Consider a s ystem with 2 parts with well defined transformation properties

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1 Direct Product of Representations Further important developments of the theory of symmetry are needed for systems that consist of parts (e.g. two electrons, spin and orbit of an electron, one electron and a vibration mode). Consider a s ystem with parts with well defined transformation properties The basis can always be chosen as if they were independent parts, and any state is a linear combination of products f (α)g (β), where f (α), f (α) m and g (β), g (β) n are bases for irreps of dimensions m and n respectively: ( α ) ( β ) Direct product basis: αβ i j = f g = αi βj i j ( αβ ) ( R) m n ( αβ α β ) kp, ij R αβ i j = k p D R D k p = direct product representation, kp ij

2 m α m n α β α β ( ) Rf Rg f g () D R D R = ( ( α ) ) i R αβ i j Rf Rg ( α ) ( ) ( ) ( ) ( ) ( = α α, β = β β ) Rf f D R Rg g D R i k ki j p pj k= p= i k p ki pj k p D ( αβ ) ( R) = direct product representation, kp ij n Direct product αβ α β kp, ij ki pj D R = D R D R αβ α β D R = D R D R

3 Direct Product of Matrices The direct product of two matrices a a b b A=, B= a a b b is defined as follows: a b a b a b a b a B a B a b a b a b a b D= A B= a B a B = a b a b a b a b a b a b a b a b A B B A Basic property of Direct Product: TrD = a b + a b + a b + a b = ( a + a )( b + b ) = TrATrB D = a b TrD = a b = TrATrB kp, ij ki pj kk jj kj

4 Direct Product of representations ( αβ ) ( α ) ( β ) ( αβ ) ( α ) ( β = ) D R D R D R D R D R D R kp, ij ki pj ( αβ ) ( αβ ) ( α ) ( β ) ( α ) ( β χ R = D R = D R D R = χ R χ ) ( R) kp, kp kk pp kp kp The character χ of the direct product of two representations is the product of the characters of the two representations. EXAMPLE Consider for instance the Group of the Square C4v I C C4 σv σ d g = 8 A A z R B x y B xy E ( xy, ) z Suppose that the two parts and belong to known irreps: their products define a representation of G and we wish to reduce it: 4

5 Direct product Irreps of parts χ are products C I C C σ σ g = 8 4v 4 v d A A z z B x y B xy E ( xy, ) A B B B B A E A E E E A A B B R Reduction by LOT n = i χ( R) χ ( R) N G R G Multiplication Table for the IRREPS () i * C A A B B E 4v A A A B B E A A A B B E B B B A A E B B B A A E E E E E E A A B B Irreps of total system 5

6 C A A B B E 4v A A A B B E A A A B B E B B B A A E B B B A A E E E E E E A A B B A in diagonal and only there. A matter of chance? NO LOT ni = χ R χ R χ R = N () i * ( A Using and ) G R G ( αβ ) ( α ) ( β ) na = χ ( R). = χ ( R). χ ( R) = δ αβ N N G R G G R G 6

7 Selection rules ( T ) For an irreducible tensor, we can find out when φ Tˆ ψ = 0, φ Γ Tˆ Γ ψ Γ ( φ) ( ψ) knowing that,. φ T ψ =invariant A integrand A ( T ) ( φ) ( ψ) Reducing Γ Γ Γ do we obtain A? If not, φ T ψ = 0 φ T ψ =invariant A requires that Γ( T ) and Γ ( φψ ) =Γ( φ) Γ( φ) have irreps in common 7

8 Example: electromagnetic transitions in square symmetry: An electron in an orbital E absorbs (or emits) a photon. What symmetries of the final orbitals are allowed? Dipole operator d=(x,y,z) E A,A,B,B allowed by x,y E E allowed by z E A = E B B forbidden because B B : = A no dipole component C I C C σ σ g = 8 4v 4 v d A A z z B x y B xy E ( xy, ) E E E E = A A B B R 8

9 Electronic ground state of molecules and clusters often closed shells A Shells are invariant subspaces for one-body states Closed shells are obtained by filling those subspaces with electrons- They are canonically conjugated to empty subspaces, that are totally symmetric. infrared selection rules Vibrational ground state=vibration vacuum: A symmetry Ground state excited state with the symmetry of dipole component (also for IR absorption, see classification of vibrations) With inversion, Ungerade modes in IR 9

10 Raman effect ε, ω ε, ω R= R ε ε Raman Tensor pq pq p q Raman tensor transforming like x p x q With inversion, Gerade modes in Raman 0

11 Reduction of the direct product representation From the m α -times degenerate irrep α and the m β -times degenerate irrep β one forms a direct product representation Γ(α) Γ(α) of dimension m α m β. For symmetry operations on the combined system, { γ r, r =,... m γ } = basis set for irrep γ of the Group. One can go, with a unitary transformation, from the basis { αiβj> } of the direct product to a basis of functions that transform according to irrep Γ(γ) of the Group G. m = αβ i j γr γr αβ i j irreps γ components r γ γr αβ i j = Clebsh Gordan coefficients of the Group

12 irreps γ m γ αβ i j γr γr αβ i j = m i components Clearly, γr αβ i j αβ i j γr = irreps αβ components i components m j r j Reduction of the D matrices ( αβ ) ( α ) ( β ) By definition, D ( R)= D ( RD ) ( R) is the direct product matrix ( αβ ) kp, ij ki pj Dkp, ij ( R) = αkβprαβ i j. Inserting basis sets αkβp R αβ i j αkβp γs γs R γr γr αβ i j = γ r s = ( α ) ( β ) ( γ ) D ( RD ) ( R) αkβpγsd ( R) γrαβ i j. ki pj sr γ r s

13 Problem: Molecule of C v symmetry To find a -e state of A symmetry with both electrons in orbitals of irrep E E E = A A B B Solution : From one-electron basis (x,y) direct product () i () * ( = i A P χ ( R) R P ) = R R We recall the matrices we used for irrep E : so two electrons of E can give A xx, xy, yx, yy 0 = = DE DC DC = 0 0 D( σ ) = D( σ ) = D( σ ) = a c b 0 R

14 irrep E : Effect of each R on (x,y) basis 0 DE = = x y y x DC Cx=, Cy=, + + = x y y x DC C x=, C y=, 0 D( σ ) = σ x= x, σ y= y, a 0 D( σ ) = x+ y y+ x σ x=, σ y =, b σ = x y y x D σ x=, σ y = c 4

15 From one-electron basis (x,y) direct product xx, xy, yx, yy A x + y x + y x+ y x + y P xx = xx + ( )( ) + x + y x + y x y x y + xx + + A Simplification gives: P xx = ( xx + yy ) P ( A ) = R R This is the wave function, which is even in the exchange on and, so it is singlet. Normalizing, A = xx + yy Clebsh Gordan coefficients αβ i j γr ExEx A = = EyEy A, ExEy A = 0. 5

16 Double Groups Spin without spin-orbit interaction trivial degeneracy Spin-orbit interaction for H atom: the degeneracy is lowered symmetry is lowered Budapest, 5 gennaio 908 Stanford, 9 settembre 00 (L + ) J=L+/ J=L-/ L+ states L states The symmetry Group is no longer O(). SU() is appropriate. Teller invented H bomb and Reagan s star wars 6

17 Spinor representation of the rotation Group The rotation around to the z axis by an angle ω is done by R ω i ω e 0 = exp[ i ω σ ] = z ω i ω χ = cos ω 0 e A similar formula for rotating Dirac s spinor. Here we assume relativistic corrections are enough R ω ω ω = cos iσ z sin and belongs to the SU() covering group of SO() (same operations occur twice). For ω = π, R ω =. 7

18 Rotation by angle ω around n = (n,n,n ) = (sinθcos φ,sinθsin φ, cos θ) x y z ω ω ω Develop R = exp[ i ( σ. n)] = i ( σ. n) + ( i ( σ. n)) +... ω! iϕ ϑ ϑe n n in z x y iϕ cosϑe sinϑ n + in n x y z sin cos ( σ. n) = = n n in n n in z x y z x y 0 = = n + in n n + in n 0 x y z x y z ( σ. n) and the series is easily summed: ω ω ω R = exp[ i ( σ. n)] = cos σ ω i(. n)sin Note: ω = π R = commutes with all other ω z ω symmetries ω ω For n= (0,0,), Tr( σ. n) = 0 and Tr exp[ i σ ] = χ = cos; this holds true (indeed, change of axis is an unitary transformation). 8

19 The π rotation as an extra symmetry (Teller 99) E= R (any n) π E= E (for a spinor) E= E (for an orbital); E = E; [ EX, ] = 0 X G: no X is altered by a π rotation. Adding XEX E X G E E to the generators of G we obtain the so-calle d double Group G'. = ' is a class, any rotation R has same character as R = R ω 4π ω ω ω 4π ω since for a spinor χ = cos is obviously = cos. ω D matrices for G ( α) ( α) ( α) ( α) E = E D ( E) =± D ( E) irrep α (in agreement with Schur lemma) χ ( ER) =± χ ( R) irrep α, R 9

20 The spinor representation i σω. χ ω D ( R ) = e ( E) = Since characters are invariant for unitary transformations, for any rotation axis we can obtain characters from D ω ( ω) = χ ( ω) = cos 0 e ω i e 0 ω i Space inversion operation i: (x,y,z) -> (-x,-y,-z) leaves angular momenta and spins invariant D(i) = D(E) and χ(i) =. Reflections = inversion*proper rotation Example: a reflection in (xyz) (xy z) in the (x,y) plane can be obtained as a rotation (xyz) ( x yz) followed by i. 0

21 Example: Structure of C v. Start from C v. In C v the square of any σ is E, and for example σ a rotation around y axis. Then in the double group is the same as a π E C C C C E C E C σ σ σ a b c σ σ σ b c a σ σ σ c a b σ σ σ a b c σ σ σ c a b σ σ σ b c a E C C C E C C C E C v multiplication table In σ σ σ a = b = c = E C v v σ = E σ = σ a a a x b c In C' σ = E, σ σ = E σ = σ = σ E a a a a a a In In C σ Cσ = σ Cσ = σ σ = C v a a a a b a thus C and C are in the same class. a C ' v σ Cσ = σ C σ E = σ σ E = CE a a a b a C C C E E σ E ' v σ v v CE CE

22 Building the G character table (fast way) α α Find the classes and append the irreps and characters of G, with ) Add spinor representation Number of irreps = number of classes Size of irreps from Burnside theorem Characters from orthogonality theorems χ ( ER = χ R Example: building the C v character table C v classes: E, C, σ Add E Inverse of C rotation: C E C I C σ g = 6 v A A z E 0 ( xy, ) Add normal irreps a v R z C C C E E E ' v σ v σv CE CE A' A' E ' 0 0

23 C C C E E E ' v σ v σv CE CE A' A' E ' 0 0 C I C σ g = 6 v A A z E 0 ( xy, ) v R z Add Spinor representation: χ π = cos * π = σ : x xy, y σ = i* R, spins are even under i χσ = χ( i)* χ( R) = *cos * π= 0 C C C E E E ' v σ v σv CE CE A' A' E ' 0 0 E classes 4 irreps irreps are missing N G = Missing irreps are - dimensional

24 C C C E E E ' v σ v σv CE CE A' A' E ' 0 0 E 0 0 By orthogonality C C C' E E E v σ v σv CE CE A' A' E ' 0 0 E Γ i i i i 6 Γ Example C I C C σ σ g= 8 4v 4 v d A A z z B x y B xy E ( xy, ) R C C C σv σd C' E E CE CE σ E σ E 4 4 4v 4 CE 4 A' A' B' B' E' E' E' v d 4

25 How does the spin-orbit interaction resolve degeneracies in crystals? To find that, classify orbitals by irreps of G build spinor representation (or find in Character Table) direct product yields spin-orbital representation Example C' E E C C E v σ v σv CE CE A' A' E ' 0 0 E Γ i i i i 6 Γ reduction by LOT case of orbital A A E/ = E/ not resolved case of orbital E E E = E Γ Γ / / 5 6 5

26 Cu [Ar]4s d Example: Cu in d configuration J =, Splitting in square planar D 4 environment Characters of reducible representation of rotations found by χ ( j) ( φ) sin ( J + ) φ = φ sin sin[ φ ] φ 4cos cos φ sin χ ( φ) = = φ χ 5 ( φ) [ ] sin φ φ φ 5φ = = (cos cos cos ) φ + + sin reflections and improper rotations= inversion x rotation 6

27 Γ C C C C' C" D' E E C E C E C E C E CE ' 4 " A' A' B' B' E' E' E' Γ Γ = E' 5 E' Γ = E' E' n = i χ( R) χ ( R) N G R G () i * 7