Assignment 3. A tutorial on the applications of discrete groups.

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1 Assignment 3 Given January 16, Due January 3, 015. A tutorial on the applications of discrete groups. Consider the group C 3v which is the cyclic group with three elements, C 3, augmented by a reflection symmetry through a vertical plane, thus the v. It can be thought of as the symmetries of a pyramid which has an equilateral triangle base, as displayed in figure 1. It has two generators, c which implements a rotation of the triangle at the base of the pyramid by angle π/3 radians, and thus obeys c 3 = e and b which implements a reflection through a vertical plane which is perpendicular to the equilateral triangle base of the pyramid and bisects one of the angles. Being a reflection it obeys b = e. The b and c transformations are clearly symmetries of the object. Note that the net effect of the b and c transformations are identical to those of the generators of D 3, the proper isometries of an equilateral triangle. Their net effect is to permute the positions A, B and C. They must therefore obey the relations of D 3, that is, (bc) = e, and in fact, then, C 3v is isomorphic to D 3 which is isomorphic to S 3, the symmetric group of permutations of three objects. This means that, for example, they have the same conjugacy classes, the same irreducible representations and the same character table. However, unlike D 3 whose operations can be thought of as proper rotations (without reflections) and can be embedded in the rotation group, SO(3), C 3v contains a reflection and there are no reflections in SO(3). It therefore cannot be embedded in SO(3). What is needed is O(3). The matrix corresponding to the reflection generator b has determinant 1. Like the embedding of D 3 in SO(3), the embedding of C 3v in O(3) forms a reducible representation of C 3v. 1. What irreducible representations of C 3v are contained in the three dimensional representation which are constructed by embedding C 3v as O(3) matrices? 1

2 Figure 1: The pyramid has a base which is an equilateral triangle with vertices A, B, C placed in the x-y plane.

3 The tetrahedral group, denoted T, is the symmetry group of a proper (without reflection) transformations of a regular tetrahedron. It is of order 1 and it consists of four C 3 subgroups, corresponding to rotations by angles π/3 and 4π/3 about the centres of the faces of the tetrahedron. The group has two generators c and b with the three relations c 3 = e, b = e, (bc) 3 = e. The latter relation can be seen by tracing the path of one of the vertices of the tetrahedron under subsequent transformations. The group elements are {e, c, c, b, bc, bc, cb, c b, cbc, cbc, c bc, c bc }. Show that the group elements which are listed in the set above are all of the independent words that can be written given the two generators c and b and the relations b = e, c 3 = e and (bc) 3 = e. ( Hint:Observe that the relation (bc) 3 = e allows us to reduce any sequence of generators with two b s to one with one b, that is beb = e, bcb = c bc, bc b = cbc.) 3. The group transformations permute the vertices of the tetrahedron. Thus, we expect that T is isomorphic to a subgroup of the symmetric group S 4 of permutations of four objects. The latter group has order 4, so T, being of order 1, cannot be isomorphic to S 4 itself. S 4 does have an order 1 subgroup, the alternating group A 4. A 4 is the set of all even permutations, that is, permutations which can be implemented by an even number of exchanges of neighbouring objects (recall the generators σ i in the presentation of S n where σ i exchanged the i th objet to the i + 1 th object). Also, recall the cycle decomposition of S 4. A cycle of even length corresponds to an odd permutation and a cycle of odd length corresponds to an even permutation. For the cycle decomposition of a given permutation, the degree of the permutation is even if there is an even number of even order cycles and it is odd of there is an odd number of even order cycles. Use this fact to show that the alternating group, A 4 has order 1. As for conjugacy classes, remember that, all rotations by the same angle about any axis are in the same conjugacy class of SO(3). Thus, since T is a subgroup of SO(3), a necessary condition for two group elements to be in the same conjugacy class of T is that they must first be in the same conjugacy class of SO(3). This sufficient condition divides the group 3

4 Figure : The regular tetrahedron, whose four corners are the points a, b, c, d is inscribed in a cube. The axis of -fold rotation symmetries are denoted by X, Y and Z. The body diagonals of the cube are four three-fold rotation axes. elements of T into three potential classes, all rotations by 0, that is {e}, rotations by π/3, that is {c, c, bc, bc, cb, c b, cbc, c bc } and rotations by π radians, that is {b, cbc, c bc}. Thus, candidates for the classes are {e}, {c, c, bc, bc, cb, c b, cbc, c bc } and {b, cbc, c bc}. However, even though the elements of these sets are linked by conjugation with rotations, that is, with elements of SO(3), they are not necessarily linked by conjugation by rotations which are themselves elements of the group T, which is a subgroup of SO(3). For example, whereas there are rotations in SO(3) which invert the direction of a body diagonal of the cube in figure, there is no element of the group T which reverses the direction of a body diagonal of the cube. We therefore expect that c and c are actually in different classes, even though both are rotations by π/3 radians. 4. Show that the classes are {e}, {c, bc, cb, c bc }, {c, bc, c b, cbc}{b, cbc, c bc} Since there are four classes, there are four irreducible representations. The dimensions squared of the irreducible representations must add up 4

5 to the order of the group, 1. Thus 1+d +d 3 +d 4 = 1, where we have taken into account that there always must be a trivial one-dimensional representation. 5. Show that the only solution of the remaining integers is that there are two more one-dimensional representations and one three dimensional representation so that in total, there are three one-dimensional representations and one three dimensional representation. We can begin constructing the character table as T [e] 4[c] 4[c ] 3[b] A A 1 A 3 1 T 3 6. Show that the only possible one-dimensional representations have the generators identified as and c = ω, b = 1 c = ω, b = 1 where ω = exp(πi/3). This allows us to fill in two more rows of the character table, T [e] 4[c] 4[c ] 3[b] A A 1 ω ω 1 A 3 1 ω ω 1 T 3 7. Show that the remainder of the character table can be filled in using orthogonality of the rows or the columns, so that 5

6 T [e] 4[c] 4[c ] 3[b] A A 1 ω ω 1 A 3 1 ω ω 1 T Consider the 7-dimensional reducible representation which is formed by taking the direct product of three three-dimensional representations, T T T. Show that it can be decomposed as the direct sum T T T = A 1 A A 3 7T Magnetic Moment A material has a magnetic moment if it responds to the magnetic field by experiencing a torque. It has a polarization, or an electric dipole moment if it experiences a torque when exposed to an electric field. The magnetic moment and electric polarization transform like a pseudovector and a vector under rotations, M i M i = det(r) R ij M j, P i P i = R ij P j If R SO(3), these transformation laws are identical. However, if the transformation contains a reflection, that if it is necessarily in O(3), these transformations differ by an overall sign. The magnetic moment is said to be a pseudo-vector and the electric polarization is a vector. In an isotropic material, all quantities should be invariant under rotations. This tells us that the electric or magnetic dipole moments should obey M i = det(r) R ij M j, P i = R ij P j 9. Show that the above equations imply P i = 0 and M i = 0 for an SO(3) invariant physical system. 6

7 10. Can M i or P i be nonzero if the symmetry of the physical system is reduced to T, D 3 or C 3v? Conductivity Tensor Now, let us ask the question as to whether the conductivity of a material with various symmetries can be zero or non-zero. The conductivity tensor is a relationship between the strength of an electric field E which is applied to a material and the electric current j which results when the electric field is applied. The current is a linear function of the applied field which has the form j i = σ ij E j Time reversal symmetry implies that σ ij is an real, symmetric matrix, σ ij = σ ji. Under a spatial rotation, it transforms as σ ij σ ij = m,n=1 R im R jn σ mn We could ask what the form of the conductivity tensor is in a material which is completely isotropic. Complete isotropy implies that the conductivity tensor does not change under a spatial rotation. That is, in an isotropic environment, σ ij = m,n=1 for any rotation, that is, for any matrix R SO(3). R im R jn σ mn (1) 11. Show that the equation (1) implies that the only possibility for σ ij is that it is proportional to the unit matrix, σ ij = σδ ij (Hint: you are allowed to use Schur s Lemma for SO(3) and the fact that the R s are an irreducible representation of SO(3), the representation with l = 1.) 7

8 Now, let us assume that the symmetry of the environment of a quantum system are reduced form SO(3) to one of the groups D 3, C 3v or T. In general, the lower degree of symmetry should pose fewer constraints than the full rotation group. We want to know whether a conductivity tensor which is more general than σ ij = σδ ij is allowed by considering an environment with less symmery. Let us begin with the tetragonal group T. To answer the question about the form of σ, we need to find the reducible representation under which the conductivity tensor transforms. Since it is a symmetric matrix, and, under a rotation, it is easy to argue that it transforms like σ ij m,n=1 1 [ D (l=1) im (g)d(l=1) jn ] (g) + D (l=1) in (g)d (l=1) jm (g) σ mn where D (l=1) ij (g) is the three-dimensional representation of SO(3), which we were calling the orthogonal matrices R ij in the above, and g is an element of T (which is also an element of SO(3), where we regard T as a subgroup of SO(3)). Let us call this representation (V V ) +. Generally, we only need the character of the above representation. We can find the character from the character formula for SO(3) group elements. To begin, we take the trace of the transformation matrix in the above formula by setting i = m and j = n and then summing over m and n. We get χ (V V ) + (g) = 1 [ (TrD (l=1) (g) ) + TrD (l=1) (g )) 1. Show that the character of the representation (V V ) + above has the characters which are summarized in the last row of the following character table: T [e] 4[c] 4[c ] 3[b] A A 1 ω ω 1 A 3 1 ω ω 1 T (V V )

9 13. Show that, studying the above character table implies (V V ) + = A 1 A A 3 T Since the identity representation only appears once on the right-handside of the equation above, this implies that there is only one way for the conductivity tensor to be invariant under the tetragonal group, if σ ij = σδ ij, as before. Now, let us consider the group D 3. Recall that the character table of D 3 is D 3 [e] [c] 3[b] A A E 1 0 (V V ) () where we have added the character of the reducible representation (V V ) How does the reducible representation (V V ) + decompose into irreducible representations of D 3? If we define the representation (V V ) as one which transforms an anti-symmetric tensor a ij m,n=1 1 [ D (l=1) im with characters defined by χ (V V ) (g) = 1 (g)d(l=1) jn ] (g) D (l=1) in (g)d (l=1) jm (g) a mn [ (TrD (l=1) (g) ) TrD (l=1) (g )) 15. Find the decomposition of (V V ) in terms of representations of T and D 3. 9

5 Irreducible representations

Physics 129b Lecture 8 Caltech, 01/1/19 5 Irreducible representations 5.5 Regular representation and its decomposition into irreps To see that the inequality is saturated, we need to consider the so-called