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420 Theory 10.31 The Lorentz Group The Lorentz group O(3, 1) is the set of all linear transformations L that leave invariant the Minkowski inner product

xy x y x0y0 = xT⌘y (10.222) ⌘ · in which ⌘ is the diagonal 1000 0100 ⌘ = 0 1 . (10.223) 0010 B 0001C B C @ A So L is in O(3, 1) if for all 4-vectors x and y

(Lx) T ⌘L y = xTLT ⌘Ly= xT⌘y. (10.224)

Since x and y are arbitrary, this condition amounts to

LT⌘L= ⌘. (10.225)

Taking the of both sides and using the transpose (1.194) and product (1.207) rules, we have

(det L)2 =1. (10.226)

So det L = 1, and every L has an inverse. Multi- ± plying (10.225) by ⌘,wefind

⌘LT⌘L = ⌘2 = I (10.227)

1 which identifies L as 1 T L = ⌘L ⌘. (10.228)

The of O(3, 1) with det L = 1 is the proper Lorentz group SO(3, 1). To find its , we consider a Lorentz matrix L = I +! that di↵ers from the identity matrix I by a tiny matrix ! and require it to satisfy the condition (10.225) for membership in the Lorentz group

I + !T ⌘ (I + !)=⌘ + !T⌘ + ⌘!+ !T! = ⌘. (10.229)

Neglecting!T!,wehave !T⌘ = ⌘! or since ⌘2 = I !T = ⌘!⌘. (10.230) This equation says (exercise 10.29) that under transposition the time-time 10.31 The Lorentz Group 421 and space-space elements of ! change sign, while the time-space and space- time elements do not. That is, the tiny matrix ! must be for infinitesimal ✓ and a linear combination

! = ✓ R + B (10.231) · · of the six matrices 000 0 0000 00 0 0 000 0 0001 00 10 R1 = 0 1 R2 = 0 1 R3 = 0 1 000 1 0000 01 0 0 B001 0C B0 100C B00 0 0C B C B C B C @ A @ A @ (10.232)A and 0100 0010 0001 1000 0000 0000 B1 = 0 1 B2 = 0 1 B3 = 0 1 0000 1000 0000 B0000C B0000C B1000C B C B C B C @ A @ A @ (10.233)A which satisfy condition (10.230). The three R are 4 4 versions of the j ⇥ generators (10.88); the three Bj generate Lorentz boosts. If we write L = I + ! as

L = I i✓ iR i iB I i✓ J i K (10.234) ` ` j j ⌘ ` ` j j then the three matrices J` = iR` are imaginary and antisymmetric, and therefore hermitian. But the three matrices Kj = iBj are imaginary and symmetric, and so are antihermitian. Thus, the 4 4 matrix L is not ⇥ unitary. The reason is that the Lorentz group is not compact. One may verify (exercise 10.30) that the six generators J` and Kj satisfy three sets of commutation relations:

[Ji,Jj]=i✏ijkJk (10.235)

[Ji,Kj]=i✏ijkKk (10.236) [K ,K ]= i✏ J . (10.237) i j ijk k

The first (10.235) says that the three J` generate the rotation group SO(3); the second (10.236) says that the three boost generators transform as a 3-vector under SO(3); and the third (10.237) implies that four canceling in- finitesimal boosts can amount to a rotation. These three sets of commutation relations form the Lie algebra of the Lorentz group SO(3, 1). Incidentally, one may show (exercise 10.31) that if J and K satisfy these commutation 422 relations (10.235–10.237), then so do J and K. (10.238) The infinitesimal Lorentz transformation (10.234) is the 4 4 matrix ⇥ 1 1 2 3 1 1 ✓3 ✓2 L = I + ! = I + ✓`R` + jBj = 0 1 . (10.239) ✓ 1 ✓ 2 3 1 B ✓ ✓ 1 C B 3 2 1 C @ a Aa b It moves any 4-vector x to x0 = Lx or in components x0 = L b x 0 0 1 2 3 x0 = x + 1x + 2x + 3x 1 0 1 2 3 x0 = 1x + x ✓3x + ✓2x 2 0 1 2 3 x0 = 2x + ✓3x + x ✓1x 3 0 1 2 3 x0 = x ✓ x + ✓ x + x . (10.240) 3 2 1 More succinctly with t = x0,thisis

t0 = t + x · x0 = x + t + ✓ x (10.241) ^ in which means cross-product. ^⌘⇥ For arbitrary real ✓ and , the matrices i✓ J i K L = e ` ` ` ` (10.242) form the subgroup of SO(3, 1) that is connected to the identity matrix I. This subgroup preserves the sign of the time of any time-like vector, that is, if x2 < 0, and y = Lx,theny0x0 > 0. It is called the proper orthochronous Lorentz group. The rest of the (homogeneous) Lorentz group can be ob- tained from it by space ,time , and space-time reflections. P T PT The task of finding all the finite-dimensional irreducible representations of the proper orthochronous homogeneous Lorentz group becomes vastly simpler when we write the commutation relations (10.235–10.237) in terms of the hermitian matrices 1 J ± = (J iK ) (10.243) ` 2 ` ± ` which generate two independent rotation groups + + + [Ji ,Jj ]=i✏ijkJk

[Ji,Jj]=i✏ijkJk + [Ji ,Jj]=0. (10.244) 10.32 Two-Dimensional Representations of the Lorentz Group 423 Thus the Lie algebra of the Lorentz group is equivalent to two copies of the Lie algebra (10.100) of SU(2). Its finite-dimensional irreducible representa- tions are the direct products

+ (j,j ) i✓ J i K ( i✓ )J ( i✓ + )J D 0 (✓, )=e ` ` ` ` = e ` ` ` e ` ` ` (10.245)

(j,0) ( i✓ )J+ (0,j ) of the nonunitary representations D (✓, )=e ` ` ` and D 0 (✓, )= ( i✓ + )J + e ` ` ` generated by the three (2j + 1) (2j + 1) matrices J and by ⇥ ` the three (2j0 + 1) (2j0 + 1) matrices J`. Under a Lorentz transformation (j,j ) ⇥ L,afield 0 (x) that transforms under the D(j,j0) representation of the m,m0 Lorentz group responds as

(j,j0) 1 (j,0)) 1 (0,j0) 1 (j,j0) U(L) (x) U (L)= D (L ) D (L ) (Lx). (10.246) m,m0 mm00 m0m000 m00,m000 Although these representations are not unitary, the SO(3) subgroup of the Lorentz group is represented unitarily by the hermitian matrices

+ J = J + J . (10.247)

Thus, the representation D(j,j0) describes objects of the spins s that can arise from the direct product of -j with spin-j0 (Weinberg, 1995, p. 231)

s = j + j0,j+ j0 1,..., j j0 . (10.248) | | For instance, D(0,0) describes a spinless field or particle, while D(1/2,0) and D(0,1/2) respectively describe left-handed and right-handed spin-1/2 fields or particles. The representation D(1/2,1/2) describes objects of spin 1 and spin 0—the spatial and time components of a 4-vector. The generators Kj of the Lorentz boosts are related to J ± by

+ K = iJ + iJ (10.249) which like (10.247) follows from the definition (10.243). + The interchange of J and J replaces the generators J and K with J and K, a substitution that we know (10.238) is legitimate.

10.32 Two-Dimensional Representations of the Lorentz Group (1/2,0) The generators of the representation D with j =1/2 and j0 = 0 are + given by (10.247 & 10.249) with J = /2 and J = 0. They are 1 1 J = and K = i . (10.250) 2 2 424 Group Theory The 2 2 matrix D(1/2,0) that represents the Lorentz transformation (10.242) ⇥

i✓ J i K L = e ` ` ` ` (10.251) is D(1/2,0)(✓, )=exp( i✓ /2 /2) . (10.252) · · And so the generic D(1/2,0) matrix is

(1/2,0) z /2 D (✓, )=e · (10.253) with =Rez and ✓ =Imz. It is nonunitary and of unit determinant; it is a member of the group SL(2,C) of complex unimodular 2 2 matrices. The ⇥ (covering) group SL(2,C) relates to the Lorentz group SO(3, 1) as SU(2) relates to the rotation group SO(3). Example 10.31 (The Standard Left-Handed Boost ) For a particle of m>0, the “standard” boost that takes the 4-vector k =(m, 0)to p =(p0, p), where p0 = m2 + p2, is a boost in the pˆ direction

0 1 B(p)=R(pˆ)pB (p ) R (pˆ)=exp(↵ pˆ B) (10.254) 3 · in which cosh ↵ = p0/m and sinh ↵ = p /m, as one may show by expanding | | the exponential (exercise 10.33). For = ↵ pˆ, one may show (exercise 10.34) that the matrix D(1/2,0)(0, ) is

(1/2,0) ↵pˆ /2 D (0,↵pˆ)=e · = I cosh(↵/2) pˆ sinh(↵/2) · = I (p0 + m)/(2m) pˆ (p0 m)/(2m) · p0 + m p = p · p (10.255) 2m(p0 + m) in the third line of whichp the 2 2 identity matrix I is suppressed. ⇥ Under D(1/2,0), the vector ( I,) transforms like a 4-vector. For tiny ✓ and , one may show (exercise 10.36) that the vector ( I,) transforms as

(1/2,0) (1/2,0) D† (✓, )( I)D (✓, )= I + · (10.256) (1/2,0) (1/2,0) D† (✓, ) D (✓, )= +( I) + ✓ ^ which is how the 4-vector (t, x) transforms (10.241). Under a finite Lorentz transformation L, the 4-vector Sa ( I,) becomes ⌘ (1/2,0) a (1/2,0) a b D† (L) S D (L)=L bS . (10.257) 10.32 Two-Dimensional Representations of the Lorentz Group 425 A massless field ⇠(x) that responds to a unitary Lorentz transformation U(L)like 1 (1/2,0) 1 U(L) ⇠(x) U (L)=D (L ) ⇠(Lx) (10.258) is called a left-handed Weyl . We will see in example 10.32 why the action density for such

(x)=i⇠†(x)(@ I ) ⇠(x) (10.259) L` 0 r· is Lorentz covariant, that is

1 U(L) (x) U (L)= (Lx). (10.260) L` L` Example 10.32 (Why Is Lorentz Covariant) We first note that the L` derivatives @b0 in `(Lx) are with respect to x0 = Lx.Sincetheinverse 1 L 1 a 1a b matrix L takes x0 back to x = L x0 or in tensor notation x = L b x0 , the derivative @b0 is a @ @x @ 1a @ 1a @b0 = b = b a = L b a = @a L b. (10.261) @x0 @x0 @x @x Now using the abbreviation @ I @ Sa and the transformation 0 r· ⌘ a laws (10.257 & 10.258), we have

1 (1/2,0) 1 a (1/2,0) 1 U(L) `(x) U (L)=i⇠†(Lx)D †(L )( @aS )D (L ) ⇠(Lx) L 1a b = i⇠†(Lx)( @aL bS ) ⇠(Lx) b = i⇠†(Lx)( @0 S ) ⇠(Lx)= (Lx) (10.262) b L` which shows that is Lorentz covariant. L` Incidentally, the rule (10.261) ensures, among other things, that the di- a vergence @aV is invariant a a 1b a c b c b (@aV )0 = @a0 V 0 = @b L a L cV = @b c V = @b V . (10.263) Example 10.33 (Why ⇠ is Left Handed) The space-time integral S of the action density is stationary when ⇠(x) satisfies the wave equation L` (@ I ) ⇠(x) = 0 (10.264) 0 r· or in momentum space (E + p ) ⇠(p)=0. (10.265) · Multiplying from the left by (E p ), we see that the energy of a particle · created or annihilated by the field ⇠ is the same as its momentum E = p | | in accord with the absence of a mass term in the action density . And L` 426 Group Theory because the spin of the particle is represented by the matrix J = /2, the momentum-space relation (10.265) says that ⇠(p) is an eigenvector of pˆ J · 1 pˆ J ⇠(p)= ⇠(p) (10.266) · 2 with eigenvalue 1/2. A particle whose spin is opposite to its momentum is said to have negative helicity or to be left handed. Nearly massless neutrinos are nearly left handed.

One may add to this action density the Majorana mass term

1 T (x)= m ⇠†(x) ⇠⇤(x)+⇠ (x) ⇠(x) (10.267) LM 2 2 2 ⇣ ⌘ which is Lorentz covariant because the matrices 1 and 3 anti-commute with 2 which is antisymmetric (exercise 10.39). This term would vanish if ⇠1⇠2 were equal to ⇠2⇠1. Since is conserved, only neutral fields like neutrinos can have Majorana mass terms. (0,1/2) The generators of the representation D with j = 0 and j0 =1/2 are + given by (10.247 & 10.249) with J = 0 and J = /2; they are 1 1 J = and K = i . (10.268) 2 2 Thus 2 2 matrix D(0,1/2)(✓, ) that represents the Lorentz transformation ⇥ (10.242)

i✓ J i K L = e ` ` j j (10.269) is

D(0,1/2)(✓, )=exp( i✓ /2+ /2) = D(1/2,0)(✓, ) (10.270) · · which di↵ers from D(1/2,0)(✓, ) only by the sign of . The generic D(0,1/2) matrix is the complex unimodular 2 2 matrix ⇥ (0,1/2) z /2 D (✓, )=e ⇤· (10.271) with =Rez and ✓ =Imz.

Example 10.34 (The Standard Right-Handed Boost ) For a particle of mass m>0, the “standard” boost (10.254) that transforms k =(m, 0)to p =(p0, p)isthe4 4 matrix B(p)=exp(↵ pˆ B) in which cosh ↵ = p0/m ⇥ · and sinh ↵ = p /m. This Lorentz transformation with ✓ = 0 and = ↵ pˆ | | 10.32 Two-Dimensional Representations of the Lorentz Group 427 is represented by the matrix (exercise 10.35)

(0,1/2) ↵pˆ /2 D (0,↵pˆ)=e · = I cosh(↵/2) + pˆ sinh(↵/2) · = I (p0 + m)/(2m)+pˆ (p0 m)/(2m) · (10.272) p0 + m + p = p · p 2m(p0 + m) in the third line of whichp the 2 2 identity matrix I is suppressed. ⇥ Under D(0,1/2), the vector (I,) transforms as a 4-vector; for tiny z

(0,1/2) (0,1/2) D† (✓, ) ID (✓, )=I + · (10.273) (0,1/2) (0,1/2) D† (✓, ) D (✓, )= + I + ✓ ^ as in (10.241). A massless field ⇣(x) that responds to a unitary Lorentz transformation U(L) as 1 (0,1/2) 1 U(L) ⇣(x) U (L)=D (L ) ⇣(Lx) (10.274) is called a right-handed Weyl spinor. One may show (exercise 10.38) that the action density

(x)=i⇣†(x)(@ I + ) ⇣(x) (10.275) Lr 0 r· is Lorentz covariant

1 U(L) (x) U (L)= (Lx). (10.276) Lr Lr Example 10.35 (Why ⇣ Is Right Handed) An argument like that of ex- ample (10.33) shows that the field ⇣(x) satisfies the wave equation

(@ I + ) ⇣(x) = 0 (10.277) 0 r· or in momentum space

(E p ) ⇣(p)=0. (10.278) · Thus, E = p , and ⇣(p) is an eigenvector of pˆ J | | · 1 pˆ J ⇣(p)= ⇣(p) (10.279) · 2 with eigenvalue 1/2. A particle whose spin is parallel to its momentum is said to have positive helicity or to be right handed. Nearly massless antineutrinos are nearly right handed. 428 Group Theory The Majorana mass term

1 T (x)= m ⇣†(x) ⇣⇤(x)+⇣ (x) ⇣(x) (10.280) LM 2 2 2 ⇣ ⌘ like (10.267) is Lorentz covariant.

10.33 The Dirac Representation of the Lorentz Group Dirac’s representation of SO(3, 1) is the direct sum D(1/2,0) D(0,1/2) of D(1/2,0) and D(0,1/2). Its generators are the 4 4 matrices ⇥ 1 0 i 0 J = and K = . (10.281) 2 0 2 0 ✓ ◆ ✓ ◆ Dirac’s representation uses the Cli↵ord algebra of the a which satisfy the anticommutation relation

a,b a b + b a =2⌘abI (10.282) { }⌘ in which ⌘ is the 4 4 diagonal matrix (10.223) with ⌘00 = 1 and ⌘jj =1 ⇥ for j = 1, 2, and 3, and I is the 4 4 identity matrix. ⇥ Remarkably, the generators of the Lorentz group

ij 0j J = ✏ijkJk and J = Kj (10.283) may be represented as commutators of gamma matrices i J ab = [a,b]. (10.284) 4 They transform the gamma matrices as a 4-vector

[J ab,c]= ia ⌘bc + ib ⌘ac (10.285) (exercise 10.40) and satisfy the commutation relations

i[J ab,Jcd]=⌘bc J ad ⌘ac J bd ⌘da J cb + ⌘db J ca (10.286) of the Lorentz group (Weinberg, 1995, p. 213–217) (exercise 10.41). The gamma matrices a are not unique; if S is any 4 4 matrix with an ⇥ inverse, then the matrices a SaS 1 also satisfy the definition (10.282). 0 ⌘ The choice 01 0 0 = i and = i (10.287) 10 0 ✓ ◆ ✓ ◆ 10.33 The Dirac Representation of the Lorentz Group 429 makes J and K block diagonal (10.281) and lets us assemble a left-handed spinor ⇠ and a right-handed spinor ⇣ neatly into a 4-component spinor ⇠ = . (10.288) ⇣ ✓ ◆ Dirac’s action density for a 4-spinor is

= (a@ + m) (@ + m) (10.289) L a ⌘ 6 in which

0 01 i † = † = ⇣ ⇠ . (10.290) ⌘ 10 † † ✓ ◆ The kinetic part is the sum of the left-handed and right-handed action L` Lr densities (10.259 & 10.275)

a @ = i⇠† (@ I ) ⇠ + i⇣† (@ I + ) ⇣. (10.291) a 0 r· 0 r· If ⇠ is a left-handed spinor transforming as (10.258), then the spinor

0 i ⇠1† ⇣ = 2 ⇠⇤ (10.292) ⌘ i 0 ⇠† ✓ ◆ 2! transforms as a right-handed spinor (10.274), that is (exercise 10.42)

z⇤ /2 z /2 ⇤ e · 2 ⇠⇤ = 2 e · ⇠ . (10.293) ⇣ ⌘ Similarly, if ⇣ is right handed, then ⇠ = 2 ⇣⇤ is left handed. The simplest 4-spinor is the Majorana spinor

⇠ 2⇣⇤ 2 = = = i ⇤ (10.294) M ⇠ ⇣ M ✓ 2 ⇤◆ ✓ ◆ whose particles are the same as its antiparticles. (1) (2) If two Majorana spinors M and M have the same mass, then one may combine them into a

(1) (2) 1 (1) (2) 1 ⇠ + i⇠ ⇠D D = M + i M = (1) (2) = . (10.295) p2 p2 ⇣ + i⇣ ⇣D ⇣ ⌘ ✓ ◆ ✓ ◆ The Dirac mass term

m = m ⇣† ⇠ + ⇠† ⇣ (10.296) D D D D D D ⇣ ⌘ conserves charge, and since exp(z /2) exp( z /2) = I it also is Lorentz ⇤ · † · 430 Group Theory invariant. For a Majorana field, it reduces to

1 1 1 T 2 m M M = 2 m ⇣†⇠ + ⇠†⇣ = 2 m ⇠†2⇠⇤ + ⇠ 2⇠ (10.297) 1 ⇣ ⌘T ⇣ ⌘ = m ⇣† ⇣⇤ + ⇣ ⇣ 2 2 2 ⇣ ⌘ a Majorana mass term (10.267 or 10.280).

10.34 The Poincar´eGroup The elements of the Poincar´egroup are products of Lorentz transformations and translations in space and time. The Lie algebra of the Poincar´egroup therefore includes the generators J and K of the Lorentz group as well as the hamiltonian H and the momentum operator P which respectively generate translations in time and space. Suppose T (y) is a that takes a 4-vector x to x + y and T (z)is a translation that takes a 4-vector x to x + z.ThenT (z)T (y) and T (y)T (z) both take x to x + y + z. So if a translation T (y)=T (t, y)isrepresented by a unitary operator U(t, y)=exp(iHt iP y), then the hamiltonian H · and the momentum operator P commute with each other

[H, P j] = 0 and [P i,Pj]=0. (10.298)

We can figure out the commutation relations of H and P with the angular- momentum J and boost K operators by realizing that P a =(H, P )isa 4-vector. Let i✓ J i K U(✓, )=e · · (10.299) be the (infinite-dimensional) unitary operator that represents (in Hilbert space) the infinitesimal Lorentz transformation

L = I + ✓ R + B (10.300) · · where R and B are the six 4 4 matrices (10.232 & 10.233). Then because ⇥ P is a 4-vector under Lorentz transformations, we have

1 +i✓ J+i K i✓ J i K U (✓, )PU(✓, )=e · · Pe · · =(I + ✓ R + B) P · · (10.301) or using (10.273)

(I + i✓ J + i K) H (I i✓ J i K)=H + P (10.302) · · · · · (I + i✓ J + i K) P (I i✓ J i K)=P + H + ✓ P . · · · · ^ Exercises 431 Thus, one finds (exercise 10.42) that H is invariant under rotations, while P transforms as a 3-vector

[Ji,H] = 0 and [Ji,Pj]=i✏ijkPk (10.303) and that [K ,H]= iP and [K ,P ]= i H. (10.304) i i i j ij By combining these equations with (10.286), one may write (exercise 10.44) the Lie algebra of the Poincar´egroup as i[J ab,Jcd]=⌘bc J ad ⌘ac J bd ⌘da J cb + ⌘db J ca i[P a,Jbc]=⌘abP c ⌘acP b [P a,Pb]=0. (10.305)

Further Reading The classic Lie Algebras in Particle Physics (Georgi, 1999), which inspired much of this chapter, is outstanding.

Exercises 10.1 Show that all n n (real) orthogonal matrices O leave invariant the ⇥ x2 +x2 + +x2 , that is, that if x = Ox,thenx 2 = x2. 1 2 ··· n 0 0 10.2 Show that the set of all n n orthogonal matrices forms a group. ⇥ 10.3 Show that all n n unitary matrices U leave invariant the quadratic ⇥ form x 2 + x 2 + + x 2, that is, that if x = Ux,then x 2 = x 2. | 1| | 2| ··· | n| 0 | 0| | | 10.4 Show that the set of all n n unitary matrices forms a group. ⇥ 10.5 Show that the set of all n n unitary matrices with unit determinant ⇥ forms a group. (j) 10.6 Show that the matrix D (g)= j, m0 U(g) j, m is unitary because m0m h | | i the rotation operator U(g) is unitary j, m U (g)U(g) j, m = . h 0| † | i m0m 10.7 Invent a group of 3 and compute its multiplication table. For extra credit, prove that the group is unique. 10.8 Show that the relation (10.20) between two equivalent representations is an . 10.9 Suppose that D1 and D2 are equivalent, finite-dimensional, irreducible representations of a group G so that D (g)=SD (g)S 1 for all g G. 2 1 2 What can you say about a matrix A that satisfies D2(g) A = AD1(g) for all g G? 2 432 Group Theory 10.10 Find all components of the matrix exp(i↵A)inwhich

00 i A = 00 0 . (10.306) 0 1 i 00 @ A i↵A i↵A 10.11 If [A, B]=B,finde Be . Hint: what are the ↵-derivatives of this expression? 10.12 Show that the tensor-product matrix (10.31) of two representations D1 and D2 is a representation. 10.13 Find a 4 4 matrix S that relates the tensor-product representation ⇥ D 1 1 to the direct sum D1 D0. 2 ⌦ 2 10.14 Find the generators in the of the group with structure constants fabc = ✏abc where a, b, c run from 1 to 3. Hint: The answer is three 3 3 matrices t , often written as L . ⇥ a a 10.15 Show that the generators (10.90) satisfy the commutation relations (10.93). 10.16 Show that the demonstrated equation (10.98) implies the commuta- tion relation (10.99). 10.17 Use the Cayley-Hamilton theorem (1.264) to show that the 3 3 ⇥ matrix (10.96) that represents a right-handed rotation of ✓ radians about the axis ✓ is given by (10.97). 10.18 Verify the mixed Jacobi identity (10.142).

10.19 For the group SU(3), find the structure constants f123 and f231. 10.20 Show that every 2 2 unitary matrix of unit determinant is a quater- ⇥ nion of unit . 10.21 Show that the as defined by (10.175) are closed under addition and multiplication and that the product xq is a if x is real and q is a quaternion.

10.22 Show that the one-sided derivative f 0(q) (10.185) of the quaternionic function f(q)=q2 depends upon the direction along which q 0. 0 ! 10.23 Show that the generators (10.189) of Sp(2n) obey commutation rela- tions of the form (10.190) for some real structure constants fabc and a suitably extended set of matrices A, A0, . . . and Sk, Sk0 ,.... 10.24 Show that for 0 <✏ 1, the real 2n 2n matrix T =exp(✏JS) in ⌧ ⇥ which S is symmetric satisfies T TJT = J (at least up to terms of order ✏2) and so is in Sp(2n, R). 10.25 Show that the matrix T of (10.198) is in Sp(2,R). 10.26 Use the parametrization (10.218) of the group SU(2), show that the Exercises 433 parameters a(c, b) that describe the product g(a(c, b)) = g(c) g(b) are those of (10.220). 10.27 Use formulas (10.220) and (10.213) to show that the left-invariant measure for SU(2) is given by (10.221). 10.28 In tensor notation, which is explained in chapter 11, the condition (10.230) that I + ! be an infinitesimal Lorentz transformation reads a !T = !a = ⌘ !c ⌘da in which sums over c and d from 0 to 3 b b bc d are understood. In this notation, the matrix ⌘ lowers indices and ⌘gh ef raises them, so that ! a = ! ⌘da. (Both ⌘ and ⌘gh are numeri- b bd ef cally equal to the matrix ⌘ displayed in equation (10.223).) Multiply both sides of the condition (10.230) by ⌘ae = ⌘ea and use the relation ⌘da ⌘ = ⌘d d to show that the matrix ! with both indices ae e ⌘ e ab lowered (or raised) is antisymmetric, that is,

! = ! and !ba = !ab. (10.307) ba ab 10.29 Show that the six matrices (10.232) and (10.233) satisfy the SO(3, 1) condition (10.230). 10.30 Show that the six generators J and K obey the commutations rela- tions (10.235–10.237). 10.31 Show that if J and K satisfy the commutation relations (10.235– 10.237) of the Lie algebra of the Lorentz group, then so do J and K. 10.32 Show that if the six generators J and K obey the commutation re- + lations (10.235–10.237), then the six generators J and J obey the commutation relations (10.244). 10.33 Relate the parameter ↵ in the definition (10.254) of the standard boost B(p) to the 4-vector p and the mass m. 10.34 Derive the formulas for D(1/2,0)(0,↵pˆ) given in equation (10.255). 10.35 Derive the formulas for D(0,1/2)(0,↵pˆ) given in equation (10.272). 10.36 For infinitesimal complex z, derive the 4-vector properties (10.256 & 10.273) of ( I,)underD(1/2,0) and of (I,)underD(0,1/2). 10.37 Show that under the unitary Lorentz transformation (10.258),the action density (10.259) is Lorentz covariant (10.260). 10.38 Show that under the unitary Lorentz transformation (10.274), the action density (10.275) is Lorentz covariant (10.276). 10.39 Show that under the unitary Lorentz transformations (10.258 & 10.274), the Majorana mass terms (10.267 & 10.280) are Lorentz covariant. 10.40 Show that the definitions of the gamma matrices (10.282) and of the generators (10.284) imply that the gamma matrices transform as a 4- vector under Lorentz transformations (10.285). 434 Group Theory 10.41 Show that (10.284) and (10.285) imply that the generators J ab satisfy the commutation relations (10.286) of the Lorentz group. 10.42 Show that the spinor ⇣ = 2⇠⇤ defined by (10.292) is right handed (10.274) if ⇠ is left handed (10.258). 10.43 Use (10.302) to get (10.303 & 10.304). 10.44 Derive (10.305) from (10.286, 10.298, 10.303, & 10.304).