PHYSICS 220 : GROUP THEORY FINAL EXAMINATION SOLUTIONS

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1 PHYSICS 0 : GROUP THEORY FINAL EXAMINATION SOLUTIONS This exam is due in my office, 5438 Mayer Hall, at 9 am, Monday, June 6. You are allowed to use the course lecture notes, the Lax text, and the character tables (link from lecture notes web page), but no other sources, and please do not discuss the exam with anyone other than me, other than to reassure your classmates that indeed this is a most fair and excellent exam. [1]C 0 isthesmallestfullerene. Ithasthestructureofadodecahedron, i.e.athreefold coordinated cage with 0 sites arranged in 1 pentagonal faces and 30 edges. Its symmetry group is I h. (a) Similar story to C 60 : The sp hybridized orbitals account for three of the four electrons in the s and p shells. (The 1s shell is completely inert.) The remaining orbital is the π orbital, oriented along a line from each carbon atom to the center of the molecule. Such orbitals transform trivially under I h. Consider a tight-binding Hamiltonian Ĥ = t ( ) i j + j i, ij where i and j are π-orbitals for neighboring carbon atoms i and j. Classify the multiplets in the tight binding eigenspectrum according to irreps of I h. For extra credit, write a computer program which diagonalizes Ĥ for t = 1 and plot the spectrum, labeling each multiplet accordingly. Solution : We have χ per (E) = 0, χ per (0C 3 ) =, and χ per (15σ) = 4. All other classes have χ per (C) = 0. Thus, with N Ih = 10, n Γ (Γ per ) = 1 ( ) χ Γ (E)+χ Γ (0C 6 3 )+3χ Γ (15σ) and using the decomposition formula we obtain, Γ elec = Γ orb Γ per = A g G g H g T 1u T u G u the total dimension of which is 0. (b) If pressure is applied to opposite pentagonal faces, the symmetry is broken from I h down to D 5d. Find how the tight-binding multiplets from part (a) split into D 5d irreps. Solution : For D 5d, we lose the C 3 (and S 6 ) operations. The order of the group is N D5d = 0, and we have the decomposition n Γ (Γ per ) = 1 0 ( ) 0χ Γ (E)+4 5χ Γ (5σ) = χ Γ (E)+χ Γ (5σ). 1

2 Decomposing in D 5d using the character table, Γ elec = A 1g E 1g E g A u E 1u E u whose total dimension is again 0. Decomposing relevant the I h irreps in D 5d, we find A 1g = A 1g T 1u = A u E 1u G g = E 1g E g T u = A u E u H g = A 1g E 1g E g G u = E 1u E u, and the decompositions agree. (c) Classify all vibrational modes of C 0 according to irreps of I h. Indicate whether each multiplet is IR active, Raman active, or Raman silent. If you really want to impress me, find the normal modes numerically for a ball and spring model, where all masses and spring constants are the same. Solution : We first construct Ψ = Γ vec Γ per, for which χ Ψ (E) = 3 0 = 60, χ Ψ (15σ) = 1 4 = 4. The decomposition formula for Ψ is thus n Γ (Ψ) = 1 χγ (E)+ 1 χγ (15σ). Appealing to the character table for I h we find Ψ = A g T 1g T g G g 3H g T 1u T u G u H u. ToobtainΓ vib,wemustsubtractfromψ thezeromodesfromγ vec = T 1u (translations) and Γ rot = T 1g (rotations). Thus, Γ vib = A g T g G g 3H g T 1u T u G u H u Only the T 1u (blue) multiplet is IR active. To check Raman activity, we need Γ sym from the symmetrized product of vector representations. It is an easy matter to decompose Γ vec Γ vec = A g T 1g H g. The total dimension is nine, corresponding to a real 3 3 matrix. We only need the symmetric part, which is of dimension six, so clearly T 1g, which is of dimension three, corresponds to Γ asy and therefore Γ sym = A g H g. The four ungerade multiplets are Raman inactive (blue/red), and among the gerade multiplets, only A g and H g are active (bold black); the others are silent (bold gray). Nota bene : Sami Ortoleva astutely discovered several errors in the Atkins, Child, and Phillips character table for I h (their I table is OK). I have corrected the error in the linked file on the Physics 0 Lecture Notes web page. Original version at

3 [] Consider the structure in Fig., which is the Shastry-Sutherland lattice with nonsymmorphic wallpaper group p4g (from Fig of the notes). (a) Let the length of each bond be a. Choose as the origin the center of the figure, and find expressions for the four basis vectors δ 1,,3,4 corresponding to the red, yellow, grey, and blue sites in the first Wigner-Seitz cell (counterclockwise starting from the right, for the color blind among you). You may assume that the smaller of the two internal angles of each rhombus is α. Solution : We have δ 1,3 = ±asin( 1 α) ˆx, δ,4 = ±acos( 1 α)ŷ (b) Find two elementary direct lattice vectors a 1,. Find the corresponding reciprocal lattice vectors b 1,. Solution : The primitive direct lattice vectors are chosen to be a 1 = a 0 (ˆx+ŷ), a = a 0 ( ˆx+ŷ) with a 0 = a ( cos( 1 α) + sin(1α)). The elementary reciprocal lattice vectors are then π π b 1 = (ˆx+ŷ), b a = ( ˆx+ŷ) 0 a 0 (c) Find the scattering form factor F(K) defined in Eqn of the notes and show there are extinctions in the Bragg pattern. Solution : The form factor is F(K) = 4 i=1 e ik δ i. We write K = n 1 b 1 +n b. Now b 1 δ 1,3 = ± πs It is convenient to define, b 1 δ,4 = ± πc θ = πs, b δ 1,3 = πs, π θ = πc,, b δ,4 = ± πc. 3

4 Figure 1: A lattice with p4g symmetry. where s = sin( 1α) and c = cos(1 α). Thus, and therefore K δ 1,3 = ±(n 1 n )θ, K δ,4 = ±(n 1 +n )(π θ), F(K) = 4 [cos [ (n 1 n )θ ] [ +( 1) n 1 +n cos (n1 +n )θ ]] So for generic θ there will be extinctions in the Bragg pattern under two circumstances: (i) n 1 = 0 and n odd, (ii) n 1 odd and n = 0 This is all consistent with the discussion in 5.4. of the Lecture Notes. If K lies along the invariant line of a mirror m for a two-dimensional point group, then if K = nβ, where β is a basis vector for the one-dimensional subset of the reciprocal lattice within that invariant line, we have extinctions whenever n is odd. In our problem, there are glide mirror lines along a 1 and a. In the former case, n = 0 and odd n 1 are extinguished. In the latter case, n 1 = 0 and odd n are extinguished. (d) When α = 1 π, the lattice becomes square. Explain the Bragg extinctions in this case. When α = 1 π we have θ = 1 π and F(K) = 16cos ( 1 πn 1)cos ( 1 πn ). There are then extinctions whenever either n 1 or n is odd, i.e. only the (even,even) Bragg 4

5 points survive, which are one quarter of all the Bragg points. The reason is that the Shastry-Sutherland lattice (SSL) for α = 1 π becomes a square lattice rotated by 1π, i.e. a Bravais lattice with a four element basis. The only Bragg peaks are those 4 corresponding to the small square lattice, and these are the surviving ones from the SSL structure. Bragg peaks with n 1 odd or n odd extinguished due to quadrupled unit cell (e) How do the basis points transform under the glide mirror? I.e. what basis color does yellow get mapped to, etc.? Also indicate which glide mirror you have chosen. Solution : There is a glide mirror parallel to a 1 which intersects the link between the red and blue basis elements at its midpoint. The glide operation has the effect of exchanging blue and red, and exchanging yellow and grey: mirror along a 1 : B R, Y G If you chose the other direction for the glide, you should find mirror along a : Y R, B G [3] Consider a V + ion in a D 4 environment. (a) The electronic configuration is [Ar]4s 0 3d 3. Hund s first two rules say S = 3 and L = 3 (F). What is the ground state term according to Hund s third rule? Solution : Hund s third rule says J = L S if an incomplete shell is not more than half filled, hence the ground state term for the isolated ion is S+1 L J = 4 F 3/ (b) Using the notation of Atkins, Child, and Phillips, the double group D 4 has two spin representations, E 1/ and E 3/, both of which are two-dimensional. Ignoring spin-orbit, decompose F into irreps of D 4 (the decomposition will be the same in D 4 ). You may find some of the results in Tab. 6.5 of the notes useful (e.g. for the characters of certain rotations in the l = 3 representation of O(3)). Then decompose 5

6 D 4 E C 4 C C C A A B B E P A E D A 1 B 1 B E F A B 1 B E Table 1: Character table of D 4 and decomposition of L = 1,, and 3 multiplets in a D 4 environment. the Γ 3/ representation for S = 3 in D 4. (Hint: it can give only some combination of the two spin irreps.) Finally, decompose the product Γ 3/ F = 4 F and find the irreps for all crystal field levels. Once again, your answer can only involve the E 1/ and E 3/ irreps, so in the end this isn t so nasty. For a template of how to proceed, see the section Co ++ in a cubic environment in 6..6 of the notes. You may find Tab helpful as well. Solution : With the help of the aforementioned tables, we construct Tab. 1, which we use to decompose F into D 4 irreps. We also must decompose Γ 3/ in terms of the spin irreps of D 4, using Tab.. Finally, we must multiply the spinless and spin irreps, which we do in Tab. 3. We arrive at the result ) Γ 3/ F = (E 1/ E 3/ ( ) A B 1 B E = 7E 1/ 7E 3/ Note that the total dimension is 8, corresponding to the product of the four-dimensional Γ 3/ and the seven-dimensional F SO(3) irreps. (c) Starting on the dominant LS coupling end, decompose 4 F into irreps of O(3), i.e. by good old addition of angular momentum. Then decompose into D 4 irreps and check that you get the same answer as in part (b). Solution : Within SO(3), we have Γ 3/ F = Γ 3/ Γ 5/ Γ 7/ Γ 9/ all of which are decomposed into D 4 irreps in Tab.. Adding up all the decompo- 6

7 C C D 4 C E Ē C 4 C C 4 C C E 1/ E 3/ Γ 3/ E 1/ E 3/ Γ 5/ E1/ E 3/ Γ 7/ E 1/ E 3/ Γ 9/ E1/ E 3/ Table : Spin irreps for D 4 and decomposition of the S = 3 multiplet. D 4 A 1 A B 1 B E E 1/ E 1/ E 1/ E 3/ E 3/ E 3/ E 3/ E 3/ E 3/ E 1/ E 1/ E 1/ Table 3: Products of spinless and spin irreps within D 4. sitions, we obtain once again Γ 3/ F = 7E 1/ 7E 3/ 7