Particles I, Tutorial notes Sessions I-III: Roots & Weights

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1 Particles I, Tutorial notes Sessions I-III: Roots & Weights Kfir Blum June, 008 Comments/corrections regarding these notes will be appreciated. My address is: kf Contents 1 Groups and representations 1.1 Groups Representations: equivalence, unitarity, ir)reducibility Schur s lemma Lie groups and algebras.1 Lie groups and algebras Jacobi identity The adjoint representation States in the adjoint U1) factors Roots & weights: Classifying irreps 7.1 Cartan subalgebra Roots & weights SU) subalgebras Example: SU) The fundamental representation of SU) The adjoint representation of SU) Weighteology The master formula of roots and weights Simple roots & Highest weight Building all the irreps, and fundamental weights Weyl reflections Another example: The 6 of SU) Product decomposition

2 1 Groups and representations 1.1 Groups A group is a set G with a multiplication such that: 1. x, y ɛ G x y ɛ G. e ɛ G such that x ɛ G : e x = x e = x. x, y, z ɛ G x y z) = x y) z 4. x ɛ G there is an inverse x 1 ɛ G such that x x 1 = x 1 x = e In what follows, I will denote xy x y. A group is finite if it has a finite number of elements. O.w. it is infinite. The order of a finite group G is the number of elements in G. A group is abelian if g 1, g ɛ G : g 1 g = g g 1. A subgroup of G is a group H which is contained in G. All elements of an invariant subgroup H of G transform among themselves under the adjoint action of any element of G g ɛ G, h 1 ɛ H, h ɛ H s.t. gh 1 = h g A simple group contains no nontrivial invariant subgroup. A semisimple group contains no nontrivial abelian invariant subgroup. 1. Representations: equivalence, unitarity, ir)reducibility A representation of G is a mapping, D of G, into linear operators such that: 1. De) = I. Dg 1 )Dg ) = Dg 1 g ) The dimension of a representation is the dimension of the vector space on which it acts. Let D be a representation of dimension N. We span the states in the vector space of this representation by the set { i }, i ɛ 1,..., N. The matrix elements of the linear operators Dg) in this representation are given by [Dg)] ij = i Dg) j. One may choose to represent the states in a more convenient way, by making linear transformations i i = S i. An invertible transformation S induces a similarity transformation on the linear operators D D = SDS 1. D and D are equivalent representations. A representation is unitary if Dg) is unitary for every g: g ɛ G : Dg) = Dg) 1. All of the representations of a finite group are equivalent to unitary representations. An invariant subspace is a subspace which transforms within itself under the action of Dg) for any g ɛ G. A representation is reducible if it contains an invariant subspace.

3 The restriction of the representation to the invariant subspace is itself a representation. A representation is completely reducible if it is equivalent to a block diagonal form: D 1 g) D g)... 1)..... where all of the D i g) are not reducible. i.e. irreducible or irrep.) Eq.1) means that a completely reducible rep. can be decomposed into a direct sum of irreps. 1. Schur s lemma Schur s lemma: Let Dg) be a finite dimensional irrep. of G. Then an operator A which commutes with Dg) for every g must be proportional to the identity: [Dg), A] = 0 g ɛ G A I ) Schur s lemma has crucial implications for QM, where the common practice is precisely to classify the eigenspaces of hermitian operators. In QM a symmetry is typically realized by putting the states of the Hilbert space in unitary representations. The transformation law is for states, and Ψ Dg) Ψ, Ψ Ψ Dg) O Dg)ODg) for operators. An invariant observable satisfies O = Dg)ODg), which, since D is unitary, means that we may use Schur s lemma. Eq.) then says that the eigenstates of O can be arranged in irreps D k of G, where each irrep corresponds to some fixed eigenvalue o k. Lie groups and algebras.1 Lie groups and algebras The elements of a Lie group depend smoothly on some continuous parameter: g = gα). We may label the operators in a representation of such group by Dα) = Dgα)). We will mainly deal with compact groups, i.e. groups for which α is confined to some finite volume parameter space. Any representation of a compact lie group is equivalent to a unitary representation. It is convenient to set gα = 0) = e Now we can use an exponential parametrization in order to represent a group element in the vicinity of the identity: Dα) = e iαaxa ) The X a, a = 1,..., N are called the generators: ) D X a = i α a α=0 4)

4 Taking N as the minimal number of independent real parameters required in order to represent any element in the group, the generators X a are linearly independent. Due to the form of ), and as long as we stick to real α, the X a of a unitary representation are hermitian. They span a linear vector space of hermitian operators. Since a representation maintains the closure property of the group, we must have e iαx e iβx = e iγx 5) Take log of both sides and expand in the vicinity of the unity iγx = ln [1 + iαx + βx) 1 αx + βx) 1 ] [αx, βx] +... = We can now define δ = α + β γ), thereby 5) becomes δx = i [αx, βx] This tells us that the vector space of the X is closed under commutation: with indices now explicit. δ is given by δ c = α a β b f abc. The relation 7) is called the Lie algebra. The structure constants f abc obey f abc = f bac iα + β)x 1 [αx, βx] ) [X a, X b ] = if abc X c 7) They are determined by the multiplication law of the group. One can verify that the ellipses on the right hand side of 6), i.e. terms of higher order in the arbitrary parameters α, β, depend again only on f abc. This means that the f abc also determine the multiplication law of the group. The structure constants can be computed in any nontrivial representation. The result will be the same every time. We may thus think of representations of the algebra as representations of the group. Typically, when people talk about groups in physics, they actually think of some representation of the group. Therefore it turns out that you may typically just swap the word group by the word algebra. Notions like reducibility, invariance or compactness can be carried to the algebra in obvious ways. An invariant subalgebra Y in the algebra X is some subset of the generators which transform within itself upon commutation with any element of the algebra x ɛ X, y ɛ Y : [x, y] ɛ Y An invariant subalgebra generates an invariant subgroup. A simple algebra contains no nontrivial invariant subalgebra. A semisimple algebra contains no nontrivial abelian invariant subalgebra. Exercise: show that if a group has a unitary representation, then the structure constants are real. 4

5 . Jacobi identity The generators X obey the Jacobi identity [X a, [X b, X c ]] + [X c, [X a, X b ]] + [X b, [X c, X a ]] = 0 8) Which is easy to prove for matrices, but true also in a more abstract sense for linear operators. A similar identity occurs for the structure constants. The adjoint representation Define the set of matrices Using the Jacobi identity in the form 9), one finds f bcd f ade + f abd f cde + f cad f bde = 0 9) T a ) bc = if abc 10) [T a, T b ] = if abc T c This means that 10) is a representation of the algebra, with dimension N. This representation is called the adjoint representation. We should define a scalar product on the linear space of the generators. Consider tr T a T b ) 11) This is a real symmetric matrix. We can diagonalize it by performing a linear transformation on the generators: X a X a = S ab X b 1) with some orthogonal matrix S. Eq.1) induces the following transformation of the structure constants f abc f abc = S ad S be f deg S 1 gc The matrices T a of the adjoint representation transform by T a T a = S ab ST b S 1 ) 1) combining a similarity transformation simultaneously with a linear transformation of the form 1). In the trace, however, only the latter transformation matters and we obtain Selecting a diagonal basis, we write tr T T ) tr T T ) = Str T T ) S T tr T a T b ) = κ a δ ab no sum!) However, we still have freedom to re-scale our generators. We shall assume that all of the κ a are positive. This will be the case if the algebra is compact 1 and semisimple. Then we can just go ahead and write tr T a T b ) = λδ ab 14) 1 Non-compact Lie algebras, in which some of the κ a 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. The requirement for the algebra to be semisimple is discussed in section.4. 5

6 with some positive λ. The basis in which 14) holds is particularly nice: In this basis, the structure constants are completely antisymmetric f abc = f bac = f acb =... This property follows because we can write f abc = i λ tr [T a, T b ]T c ) The T a matrices are therefore antisymmetric and pure imaginary, and so hermitian. Thus, in this basis, the adjoint representation is unitary...1 States in the adjoint The states in the adjoint correspond to the generators themselves X a X a We can find the action of the generators on the states by inserting a complete set of states: X a X b = X c X c X a X b = X c T a ) cb = if acb X c = [X a, X b ] 15) The scalar product is defined on this space by X a X b = 1 λ tr T at b ) 16) The dagger is there because we will sometimes work with complex linear combinations of generators..4 U1) factors In order to write 14), we had to assume that our algebra was compact and semisimple. We now discuss the notion of a semisimple algebra. An algebra is semisimple if it does not contain a non-trivial abelian invariant subalgebra. An abelian subalgebra follows from an abelian subgroup. Such invariant subgroup is called a U1) factor. In the algebra, it corresponds to a specific generator X which commutes with all of the other generators. The structure constants give us no information about this subalgebra. However, Schur s lemma tells us that the existence of such a U1) factor is not really much trouble. The representation of the group will split into irreps of the semisimple part of the algebra, each irrep characterized by some fixed eigenvalue under the generator X of the U1) factor. In order to classify the algebra representations, we may therefore keep to semisimple algebras. These are built by factoring simple algebras to each other. For example: SU) : simple. SU) SU) : semisimple. U) U1) SU) : neither simple nor semisimple. 6

7 Roots & weights: Classifying irreps The useful thing about the adjoint representation is that it is a representation of the algebra on itself. This means that every structure we expose in the adjoint representation is in fact inherent to the algebra: studying the adjoint representation is studying the algebra. This notion gets exploited to a great extent in the analysis of Roots & Weights..1 Cartan subalgebra Suppose the X a, a = 1,..., N are the generators of a given irrep of dimension D. The maximum subset of generators which can be diagonalized simultaneously is called the Cartan subalgebra. Denote these generators by H i, i = 1,..., M. They satisfy M is called the rank of the algebra. We can choose a basis in which H i = H i, [H i, H j ] = 0 i, j 17) tr H i H j ) = λ D δ ij After diagonalizing all of the H i, we have for the states of the representation H i µ = µ i µ µ is a real vector of length M; We have D of those. The vector µ is called the weight vector. The entries µ i are the weights. In different irreps we will get different weight vectors.. Roots & weights We now return for a moment to the adjoint representation. The weights of the adjoint representation are called roots. Using the conventions of section.1 we have, in the adjoint, D = N. Due to 15,17) we know that H i H j = 0 18) for any H i, H j in the cartan. We can label the other N M states by E α, where This implies It follows that And, therefore H i E α = α i E α 19) [H i, E α ] = α i E α 0) [H i, E α] = α i E α 1) E α = E α ) The E α are not hermitian. They should remind you of the raising and lowering operators in the analysis of the irreps of SU). Indeed, one sees H i E ±α µ = [H i, E ±α ] + E ±α H i ) µ = µ i ± α i ) E ±α µ ) Make sure you understand the difference between the dimension of the group N) and the dimension of the representation D). 7

8 The convention 16) ensures us of the normalization H i H j = δ ij E α E β = δ αβ 4) You can easily check that this is so, by noting that the similarity transformation connecting the E α s to the X a s is unitary: it is the diagonalizing transformation for the hermitian operators H i. Notice that the state E α E α 5) has weight zero. This implies that the operator [E α, E α ] is in the cartan: [E α, E α ] = β i H i We can compute β by projecting the state 5) onto a cartan state: β i = H i E α E α = 1 λ tr H i[e α, E α ]) = 1 λ tr E α[h i, E α ]) = α i 6) so [E α, E α ] = α i H i 7) Now comes an important point. Equations 0-) and 7) that we have found are operator equations. These equations, together with the conclusion ), also hold in an arbitrary representation!. SU) subalgebras Putting equations,7) together, we can make explicit the relation between the E ±α s and SU) irreps. For every non-zero root vector α there exists an SU) subalgebra, with the generators: such that E = α ih i α, E ± = E ±α α [E, E ± ] = ±E ±, [E +, E ] = E 8).4 Example: SU).4.1 The fundamental representation of SU) The fundamental representation of the group SU) consists of unitary matrices with determinant 1, acting on -dimensional complex vectors. It is also called the. These matrices are generated by hermitian traceless matrices, which form an 8-dimensional algebra. A basis for the algebra is i λ 1 = λ = 1 i 0 0 λ = λ 4 = ) 0 0 i λ 5 = λ 6 = λ 7 = i λ 8 = i i

9 This basis is normalized such that The cartan consists of tr λ i λ j ) = 1 δ ij H 1 = λ, H = λ 8 The weights are: ) 1, 1, ), 1, , 1 ) Figure 1 shows a geometrical diagram of the weights. Since the cartan is two dimensional, we can draw the weights on a two dimensional plain. There is no degeneracy: Each weight vector corresponds to a unique state..4. The adjoint representation of SU) We now continue to construct the adjoint representation of SU). The vector space on which this representation acts is eight dimensional - this representation is also simply called the 8. However, using our weight diagram from the previous section, we don t even need to write down explicitly the eight-by-eight matrices which act on this space. We know that the six generators which are not in the cartan can be rearranged as raising and lowering operators of SU) subalgebras. These subalgebras appear as SU) irreps which are contained in the SU) representation we have found. The arrows connecting the states in figure 1 reveal these SU) irreps: We have three of those, each of them spin half. The weight vectors we have found are separated by the vectors: I chose the labels for later convenience) ) ) ±α ) = ± 1, 0), ±α 1) = ± 1,, ±α ) = ± 1, The α i) s are precisely the root vectors of the non-cartan generators in the adjoint. Indeed, the E α s are given by E α ) = 1 λ 1 + iλ ), E α 1) = 1 λ 4 + iλ 5 ), E α ) = 1 λ 6 iλ 7 ) ) The two cartan generators will correspond to two vanishing root vectors: All together, we have eight roots corresponding to eight states. We can now go ahead and draw a weight diagram for the adjoint representation. This diagram is depicted in figure. The non vanishing roots form a regular hexagon. The states corresponding to the cartan generators are degenerate: These are two states with the same root. I have marked with arrows specific SU) irreps all corresponding to the same SU) subalgebra) which are hiding in this SU) representation..5 Weighteology It may be the case that several states in a representation will have the same weight vector. The adjoint representation is an example: All of the states corresponding to cartan generators have zero weight vectors. We would like to have a systematic method to label all of the states in a given representation by unique labels. In order to do this, let s start working with the tools of section.. 0) 1) 9

10 .5.1 The master formula of roots and weights Define: Now compute: However, notice that so that 4) implies E ±α µ = N ±α,µ µ ± α ) µ [E α, E α ] µ = µ α i H i µ = α µ = µ E α E α µ µ E α E α µ = N α,µ N α,µ 4) N α,µ = µ α E α µ = µ α E α µ = µ E α µ α = N α,µ α α µ = N α,µ α N α,µ 5) Because the E ±α are raising and lowering operators of SU) subalgebras, we know that, for any state µ, there exists p, q ɛ N such that E α µ + pα = E α µ qα = 0 In other words, the N ±α,µ vanish for high enough or low enough µ: N α,µ+pα = 0, N α,µ qα = N α,µ q+1)α = 0 We can stack evaluations of 5) to write the following tower of equations: N α,µ+p 1)α 0 = α µ + pα) N α,µ+p )α N α,µ+p 1)α = α µ + p 1)α).. Summing both sides of the tower, we obtain N α,µ qα N α,µ q 1)α = α µ q 1)α) 0 N α,µ qα = α µ qα) 0 = p + q + 1) ] [α µ + α p q) We conclude: α µ α = q p = integer 7) This is a surprising constraint on the weight vectors! We will call 7) the master formula. Specifically, in the adjoint α β α = m but, just the same β α β = m We can write: α β) α β = cos θ = mm 8) 4 There are only very limited choices to the relative orientation of root vectors. I wrote them in table 1. 6) 10

11 Table 1: relative root orientations mm θ , 10 45, 15 0, Simple roots & Highest weight A weight vector α is positive if its first non-zero entry is positive. We say α > β if α β is positive. A highest weight within an irrep is larger than any other weight in the irrep. The highest weight of an irrep is unique. In the adjoint, a simple root is a positive root which cannot be written as a sum of positive roots. Some facts concerning simple roots: 1. If α, β are simple, than α β is not a root. The master formula 7) then implies that Similarly, E α E β = 0 α β α = 1 p, i.e., q = 0) 9) E β E α = 0 β α β = 1 p, i.e., q = 0) 40) This restricts the length and the relative orientation of any two simple roots: β α = p p, pp cos θ = Note also that π θ < π This follows because, by 9-40), the cosine cannot be positive.. It follows that the simple roots are linearly independent. In fact, they are also a complete set: Any positive root µ can be decomposed as a sum of simple roots: µ = r i α i), with r i = integer 0.. The number of simple roots is equal to the rank of the group, and so to the number of generators in the cartan. Knowing the p s of all the simple roots, we can construct the whole algebra. This is done by finding all the roots, through recursive application of the master formula. 11

12 Example: Take the adjoint representation of SU). The two simple roots are ) ) α 1) 1 =,, α ) 1 =, They give α 1) α = ) = 1, α 1) α ) = 1 p = p = 1 Since both α 1), α ) are simple roots, we know that the q s are zero. Thus α 1) can raise α ) once and vice verse), giving the third positive root α ) = α 1) + α ). Note that α ) cannot be raised further, because p = p = 1. So α 1) + α ), for example, is not a root. Since roots come in pairs ±α k), we now know all of the negative roots, too..6 Building all the irreps, and fundamental weights Let α k), k = 1,..., m be the simple roots. Consider a weight µ in some arbitrary irrep. µ is the highest weight if and only if µ + α k) is not a weight for any k. This means that E α k) µ = 0 k, so that α k) µ α k) = q k 0, i.e. p k = 0) 41) Since the α k) are independent and since m is the rank of the group, the q k s determine the highest weight µ completely. Because the highest weight of an irrep is unique, we arrive at an important result, instructing us how to build all of the irreps of a Lie algebra: Given a set of integers q k, k = 1,..., m, we have an irrep with highest weight µ, defined uniquely through 41). The q k s are called the Dynkin indices. We define the fundamental weights by the set of weights µ i) such that µ l) : q k = δ kl Example: SU)... The two simple roots are α 1) = 1, ) ), α ) 1 =, We have two fundamental weights, corresponding to the two different choices: q = 1, 0) or q = 0, 1) 1

13 The first choice, which we will label by µ 1), gives us α 1) µ 1) = 1 α ) µ 1) = 0 µ 1) = ) 1, 1 The fundamental weight µ 1) is just the highest weight of the fundamental representation, which we already know. We will also call this representation 1, 0). In a similar way the second choice, labeled µ ), gives us α 1) µ ) = 0 α ) µ ) = 1 µ ) = 1, 1 ) The fundamental weight µ ) is the highest weight of another three-dimensional irrep of SU), called the anti-fundamental representation, or. We will also call this representation 0, 1). We can continue to construct this representation: Since q 1 = 0, we can not lower µ ) by α 1). We can, however, lower it once with α ). So the weight µ ) α ) = 0, ) 1 does belong in the irrep. The master formula dictates: α 1) µ ) α )) α 1) = 1 = q p 4) However, we know that [E α 1), E α )] = 0 because α 1) α ) isn t a root. HW!) Therefore we know that E α 1)E α ) µ ) = E α )E α 1) µ ) = 0 telling us that p = 0. Eq.4) than implies q = 1. We have found the last missing state of the irrep: It is µ ) α ) α 1) = 1, 1 ) Comparing the three weights that we have found to the weights 0) of the, we see that the weight diagram for the is obtained by reflecting the diagram of figure 1 around the µ 1 axis. It is true in general that if the set of matrices X a form a representation D of the algebra, than the matrices X a obtained from the X a by X a = X a also form a representation of the algebra, called D. The and the are an example for a situation in which the D and the D are not equivalent: There is no similarity transformation which maps the set of matrices λ a of 9) into λ a. If D is equivalent to D, than D is called a real representation. 1

14 .6.1 Weyl reflections The fact that roots come in pairs ±α which generate SU) subalgebras together with the cartan element α H) is manifest in weight diagrams through Weyl reflection symmetry. A Weyl reflection operation on the weight diagram corresponds to a reflection of the weights around an axis perpendicular to one of the roots. Weyl reflection symmetry means that the weight diagram of any representation is invariant under such operation. Multiple reflections are also a symmetry. Weyl reflection symmetry is easily verified for the SU) irreps we have studied so far. All are invariant for reflection around the axes perpendicular to any of α 1), α ) or α ) of 1). This property is responsible for the regular triangular and hexagonal patterns in the irreps of SU). Weyl reflection symmetry is very useful when constructing irreps from highest weights, saving much computing time..6. Another example: The 6 of SU) Let s construct another irrep of SU): The one with highest weight defined by q =, 0) Once again, we will not need to write explicitly the matrix realization of the generators. All we will need is Weighteology. The highest weight is just µ = µ 1) = ) 1 1, Two states we immediately recognize are those corresponding to the weights ) µ α 1) 1 =, 1 ) µ α 1) = 0, We start by checking the possibility to raise or lower the first of these weights by α ), using the master formula: ) α ) µ α 1)) 1 1 =,, 1 ) = 1 = q p I leave it as an exercise to show that, in fact 4, p = 0. This gives q = 1, and so a forth weight: ) µ α 1) α ) 1 = 0, Obviously, this weight cannot be raised by α 1), since this would contradict µ being highest weight. So p = 0. It can, however, be lowered by α 1) : α 1) µ α 1) α )) = 1 = q 4 The proof follows from a two-step logic: A. Assume that there is a state with the weight µ α 1) + α ). B. Weyl reflection around the vector perpendicular to α 1) then implies that there exists a weight which is larger than µ. This contradicts with µ being highest weight. 14

15 So we have a fifth weight µ α 1) α ) = 1, 1 ) We proceed by checking the possibility to raise or lower the weight µ α 1) by α ), using the master formula: ) α ) µ α 1)) 1 =, 0, ) = 1 = q p You can show that the consistent solution is q =, p = 0. The weight corresponding to lowering once by α ) is just the fifth weight we ve found above. The weight corresponding to lowering twice by α ) is the sixth and last of this irrep: ) µ α 1) α ) 1 = 1, It is indeed the last weight, since it cannot be neither lowered nor raised by α 1) : α 1) µ α 1) α )) = 0 = q p and you can show that the correct solution is q = p = 0. The weight diagram we have found is depicted in figure..7 Product decomposition The product of irreps is usually not an irrep. However, there s a very painless way to decompose a product of irreps into a direct sum of irreps. It consists of writing down all of the combinations of weight sums, arranging these new summed) weights in degenerate bunches, and then removing the revealed irreps one by one until there s nothing left. I will demonstrate the method with SU). By now, we are familiar with five inequivalent irreps of SU): These are the 8, 6,, and the trivial rep 1. Let us find how the product is decomposed into a direct sum. Putting together what we ve learned in sections.4-.6, we construct new weights by adding all the weights of the to each weight of the, keeping track of the multiplicity. I have collected the result in table. There are nine states, as the sum over multiplicities confirms. Substituting for the vectors µ k), α k) enables us to plot the weight diagram. The three-fold degenerate state has weight zero. Subtracting the diagram of the 8, we are left with a simple irrep consisting of one zero weight. This is of course just the trivial representation. We conclude that: = ) Its convenient to do what we have just done pictorially, with weight diagrams. To re-derive the result above, take the following steps: 1. Draw the weight diagram of the, the triangle. 15

16 Table : States of weight multiplicity µ 1) + µ ) 1 µ 1) + µ ) α 1) 1 µ 1) + µ ) α 1) α ) µ 1) + µ ) α ) 1 µ 1) + µ ) α 1) α ) 1 µ 1) + µ ) α 1) α ) 1 µ 1) + µ ) α 1) α ) 1. For each vertex of the triangle, draw the reversed triangle) weight diagram of the, centered around it. The result is the diagram of 4). I tried to illustrate this process in figure 4. 16

17 Figure 1: SU) fundamental Figure : SU) adjoint 17

18 Figure : SU) 6 Figure 4: =

19 References [1] H. Georgi, Lie Algebras in Particle Physics [] B. C. Hall, Lie Groups, Lie Algebras, and representations 19