Landau & Lifshits, Quantum Mechanics, Ch. 12. Tinkham, Group Theory and Quantum Mechanics

Transcription

1 Suggested reading: Landau & Lifshits, Quantum Mechanics, Ch. 2 Tinkham, Group Theory and Quantum Mechanics Dresselhaus, Dresselhaus, Jorio, Group Theory: Applications to the Physics of Condensed Matter Ramond, Group Theory: a Physicist s Survey

2 A (finite or infinite) sequence of elements A,B,C form a group, if the following four conditions are satisfied. CLOSURE: If A and B are belong to the group, then A B also belongs to the group. 2. ASSOCIATIVITY: If A, B and C belong to the group, then (A B) C = A (B C). 3. IDENTITY: There is an element e of the group such that for any element a of the group A I = E I = I. 4. INVERSE: For any element A of the group there is an element A such that A i A =A i A = I

3 D3

4 2 2 3 A 2 3 C 3 B

5 2 D F 2

6 Closure property 2 2 AB 3 2 =A = =D 3 Six elements: identity, three filps, two rotations Group of order 6

7 Left classes I 3C 2 2C 3 I A B C D F I I A B C D F Right A A I D F B C B B F I D C A C C D F I A B D D C A B F I F F B C A I D Three classes: ) identity (E), 2) three 8 rotations (A,B,C), 3) 2 rotation (D) and 24 rotation (F)

8 Two groups G and G are called isomorphic, if there is one-to-one correspondence between their elements { } { } G = A, B,C...P... G ' = A', B',C '...P'... A A' B B'... AB = P A' B' = P' Example: D 3 is isomorphic to C 3v (rotations by 2 + reflections in three vertical planes)

9 ( ) is an arbitrary (single-valued) function of x. Let ψ x Take an element R of group G (order g) Apply the operator P R P( R)ψ ( x) ψ R x ( ) to ψ ( x) defined as ( ) ( ) Φ R x Operators P(R) form a group which is isomorphic to G : P(S)P(R) = P(SR) Proof: P( S)P( R)ψ ( x) = P( S)Φ R ( x) = Φ R S x ( ) =ψ ( R S x) =ψ SR (( ) x) = P SR ( )ψ x Applying all symmetry operations to ψ, we get a set of r linearly indepedent functions In general, r g. { } ψ...ψ r BASIS ( ) NB :choice of ψ is arbitrary.

10 Applying a symmetry operation to the basis function, we get a linear superposition of basis functions P( S)ψ i = G ki ( S)ψ k r k= Matrices G S ( ) form a representation of the group. Representation of a group is as arbitrary as the choice of the basis function. If a matrix of particular representation cannot be reduced to a block-diagonal form by any similarity transformations, such a representaton is called irreducible.

11 D 3 ) consider a function which does not change either upon rotations or flips This function generates a trivial D representation ( ) = G( A) = G(B) = G(C) = G(D) = G(F) = G I f ( x, y,z) =, x 2 + y 2,z 2,... 2) consider a function which is invariant with respect to 2 rotations but changes its sign upon flips ( ) = z f x, y,z This function generates another D representation ( ) = G(D) = G(F) = ( ) = G(B) = G(C) = G I G A

12 D 3 3) 2D representations are formed by two basis functions which transform as elements of a vector (x,y) x 3 y ( ) = G B G I ( ) = identity ; Tr=2 / 2 3 / 2 3 / 2 / 2 G( A) = ; Tr= G C ( ) = ; Tr= 22 flip: x -x,y y 33 flip flip / 2 3 / 2 3 / 2 / 2 ; Tr= G( D) = / 2 3 / 2 3 / 2 / 2 ; Tr=- ( ) = / 2 3 / 2 G F 3 / 2 / 2 ; Tr=- 2 rotation 24 rotation

13 Character=trace of an irreducible representation matrix Traces are invariant characters do not depend on the choice of basis functions

14 Group A,B: D representations (A is even upon rotation, B is odd) E: 2D representation (not to be confused with identity!) F: 3D representation Class D 3 I 2C 3 3C 2 A Basis function z A 2 (x, y) E 2 irrep Trace of irrep

15 Take trace 2, multiply by the number of the elements in the class, and sum over classes z (x, y) D 3 A A 2 E I 2C 3 3C 2 2 A and A 2 : ( ) + 2 ( ) + 3 ( ( )) = A and E : ( 2) + 2 ( ( )) + 3 ( ) = C different irreps N C number of elements in C χ β (C) * χ α ( C ) = gδ αβ class same irrep = 6 = g ( ) 2 = 6 = g ( ) = 6 = g C NC number of elements in C Also, = 6 = g f 2 α = g α : over irreps dim of irrep χ α ( C ) 2 class = g

16 Van Vleck orthogonality theorem for irreps R: symmetry elelements G ik α * β ( R) Glm (R) = g δ αβ δ il δ km f α Set i = k,l = m and take a trace R: symmetry elelements χ α * ( R) χ β ( R) = gδ αβ All elements of the same class (C) have the same characters C: classes N C χ α * ( C) χ β ( C) = gδ αβ

17 Let G be a reducible representation of dim f with character χ R. A reducible representation can be expanded over irreps How many times an irrep G α is contained in G? Using orthogonality of characters, a α = g G = α N C χ R χ α * ( C) C:all classes of G α a α G α or, since dims of G may be different, G = a G a 2 G 2... A B = Applying trace, χ R = a α χ α α A B direct sum

18 Ĥψ = Eψ Wavefunctions must obey all symmetry properties of the Hamiltonian. A proper description of a degenerate state is a linear superposition of wavefunctions. Basis functions of a given irrep are transformed into each other under group operations Degenerate states form a basis of a given irrep Dimensionality of a given irrep gives us immediately degeneracy of the corresponding energy level A,B: D representations non-degenerate levels E: 2D representation two-fold degeneracy F: 3D representation three-fold degeneracy

19 Ĥ = Ĥ + Ĥ ' Symmetry of Ĥ ' < Symmetry of Ĥ Representations of Ĥ ' are contained in Ĥ In general, a representation of Ĥ ' is a reducible representation of Ĥ Decomposing representations of Ĥ ' into irreps of Ĥ, we find which degeneracies are lifted.

20 Ĥ ( O) : Rotational group of a cube (without inversion and reflection symmetries) 4 C 3 axes 8C 3 (4C + 4C 2 3 ) 3 Classes: O I 8C 3 3C 2 6C 2 6C 4 A (Γ ) A 2 (Γ 2 ) ( ) 2 2 ( ) 3 ( ) 3 E Γ 2 F Γ 5 F 2 Γ 25 6 C 2 axes 6C C 4 axes 6C 4 (3C 4 + 3C 4 3 ) 3C 2 (=3C 4 2 ) } } } non-degen. 2-fold 3-fold Ĥ '( D ) 3 : A strain is applied along the main diagonal How does the strain split the degenerate levels?

21 Group O contans all the elements of D 3 [E,2C 3,3C 2 ] For example, irrep F 2 of O is a reducible representation of D 3 O I 8C 3 3C 2 (= 3C 2 4 ) 6C 2 6C 4 A A 2 E 2 2 F 3 F 2 3 ( ) D 3 I 2C 3 3C 2 A A 2 E 2 Decomposition formula: a α = N C χ R C χ α * g ( C) C:all classes of G α a( A ) = I 2C 3 3C N C χ( I ) χ A ( I ) N C χ( 8C 3 ) χ A ( 2C 3 ) N C χ( 6C 2 ) χ A ( 3C 2 ) = ( ) = ( ) a A 2 = a( E) = [ ( ) + 3 ] = F 2 ( 3) A ( ) E( 2)

22 Rotational symmetries of building blocks (polygons) must be consistent with translational symmetry crystallographic restriction theorem: lattice can have only 2, 3, 4, and 6- fold rotational symmetries

23 Crystal structure can be obtained by attaching atoms or groups of atoms --basis-- to lattice sites. Crystal Structure = Crystal Lattice + Basis Crystal Structure 23 Partially from Prof. C. W. Myles (Texas Tech) course presentation 23

24 Bravais lattices: monoatomic basis Non-Bravais lattices: polyatomic basis Graphene: Honeycomb

25 8 2π = 2 Oblique 2π 8 = 2 fold axis 2 Rhombohedral Orthorhombic (Rectangular) 2 8 = π 2 2π 6 = 6-fold axis 6 2π 9 = 4-fold axis 4 Tetragonal (Square) Hexagonal (Triangular) 25

26 8 Oblique Elements of symmetry: C 2 rotations Group: C 2

27 Ohm s law on a lattice: j i = ij E j Conductivity tensor ˆ = xx yx xy yy C A () How many independent components does an oblique lattice have? The only symmetry operation is the 8 rotation about z. Matrix of rotation about z by : U = U 8 = cos sin C A ) (2) sin cos C A = (3) Transformation of the Ohm s law under rotation Û~j = Û ˆÛ Û ~ E ) ˆ! Û ˆÛ (4) For rotations, Û ˆ= U ) ˆ = Û ˆÛ (5) 2

28 3 xx yx xy yy C A = Û8ˆÛ8 =( ) = xx yx xy yy xx yx xy yy C A ( ) C A (6) ) lattice symmetry imposes no constrains on ˆ. Time-reversal symmetry (symmetry of Onsager kinetic coe cients): ij = ji ) (7) ˆoblique = xx xy xy yy C A (8)

29 Rhombohedral Orthorhombic (Rectangular) D 2 D 2

30 Equivalently, one can do reflections in vertical planes 8 D 2 =C 2v (=means isomorphic )

31 D 2 =(C 2v ) group (rhombohedral and orthorhombic lattices) We already know that Û8 about z axis imposes no constraints. 8 rotation about the x axis: x! x, y! y Û x 8 = C A (9) 4 xx yx xy yy C A = Û 8ˆÛ x 8 x = = = xx yx C A yy xx yx xy xy yy C A C A xx yx xy yy C A C A C A ) xy =, yx = () ˆrhombo/ortho = xx C A () yy

32 Tetragonal (Square) 9 Symmetry operations: 3 9 rotations 8 rotations about 4 horizontal axes D 4 Symmetry operations: 3 9 rotations Reflections in 4 Vertical planes C 4v D 4 =C 4v

33 D 4 = C 4v (tetragonal) D 4 (C 4v )alreadycontainsallsymmetriesofd 2 (C 2v ). ) At least, we must have ˆtetra = xx C A (2) yy However, we have additional symmetries: 8 rotations about diagonals (or reflections in the diagonal vertical planes). Reflection in a diagonal vertical plane: x! y, y! x ˆR x y ˆR = C B A C A C B x y C A = 5 y C A (3) x xx C A = ˆRˆ ˆR = yy B C B xx C B A yy B C B xx C B yy C A A yy xx ) xx = yy (4) ˆtetra = xx C A (5) xx

34 Any tensor describing physical properties of the lattice 6 dielectric permittivity ˆ" elastic moduli electron e ective mass ˆm... have the same symmetries. Exercise: work out symmetries of ˆ for the hexagonal lattice.

35 Vibrational modes of the H 2 O molecule System of N particles (not on the same line): 3N degrees of freedom 3 translational 3 rotational # of vibrational modes: N v = 3N-3-3=3N-6 For H 2 O: N=3 N v =3 What are those 3 modes?

36 H 2 O σ' v σ v C 2 axis+2 vertical planes (σ v and σ v ) C 2v group

37 Symmetry operations of C 2v group: I, C 2, v, and v. g =4. Special property: C 2v is cyclic group. Cyclic group: Take one symmetry element, S. All other elements are given by S k, k =, 2... Cyclic groups have only D irreps. C 2v is particularly simple: applying any element twice, we get I. C 2 2 = I, 2 v = I, ( v) 2 = I. Let be any basis function. S(S ) =S 2 = Irreps are numbers such that G 2 =) G = ±. X irreps X irreps dim2 (irrep) = g =4) #ofirreps=4 7

38 How do the group elements act on coordinates? C 2 : x! x, y! y v : x! x, y! y v : x! x, y! y For all operations, z! z. Table of characters C 2v I C 2 v v z, A xy, A x, B - - y, B A,2 even under C 2, B,2 odd under C 2. We know that N v =3butwehave4irreps. Some of the irreps do not correspond to molecular vibrations! Need to get rid of irreps that correspond to translations and rotations rather than to vibrations. 8 σ' v σ v

39 Equivalence ( atomic site ) representation (G eq )isformed by vibrational displacements of all atoms consistent with the symmetry operations of the group. In general, G eq is reducible. Expand G eq over irreps. C 2v : G eq = a A + a 2 A 2 + a 3 B + a 4 B 2 Some of the coe cients should be zero. Consider rotation by angle about some symmetry axis of the molecule in 3D U = cos sin sin cos Its character: C A (U )=TrU =+2cos If there are N a atoms on the same axis, (U )=N a ( + 2 cos ) Each of the N a atoms was subjected not only to vibrations but also to translations and rotations ) (U )containsextradegreesoffreedom. 9

40 Translation is a vector ~u =(u x,u y,u z ). U transforms its as any other vector. The corresponding character is +2cos Rotation by an infinitesimally small angle is described by vector ~ of magnitude and along the axis of rotation. ~ is a polar vector but, under rotations, it transforms as a polar vector. Its transformation adds another + 2 cos term to (U ). Subtracting 2( + 2 cos )from (U ), we obtain a character of purely vibrational degrees of freedom v = N a ( + 2 cos ) 2( + 2 cos ) =(N a 2)( + 2 cos )

41 Equivalence representation of C 2v v =(N a 2)( + 2 cos ) Identity (I): N a = N, =) v(i) =3. C 2 contains oxygen: ) N a =, =8) v(c 2 )= ( 2) = v leaves all atoms intact) v( v )= v (I) =3 v is equivalent to C 2 ) v( v)= v (C 2 )= C 2v I C 2 v v G eq 3 3 σ ' v σ v

42 2 Last part: decompose G eq into irreps of C 2v C 2v I C 2 v v z, A xy, A x, B - - y, B C 2v I C 2 v v G eq 3 3 G eq = a A + a 2 A 2 + a 3 B + a 4 B 2 Decomposition formula a = X g N C v(c)[ (C)] C g =4,N C = a = ( ) = 2 4 a 2 = (3 + 3 ) = 4 a 3 = (3 +3 ) = 4 a 4 = (3 3 + ) = 4 G eq =2A + B ) Two vibrational modes with symmetry A,onewith symmetry B,nonewithA 2 and B 2.

43 A χ ( C 2 ) = + A χ ( C 2 ) = B

44 3 Selection Rules Consider a basis function i (i =...f) oftheanf dimensional representation i. Theorem : Theintegralof i over an entire configurational space of the system ˆ i dq is non-zero is non-zero if and only if is the identical (or symmetric) representation,. Proof: Assumethat 6=.Sincetheintegralmustbe the same in any coordinate system (it is over all space), we can say ˆ ˆ ˆ X i dq = ˆ i dq = ki k where ˆ transforms the function by acting as a matrix product. Now we sum the last equation over all symmetry elements of group G: ˆ ˆ X G i dq = X G k ki k dq On the LHS, we simply get the group order (g) timesthe original integral, ˆ ˆ X g i dq = ki k dq G

45 Recall the orthogonality theorem X ik jl = g ij kl f G X G If =,then jl = g = ( g, =, 6= Because of this, it immediately follows that ˆ i dq = ˆ X ki k dq = g if 6=.Hence, =. Consider the Hamiltonian G H = H o + H 4 where H o has group G and H has some lower symmetry group. Define the matrix elements ˆ M = i dq k H The direct product (or tensor product or Kronecker product) of representations is denoted by C = A B, C sr = A ij B kl for apple (i, j, k, l) apple (m, n, p, q) andso

46 apple (s, r) apple (mp, nq). For example, b a a b 2 b 3 A B = b a 2 a 2 b 22 b 23 A a Ba = 2 B 22 a b 3 b 32 b 2 Ba 22 B 33 b b 2 b 3 b b 2 b 3 b 2 b 22 b 23 A a b 2 b 22 b 23 A b 3 b 32 b 33 b 3 b 32 b 33 = (6) b B b 2 b 3 b b 2 b a b 2 b 22 b 23 A a b 2 b 22 b 23 A C A b 3 b 32 b 33 b 3 b 32 b 33 Tr {A B} = X i Notice that X a ii b jj =Tr{A} Tr {B} Suppose we have two representations ij and kl.thedirect product of two representations j = is a matrix given by the direct product of matrices corresponding to and. Going back to the matrix element, ˆ M = i dq k H we note that M 6= ifandonlyiftheintegrandtransforms as. SupposethatH transforms as some representation 5

47 H which, in general, is a reducible representation of the Hamiltonian group G. Theintegrandtransformsasatriple direct product H According to Theorem, the matrix element is non-zero if and only if this triple product contains an identical representation H =... Now we have practical way to find if the matrix element is non-zero: we must decompose the triple product into irreps of G, and see if occurs in the decomposition. If it does, then M 6= ;ifitdoesnot,m =. This procedure can be simplified further if we observe that the direct product of some representation with itself,, must necessarily contain.indeed,letusdecompose into irreps = a a 2 2 The weight of in this decomposition is found as a = X (G) g (G) G For a direct product of two matrices, Tr (A B) =Tr(A)Tr(B). (To convince yourself in the validity of this identity have a look at the example of the direct product of 2 2and3 3matrices.)Therefore, 6

48 = 2 while, by definition, =. Bythe orthogonality property of characters, X 2 =. g a = g X G G Therefore, = X g G 2 = g g =, which means that contains in once. Thus the condition H =... can be replaced by an equivalent one H =... H Indeed, in this case, a must necessarily contain. 7 A. Dipole selection rule Consider am electron system subject to a weak electromagnetic field. In the transverse gauge, r ~A =, H = ~p + e A 2m c ~ 2 p 2 e = + A {z} 2m mc ~ ~p +O(A 2 ) {z } H o H The matrix element of the transition is h,j H,ii = e mc ~ A h,j ~p,ii Vector ~p is a polar vector which transforms as the radial vector, ~r =(x, y, z). A representation of a polar vector is

49 denoted by.weneedtodecompose into irreps of H. Recall the character table of the C 2v group which described the H 2 Omolecule.Thex, y, andz components of ~r transform as B, B 2,andA.Therefore, (C 2v )=A B B 2 On the other hand, for the D 3 group (D 3 )=A 2 E Suppose that the initial state of the H 2 OmoleculeisA. Then we need to find a = g A =(A B B 2 ) A Here is one more useful property: =, 8 Proof: Decompose into irreps X G ( a = X a where ) = X g G {z} = ( ) = Then, (A B B 2 ) A = A B B 2. Vibrational normal modes transform as A and B. This means that the final state can be either A or B :each of these modes has a non-zero overlap with A. In other words, all vibrational modes of H 2 Oare infrared-active. 4

50 5 I. EXAMPLE: DIPOLE TRANSITIONS IN A D 3 MOLECULE Consider a (hypothetical) molecule with D 3 symmetry. Determine possible transitions between electronic states of this molecule. D 3 I 2C 3 3C 2 A z, A 2 - (x, y), E 2 - Apolarvectortransformsas = A 2 E. Suppose that the initial state is A. A = = A 2 E Allowed transitions: A! A 2 and A! E. Initial state: A 2. A 2 =(A 2 E) A 2 = A 2 A 2 E A 2 A 2 A 2 = 2 A 2 = for all symmetry classes ) A 2 A 2 = A

51 6 D 3 I 2C 3 3C 2 A z, A 2 - (x, y), E 2 - E A ) E A 2 = E A 2 = A + E Allowed transitions A 2! A,A 2! E Initial state: E E =(A 2 E) E = A 2 E {z } =E E E = E E E

52 Decomposing E E 7 D 3 I 2C 3 3C 2 A z, A 2 - (x, y), E 2 - E E 4 E E = X a G a = 6 X G E E a A +2 A = 6 {z } {z } I 2C 3 a A2 +2 A = 6 {z } {z } I 2C 3 a E = 2 {z } I 2 A = {z } 2C 3 E E = A A 2 E E = E A A 2 E = A A 2 2E Transitions E! A,E! A 2,E! E

53 Jahn-Teller Effect

54 From Prof. Hancock Chemistry, UNC-Wilmington All six Ni-O bonds equal at 2.5 Å two long axial Cu-O bonds = 2.45 Å [Ni(H 2 O) 6 ] 2+ [Cu(H 2 O) 6 ] 2+ four short in-plane Cu-O bonds = 2. Å

55 High-spin Ni(II) only one way of filling the e g level not degenerate, no J-T distortion e g d 8 Cu(II) two ways of filling e g level it is degenerate, and has J-T distortion energy e g e g d 9 Ni(II) t 2g t 2g t 2g

56 Jahn-Teller Theorem (937) Symmetric configurations of molecules with degenerate electron states are unstable 8 )The nuclei in the molecule will distort to remove the degeneracy.

57 Let qi be the normal modes of molecular vibrations (i = x, yz, labels the irreducible representation) We are interested only in asymmetric positions of ions: is not considered Electron Hamiltonian: H Suppose that the electrons are in one the multidimensional representations, dim( ) > b = E,F,T,... Suppose that the molecule get distorted H = H! H = H + X V i q i + X W i, j q i q j +...,i,,i,j Is the matrix element of the linear in q term finite? If it is, the energy will always be lowered (the sign of q can always be chosen negative) for small enough q. (Higher-order terms non-linear JT-e ect) Coe cients V transform as one of the equivalence representations of the symmetry group V / Reminder: an equivalence representation describes purely vibrational modes and exclude translations and vibrations eq H 2 O(C 2v ) C 2v I C 2 v v eq eq =2A B

58 If the electron state is degenerate, the energy splitting due to adeformationis 2 E = h,b V,i,ai where,ai (a =...f )isthebasisfunction E 6= if and only if eq Equivalently, Non-degenerate electron configuration )no JT distortion JT distortion: an asymmetric configuration of nuclei ) eq 6= If the electron configuration is non-degenerate, is a D representation and eq = (G) =± ) B = 2 (G) =) = ) E = If the electron configuration is degenerate, E 6=

59 2 Td symmetry FIG.. Methane molecule Example: CH 4 (methane) Vibrational modes eq = A E 2T 2 Excluding the symmetric mode eq = E 2T 2 Degenerate electron configurations: E,T,T 2 E E = A E E This document is provided by the Chemical Portal Character table for T d point group E 8C 3 3C 2 6S 4 6σ d linear, rotations quadratic A x 2 +y 2 +z 2 A E 2-2 (2z 2 -x 2 -y 2, x 2 -y 2 ) T (R x, R y, R z ) T (x, y, z) (xy, xz, yz) You may print and redistribute verbatim copies of this document. T T = T 2 T 2 = A E T 2 T 2 ) symmetric molecule is unstable

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