirac s H i. k ti H and T = e, with k k k ik x k, k, k, i k (x) = e u (x ), u (x ) lattice periodic spinor. k 11

12 i. k ti H and T = e, with k k k ik x k, k, k, i k (x) = e u (x ), u (x ) lattice periodic spinor. HSO 1 ' V. p 4mc k The 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 ( 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 1

13 Space Inversion operation (0) P x x Stereoisomers that are mirror images are callend enantiomers derived from 'ἐνάντιος', opposite, and 'μέρος', part or portion. Two enantiomers of a generic aminoacid Lactic acid enantiomers are mirror images of each other. Mirror: P (0) C P (0) =inversion p Let H ( x) ( x), spinor index, H V( x) H ' describe k, k, k, SO m Then, 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, 13

14 (0) P x x Energy eigenvalues are the same (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, x) P ( x), is the same for enantiomers. k, k, But what happens if the enantiomers are identical, and P is a symmetry? 14

15 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-abelian 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 15

16 Crystals with Space Inversion symmetry Question: u ( x) u ( x)? k, k, Yes ( x) ( x) ( x) e u ( x) ikt k, k, k, k, ikt belongs to -k therefore coincides with e u ( x) u ( x) u ( x) k, k, k, 16

17 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) Are 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, 17

18 Crystals with Space Inversion symmetry Are the degenerate ( x) and ( x) orthogonal? k, k, Yes, because ( 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, They belong to different eigenvalues of the unitary translation operator. Then H ( x) ( x) k, k, k, k, k, (in simple tight binding chain,too, it is P which implies it). 18

19 Time 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 T such that '( t ') T( t). and obtain Time-reversed operators: Aˆ ' TAT ˆ 1 Introduce Kramers operator : t t t ' K K Ki i i K K 19

20 Schroedinger equation at time ( ) i H( ) ( ) Apply 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( t) with t'=-t T Time-reversed operators: p' TpT 1 KpK p. K 0

21 Time 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 A ( t) A( 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) Time-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. Thus, B =-B, A =-A: Therefore the imaginary part of H 0 (term in Ap) changes sign. '( t) time-reversed dynamics i [ H ( t). B( t)] '( t) 0 t We wish T such that '( t') T( t) 1

22 e Original problem: i [ H ( t). B( t)] ( t), = 0 t mc '( t) time-reversed dynamics i [ H ( t). B( t)] '( t) 0 t (we seek relation of ' to ) K must obviously occur in T to change sign of t. Write for t ( ) Original problem: i [ H ( ). B( )] ( ); apply K, 0 ( ) i [ H ( ). B( )] ( ). Now set t, 0 inverted-time complex-conjugate equation reads ( t) i H ( t) ( t). B( t) ( t) 0 t., Compare to time-reversed dynamics '( t) i [ H ( t). B( t)] '( t) : 0 t almost O K, but ( t) does not work because of instead of -

23 One should expect that the spin must be reversed by time reversal. Next, I show that time reversed spinor '( t) i ( t) Note: i, flips spin =, ( 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

24 i ( i ) ( t) H ( t) ( i ) ( t) ( i ) ( i ). B( t) ( i ) ( t). 0 t This is what we want. To 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 4 '( t) i ( t) (Not only t -t, not only K, but also reverse spin).

25 Thus, the time reversal operator is T i yk Time-reversal of matrix elements: for spinors ( ) K K 0 1 Square of T i K K : y T K K T T 0 1 i 1 0 Who could have predicted this - sign? 5

26 Time-reversal of dynamical variables: p' TpT Kp( ) K KpK p L' TLT KL( ) K KLK L ( i ) ( i ) TST S, 1 TL. ST L. S L. S invariant Spin-orbit interaction is time-reversal invariant: Dirac s Theory is. 6

27 Question: In the case of P, we found that if it is a symmetry, it bears degeneracy. What about T?, when is it a symmetry? If it is, does it imply degeneracy? Answer: T 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

28 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. The time-reversed spinor has the same energy just because T is a symmetry: T i K [ T, H ] 0 H THT y H E TH ET HT ET 0 1 T i K K y 1 0 _ 1 Proof that T : 0 1 T T (note: T flips spin, indeed). This implies degeneracy. Note: nothing to do with different irreps. This is spin degeneracy. However in solids time-reversed Bloch states also belong to opposite k. T ( x) ie u ikx k y k belongs to k and has the same energy 8

29 T ( x) ie u ikx k y k belongs to k and has the same energy reverses spin Next, I show that z reverses under T. Note that T i K implies T T since [, ] 0 y z z y z and that T T i K i K K K K K. y y y 9

30 Proof that time reversal reverses spin k z k k z k. Use T T. obtain T T z k k Use T z T z obtain T T T T by z k k k z k Hermiticity of z i.e. time reversal reverses spin In conclusion : T ( x) ( x) k, k, [ T, H] 0 also with spin-orbit. k, k,

31 Summary on symmetry and degeneracy: P x x P H 0 0 Parity k ( ) k ( ), 0 k, k, -k k Time reversal T ( x) ( x) T, H 0 k k k k -k k 31

32 Coniugation In molti cristalli coesistono P 0 e T C 0 P T 0 [ P, H] 0, [ T, H] 0 [ C, H] 0 C x P T 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 3