Physics 129B, Winter 2010 Problem Set 5 Solution

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1 Physics 9B, Winter 00 Problem Set 5 Solution Chan Y. Park March 3, 00 Problem Four equal masses m are connected by six springs of spring constant k in such a way that the equilibrium positions of the masses are at the corners of a regular tetrahedron. Find the frequencies of all the normal modes. Solution: If we use the coordinate system such that the vertices of the tetrahedron are at v = (0, 0, 0), v = (,, 0), v 3 = (, 0, ), and v 4 = (0,, ), the potential energy of the system is given as where k i,j= η i U ij η j, () η = ( r, r, r 3, r 4 ) () = (x, y, z, x, y, z, x 3, y 3, z 3, x 4, y 4, z 4 ), (3) when r i is the displacement of a mass m i at vertex v i, and U = (4) Then when m i = m, the equation of motion is m η i + k j U ij η j = 0. (5)

2 We want to find out the normal modes of the system. equation of motion of the form A normal mode is a solution of the η(t) = ξe iωt, (6) that every component oscillates with the same frequency. If we substitute this back into the equation of motion, we get a matrix equation (U λi)ξ = 0, (7) where λ = ω /ω0 and ω 0 = k/m, the characteristic frequency of the system. Therefore to find out the normal modes, we need to solve the above equation to find out eigenvalues and eigenvectors, which will correspond to the frequencies and the amplitudes of the normal modes. A straightforward calculation would also give us the answer, but we can use the symmetry of the system to do a better job. Because every mass sits on a vertex of a tetrahedron, the Lagrangian of this system has the symmetry of S 4. Therefore the eigenspace of the normal modes should be a -dimensional representation of S 4. Let s assume that we found the right basis of the eigenspace. Then by a similarity transformation that corresponds to a suitable coordinate transformation from (3) to the basis, U will be transformed into V = P UP, where P is a matrix that gives the coordinate transformation, and λ... λ λ... V = λ, (8)... λn... λn where each λ i repeats n i times. Considering decomposing the -dimensional (reducible) representation into the irreps of S 4, these n i should be the dimension of the corresponding irrep, and the subspace of the normal modes that have the same eigenvalue λ i consist of a vector space of the irrep. How does this help us finding the eigenvalues? For any g S 4, if we call the -dimensional representation of the element as D(g), then by applying the same similarity transformation D(g) can be made into a block-diagonal form, which is the same as decomposing it into the irreps of S 4. P D(g)P = i D i (g). (9) Here D i is a irrep of S 4 that corresponds to the i th subspace of the normal modes. Then by taking the trace of D(g)U, we get tr(d(g)u) = tr(p D(g)P P UP ) = i tr(λ i D i (g)) = i λ i χ i (g). (0)

3 Therefore we got the equations for λ i. If the number of equations is not enough to solve for every λ i, or if we have any other reason for more equations of λ i, we can use the equations like tr((d(g)u) ) = i λ i tr(d i (g) ), () and so forth. Therefore what we have to do now is first decomposing the -dimensional representaion into the irreps of S 4, and then setting up the equations for λ i. To decompose the -dimensional representation, we need the character table of S 4. C 0 C C C 3 C 4 χ 0 χ χ 0 0 χ χ () Here C i represents a conjugate class of S 4, C 0 = {e}, (3) C = {( )}, (4) C = {( )}, (5) C 3 = {( )}, (6) C 4 = {( )( )}. (7) We also need to find the character table of the -dimensional representation D(g) for g S 4. To build the table we need some explicit representation of D(g) for each g C i. For e C 0, D(e) = I, (8) where I is a -by- identity matrix. For p = (34) C, P D(p) = 0 P P, P = 0 0. (9) P 0 For r = (34) C, For q = (34) C 3, R D(r) = R R 0 0, R = 0 0. (0) R Q D(q) = Q Q 0 0, () 0 0 Q 0 3

4 where we don t need to know what Q is to get the character of D(q). For s = ()(34) C 4, 0 S 0 0 D(s) = S S, S = 0 0. () S 0 Now we can build the character table of D. C 0 C C C 3 C 4 χ (3) When we denote the decomposition of D into D i as the orthogonality relation of the characters a i = 4! D = 4 a i D i, (4) i=0 4 N k χ i (C k )χ(c k ), (5) k=0 where N k is the number of the elements in class C k, gives D = D 0 D D 3 D 4, (6) from which we can deduce that there are five eigenvalues, each one corresponding to one irrep. We denote with λ 0 the eigenvalue for D 0, λ for D, λ 3 for D 3, and λ 4, λ 4 for two D 4 s. Then the trace equations are given as tr(d(e)u) = = λ 0 + λ + 3λ 3 + 3(λ 4 + λ 4 ), (7) tr(d(p)u) = 6 = λ 0 λ 3 + (λ 4 + λ 4 ), (8) tr(d(r)u) = 3 = λ 0 λ, (9) tr(d(s)u) = 4 = λ 0 + λ λ 3 (λ 4 + λ 4 ), (30) tr((d(e)u) ) = 30 = λ 0 + λ + 3λ 3 + 3λ 4 + 3λ 4, (3) where the first equalities of (7)-(30) are from directly evaluating the trace of D(g)U, and the second equalities are from (0). Instead of considering D(q)U, we calculated the trace of (D(e)U), which is easier to calculate because we don t know the exact form of D(q) yet. This gives us (3). These are equations of enough number to get λ s. By solving (7)-(3), we get λ 0 = 4, λ =, λ 3 = 0, λ 4 =, λ 4 = 0, (3) from which we can get the frequencies of the normal modes using ω = λω0. We can try to simplify the problem using a subgroup of S 4, for example A 4, a group of even permutations. If we do that, we get the result that the -dimensional representation should be decomposed into three -dimensional irreps and three 3-dimensional irreps. In principle each irrep has different eigenvalue, because if two irreps have the same eigenvalue than their linear combination does too, which implies there is a -dimensional irrep instead of two -dimensional irreps. This means that the chosen group is not a maximal symmetry group of the system. Indeed this is the case when we use A 4, where two of the three -dimensional irreps have the same eigenvalue. Therefore A 4 does not see the symmetry that S 4 can, and it is the reason we started with S 4 instead of A 4. 4

5 Problem Show that there are only two groups of order 4, and both are Abelian. Find all the irreducible representations for both groups. Solution: By building group multiplication tables, we can see there are only two finite groups of order four. Let s start with (33). e a b c e e a b c a a? b b c c (33) Now we have a choice. One is setting a a = e, the other a a = b, because a a = c can be considered as the same as the second choice after renaming b as c. If we make a choice, the rest of the multiplication table is fixed by that. For a a = e, we have (34), e a b c e e a b c a a e c b b b c e a c c b a e (34) which is so-called Klein four-group, isomorphic to Z Z. And for a a = b, we have (35), e a b c e e a b c a a b c e b b c e a c c e a b (35) which is isomorphic to Z 4. As the order of a group is the same as the sum of the squares of the irreps, the only possibility for groups of order 4 is to have four one-dimensional irreps, because every group have trivial onedimensional irrep which maps every group element to. These two groups are abelian, and every element of an abelian group makes a class by itself. Therefore if we build a character table, it gives us all the one-dimensional irreps because a trace of a one-dimensional irrep is the irrep itself. Using the orthogonality of characters, we can find out that the character table of Z Z is (36). (0, 0) (, 0) (0, ) (, ) χ 0 χ χ χ 3 (36) 5

6 where χ 0 is the character for a trivial representation. In a similar way, we can build the character table of Z 4 as (37). 0 3 χ 0 χ i i χ χ 3 i i (37) Problem 3 Suppose M is a square matrix. Show Solution: det(exp M) = exp(trm). Every square matrix can be made in a Jordan canonical form, which is a block diagonal form J = J... J n, (38) where each block M i is of the following form λ i. J i = λ.. i.... (39) λi In other words, there is an invertible matrix P such that P MP = J, where J = n J i, (40) i= and each J i = λ i I i + N i, where N i is a nilpotent matrix defined as (N i ) j,k = δ j,k. This is as closest as we can get to a diagonal form for a general square matrix. Note that when J i is an n i n i matrix N n i = 0, therefore the name nilpotent. Also, it is easy to see that which leads to tr(n i ) = 0, (4) tr(m) = tr(p JP ) = tr(j) = i tr(λ i I i + N i ) = i tr(λ i I i ) + tr(n i ) = i n i λ i, (4) where in the second equality we used the cyclic property of a trace. As N i commutes with I i, exp(j i ) = exp(λ i I i ) exp(n i ). (43) 6

7 Using N n i = 0, we can get exp(n i ) = I i + N i + N i + + (n i )! N n i i = (n i )!, (44) from which we can easily see that det(exp(n i )) =. (45) Now we gathered all the information we need. Let s prove the equality given in the problem. det(exp(m)) = det(p exp(j)p ) (46) = det(exp(j)) (47) = det(exp(j i )) i (48) = i = i = i = i det(exp(λ i I i ) exp(n i )) (49) det(exp(λ i I i ) det(exp(n i ))) (50) det(exp(λ i )I i ) (5) exp(n i λ i ) (5) ( ) = exp n i λ i = exp(trm). (53) i (54) Problem 4 The Baker-Campbell-Hausdorff formula enables you to express e ɛx e ɛy as in the form e ɛz, where X, Y, and Z are matrices and ɛ is some constant. In the class, the formula for Z to all order in the ɛ expansion was given without a proof. Derive the formula for Z to the order ɛ 3 by expanding e ɛx and e ɛy in powers of ɛ. Solution First let s expand e ɛx e ɛy up to ɛ 3. ( e ɛx e ɛy = I + ɛx + ɛ X + ) ( 6 ɛ3 X 3 + I + ɛy + ɛ Y + ) 6 ɛ3 Y 3 + (55) = I + ɛ(x + Y ) + ɛ (X + XY + Y ) + 6 ɛ3 (X 3 + 3X Y + 3XY + Y 3 ) + (56) 7

8 Now using which will work if ɛ is small, we get, up to ɛ 3, Problem 5 log(i + ɛm) = ɛm ɛ M + 3 ɛ3 M 3 +, (57) Z = log(e ɛx e ɛy ) (58) = ɛ(x + Y ) (59) + ɛ (X + XY + Y ) ɛ (X + Y ) (60) + 6 ɛ3 (X 3 + 3X Y + 3XY + Y 3 ) (6) ( ɛ3 (X + Y )(X + XY + Y ) + ) (X + XY + Y )(X + Y ) (6) + 3 ɛ3 (X + Y ) 3 (63) = ɛ(x + Y ) (64) + ɛ (XY Y X) (65) + ɛ3 (X Y + XY XY X + Y X + Y X Y XY ) (66) = ɛ(x + Y ) + ɛ ɛ3 [X, Y ] + ([X, [X, Y ]] [Y, [X, Y ]]). (67) Consider the tensor product of the spin / and the spin 3/ representations of SU(). Decompose the product into irreducible representations of SU(). Solution: When we see the (j + )-dimensional vector space of a spin j representation of SU() as the Hilbert space of a quantum mechanical system with spin-j degree of freedom, decomposing the tensor product of two irreps of SU() with spin j and j into the direct sum of the irreps of SU() is the same as considering the addition of angular momenta j and j. We represent the vector space of a irrep of spin j as j, m (68) where m shows which component of the vector space of spin j representation the ket represent, and has j + values from j to +j. Then the tensor product of two irreps of spin j and j is j, m j, m = j, j ; m, m, (69) where the vector space of the tensor product of two irreps has dimension (j + )(j + ). What we want to do is to find out how this vector space decompose into the irreps of SU(), which we will denote as J, M; j, j. (70) 8

9 To show from what tensor product we get the irreps, we embellish the usual ket with the value of j and j. With this language, we can reformulate our original question. By using the fact that +J J=0 M= J J, M; j, j J, M; j, j (7) is an identity operator in the representation of SU(), we can say that j, j ; m, m = = +J J=0 M= J j +j J= j j M= J J, M; j, j J, M; j, j j, j ; m, m (7) +J ( J, M; j, j j, j ; m, m ) J, M; j, j j, j ; m, m, (73) where we used the fact that J should take value between j j and j + j, which can be seen either from a vector addition of angular momenta or from the argument that the vector space of J, M; j, j should have dimension (j + )(j + ), the same dimension as the tensor product of two original irreps. Therefore the problem boils down to finding out the coefficients J, M; j, j j, j ; m, m (74) for every possible value of J, M, m, m. These coefficients are known as Clebsch-Gordan coefficients. To calculate the coefficients, we identify the top states in the two different representation, fixing the phase factor between the two by a convention, then get the rest by applying lowering operators. The details of how to do it can be found in many quantum mechanics textbooks, for example see chapter 5 of Shankar(994). It is worth studying the principle underlying the calculation of Clebsch-Gordan coefficients, but the hard work of actual calculation can be spared by looking up good references, for example Particle Data Group s Review of Particle Physics, paying attention to what conventions the reference of your choice are using. From this page of PDG s book, we can read off the coefficients we want to find out., + + 3/, +/ =, (75), + + 3/, / = /4, (76), + + /, +/ = 3/4, (77), 0 + /, / = /, (78), + + 3/, / = 3/4, (79), + + /, +/ = /4, (80), 0 + /, / = /, (8) where we omitted j = 3/ and j = / for brevity. The rest of the coefficients can be obtained by using J, M; j, j j, j ; m, m = ( ) j j j J, M; j, j j, j ; m, m, (8) or by looking up the coefficients from the reference as we did previously. 9