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Tutorial #5: Characters and Orthogonality Relations April 12, 2016 1 Schur's lemma in quantum mechanics 1.1 Particle on a pyramid Consider a particle hopping between the verticies of a pyramid whose base is an equilateral triangle. Figure 1: Pyramid The particle hops between the base verticies with amplitude amplitude t1 ; that is, the Hamiltonian is 0 t1 H t1 t1 where on site 1 2 0 0 0 T t1 t 0 t and from the base to the top of the pyramid with t1 t t 0 is the state corresponding to the particle on site and so on. The Hamiltonian is obviously basis of the form t1 0 t t t, C3v (1) 1 (head of the pyramid), 0 1 0 0 T is the state symmetric; as you saw in class, we can construct an orthonormal |a, l, xy, with a denoting irreducible representation, l the state within the representation, and x is some extra index. In order to do that, we need to know what are the relevant irreducible representations. Our Hilbert space is four-dimensional, so it must transform according to some reducible representation (because the irreducible ones are only one- and two-dimensional). First, we notice that action of the group only exchange the base verticies, and leave the pyramid head in place, so 1 |1, 1, 1y |1y 0 0 (2) 0 is an invariant state - that is, it transform according to the trivial representation. to denote the rst irreducible representation of C3v , Mind the notation: the rst 1 is the trivial one; the second 1 is for the state itself (but the trivial representation is one-dimensional, so this index cannot obtain any other value); the third 1 is for dierent copies of the trivial representation. And indeed, there is another copy: if we consider the representation of the group that act on the Hilbert space: 1 0 D pRq 0 0 0 0 0 1 0 1 0 0 0 0 , 1 0 1 0 D pσ1 q 0 0 1 0 1 0 0 0 0 0 1 0 0 , 1 0 etc. (3) Tutorial 5: Characters and Orthogonality Relations 3 CHARACTERS we can notice that |1, 1, 2y |2y 0 1 1 ? 3 1 1 (4) is also invariant under the action of the group. Finally, since the Hilbert space representation is faithful, we must have the (only) faithful representation of orthogonal to both C3v , the two-dimensional one. This means that the remaining two dimensional space |1y and |2y transform according the the representation Dp3q , and we are free to choose a basis, for example: 0 1 ? 11 , 6 2 |3, 1, 1y |3y |3, 2, 1y |4y 0 1 ? 11 2 0 (5) Now, recall that according to Schur's lemma the action of the Hamiltonian in this basis is in the form H |a, l, xy ¸ paq |a, l, yy fxy (6) y we can see this explicitly: ? 3t1 ?2 t ? H |2y ?23 3t1 |1y 3t ?23 t 0 t1 ? 1 H |1y t1 3t |2y , t1 that is, 2t |2y (7) p1q f p1q ?3t1 , f p1q 2t, so we now only have to diagonalize a 2 2 matrix, rather then a 4 4: 22 21 ? 1 0 3t f p1q ? 1 3t 2t f12 (8) for the other two states the situation is even better, since they are already eigenstates: p3q where 2 p3q f11 H |3, l, 1y f11 |3, l, 1y is just a number (that happens to be t), because the representation Dp3q only appear once. Schur orthogonality relation Reminder: Schur orthogonality relation is the equation ¸ P Dpaq g 1 • note that there is a small caveat: if P Dpaq pg q P 1 for some xed P, Dpaq mn g G and Dpbq |dG| δab δni δmj Dpbq pg q ij (10) a are equivalent representations, but not identical, namely Dpbq pg q the orthogonality relation reads ¸ P Dpaq g 1 mn g G 3 (9) Dpbq pg q |G| P P 1 in mj d ij (11) a Characters Reminder: The (rst) orthogonality relation for characters is ¸ kα χa pCα q χb pCα q |G| δab (12) Cα (where a, b are indecies for irreducible representations, and relation (or completeness relation) is ¸ α, β for conjugacy classes). The seconed orthogonality χa pCα q χa pCβ q a |G| δ kα αβ (13) As you saw in the lecture, we can construct the following vectors: d vaα using dirac notation, the rst relation is Group Theory in Physics |kGα| χa pCα q (14) xva |vb y δab , while the seconed one is °a |va y xva | 1. page 2 of 3 Spring 2016 Tutorial 5: Characters and Orthogonality Relations 3.1 Alternating group A4 Recall the alternating group Construct the character table of A4 1 - the group of even permutations on four letters. It consists of four conjugacy classes : A4 d21 is a normal subgroup of d1 A4 d22 d23 d24 |A4 | 12 G H of a group from the irreducible representations of G H : D pg q D pgH q { In our case, A4 H is a group of order 3, so it must be isomorphic to Dp1q Dp2q Dp3q e 1 1 1 R 1 ω ω2 2πi 3 { G, you can immediately construct (18) C3 , for which the character table is R̄ 1 ω2 ω Table 1: Characters of e (17) (you can verify directly that it is closed under multiplication, and thus a subgroup, and it irreducible representations of ω (16) d2 d3 1, d4 3. Now, notice that H tid, p12q p34q , p13q p24q , p14q p23qu is normal it is a union of conjugacy classes). Given a normal subgroup, where (15) has four conjugacy classes, it also has four irreducible representations. The dimensionality of the representations must satisfy The only possible choice id CHARACTERS A4 C1 tidu , Cπ tp12q p34q , p13q p24q , p14q p23qu tp123q , p142q , p134q , p243qu , Cα1 tp132q , p124q , p143q , p234qu Cα Solution: First, Since 3 C3 . This gives us all the one-dimensional representations of A4 , and we can use orthogonality to nd the character for the three-dimensional representation. C1 1 1 1 3 Dp1q Dp2q Dp3q Dp4q 3Cπ 1 1 1 1 4Cα 1 ω ω2 0 Table 2: Characters of 3.2 4Cα 1 ω2 ω 0 1 A4 Quaternion group Construct the character table for the Quaternion group. Solution: As you recall, Q has 5 conjugacy classes: teu , Cw twu , Cx tx, x̄u , Cy ty, ȳu , Cz tz, z̄u (19) ° 2 So it also has ve irreducible representations. From the constraint a da |Q| 8 we obtain d1 d2 d3 d4 1, d5 2. Consider now the Subgroup X te, w, x, x̄u. The quotient group Q{X is isomorphic to Z2 , so it induces a sign representation: 1 for the elements contained in X , and 1 for elements of QzX (elements contained in Q C1 but not in X ). The same argument can be also applied to the subgroups Y and Z, and thus we obtain the three non-trivial one-dimensional representations. The characters for the two-dimensional representation can be obtained using orthogonality. Dp1q Dp2q Dp3q Dp4q Dp5q C1 1 1 1 1 2 Cw 1 1 1 1 2 2Cx 1 1 1 1 0 2Cy 1 1 1 1 0 Table 3: Characters of 1 2Cz 1 1 1 1 0 Q The derivation of this partition appears in tutorial #3, even though we haven't done it in class Group Theory in Physics page 3 of 3 Spring 2016