Show, for infinitesimal variations of nonabelian Yang Mills gauge fields:

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1 Problem. Palatini Identity Show, for infinitesimal variations of nonabelian Yang Mills gauge fields: δf i µν = D µ δa i ν D ν δa i µ..) Begin by considering the following form of the field strength tensor for a nonabelian Yang Mills field: F i µν = µ A i ν ν A i µ + gf ijk A j µa k ν,.) take the infinitesimal variation of this to first order), noting the use of the Leibniz rule to the last term: δf i µν = µ δa i ν ν δa i µ + gf ijk δa j µ A k ν + gf ijk A j µ δa k ν..3) Next, use the following definition for the covariant derivative: where, the standard convention utilized in physics is applied: D µ A i ν = µ A i ν igt ijk adj Aj µa k ν,.4) T ijk adj and, the using antisymmetry of the structure constants: = if ijk adj,.5) f ijk = f ikj gf ijk δa j µ A k ν = gf ijk A j ν δa k µ,.6) Eq..3) may be rearranged as: δf i µν = µ A i ν it ijk adj Aj µ δa k ν ν A i µ it ijk adj Aj ν δa k µ = D µ δa i ν D ν δa i µ..7). NonAbelian Bianchi Identity Show, for a nonabelian Yang Mills field strength tensor: where, F µν = εµναβ F αβ. First, take a generic result about antisymmetric tensors: D µ F µν =,.8) T µνρ = T µνρ T µνρ = 6 T µνρ T µρν + T ρµν T ρνµ + T νρµ T νµρ ) = 3 T µνρ + T ρµν + T νρµ )..9) Here, since ε µναβ = ε µναβ and T µν S µν = T µν S µν, D µ F µν = D µ F νρ = D µ F νρ + D ν F ρµ + D ρ F µν =..) This already looks much like the cyclic combination which appears in the Jacobi identity. Therefore consider: D µ, D ν, D ρ,.) These covariant derivatives are defined by their action on fields for instance, a general field φ), however:

2 µ, D ν, D ρ φ = µ Dν, D ρ φ D ν, D ρ µ φ = µ Dν, D ρ φ + D ν, D ρ µ φ Dν, D ρ µ φ = µ Dν, D ρ φ..) Here, means take the commutator with respect to the Lie algebra, and O means apply the differential operator O. The second equality is due to the Liebniz rule, and the full statement means that µ, D ν, D ρ φ is independent of derivatives of φ, or acts only through multiplication. Similarly: Putting Eqs..) and.3) together, find: iga i µt i, D ν, D ρ = ig A i µt i, igf j νρt j = g A i µf j νρt i, T j = ig A i µf j νρf ijk T k = g A i µt ijk adj F j νρt k..3) D µ, D ν, D ρ = µ ig A µ, D ν, D ρ.4) = µ, D ν, D ρ + iga i µt i, D ν, D ρ.5) = ig µ δ ik + igt ijk adj Aj µ)fνρt k i.6) = igd µ F νρ..7) Hence each of the left hand side terms in Eq..) is proportional to:.3 F F D µ, D ν, D ρ + D ρ, D µ, D ν + D ν, D ρ, D µ =..8) Show the infinitesimal veriation of F µν F µν is a total derivative. Therefore such terms do not contribute to the classical field equations. δ F µν εµναβ F αβ ) = δ F µν εµναβ F αβ ) + F µν δ εµναβ F αβ ),.9) by the Leibniz rule, δ ε = and ε µναβ = ε αβµν, therefore: δ F µν F µν ) = δ F µν εµναβ F αβ ) + δ F αβ εαβµν F µν ) = δ F µν F µν..) Using the Paletini identity: Now applying Leibniz again: δ F µν F µν ) = D µ δ A ν D ν δ A µ ) F µν = 4D µ δ A ν F µν..) δ F µν F µν = 4D µ δ A ν F µν 4δ A ν Dµ F µν..) The first term of the above is a gauge invariant. That is the term as it appears in the Lagrangian: { tr δ µν F µν F } d 4 x = δ D µ δ A i ν F µν i d 4 x,.3) is invariant transforms trivially) under the action of the gauge group. Hence: δ F µν F µν = 4D µ δ A ν F µν 4δ A ν Dµ F µν = 4 µ δ A ν F µν..4) Therefore this term is a total derivative and does not contribute to the classical equations of motion.

3 Problem. Converting Complex Representations to Real Representations Defining real and imaginary parts of a scalar may be done through the use of the non-holomorphic) R and I functions: Defining φ, so that. Rz = z + z Iz = z z i,. that is: φ = φ + iφ,.) φ = Rφ, φ = Iφ..) Then the terms of the action of the Lie algebra generators on φ is: it k φ = irt k + iit k )φ + iφ ).3) = irt k φ iit k φ IT k φ RT k φ,.4) Defining a new real) scalar φ) that transforms under a real representation of the same algebra: expiθk T k )φ = it k φ,.5) θ k expiθ k T k ) θ φ = i T k φ..6) k Note, T must be imaginary cancel the i) and antisymmetric i.e. Hermitian and Imaginary to cancell the i): T ) k T = T k.7) for imaginary T k. T k is a matrix, With this notation: ) T T k k k = UL T UR T LL k T LR k,.8) ) φ φ =..9) φ i T k i k T φ = UL φ + i T UR k φ ) ) IT i T LL k φ + i T LR k φ k φ RT k φ i ) irt k φ iit k,.) φ therefore taking Eqs..4) and.) together implies: The above may be written: i T k UL = IT k,.) i T k UR = RT k,.) i T k LL = RT k,.3) i T k LR = IT k..4) ) T k IT = i k RT k RT k..5) IT k 3

4 . Symmetry of Mass Matrices The inner product of the vacuum after the action of Lie algebra generators is: S ij = T i φ, T j φ),.6) Since these generators are Hermitian with respect to the, ) inner product, and the operator is symmetric: = T j T i φ, φ) = φ, T j T i φ).7) = T i T j φ, φ) = T j φ, T i φ) = S ji..8) Furthermore, considering any vector annihilated by the mass matrix: M ABT i BC φ C =. ϕ i B = T i BC φ C M ABϕ i B = = ϕ i. That is if a matrix annihilates a vector, that vector is an eigenvector. In this case, a fluctuation about the vacuum with zero mass, i.e., a Goldstone boson. Additionally, S ij is real, since it is the product of two purely imaginary vectors. Furthermore, this may be diagonalized using an orthogonal transformation: S ij = O T SO) ij,.9) where O are orthogonal matrices. The nonzero diagonal components of S correspond to independent non vanishing T i φ. Therefore, all non vanishing T i φ correspond to different eigenvectors of equation: with eigenvalue. M ABT i BC φ C =,.) 3 Problem 3 3. The Real Representation Considering how the SU) U) term appears int the Lagrangian: L eff = i g W iµσ i i ) g B µ ϕ C, 3.) the desired matrices transform as: Expanding out the effective term: it C ϕ C it 4R φ 4R. 3.) ) ) i g B µ + g Wµ 3 g Wµ + iwµ φ4r3 + iφ 4R4 g Wµ ig Wµ g B µ + g Wµ 3 φ 4R + iφ 4R. 3.3) Then taking the above expression field by field: Therefore: ) g Wµ iφ4r φ 4R iφ 4R3 φ 4R4 g Wµ φ 4R4 φ 4R3 φ 4R φ 4R = ig W µ T φ 4R φ 4R φ 4R3 φ 4R4. 3.4) 4

5 Similar calculations yield each T i 4R : T4R = i. 3.5) T4R = i T 4R 3 = i T 4R 4 = i. 3.6) 3. Mass Squared Matrix Next, calculating M ) ij is straightforward, for example: M ) 34 = g g 4 = g g 4 v ) v 3.7) ) v v = 4 g g v. 3.8) This calculation may be sped through caching, either the general form of S ij as described in the problem sheet, or the general form of T i φ 4R which are: T φ4r = i v, T φ4r = i v, T 3 φ4r = i v, T 3 φ4r = i v. 3.9) Taking the possible products of these find the mass squared matrix is: M ) gv = gv 4 gv g g v. 3.) g g v gv 3.3 Custodial Symmetry In general, however, simply requiring that there be only one uneaten mode means that some linear combination of generators annihilates the vacuum gives a different mass squared matrix. In the current representation with the current Lagrangian: g i g j φt i T j φ = 3.) g φ ij g g ac + bd) g φ g g bc ad) 4 g φ g g c + d a b ) g g ac + bd) g g bc ad) g g c + d a b ) g φ 3.) Here φ = a b c d ). Breaking custodial symmetry with this Lagrangian is impossible, a more complicated scalar sector specifically a higher representation of SU) U)) is required. 5

6 However, insisting that an arbitrary, for simplicity sake T 3 + T 4 ) linear combination of SU) generators and U) generators remains unbroken yields: T 3 φ, T 3 + T 4 ) φ) = T 3 φ, T 3 φ) = T 3 φ, T 4 φ) 3.3) multiplying by the relevant coupling constants: g S 33 = g g g g S 34 M ) 33 = g g M ) ) Furthermore this real symmetric matrix must have three non-zero eigenvalues, therefore, diagonalize the further terms to get: a b M ) = a c 4 d g g d. 3.5) b c g g d g g d Given that no linear combination of the remaining generators may be unbroken, that is: for i =, and for all k, therefore: T i φ, T i + kt 4 ) φ) 3.6) S ii + ks i4 S i4 =, 3.7) for i =,. Therefore choosing the correct normalizations of a, and d, get: gv M ) = gv 4 gu g g u. 3.8) g g u gu With broken custodial symmetry. 4 Problem 4 Setup are generators of SO3) algebra in the adjoint representation: T = i, T = i, T ijk = iɛ ijk 4.) T = i. 4.) 4. VEV a Let then φ =, 4.3) T k φ = 4.4) for all k. This means that no generator is broken and the stability group is H = G = SO3). Expanding φ around φ: where φ, φ, φ 3 are massive fields. φ = φ + φ + φ + φ 3, 4.5) 6

7 4. VEV b φ = v, T φ =, 4.6) so T is unbroken. Let a generic linear combination of T, T 3 act on φ: at + bt 3 ) φ = ia + ib v =. 4.7) v For this last equality to be true, we need a = b =, therefore no linear combination of T, T 3 annihilates φ, and T, T 3 are both broken generators. Since there is only one unbroken generator, the stability group is H = SO). If we expand the field φ around φ: φ = v + φ + φ + φ ) φ is massive since it is perpendicular to the SO3) orbit of φ = v, an S with radius v. φ and φ 3 are massless since they go in the two tangent directions to the orbit. 4.3 VEV c φ = v. 4.9) Let a linear combination of the generators act on φ: at + bt + ct 3 ) φ = i v c b c a 4.) b a = i v c c =. 4.) a b This requires c = and a = b. Therefore, the generator T +T is unbroken and the generators T T and T 3 are broken. The stability group is H = SO): d rotations in the plane perpendicular to φ = i v. Let φ = φ + φ + φ + φ 3. 4.) φ is massive, φ and φ 3 are massless. 4.4 VEV d Let φ = v. 4.3) at + bt + ct 3 ) φ = i v c b c a 4.4) b a = i v c c =. 4.5) a b 7

8 This is satisfied for c =, a = b. Hence the generator T T is unbroken and the generators T + T and T 3 are broken. The stability group is H = SO), the d rotations of the plane perpendicular to φ. Expanding the field φ: φ = φ + φ + φ + φ ) φ is massive, φ and φ 3 are massless. 4.5 Generic VEV In the general case we have two options:. φ =, discussed in part a, three unbroken generators.. φ = v n, where n is a 3 vector of length, n n = n, n + n + n 3 =. 4.7) n 3 The linear combination of generators a i T i that generates the SO) group of d rotations of the plane perpendicular to n is unbroken. Two other linearly independent combinations b i T i and c i T i will be broken. Let a denote the vector then a = a a a 3 This is true when a and n are om the same direction. 5 problem 5 5. Hermitian Diagonalization M M is Hermitian since:, 4.8) a i T ijk φk = 4.9) ia i ɛ ijk φk = 4.) ia i ɛ ijk vn k = 4.) iv a n) i =. 4.) M M) = m M ) = M M. 5.) Therefore prove by using the inner product) it must have real eigenvalues: ψ, M Mψ) = λψ, ψ) = M Mψ, ψ) = λ ψ, ψ) λ = λ, 5.) its eigenvectors with different eigenvalue) must be orthogonal: ψ, M Mφ) = M Mψ, φ) = λ ψ ψ, φ) = λ φ ψ, φ) ψ, φ) = when λ ψ λ φ, 5.3) therefore its eigenvectors may be chosen to) form an orthonormal basis: Σψ ψ = ˆ, 5.4) 8

9 therefore M M is defined by its action on its eigenvectors: M Mφ = ˆM Mˆφ = ΣΣψ ψm Mψ ψφ = Σψ λ ψ ψφ = UD U φ, 5.5) where D is the diagonal with D ) i i = λ ψ, the eigenvalue corresponding to the ith column of the matrix U ) i j = ψ j, which is composed of eigenvectors U is unitary since the basis was orthornormal). Alternately: D = U M MU. 5.6) To show that the entries of D are real guaranteed by Hermiticity) and positive consider eigenvector ψ with eigenvalue λ: Multiplying with ψ from the left: Now define vector φ = MUψ. Hence: D ψ = λψ. 5.7) ψ D ψ = λψ ψ. 5.8) φ φ = λψ ψ. Since φ φ and ψ ψ are manifestly real and positive, λ must be real and positive as well. 5. General Diagonalization We define H = UDU, hence: U is unitary, U = U, so H = UDU ) = U ) D U. 5.9) Furthermore D is Hermitian and therefore: H = UD U. 5.) is Hermitian. Ergo: H = U ) D U = UDU = H, 5.) Finally: Ũ Ũ = H ) M MH = UD U UD U UD U = ˆ. 5.) V MU = U Ũ MU = U Ũ MU = U HM MU = U UDU U = D. 5.3) Therefore M is diagonalizable. 6 Problem 6 Given three complex numbers A, B, C satisfying A + B + C =, show that the area of the triangle with vertices at the two dimensoinal points, A, A + B is given by Let: IAB ) = IAC ) = IBC ). 6.) A = expiα)a, 6.) B = expiβ)b, 6.3) C = expiγ)c, 6.4) where α, β, γ, π and a, b, c R +. 9

10 a) b) c) We can divide the triangle with base C in two halves, then compute height h: sinθ) = h b h = b sinθ). 6.5) From the figure we see And we know that θ = β γ π) = β γ + π. 6.6) Area = base) height) = c h = c b sinβ γ + π). 6.7) Therefore: IBC ) = b c expiβ) exp iγ) = b c expiβ γ)) 6.8) = b c sinβ γ) = b c sinβ γ + π). 6.9) If we arrange A, B, C as in the figure, then θ is between and π, so sinθ) = sinθ). Hence: IBC ) = b c sinβ γ + π) = Area. 6.) Similarly we can choose A or B to be the base and obtain: Area = IAB ) = IAC ). 6.)