Symmetries, Fields and Particles 2013 Solutions

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1 Symmetries, Fields and Particles 013 Solutions Yichen Shi Easter (a) Define the groups SU() and SO(3), and find their Lie algebras. Show that these Lie algebras, including their bracket structure, are isomorphic. SU() = { by complex matrices U with U U = I, det(u) = 1}. Consider a curve U(t) := I + tz + O(t ) SU() with U(0) = I. U U = I gives Also, in matrix form, det U = 1 gives Hence d dt [U U] t=0 = [ U U + U U] t=0 = U + U = Z + Z = 0. ( 1 + tz11 tz U(t) = 1 tz tz ). 1 + t(z 11 + Z ) + O(t ) = 1 + ttr(z) + O(t ) = 1 Tr(Z) = 0. L(SU()) = {vector space of 3 by 3 complex antihermitian traceless matrices}. Let L(SU()) have bases T a = 1 iσ a where 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 ] = 1 4 (σ aσ b σ b σ a ) = 1 iɛ abcσ c = ɛ abc T c Now, SO(3) = { by real matrices O with O T O = I, det(o) = 1}. Now consider a curve O(t) = I + tx + O(t ) SO(3) with O(0) = I. O T O = I gives d dt [OT O] t=0 = [ȮT O + O T Ȯ] t=0 = ȮT + Ȯ = 0. det O = 1 gives no extra constraint since all elements in SO(3) near the identity have det= 1. Hence L(SO(3)) = {vector space of 3x3 real antisymmetric matrices} SU() is the universal covering group of SO(3) hence their Lie algebras are isomorphic. The isomorphism is induced by the universal covering map. 1

2 For an explicit computation, let L(SO(3)) have bases T 1 = , 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 [ T a, T b ] = ɛ abc Tc Since their bracket structures are the same, we have L(SU()) L(SO(3)). (b) Define the group SU(3). Identify an SU() subgroup and an SO(3) subgroup of SU(3). SU(3) = {3 by 3 complex matrices U with U U = I, det(u) = 1}. K is an SU() subgroup of SU(3) if K = H is an SO(3) subgroup of SU(3) if {( B ) } : B SU(). H = {A SU(3) : A ij R}. (c) Consider the standard action of SU(3) on C 3, and let z = z 1 z z 3 denote a general vector in C 3. For the vector 1 0 0, determine the orbit and the isotropy group. Note that the orbit of a point m M is the set G(m) = {g(m) : 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 : h(m) = m}. Here M = C 3, m = v 0 := (1, 0, 0). The orbit of the vector v 0 under the action of SU(3), i.e., is then given by [SU(3)](v 0 ) = {A(v 0 ) : A SU(3)}, Orb = {v C 3 : v is the first column of some A SU(3)} = {v C 3 : v = 1}.

3 Proof : Let v Orb. The statement that the matrix is an element in U(n) implies that the columns are orthonormal. Hence v = 1, and Orb { v C 3 : v = 1 }. Conversely, let v C 3 be of norm 1. Then, we can find two vectors u and w of norm 1 which are orthogonal to v. Then, the matrix with columns v, u, and w will be an element of U(3) whose first column is v. We can then scale u or w by e iθ with an appropriate θ to ensure det = 1. Hence v Orb. ( ) 1 0 The isotropy group is the set of matrices of the form where U SU(). 0 U (d) The equation z 1 + z + z 3 = 1 defines a subset M of C 3. Show that M is not an orbit of SU(3). What can you say about the action of the real SO(3) subgroup on M? Set A := i i SU(3). Then z := Av 0 = (i, 0, 0) is an element of the orbit of v 0 under the action of SU(3). However, it does not satisfy the condition z 1 + z + z 3 = 1. Hence M is not an orbit of SU(3). M is invariant under the action of the real SO(3) 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 we are in -dimensions and F xy is antisymmetric, it automatically only has one independent component. For a general gauge group G := U(n), introduce a fundamental scalar group Φ(x) = (Φ 1 (x),..., Φ n (x)). We require the theory to be invariant under the transformation Introduce covariant derivative D µ such that Φ(x) g(x)φ(x). D µ Φ(x) = ( µ + A µ )Φ(x) (where we postulate that the gauge potential A µ transforms as A µ ga µ g 1 ( µ g)g 1 ). 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 ]. 3

4 (b) Let g(x, y) be a G-valued function, and suppose A x = α( x g)g 1, A y = α( y g)g 1 where 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 g(x, y). Note that a G-valued function g has i g in the tangent space to G at g, hence ( i g)g 1, g 1 i g L(G). F xy = x A y y A x + [A x, A y ] = α( x y g)g 1 α( y g)g 1 ( x g)g 1 α( y x g)g 1 + α( x g)g 1 ( y g)g 1 + α [( x g)g 1, ( y g)g 1 ] = (α + α )[( x g)g 1, ( y g)g 1 ]. Hence if α = 0 or 1, then F xy = 0. This makes sense since A µ is gauge equivalent to 0 when α = 0 or -1, according to A µ ga µ g 1 ( µ g)g 1. (c) Suppose now that G = SU() and that g(x, y) = exp ( 1 ) i(xσ 1 + yσ ) where 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, g(0, 0) = I. Now, using polar coordinates x = r cos θ, y = r sin θ, g(x, y) can be written using Taylor series expansion, or better, the formula exp(iv σ) = cos v I + i v σ v sin v, where we have v = r (cos θ, sin θ) and v = r. Hence, ( ) cos r g(r cos θ, r sin θ) = i(cos θ i sin θ) sin r i(cos θ + i sin θ) sin r cos r := cos r I + A sin r. Hence g = ±I on an infinite number of circles in the plane, as long as the circles have r = nπ, n 0, n Z. To find the value of F xy at the origin, notice that ( 1 g 1 x=y=0 = exp i(xσ 1 + yσ )) = 1 x=y=0 and x g x=y=0 = 1 iσ 1, y g x=y=0 = 1 iσ. From the previously found expression F xy = (α + α )[( x g)g 1, ( y g)g 1 ], we see that F xy x=y=0 = 1 4 (α + α )[σ 1, σ ] = 1 i(α + α )σ 3. 4

5 Now, using polar coordinates for the circles case, det g = cos r + sin r (cos θ + sin θ) = 1. g 1 r=nπ = cos r I A sin r r=nπ = ( 1) n I And using x = cos θ r 1 r sin θ θ, y = sin θ r + 1 r cos θ θ, we have for n 0, x g r=nπ = 1 ( cos θ sin r I + A cos r ) 1 ( ) r sin θ 0 (i sin θ cos θ) sin r (i sin θ + cos θ) sin r 0 = 1 cos θ( 1)n A r=nπ and similarly, y g r=nπ = 1 sin θ( 1)n A. From the previously found expression F xy = (α + α )[( x g)g 1, ( y g)g 1 ], we see that [( x g)g 1, ( y g)g 1 ] r=nπ = 1 [cos θa, sin θa] = 0. 4 Hence F xy vanishes at all points on the circles. 3. (a) What are meant by the weights of a representation of SU(3). Calculate, from the definitions, the weights of 3, the fundamental representation of SU(3), and 3, its complex conjugate. Calculating with weights, show that 3 3 = 8 1. and describe briefly what the representations 8 and 1 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 L(SU(3)). First note that a Cartan subalgebra of L(G) is a maximal abelian subalgebra of L(G). In the case of L(SU(3)), the Cartan subalgebra is spanned by h = (h 1, h ), where h 1 := 0 1 0, h := , which are two of the eight Gell-Mann matrices that form a basis for L(SU(3)). 5

6 Now, suppose d is a representation of L(G) acting on V. The weights λ i of d are defined to be the eigenvalues of some set of maximally commuting generators d(h i ). Hence h above allows us to find weights λ i of L(SU(3)) through d( h)v = λv. Note that the weight space V λ = {v V : d( h)v = λv}. Now, the fundamental representation has d( h) = h, so we can simply read off weights from the diagonal entries of h. Hence the weights of 3 are ( 1, 1 ), ( 1 3, 1 ), and (0, ) and those of 3 are ( 1, 1 ), 3 ( 1, 1 ), and (0, 3 1 ). 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 L(SU(3)) is antihermitian. Note that tensor products add. By adding the two diagrams, we see 8 positions where weights are present. At the origin, there are three weights, out of which one belongs to the trivial representation. The results agree with direct computation of 8, using (ad h)t i = [ h, T i ] = (root i)t i, where T i are the bases of L(SU(3)). 1 obviously has weight 0. Hence 3 3 = is the trivial representation, i.e., d(x) = 0, and all trivial representations (and one-dimensional representations) are irreducible. 8 is the adjoint representation, i.e., d(x) = ad X where ad X (Y ) = [X, Y ]. Since L(SU(n)) is simple, its adjoint representation must be irreducible. (Simple means that there are no invariant ideals of the algebra, i.e. subspaces closed under the action of the whole algebra by commutation, which is just the adjoint action.) (b) In the context of the quark model with approximate SU(3) 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 SU(3) are the quarks (u, d, s), e.g., (1, 0, 0) is identified with a u quark state. Hadrons (baryons and mesons) are multi-quark states in the SU(3) flavour multiplets (represented as tensor products). Note that particles in irreducible representations have approximately the same mass, and related strong interactions. 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 1 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 1 corresponds to η. The three meson states at the centre are η = 1 (uū + d d + s s), 6 η = 1 6 (uū + d d s s), and π 0 = 1 (uū d d), which all have 0 charge. 6

7 4. Give an account of how the irreducible representations of SO(4), and SO(1,3), the connected component of the Lorentz group, may be constructed using irreducible representations of SU(). Spin(4) = SU() SU(). Now, note that Spin(n) is the universal covering group of SO(n) hence they have the same Lie algebra. Thus, L(SO(4)) L(Spin(4)) L(SU()) L(SU()) It follows that the irreducible representations of L(SO(4)) are in one-to-one correspondence with pairs of irreducible representations of L(SU()). The Lie algebra of SO(1, 3) is L(SL(, C)) (regarded as a real Lie algebra) since SL(, C) is the universal covering group of SO(1, 3). The complexification of L(SL(, C)) is L(SL(, C)) L(SL(, C)), both regarded of course as complex Lie algebras. However, the complexifcation of L(SU()) is also L(SL(, C)). That is, L(SO(1, 3)) R C [L(SU()) R C] [L(SU()) R C]. It follows that the irreducible representations of L(SO(1, 3)) are in one-to-one correspondence with pairs of irreducible representations of L(SU()). (Weinberg, Quantum Field theory vol.1 section 5.6 has examples.) Appendix The proof of SU() being the universal covering group of SO(3): Let v R 3. Define V := v 1 σ 1 + v σ + v 3 σ 3. Check that det(v ) = v. Furthermore, the map v V is linear. V is traceless. So this is a linear map from R 3 to L(SL(, C)). Furthermore, it is an isomorhphism of real vector spaces. Let U SU() and let A L(SL(, C)). Define U.A := UAU 1. This is a linear map from L(SL(, C)) to itself that preserves the determinant. And hence the corresponding map on R 3 (obtained by the isomorphism mentioned before) preserves norm. Thus, A U.A corresponds to a norm-preserving linear map of R 3. Call this map c(u) (c for covering). Thus, c(u) O(3). However, c is continuous and SU() is connected, and so c(u) SO(3). c is also a group homomorphism with kernel ±1. Hence, c is a covering map (see Lemma of Thus, SU() is a covering group of SO(3). It is universal because SU() is diffeomorphism to S 3 which is simplyconnected. 7