Symmetries, Fields and Particles 2013 Solutions

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1 Symmetries, Fields and Particles 03 Solutions Yichen Shi July 9, 04. a Define the groups SU and SO3, and find their Lie algebras. Show that these Lie algebras, including their bracket structure, are isomorphic. SU = {U U : detu = } U = {U GL, C : U U = }. Consider a curve Ut SU with U0 = I. U U = I gives Let Ut := I + tz +, i.e., det U = gives Hence d dt [U U] t=0 = [ U U + U U] t=0 = U + U = 0. + tz tz Ut = tz + tz tz + Z + + = + ttrz + = TrZ = 0 LSU = {vector space of 3x3 complex antihermitian traceless matrices}. Let LSU have bases T a = iσ a σ a are the Pauli matrices, which have the property σ a σ b = δ ab I + iɛ abc σ c. [T a, T b ] = 4 σ aσ b σ b σ a = iɛ abcσ c = ɛ abc T c Now, SO3 = {O O3 : deto = } O3 = {O GL3, R : O T O = }. Now consider a curve Ot SO3 with O0 = I. O T O = I gives d dt [OT O] t=0 = [ȮT O + O T Ȯ] t=0 = ȮT + Ȯ = 0.

2 det O = gives no constraint since all elements in SO3 near the identity have determinant. Hence LSO3 = {vector space of 3x3 real antisymmetric matrices} SU is the universal covering group of SO3 hence their Lie algebras are isomorphic. The isomorphism is induced by the universal covering map. For an explicit computation, let LSO3 have bases T = 0 0, T = T a bc = ɛ abc., T3 = [ T a, T b ] cd = T a ec T b de T b ec T a de = ɛ ace ɛ bed ɛ bce ɛ aed = δ ad δ cb δ ab δ cd δ bd δ ca + δ ba δ cd = ɛ eab ɛ edc = ɛ abe ɛ ecd = ɛ abe T e cd Hence LSU LSO3. [ T a, T b ] = ɛ abc Tc b Define the group SU3. Identify an SU subgroup and an SO3 subgroup of SU3. K is an SU subgroup of SU3 if SU3 = {U U3 : detu = } U3 = {U GL3, C : U U = I}. K = H is an SO3 subgroup of SU3 if { X 0 0 } : X SU. H = {A SU3 : A ij R}. c Consider the standard action of SU3 on C 3, and let denote a general vector in C 3. For the vector z = 0 0 z z z 3, determine the orbit and the isotropy group. Note that the orbit of a point m M is the set Gm = {gm : g G}. If G acts on M transitively, then the isotropy subgroup at m is the subset H of G that leaves m fixed, i.e., H = {h G : hm = m}.

3 The orbit O of the vector v 0 :=, 0, 0 under the action of SU3, i.e., is given by We show below that O := [SU3] v 0 = {Av 0 : A SU3}, O = {v C 3 : v is the first column of some A SU3}. O = { v C 3 : v = }. Note that the statement that the matrix is an element in Un or On is equivalent to the statement that the columns or equivalently, rows, are orthonormal. Now, let v O. Then, by the note above, v =, so O { v C 3 : v = }. Conversely, let v C 3 be of norm. Then, we can find two vectors u and w of norm which are orthogonal to v pick a basis of C 3 containing v and apply the Gram-Schmidt process of orthonormalisation. Then, the matrix with columns v, u, and w will be an element of U3 also by the note above whose first column is v. The determinant of this matrix will not necessarily be, however. To remedy this, scale w by e i θ : the matrix obtained thereby will still have orthonormal columns and hence be an element of U3 still with first column v; however, now the determinant will be a factor of e i θ different from the previous. Choosing θ appropriately, we obtain a matrix in U3 whose first column is v and has determinant. This shows that v O, and hence O = { v C 3 : v = }. 0 The isotropy group is the set of matrices of the form X SU. 0 X d The equation z + z + z 3 = defines a subset M of C 3. Show that M is not an orbit of SU3. What can you say about the action of the real SO3 subgroup on M? Set A := i i Then z := Av 0 = i, 0, 0 is an element of the orbit of v 0 under the action of SU3. However, it does not satisfy the condition z + z + z 3 =. Hence M is not an orbit of SU3. M is invariant under the action of the real SO3 subgroup, and is thus a disjoint union of orbits.. a The gauge potential of a gauge theory in the x, y plane, with gauge group G, has components A x, A y. Show that the field tensor has only one independent component, F xy, and give the formula for this. Because F xy is antisymmetric, it automatically only has one independent component. For a general gauge group G := Un, introduce a fundamental scalar group Φx = Φ x,..., Φ n x. We require the theory to be invariant under the transformation Introduce covariant derivative D µ such that Φx gxφx. D µ Φx = µ + A µ Φx 3

4 we postulate that the gauge potential A µ transforms as A µ g A µ g µ gg. Hence if A µ = A x, A y only, then [D µ, D ν ]Φ = µ + A µ ν + A ν Φ + A µ A ν µ ν = µ ν Φ + µ A ν Φ + A µ ν Φ + A µ A ν µ ν = µ A ν Φ + A ν µ Φ + A µ ν Φ + A µ A ν µ ν = µ A ν ν A µ + [A µ, A ν ]Φ := F µν Φ. F xy = x A y y A x + [A x, A y ]. b Let gx, y be a G-valued function, and suppose A x = α x gg, A y = α y gg α is a real constant. Evaluate F xy in terms of g and its derivatives. Explain why, for certain values of α, F xy vanishes for all gx, y. Note that a G-valued function g has i g in the tangent space to G at g, hence i gg, g i g LG. F xy = x A y y A x + [A x, A y ] = α x y gg α y gg x gg α y x gg + α x gg y gg + α [ x gg, y gg ] = α[ x gg, y gg ] + α [ x gg, y gg ] Hence if α = 0 or, then F xy = 0. This makes sense since A µ is gauge equivalent to 0 when α =0 or -, according to A µ g A µ g µ gg. c Suppose now that G = SU and that gx, y = exp ixσ + yσ σ a are the Pauli matrices. Show that g = ±I at the origin and on an infinite number of circles in the plane. Calculate F xy at the origin, and show that F xy = 0 at all points on the circles. Obviously, g0, 0 = I. Now, gr cos θ, r sin θ = exp ircos θσ + rsin θσ = exp ir 0 cos θ i sin θ cos θ + i sin θ 0 := exp A = I + A + A + 3! A3 + 4! A4 + = I r 4 + 4! r A 3! = I r + 4! r A r r 4 + 5! r 4 + 3! r 3 + 5! r 5 + = cos r + r A sin r cos r = i sin r cos θ i sin θ i sin r cos θ + i sin θ cos r 4

5 since A = r 4. Hence for g = ±I on an infinite number of circles in the plane, we require r = nπ, n Z. Note that we could have used the formula Now, we have previous found that expiv σ = cos v I + i v σ v sin v. F xy = α + α [ x gg, y gg ] x g = 3. a What are meant by the weights of a representation of SU3. Calculate, from the definitions, the weights of 3, the fundamental representation of SU3, and 3, its complex conjugate. Calculating with weights, show that 3 3 = 8. and describe briefly what the representations 8 and are, and why they are irreducible. Firstly, physicists don t distinguish between representations of Lie groups and those of Lie algebras. Rigorously, this and the next questions are about the representation of LSU3. In general, the weights of a representation are defined to be the eigenvalues of some set of mutually commuting generators dh i. The dimension of the representation is the number of weights. Define 0 0 h := 0 0, h := which are two of the eight Gell-Mann matrices that form a basis for LSU3. Note that all Lie algebras of are vector spaces over the real numbers, with the extra structure of brackets. LSU3 consists of traceless antihermitian matrices. So h and h are not antihermitian, so they aren t in LSU3. However, you can write them as complex scalar multiple of something in the Lie algebra, e.g., h = ix, h = ix, X, X are in the Lie algebra. This shows that h, h are in the complexification. Indeed, complexification is anything that can be obtained by taking a complex scalar multiple of something in the original lie algebra. Obviously h, h the complexification of LSU3. Furthermore, [h, h ] = 0 and h, h are linearlyindependent, hence C := span{h, h } is a two-dimensional commuting subalgebra. Note that two elements h and h span LSU3, so it has rank. Hence C is a Cartan subalgebra of LSU3. Hence the fundamental weights of 3 are,, 3,, and 0, 3 3 and those of 3 are,, 3,, and 0, 3. 3 The weights of 3 can be shown on weight diagrams omitted and the weights of 3 are simply those of 3 reflected in the origin since LSU3 is antihermitian. Note that tensor products add. By adding the two diagrams, we see 8 positions weights are present. At the origin, there are three weights, out of which one belongs to the trivial representation. Hence 3 3 = 8. is the trivial representation, i.e., dx = 0, and all trivial representations and one-dimensional representations are irreducible. 8 is the adjoint representation, i.e., dx = ad X ad X Y = [X, Y ], and if the Lie algebra is simple, then its adjoint representation must be irreducible. 5

6 Note that a Lie algebra L is simple iff whenever there are two other Lie algebra L and L with L L L, one of L and L is trivial. b In the context of the quark model with approximate SU3 flavour symmetry, discuss how the above relation leads to a classification of meson states with spin/parity 0, including the pions. Briefly discuss the quark content and some physical properties of the meson states at the centre of the weight diagram. The particles transforming under the fundamental representation 3 of flavour SU3 are the quarks u, d, s, e.g.,, 0, 0 is identified with a u quark state. Hadrons baryons and mesons are multi-quark states in the SU3 flavour multiplets represented as tensor products. We can correspond the three weights on the weight diagram of 3 with u top right, s bottom, and d top left, and the corresponding weights of 3 with ū, s and d. It leads to the fact that the 8 in the relation 3 3 = 8 corresponds to the weight diagram with weights at the origin representing η 0, π 0 singlet, and the surrounding weights representing K 0, K +, K, K0, π, π + ; and the weight at the origin in corresponds to η. The three meson states at the centre are η = uū + d d + s s, 6 η = 6 uū + d d s s, and π 0 = uū d d, which all have 0 charge. 6