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We can now add the entries for this two-dimensional irreducible representations to the character table of C6v : C6v Γ1 Γ1 Γ1 Γ1 Γ2 Γ2 E 1 1 1 1 2 2 2C6 1 1 1 2C3 1 1 −1 C2 1 1 −2 3σv 1 −1 3σd 1 −1 0 0 (c) The one-dimensional irreducible representations must obey the multiplication table, since they themselves are representations of the group. In particular, given the products C3 C32 = E, C33 = E , if we denote by α the character of the class 2C3 = {C3 , C32 }, then these products require that α2 = 1, α3 = 1 , 9 respectively.

Use the Great Orthogonality Theorem to show that two functions which belong to different irreducible representations or are different partners in the same unitary representation are orthogonal. 7. Consider a particle of mass m confined to a square in two dimensions whose vertices are located at (1, 1), (1, −1), (−1, −1), and (−1, 1). The potential is taken to be zero within the square and infinite at the edges of the square. The eigenfunctions ϕ are of the form cos(kp x) cos(kq y) ϕp,q (x, y) ∝ sin(kp x) sin(kq y) where kp = 12 pπ, kq = 12 qπ, and p and q are positive integers.

We have two groups Ga and Gb with elements Ga = {ea , a2 , a3 , . . , a|Ga | } and Gb = {eb , b2 , b3 , . . , b|Gb | } , such that ai bj = bj ai for all i and j. We are using a notation where it is understood that a1 = ea and b1 = eb . The direct product Ga ⊗ Gb of these groups is the set obtained by forming the product of every element of Ga with every element of Gb : Ga ⊗ Gb = {e, a2 , a3 , . . , ana , b2 , b3 , . . , bnb , . . , ai bj , . } . To show that Ga ⊗ Gb is a group, we must demonstrate that these elements fulfill each of the four requirements in Sec.

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