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1 Lie Algebra and Adjoint Representation 20191051 Quantum Mechanics II Tutorial 7 Teaching Assistant: Oz Davidi December 23, 2018 1 Lie Algebra and Adjoint Representation 1.1 Structure Constants Since the generators are closed under Lie bracket, we can define the structure constants fabc in the following way c [Xa;Xb] = ifabcX : (1.1) Note that by definition fabc = −fbac. It is remarkable that fabc contains the whole information that we need in order to maintain the group multiplication law. It is important to emphasize that the actual value of fabc is basis dependent. 1.2 The Adjoint Representation Let us first define a Lie algebra representation, and then look again at the adjoint representation which you already saw in class. 1 1.2 The Adjoint Representation 1 LIE ALGEBRA AND ADJOINT REPRESENTATION Definition. A representation of a Lie algebra g is a linear map π : g −! gl (n; C) that preserves the Lie product 8X; Y 2 g : π ([X; Y ]) = [π (X) ; π (Y )] : (1.2) A linear map between two Lie algebras that preserves the Lie product is called a Lie algebra homomorphism. If this map is invertible, it is called an isomorphism. Definition. A Lie algebra representation π : g −! gl (V ) is irreducible if there is no nontrivial subspace ;= 6 W ⊂ V such that 8X 2 g : π (X) W ⊆ W: (1.3) One of the most important representations is the adjoint. There are two equivalent ways to defined it. 1.2.1 Georgi (p. 48) We plug Eq. (1.1) into the Jacobi identity, and find 0 = [[Xa;Xb] ;Xc] + [[Xc;Xa] ;Xb] + [[Xb;Xc] ;Xa] (1.4) = ifabd [Xd;Xc] + ifcad [Xd;Xb] + ifbcd [Xd;Xa] (1.5) = −fabdfdceXe − fcadfdbeXe − fbcdfdaeXe : (1.6) This gives us the condition on the structure constants fabdfdce + fcadfdbe + fbcdfdae = 0 : (1.7) We define a set of matrices Ta such that [Ta]bc ≡ −ifabc : (1.8) 2 1.2 The Adjoint Representation 1 LIE ALGEBRA AND ADJOINT REPRESENTATION Then, by using the relation of the structure constants, Eq. (1.7), we get [Ta;Tb]ce = [Ta]cd [Tb]de − [Tb]cd [Ta]de (1.9) = −facdfbde + fbcdfade (1.10) = −fcadfdbe − fbcdfdae (1.11) = fabdfdce (1.12) = ifabd [Td]ce : (1.13) We see that [Ta;Tb] = ifabcTc : (1.14) Therefore, the structure constants themselves generate a representation. This representation is called the adjoint representation. We mentioned that the structure constant are basis dependent, which is obvious, remembering that the generators span a vector space. Let us see it explicitly. We define a scalar product on the linear space of the generators in the adjoint representation as Tr [TaTb]. This is a real symmetric matrix, and can hence be diagonalized with orthogonal matrix. By performing a linear transformation on the generators 0 Xa −! Xa = SabXa ; (1.15) with some orthogonal matrix S, Eq. (1.15) induces the following transformation of the structure constants 0 −1 fabc −! fabc = SadSbefdegSgc : (1.16) The matrices Ta of the adjoint representation transform by −1 Ta −! Sab STbS : (1.17) This is a combination of a similarity transformation and, simultaneously, a linear transformation of the form of Eq. (1.15). In the trace, however, only the latter transformation matters, and we obtain 0 0 T Tr [TaTb] −! Tr [TaTb] = SacSbdTr [TcTd] = SacTr [TcTd] S db : (1.18) Selecting a diagonal basis, we write (no sum on repeated indices) Tr [TaTb] = κaδab : (1.19) However, we still have the freedom to rescale the generators. We shall assume that all of the κa 3 1.2 The Adjoint Representation 1 LIE ALGEBRA AND ADJOINT REPRESENTATION are positive. This will be the case if the algebra is compact 1 and semisimple.2 We can thus go ahead and write Tr [TaTb] = λδab ; (1.20) for some positive λ. A nice feature of this basis comes from the following Tr [[Ta;Tb] Tc] = ifabdTr [TdTc] = iλfabdδdc = iλfabc ; (1.21) meaning 1 f = Tr [[T ;T ] T ] : (1.22) abc iλ a b c We find that the structure constants in this basis are completely anti-symmetric! The Ta matrices are therefore anti-symmetric and pure imaginary, thus Hermitian. In this basis, the adjoint representation is unitary. 1.2.2 The Generators as States Here, we would like to generalize our construction of the regular representation to semisimple3 Lie groups. For this, we move to a basis in which the structure constants tensor is fully anti-symmetric. 1. For each generator Xa 2 g, associate a vector jXai, such that they form an orthonormal basis, hXa j Xbi = δab. 2. Define the adjoint representation on this vector space as ad (Xa) jXbi = j[Xa;Xb]i. Exercise. Check that this is a representation, i.e. [ad (X) ; ad (Y )] = ad ([X; Y ]). 3. The components of the matrices are given by [ad(Xa)]bc =hXb j ad(Xa) j Xci=hXb j [Xa;Xc]i=ifacd hXb j Xdi=ifacb =−ifabc =[Ta]bc :(1.23) It is sometimes said that the adjoint representation is the action of the algebra on itself. In this language of states, we define the inner product as 1 hX j X i = Tr T yT : (1.24) a b λ a b 1 Non-compact Lie algebras, in which some of the κa are negative, do not have non-trivial finite-dimensional unitary representations. They still, however, play an important role in physics. The Lorentz algebra is an example. 2A semisimple Lie algebra contains no non-trivial Abelian ideals. 3A semisimple Lie group has a trivial center. Equivalently, its algebra is semisimple. 4 2 SUBALGEBRA, CARTAN, AND CASIMIR The dagger is there because we will sometimes work with complex linear combinations of genera- y tors, but the generators themselves are still Hermitian: Ta = Ta. 2 Subalgebra, Cartan, and Casimir Definition. For a given Lie algebra g, a subspace h ⊆ g is a Lie Subalgebra if it is closed under the Lie product 8X; Y 2 h : −i [X; Y ] 2 h : (2.1) Let us consider as an example su (3) and su (2). You saw in class that the Gell-Mann matri- ces are a good basis for su (3). The generators are conventionally defined as 0 0 1 00 0 11 00 0 −i1 λ 1 σj λ 1 λ 1 X = j = B 0 C ;X = 4 = B0 0 0C ;X = 5 = B0 0 0 C ; j 2 2 @ A 4 2 2 @ A 5 2 2 @ A 0 0 0 1 0 0 i 0 0 (2.2) 00 0 01 00 0 0 1 01 0 0 1 λ6 1 λ7 1 λ8 1 X6 = = B0 0 1C ;X7 = = B0 0 −iC ;X8 = = p B0 1 0 C : 2 2 @ A 2 2 @ A 2 2 3 @ A 0 1 0 0 i 0 0 0 −2 It’s easy to see that [Xj;Xk] = ijklXl, for j; k; l 2 f1; 2; 3g. This is also the algebra of su (2). Therefore su (2) is a subalgebra of su (3). In fact, there are more copies of su (2) in su (3): p p n 3 1 o n 3 1 o X4;X5; 2 X8 + 2 X3 and X6;X7; 2 X8 − 2 X3 form an su (2) algebra. We also notice that λ2, λ5, and λ7 are the generators in so (3), hence so (3) is a subalgebra of su (3) as well. 2.1 Cartan Subalgebra Definition. A Cartan Subalgebra h, is the maximal commuting subalgebra of a Lie algebra g. Namely: 1. 8H1;H2 2 h :[H1;H2] = 0. 2. If X 2 g and [H; X] = 0 for all H 2 h, then X 2 h. All Cartan subalgebras of g have the same dimension. We define the rank of an algebra to be the dimension of its Cartan subalgebra. The importance of the Cartan subalgebra is that all its members can be diagonalized simultaneously. This means that they label states in our system by their eigenvalues (called weights). These correspond to the m quantum number of angular 5 2.2 Casimir Operator 2 SUBALGEBRA, CARTAN, AND CASIMIR momentum. Exercise. What is the rank of su (n)? The Cartan subalgebra of su (3), for example, can be chosen to be λ3 and λ8. The rank of su (3) is thus 2. 2.2 Casimir Operator A Casimir Operator is an operator (in the enveloping algebra) that commutes with all members of the algebra. The number of Casimirs is equal to the rank of the algebra. The importance of the Casimirs is that they can be brought to a diagonal form simultaneously with the Cartan. These operators allow us to differentiate between different representations. They correspond to the j quantum number of angular momentum. Let us look at su (3) again. Two Casimir operators can be constructed. One of them is quadratic C1 = XaXa ; (2.3) where the other one is cubic C2 = dabcXaXbXc : (2.4) Here, dabc is given by 1 fX ;X g = δ + d X : (2.5) a b 3 ab abc c Exercise. Prove that indeed C1 is a Casimir, namely 8X 2 su (3) : [X; C1] = 0 : (2.6) Let us now prove that C2 is a Casimir. First, we can rewrite it as 1 1 1 C = X X fX ;X g − δ = (X X )2 + X C X − C = (X X )2 + C2 − C ; (2.7) 2 a b a b 3 ab a b a 1 a 3 1 a b 1 3 1 where we have used [Xa;C1] = 0 for all Xa 2 su (3). Because the Casimir operators commute with the algebra elements regardless of basis, we work in the basis defined by Eq.
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