Group Theory. Problem Set 3, Solution

Transcription

1 Problem Set 3, Solution Symmetry group of H From O b φ n = m T mnbφ m and O a φ q = p T pqaφ p we have O b O a φ q = O b T pq aφ p = T pq a T mp bφ m p p m =! T mp bt pq a φ m = O ba φ q m p = m T mq baφ m T mq ba = p T mp bt pq a T ab = T a T b 2 The improper rotation operator The operator representing proper and improper rotation in a function space {ψ r} is defined by O R ψ r = ψr r a Applying the rotation operator on the function space {ψ r} one shows straight forward the first relation: O R OR ψ r = O R ψr r y=r r = O R R r y ψ y = O R R y ψ y = R ψ r O R OR = R b With the help of the relation a one shows directly the second relation: O R O R = O R OR O R OR = R R =, R T R =

2 3 Symmetry adapted vectors a In order to test the irreducibility for the two dimensional representation of C 3v and C 4v one can use the orthogonality theorem for characters: C 3v E 2C 3 3σ Λ 3 2 C 4v E 2C 4 C 2 2σ 2σ η a χ α a 2 = = η a χ α a 2 = = b The character tables of C 2v, C 3v and C 4 are: C 3v E 2C 3 3σ Λ Λ 2 Λ 3 2 C 4 E C 4 C4 2 C4 3 Γ Γ 2 i i Γ 3 Γ 4 i i C 2v E C 2 m x m y Σ Σ 2 Σ 3 Σ 4 c With respect to the C 2v group the two dimensional representation reads: y i i x E χ E = 2 C 4 χ C4 = C 2 4 χ C 2 4 = 2 C 3 4 χ C 3 4 = {E} {C 4 } {C 2 4} {C 3 4} Γ xy 2 2 Γ 2 Γ 4 Choosing the trial vector v =, one can easily find the symmetry adapted basis vectors using the projector technique P Γ2 v = E v + i C 4 v + C4 v 2 + i C4 v 3 = + i + + i 2 2i i i = Γ 2 + 2i + i + i 2 P Γ4 v = E v + i C 4 v + C4 v 2 + i C4 v 3 = + i + + i 2 + 2i + i + i = Γ 2 2i i i 4

3 Also for the C 4 group one can repeat the previous calculation: y x E χ E = 2 C 2 χ C2 = 2 m x χ mx = m y χ my = {E} {C 2 } {m x } {m y } Γ xy 2 2 Σ 2 Σ 4 Respectively, choosing the trial vector v =, one can find again the symmetry adapted basis vector P Σ2 v = E v + C 2 v + m x v + m y v = = Σ 4 2 The chosen trial vector is already a symmetry adapted basis vector, therefore one chose new trial vector v =, and repeat the calculation with projector technique P Σ4 v = E v + C 2 v + m x v + m y v = = Σ 4

4 4 Invariant subspaces a Recalling the Tower representation of the C 3v group calculated in the Problem Set 2, we can calculate the three dimensional matrix representation of C 3v E = σ = C 3 = σ 2 = from which we obtain the following character table: C 3 = σ 3 = C 3v E 2C 3 3σ Λ Λ 2 Λ 3 2 Λ T 3 Λ Λ 3 Similar to the previous exercise one chooses a trial vector and calculates the symmetry adapted vectors with the projector technique P Λ Λ P Λ3 = 2 E + C 3 + C3 + σ + 2 = Λ 3 2 C 3 = 2 2 Λ 3 Those vectors are the basis of the invariant subspaces b Choosing x = 6 we can construct the regular representation, which certainly contains all irreducible representation of C 3v

5 5 Ring-lie molecule a C is a cyclic group consisting of the elements { C }, =,, Because of the property of e cyclic group one nows that there will be one dimensional irreducible representations Therefore, one can directly write the character table: Γ n : n =,, n Γ n 2π = e i 2π C φ = φ = 2π φ = 2π φ = 2π Γ Γ e i 2π e i 2π e i 2π Γ n e i 2πn e i 2πn e i 2πn Γ e i 2π e i 2π e i 2π b Again, in order to find the symmetry adapted vectors we choose the trial vector v T =,,,, and we appliy the projector technique: P Γ v T = E v T + C v T + + C v T + + C v T =,,,, +,,,, + +,,,,,,,, P Γ v T =,,,, + e i 2π,,,, + + e i 2π,,,,, e i 2π,, e i 2π P Γ n v T =,,,, + e i 2πn, e i 2πn,, e i 2πn,, e i 2π = e i 2π e,,,, + + e i 2πn,,,,,, e i 2πn = e i 2π n e c As a discretization of the differential operator d 2 /dψ 2 on a ring with discrete lattice points we tae: -2 2u + u + + u ie we associate to the site the matrix element 2, the the site + and the matrix element +

6 Accordingly, the operator d 2 /dψ 2 becomes a matrix of the form: M = One can easily chec that this matrix commute with the tower representation of C, ie with all representation matrices Thus, the matrix M has common eigenspaces with the tower representation of C These invariants space we have just found in b and can be used to find the eigenvalues of M M n e = E n e i 2π n e e i 2π E n = 2 + e i 2π n + e i 2π n = 2 + cos 2π n n =,,, E n /2 n -4

7 6 An eigenvalue problem Consider the function space of periodic function fϕ defined on the unit circle with an angle ϕ and also consider the operator M = i d dϕ a According to the previous exercise the characters of the irreducible representation of the symmetry group of the circle, C are: C α Γ m = Γ e ±iα m = ± Γ m e imα ±m b A possible representation of the group C within the function space of periodic function fϕ is: O α fϕ = fϕ α O β O α fϕ = O β fϕ α = gϕ α β = O α+β fϕ homomorphism c Starting from a general function fϕ, the symmetry adapted wave function transforming according to the irreducible representation of C are: P Γ mfϕ = 2π e imα fϕ α dα = e imϕ 2π e imα fα dα 2π 2π P Γ mfϕ e imϕ d From: O α i d dϕ O α d fϕ = io α dx fx ϕ α = i d dx fx d ϕ = io α dϕ fϕ it follows that M commute with C Therefore M and the irreducible representation of C have common eigenspaces {e imϕ Am} Applying M onto these eigenfunctions we get the eigenvalues of M: i d dϕ e imϕ Am = mame imϕ m =, ±, ±2,

8 7 The point group O h According to the table for transformation of the coordinate xyz under the operations of the group O h, given in the lecture The three dimensional representation of the point group O h reads: Γ 5 of O h E xyz 3 I xȳ z 3 6C 4 ȳxz I6C 4 y x z 3C4 2 xȳz I3C4 2 xy z 6C 2 yx z I6C 2 xȳz 8C 3 zxy I8C 3 z xȳ In order to chec the irreducibility of Γ 5 we use the orthogonality theorem for characters: O h = 48 η a χ Γ 5 2 = = η a χ Γ 5 = =