Quantum Field Theory-I Problem Set n. 3 - Solutions UZH and ETH, HS-2015 Prof. G. Isidori Assistants: K. Ferreira, A. Greljo, D. Marzocca, A. Pattori, M. Soni Due: 8-10-2015 http://www.physik.uzh.ch/lectures/qft/index1.html 1 The Lorentz group Generic Lorentz transformations are defined by µ 0µ µ ν µ ν x ! x = Λ ν x ; Λ λΛ ρgµν = gλρ : (1) I. Show that these transformations form a group (see the definition of \group" in the lecture notes). Solution: In order to show that a set forms a group, we need to define (i) group element, (ii) multiplication operation and (iii) prove the group axioms are satisfied. Element of the Lorentz µ µ ν group is Lorentz transformation matrix Λ ν that satisfies Λ λΛ ρgµν = gλρ. Group multiplication µ ν operation (Λ1 ·Λ2) is defined as matrix multiplication, Λ1νΛ2ρ. Let us prove that the group axioms hold: µ µ ν (a) Let Λ1 and Λ2 be two arbitrary group elements. We define Λ3 such that Λ3ρ ≡ Λ1νΛ2ρ. By explicit computation, µ ν µ α ν β Λ3λΛ3ρgµν = Λ1αΛ2λ Λ1βΛ2ρ gµν ; = (Λµ Λν g )Λα Λβ ; 1α 1β µν 2λ 2ρ (2) α β = gαβΛ2λΛ2ρ ; = gλρ ; we find that Λ3 is also a group element. (b) (Λ1 · Λ2) · Λ3 = Λ1 · (Λ2 · Λ3) trivially holds from the properties of matrix multiplication. µ (c) The unit element is the identity matrix, g ν. The identity matrix is a group element, µ ν g λg ρgµν = gλρ ; (3) and it satisfies the following equation µ ν µ ν µ g νΛ ρ = Λ νg ρ = Λ ρ ; (4) where Λ is an arbitrary group element. −1 µ α βµ (d) For an arbitrary group element Λ there is an inverse element (Λ ) ν = gναΛ βg , that satisfies −1 µ ν α βµ ν (Λ ) νΛ ρ = gναΛ βg Λ ρ ; = (Λν Λα g ) gβµ ; ρ β να (5) βµ = gρβg µ = g ρ : 1 In the matrix notation Λ−1 = gΛT g. The inverse is the element of the group (Λ−1)µ (Λ−1)ν g = g Λα gβµ g Λα1 gβ1ν g ; λ ρ µν λα β ρα1 β1 µν = g Λα Λα1 g (gβµgβ1νg ) ; λα β β1 ρα1 µν = g Λα (g Λα1 gββ1 ) ; (6) λα β ρα1 β1 α −1 β = gλα (Λ β(Λ ) ρ) ; = gλρ : 2 0 2 II. Show that det(Λ ) = 1 and (Λ 0) ≥ 1. These results imply that the entire Lorentz group is not continuous: show 0 that the transformations with det(Λ) = 1 and (Λ 0) ≥ 1 form a subgroup; the latter is called continuous or proper Lorentz 0 group. Show that the other three discrete possibilities, obtained by setting det(Λ) = −1 and/or (Λ 0) ≤ 1, do not form separate subgroups. µ ν T Solution: The defining equation, Λ λΛ ρgµν = gλρ, can be written in the matrix form, g = Λ gΛ, µ where g and Λ are 4 × 4 matrices with µν entry gµν and Λ ν, respectively. Using the identities for the determinant, det(AB) = det(A) det(B) and det(AT ) = det(A), we find det(ΛT gΛ) = det(g) ; det(ΛT ) det(g) det(Λ) = det(g) ; (7) det(Λ)2 = 1 : µ ν Explicitly writing the λρ = 00 component of the defining equation, Λ λΛ ρgµν = gλρ, we find 0 2 X i 2 1 = (Λ 0) − (Λ 0) ; (8) i 0 2 which implies that (Λ 0) ≥ 1. 0 Let us show that the continuous Lorentz transformations (det(Λ) = 1 and (Λ 0) ≥ 1) form a subgroup, that is, a subset of the full Lorentz group which is a group itself. (I) If Λ1 and Λ2 are the elements of the continuous Lorentz group, than Λ3 = Λ1·Λ2 has determinant det(Λ3) = det(Λ1Λ2) = det(Λ1) det(Λ2) = 1. Let us also show that Λ3 is orthochronous Lorentz transformation. Let the general form of transformation matrix be a bT Λ = ; (9) c R where a is a real number, b and c are 3 × 1 column matrices, while R is 3 × 3 matrix. From the defining identity ΛT gΛ = g, we find a cT 1 0 a bT a2 − cT c : : : 1 0 = = ; (10) b RT 0 −1 c R :::::: 0 −1 that is, a2 − cT c = 1. Similarly, from (Λ−1)T gΛ−1 = (gΛT g)T ggΛT g = g which implies ΛgΛT = g we find a2 − bT b = 1. Finally, T T T a1 b1 a2 b2 a1a2 + b1 c2 ::: Λ3 = Λ1Λ2 = = : (11) c1 R1 c2 R2 :::::: 2 T In other words, a3 = a1a2 + b1 c2. From the definition of the dot product q q T T T b1 b1 c2 c2 ≥ jb1 c2j ; (12) that is q q 2 2 a3 ≥ a1a2 − a1 − 1 a2 − 1 ; (13) which for a1; a2 ≥ 1, implies a3 ≥ 1. Thus, Λ3 is also the element of the continuous Lorentz group. (II) Associativity trivially holds. (III) The unit element of the full Lorentz group is trivially the unit element of the continuous Lorentz group. (IV)The inverse of the element Λ of the continuous Lorentz group Λ−1 has determinant 1 since −1 −1 −1 0 0 det(Λ Λ) = det(Λ ) det(Λ) = 1 and (Λ ) 0 = Λ 0 ≥ 1. With this we conclude that the proper Lorentz transformations indeed form a group. On the other hand, it follows from the group axioms that the unit element is unique. Therefore, the other three discrete possibilities obtained by space inversion and (or) time reversal transformations, do not form a group since they do not contain the unit element. 2 Representations of the Lorentz group Combing a generic infinitesimal boost with a generic infinitesimal rotation we obtain the most general infinitesimal proper Lorentz transformation. Writing µ µ µ Λ ν ≈ g ν + ! ν ; (14) we have 0 1 0 −β1 −β2 −β3 B −β1 0 −α3 α2 C X ! = B C = (αiXi + βiYi) ; (15) −β2 α3 0 −α1 @ A i −β3 −α2 α1 0 and the six matrices Xi and Yi can be chosen as generators of the proper Lorentz group in this 4 × 4 matrix representation µ (the proof that this is the most general infinitesimal transformation follows from the fact that !σν = gσµ! ν is a generic antisymmetric matrix, as already seen in the exercise set n.2). A generic element of the entire (non-continuous) group can be obtained by combining an element of the proper group with the discrete transformations of parity (P ) and time-reversal (T ). Within the 4 × 4 matrix representation: 0 1 0 0 0 1 0 −1 0 0 0 1 B 0 −1 0 0 C B 0 1 0 0 C P = B C T = B C PT = −1 : (16) @ 0 0 −1 0 A @ 0 0 1 0 A 0 0 0 −1 0 0 0 1 I. Derive the commutation relations of the generators Xi and Yi. Show that the matrices 1 1 A = (X + iY ) ;B = (X − iY ) ; (17) i 2 i i i 2 i i satisfy the following commutation relations: [Ai;Aj ] = "ijkAk ; [Bi;Bj ] = "ijkBk ; [Ai;Bj ] = 0 : (18) Solution: The infinitesimal transformation of the four-vector V µ under the proper Lorentz group is ρ ρ σ δV = ! σV : (19) 3 We define the generator representation µν ρ µρ ν νρ µ (J ) σ = g g σ − g g σ ; (20) such that 1 1 ! (J µν)ρ = ! (gµρgν − gνρgµ ) ; 2 µν σ 2 µν σ σ 1 1 = gρµ! gν + gρν! gµ ; (21) 2 µν σ 2 νµ σ ρ = ! σ : Note that J µν = −J νµ by construction. By inspection, one can show 0i ρ 0ρ i iρ 0 ρ (J ) σ = g g σ − g g σ = −(Yi) σ ; (22) and ij ρ iρ j jρ i ijk ρ (J ) σ = g g σ − g g σ = (Xk) σ : (23) ijk k Here, i; j and k are the spatial indices (1; 2; 3). Using the identity ijm = 2δm, we find Xk = 1 ij 2 ijkJ . Let us derive the algebra of the Lorentz group µν ρσ α µν α ρσ γ ρσ α µν γ [J ;J ] β = (J ) γ (J ) β − (J ) γ (J ) β ; µα ν να µ ργ σ σγ ρ ρα σ σα ρ µγ ν νγ µ = (g g γ − g g γ )(g g β − g g β ) − (g g γ − g g γ )(g g β − g g β ) ; (24) νρ µα σ σα µ µρ να σ σα ν νσ µα ρ ρα µ µσ να ρ ρα ν = g (g g β − g g β ) − g (g g β − g g β ) − g (g g β − g g β ) + g (g g β − g g β ) ; that is [J µν;J ρσ] = gνρJ µσ − gµρJ νσ − gνσJ µρ + gµσJ νρ : (25) More explicitly, in terms of Xi and Yi, we get 0i 0j i0 0j 00 ij ij 00 0j i0 ij [Yi;Yj] = [J ;J ] = g J − g J − g J + g J = −J ; ijk (26) = − Xk ; 1 1 [X ;Y ] = [ J lm; −J 0j] = − (gm0J lj − gl0J mj − gmjJ l0 + gljJ m0) ; i j 2 ilm 2 ilm 1 = − (δmJ l0 − δl J m0) ; (27) 2 ilm j j ijk = Yk ; 1 1 1 [X ;X ] = [ J lm; J pq] = (gmpJ lq − glpJ mq − gmqJ lp + glqJ mp) ; i j 2 ilm 2 jpq 4 ilm jpq 1 1 1 1 = J lq + J mq + J lp + J mp ; (28) 4 ilp jqp 4 imp jqp 4 ilm jpm 4 iml jpl ijk = Xk ; 4 i j i j ij ijk where in the last derivation we used ijklmk = δl δm − δmδl and J = Xk.