1 the Lorentz Group

Home , Lorentz group

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 ≥ |b1 c2| , (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 + ω ν , (14) we have   0 −β1 −β2 −β3  −β1 0 −α3 α2  X ω =   = (αiXi + βiYi) , (15) −β2 α3 0 −α1   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 ωσν = gσµω 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:  1 0 0 0   −1 0 0 0   0 −1 0 0   0 1 0 0  P =   T =   PT = −1 . (16)  0 0 −1 0   0 0 1 0  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 ] = εijkAk , [Bi,Bj ] = εijkBk , [Ai,Bj ] = 0 . (18)

Solution: The inﬁnitesimal transformation of the four-vector V µ under the proper Lorentz group is ρ ρ σ δV = ω 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 ω (J µν)ρ = ω (gµρgν − gνρgµ ) , 2 µν σ 2 µν σ σ 1 1 = gρµω gν + gρνω gµ , (21) 2 µν σ 2 νµ σ ρ = ω 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. With this being done, let us ﬁnally compute the commutation relations for Ai and Bi 1 1 [A ,A ] = [ (X + iY ), (X + iY )] i j 2 i i 2 j j 1 i i 1 = [X ,X ] + [Y ,X ] + [X ,Y ] − [Y ,Y ] , 4 i j 4 i j 4 i j 4 i j (29) 1 i i 1 = ijkX − jikY + ijkY + ijkX , 4 k 4 K 4 K 4 k = ijkAk , 1 1 [B ,B ] = [ (X − iY ), (X − iY )] i j 2 i i 2 j j 1 i i 1 = [X ,X ] − [Y ,X ] − [X ,Y ] − [Y ,Y ] , 4 i j 4 i j 4 i j 4 i j (30) 1 i i 1 = ijkX + jikY − ijkY + ijkX , 4 k 4 K 4 K 4 k = ijkBk , 1 1 [A ,B ] = [ (X + iY ), (X − iY )] i j 2 i i 2 j j 1 i i 1 = [X ,X ] + [Y ,X ] − [X ,Y ] + [Y ,Y ] , 4 i j 4 i j 4 i j 4 i j (31) 1 i i 1 = ijkX − jikY − ijkY − ijkX , 4 k 4 K 4 K 4 k = 0 .

II. Since the commutation relations of Ai and Bi are identical to those of the generators of SO(3) (or the rotations in 3 dimensions), and the two sets commute, it follows that the irreducible representations of the proper Lorentz group can be classiﬁed in terms of two semi-integer numbers, j and j0, which labels the eigenvalues of the two Casimir operators: 2 P 2 2 P 2 A = i Ai and B = i Bi (i.e. as direct product of two representations of the angular momentum operator in 3 dimensions).

0 1 1 Show that the 4 × 4 representation of the Λ matrices corresponds to the (j, j ) = ( 2 , 2 ) representation. Suggestion: construct the two Casimir operators in this representation, starting from the explicit expressions of the generators Xi and Yi.

Solution: Let us ﬁrst explicitly compute Ai and Bi generators in the vector representation  i   i   i  0 − 2 0 0 0 0 − 2 0 0 0 0 − 2 i 1 1  − 2 0 0 0   0 0 0 2   0 0 − 2 0  A1 =  1  ,A2 =  i  ,A3 =  1  , (32)  0 0 0 − 2   − 2 0 0 0   0 2 0 0  1 1 i 0 0 2 0 0 − 2 0 0 − 2 0 0 0 and  i   i   i  0 2 0 0 0 0 2 0 0 0 0 2 i 1 1  2 0 0 0   0 0 0 2   0 0 − 2 0  B1 =  1  ,B2 =  i  ,B3 =  1  . (33)  0 0 0 − 2   2 0 0 0   0 2 0 0  1 1 i 0 0 2 0 0 − 2 0 0 2 0 0 0

5 By means of explicit computation it can be shown 1 A A = A A = A A = B B = B B = B B = − 1 , (34) 1 1 2 2 3 3 1 1 2 2 3 3 4 where 1 is 4 × 4 identity matrix. Finally, we get 3 A2 = B2 = − 1 . (35) 4 The Casimir operator for SO(3) group J2 takes the form

2 X J = JiJi = −j(j + 1)1 , (36) i where j is the quantum number of the representation. Thus, vector representation is (j, j0) = (1/2, 1/2).

3 Higher-dimensional tensors

A tensor with n indices is an object transforming as

µ1µ2...µn 0 µ1µ2...µn µ1 µ2 µn ν1ν2...νn T → (T ) = Λ ν1 Λ ν2 ... Λ νn T . (37) A notable example is the metric tensor gµν . Applying the general rule (37) to the case of gµν it is easy to realize that gµν is an invariant tensor: (g0)µν = gµν . Another important example is the Levi-Civita tensor, εµνρσ. This is deﬁned by the condition of being completely antisymmetric in the exchange of any of the two indices, and by ε0123 = 1. Applying the general rule (37) to the Levi-Civita tensor one ﬁnds εµνρσ → det(Λ)εµνρσ . (38) The Levi-Civita tensor is therefore invariant under proper Lorentz transformations, while it is an odd quantity under parity transformations. It can be shown that there are no other invariant tensors beside g, and their possible combinations. The transformation law of a generic tensor in Eq. (37) provides a higher-dimensional representation of the Lorentz µ 1 1 group. Since Λ ν belongs to the ( 2 , 2 ) irreducible representation, the generic transformation in Eq. (37) is the direct product of n of such representations. In general, this product is not irreducible can be decomposed into irreducible representations of lower dimension. The following tools are quite useful to obtain this decomposition: 1. The property of symmetry or antisymmetry under the exchange of two indices is invariant under Lorentz transformations:

if T µ1...µi...µj ... = ±T µ1...µj ...µi... then (T 0)µ1...µi...µj ... = ±(T 0)µ1...µj ...µi...

µ µ 2. The multiplication with gµν or g ν = δ ν let us to construct a tensor with (n − 2) indices starting from a n-indices one. Similarly, the multiplication with εµνρσ can be used to decrease the number of indices.

I. A generic tensor with 2 indices, T µν (16 independent components), can be decomposed as follows 1 T µν = Sµν + Aµν + Cgµν , (39) 4 µν µν µ µν where S is a symmetric tensor with null trace (gµν S = Sµ = 0), A is an antisymmetric tensor and C is a scalar. The transformation properties of these three elements provide irreducible representations of the full Lorentz group. Derive the expressions of Sµν , Aµν and C in terms of the original tensor T µν .

6 Solution: Starting with 16-component tensor T µν, we can deﬁne an antisymmetric tensor 1 Aµν = (T µν − T νµ) , (40) 2 and a symmetric tensor 1 Sµν = (T µν + T νµ) , (41) 1 2 µν µν µν such that T = A + S1 . The two tensors indeed span the invariant subspaces under the µν 0µν µ ν αβ Lorentz transformations. In particular, for A → A = Λ αΛ βA , the transformed ﬁeld remains antisymmetric

0νµ ν µ αβ A = Λ αΛ βA , ν µ βα = −Λ αΛ βA , (42) = −A0µν

µν 0µν µ ν αβ and similarly for S1 → S1 = Λ αΛ βS1 , the transformed ﬁeld remains symmetric

0νµ ν µ αβ S1 = Λ αΛ βS1 , ν µ βα = Λ αΛ βS1 , (43) 0µν = S1 . The symmetric tensor can further be decomposed 1 Sµν = Sµν + Cgµν , (44) 1 4 µν where gµνS = 0. Contracting the above equation with gµν, we ﬁnd 1 C = g Sµν = g (T µν + T νµ) , µν 1 µν 2 (45) µν = gµνT ,

µν where we used gµνg = 4. The invariance of the scalar C is trivial. Let us ﬁnally prove that the traceless symmetric tensor spans the invariant subspace

0µν µ ν αβ gµνS1 = gµνΛ αΛ βS1 , αβ (46) = gαβS1 , = 0 .

II. Looking at the number of independent components of Sµν , Aµν and C, try to identify them with the irreducible representations obtained by the direct product 1 1 1 1 ( , ) ⊗ ( , ) = (1, 1) ⊕ (1, 0) ⊕ (0, 1) ⊕ (0, 0) . (47) 2 2 2 2

7 Why we have only three tensors in Eq. (39) and not four, as one could guess from the above decomposition?

1 1 1 1 Solution: The direct product of the two vector representations ( 2 , 2 ) ⊗ ( 2 , 2 ) spans 4 × 4 = 16 dimensional vector space (T µν) and it can be decomposed as the direct sum of the irreducible representations. The (1, 1) irreducible representation is 3 × 3 = 9 dimensional and it corresponds to the traceless symmetric tensor Sµν, while the singlet representation (0, 0) corresponds to the scalar function C. The antisymmetric tensor Aµν (6-dimensional) is the direct sum of the (1, 0) ⊕ (0, 1) representations. Given the antisymmetric tensor, one can construct self-dual and anti-self-dual parts

µν 1 µν i µνρσ G = A − Aρσ , 2 4 (48) 1 i G†µν = Aµν + µνρσA , 2 4 ρσ and these do not mix under Lorentz transformations 0µν µ ν 1 αβ i αβρσ G = Λ αΛ β A − Aρσ , 2 4 (49) 1 i = A0µν − µνρσA0 . 2 4 ρσ

4 Poincar´ealgebra

Given the following general deﬁnition of the Poincar´egenerators, J µν and P µ, i U(1 + ω, ) = 1 − ω J µν − i P µ + ... (50) 2 µν µ and the composition law U(Λ2, a2)U(Λ1, a1) = U(Λ2Λ1, Λ2a1 + a2) (51)

I. Show that U −1(Λ, a)U(1 + ω, )U(Λ, a) = U(1 + Λ−1ωΛ, Λ−1 + Λ−1ωa) (52) and, expanding this expression, show that this implies the following commutation relations

[J µν ,J ρσ] = i (gνρJ µσ − gµρJ νσ − gνσJ µρ + gµσJ νρ) (53) h i J µν ,P λ = i gνλP µ − gµλP ν (54)

Solution: First, let us show that U −1(Λ, a) = U(Λ−1, −Λ−1a) . (55) Namely, U(Λ−1, −Λ−1a)U(Λ, a) = U(Λ−1Λ, Λ−1a − Λ−1a) , (56) = U(1, 0) .

8 Applying the composition law in two steps

U −1(Λ, a)U(1 + ω, )U(Λ, a) = U(Λ−1, −Λ−1a)U(Λ + ωΛ, a + ωa + ) , (57) = U(1 + Λ−1ωΛ, Λ−1 + Λ−1ωa) .

Expanding this result to ﬁrst order in ω and , one gets i i U −1(Λ, a)(1 − ω J αβ − i P α)U(Λ, a) = 1 − ω Λα Λβ J µν − iΛα ( + ω aβ)P µ . (58) 2 αβ α 2 αβ µ ν µ α αβ Equating the corresponding terms

−1 αβ α β µν α β µ β α µ U (Λ, a)J U(Λ, a) = Λ µΛ νJ + Λ µa P − Λ µa P , (59) −1 α α µ U (Λ, a)P U(Λ, a) = Λ µP , (60) Treating Λ ≡ 1 + ω and a ≡ as inﬁnitesimal and expanding the above equations i ω J µν + i P µ,J αβ = ωα J ρβ + ωβ J αρ + δα βP ρ − δβ αP ρ , 2 µν µ ρ ρ ρ ρ (61) i ω J µν + i P µ,P α = ωα P ρ , 2 µν µ ρ one can derive the following commutation relations

[J µν,J ρσ] = i (gνρJ µσ − gµρJ νσ − gνσJ µρ + gµσJ νρ) , (62) [P µ,P ρ] = 0 , (63) J µν,P λ = i gνλP µ − gµλP ν . (64)