Physics 9B, Winter Problem Set 4 Solution H-J Chung March 8, Problem a Show that the SUN Lie algebra has an SUN subalgebra b The SUN Lie group consists of N N unitary matrices with unit determinant Thus, it has a natural N-dimensional representation How does it decompose into irreducible representations of the SUN? Solution: a Let s think about the fundamental representation which is also called defining representation Fundamental reprsentation of SUN Lie algebra T N is N N traceless hermitian matrix where SUN group U is U = expiαt N with α being real Among these N matrices, we can consider of following form of N N matrices; M N A = T N, where M N are traceless and hermitian matrices and they are elements of fundamental representation of SUN Lie algbera More concretely, for SUN traceless and hermitian, we can choose T N in following way Firstly, we put i above the main diagonal and corresponding i below so that they become hermitian There are NN number of these Secondly, we put above the main diagonal and a corresponding below There are NN number of these Finally, we put n s along the main diagonal, followed a single entry n, followed by zeroswith overall normalization constant to satisfy orthonormal condition There are N number of these In sum, we have N traceless hermitian matrices In this choice of generators, we choose set whose elements are of form as in Due to the form of the matrices, commutator of them is also form of A Therefore algebra is closed under Lie bracket Hence, SUN Lie algebra has an SUN subalgebra Above argument can be made in terms of roots In above construction, the weights look like as follows; for detail, please refer to the book by Georgi
ν = ν = ν 3 =,,, mm + N N mm + mm + N N N N ν m+ m =,, }{{} mm + m ν N N =, }{{} N N N m + m + N N With the convention that a positive weight is one in which the last non-zero component is positive, the weight satisfy Simple roots are for i =, N, and they satisfy ν > ν > ν N > ν N α i = ν i ν i+ 3 α i α j = δ ij δ i,j± 4 More explicitly, simple roots are α =, }{{} N α =, 3, }{{} N α m m m + =,,, }{{} m m mm + }{{} m N m α m N =,, }{{} N N N 3 N N N
They are linearly independent and complete, so span N dimensional Euclidean space Once we know the simple roots, then we can construct whole algebra from them and we can construct states from the highest weight state by acting lowering operators By construction, if we have only α α N, they are simple roots for SUN Therefore, SUN is subalgebra of SUN b After exponentiating A, we get following form; M N 5 Thus we have block-diagonalized form, so N is decomposed into N fundamental representation and a singlet of SUN, ie N = N- under SUN Indeed from matrix 5, M N only acts on N vector space, and at N, N component of matrix 5 is invariant under M N We can do this with the highest weight state and lowering operators From the set-up, the fundamental weights are µ j = j ν k, 6 k= and they satisfy αi µ j = µ α i ij µ = ν is the highest weight of the fundamentaldefining representation N weights, ν ν N note that the weight ν i corresponds to the eigenvalues of the eigenvector of Cartans, are gone through by acting lowering operators associated to α α N, and ν N cannot be accessed and go to other weights by those raising and lowering operators Therefore, N = N- under SUN Though above argument is true in general, there is another possibility for the case of N = 3 It is possible that 3 of SU3 is reduced to 3 of SU adjoint representation or spin- representation under certain SU subalgebra Notice that except for N = 3 case, this kind of thing doesn t happen, because, for other cases, dimension of irreducible representation larger than N of SUN is not smaller than or equal to N This case of N = 3 is given in problem [3] in this problem set }{{} i th T Problem Let r a a =,, 3 be a set of real numbers Calculate expi a=,,3 r aλ a, where λ a are the first 3 Gell-Mann matrices 3
Solution: Let X = a=,,3 r aλ a, then expix = n= n! Xn Since up and left matrices of λ,,3 are Pauli matrices, we have {λ a, λ b } = δ ab By using this relation, a=,,3 r a λ a = r a r b {λ a, λ b } 7 raλ a + a=,,3 a>b = ra 8 a=,,3 Let R = a=,,3 r a /, then expix = n! Xn = + X! X i 3! X3 + 4! X4 + 9 n= = + r a λ a! R i 3! R r a λ a + 4! R4 + a=,,3 a=,,3 = cos R + ir aλ a sin R R cos R + ir 3 R sin R ir +r R sin R = ir r R sin R cos R ir 3 R sin R Problem 3 Show that the Gell-Mann matrices λ, λ 5, λ 7 make a subalgebra of SU3 What is it? How does the 3 representation of the SU3 decompose into irreducible representation of SU3 Solution: The Gell-Mann matrices λ, λ 5, λ 7 are given as follows; i i λ = i, λ 5 =, λ 7 = i 3 i i They satisfy the commutation relation of SU Lie algebra, so they are SU subalgebra of SU3; [λ i, λ j ] = iɛ ijk λ k, 4 where i, j, k is cyclic permutation of,5,7 In this case, 3 of SU3 is reduced to 3 of SU, adjoint representation of SU, under SU as mentioned in problem [] Let s check this more carefully In this problem, we take Cartan subalgebra as λ, and rasing and lowering operator as λ ± = 4
λ 5 ± iλ 7 / satisfying [λ, λ ± ] = ±λ ±, [λ +, λ ] = λ 5 The eigenvectors of λ, and associated weights are / v = i/ + 6 v = 7 i/ v 3 = / 8 v is the highest weight state; λ + v = By taking lowering operator subsequently, we see that there is no invariant subspace, and it is 3 dimensional vector space; λ v = v, λ v = v 3, and λ v 3 = We know that T,,3 = λ,,3 / also forms SU subalgebra of SU3 We set T 3 as Cartan subalgebra and raising and lowering operator as T ± = T ± it / The eigenvectors of T 3, and associated weights are / v = / +/ 9 v = v 3 = / v is the highest weight state; T + v = By taking lowering operator T, we can check that T v = v 3, T v 3 = In addition, T ± v = Therefore v,3 form -dimensional invariant subspace of 3-dimensional vector space Thus in this case, fundamental representation of SU3 is decomposed to fundamental representation and singlet of SU subalgebra, 3 = Problem 5 Let J be the n n skew-symmetric matrix, J i+n,j = δ i,j, J i,j+n = δ i,j, J i,j = J i+n,j+n =, i, j =, n Show that the set of all n n real matrices S satisfying S T JS = J forms a group, the so-called symplectic group, under matrix multiplication Compute the dimension of the symplectic group 5
Solution: closure Let S, be two elements of the set G whose elements are n n real matrices satisfying S T JS = J Then S S G, because Identity Identity I is in G, because I T JI = J S S T JS S = S T S T JS S = S T JS = J 3 Inverse From S T JS = J, det S T JS = det S det J = det J, so det S Therefore S is invertible If S G,then its inverse S is also in G Since S T = S T, from S T JS = J, we get Therefore, S G J = S T JS = S T JS 4 Associativity Group multiplication is matrix multiplication here Since we know that associativity holds for matrix multiplication, S S S 3 = S S S 3, associativity holds here Since axiom of group is satisfied, the set of all n n real matrices S satisfying S T JS = J forms a group Consider the following form of n n matrix; A B S = C D 5 Then, from S T JS = J, A T C T B T D T I I A B C D C T A A T C C T B A T D = D T A B T C D T B B T D = I I 6 Therefor, we have four equations for the constraints for the elements of S C T A A T C = 7 D T B B T D = 8 C T B A T D = I 9 D T A B T C = I 3 Transpose of eq9 is eq3, so they are same conditions, which give n conditions LHS of eq7 is antisymmetric matrix Therefore eq7 gives nn conditions, and same argument is applied to eq8 and it gives another nn In sum, we have n n constraints on n variables Therefore dimension is n + n 6