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18 In this way we will obtain all possible (finite dimensional) representations of the Lorentz group. In order to obtain the transformation law for spinors we will consider the set of 2 2 complex matrices with determinant =1. These form the group called SL(2,C). We will represent the Lorentz group by the action of these matrices on two component complex spinors. Equivalently we could build the spinor transformation law from the spin 1 angular momentum representation 2 matrices familiar from quantum mechanics, the Pauli matrices. Once we know the action of N and N on the spinors we can reconstruct that of M μν. However, it is more useful and to the point to proceed by considering directly SL(2,C). To obtain the relation of the Lorentz group to SL(2,C) we must first recall that there exists a one to one correspondence between 2 2Hermitian matrices and space-time points. The Pauli matrices ( ) 1 0 (σ 0 ) α α 0 1 α α ( ) 0 1 (σ 1 ) α α 1 0 α α ( ) 0 i (σ 2 ) α α +i 0 α α ( ) 1 0 (σ 3 ) α α, (II.1) 0 1 α α where α =1, 2labelsthetworowsand α =1, 2 labels the two columns, form abasisfor2 2 Hermitian matrices. Let X α α be a Hermitian matrix, that is, X = X (X ) αα =(X) α α. (II.2) It has the general form X α α = ( (x0 + x 3 ) ) (x 1 ix 2 ) (x 1 + ix 2 ) (x 0 x 3 ) α α = x μ (σ μ ) α α x α α (II.3) for x μ real with x called x slash. Thus corresponding to any 4 vector x μ we associate a 2 2 Hermitian matrix X α α by equation (II.3). Using the trace relation for the product of two Pauli matrices (σ μ ) α α (iσ 2 ) α β(σ ν T ) ββ (iσ 2 ) βα = 2g μν, 203

19 or more succinctly written Tr[σ μ (iσ 2 )σ ν T (iσ 2 )] = 2g μν, (II.4) we have for every Hermitian matrix X α α an associated four vector x μ x μ = 1 2 Tr[X(iσ2 )σ μt (iσ 2 )]. (II.5) This correspondence is one to one (we will use X = x in what follows to underscore this correspondence). Simplifying the notation, since we would like to keep our dotted and undotted indices separate that is when we sum over indices we would like them to be of the same type in order to avoid extra confusion, we introduce an antisymmetric tensor ɛ αβ,thatis,ɛ αβ = ɛ βα with ɛ 12 =+1andwith lowered indices ɛ αβ = ɛ αβ = ɛ βα, that is ɛ 12 = ɛ 12 = 1. Note that the matrix is the same when we use dotted indices, that is, Also note that ɛ αβ =(iσ 2 ) αβ = ɛ α β =(iσ 2 ) α β = ɛ αβ ɛ βγ = δ α γ ɛ α βɛ β γ = δ α γ. ( 0 ) αβ Then we can define the Pauli matrices with upper indices ( ) 0 1. (II.6) 1 0 α β ( σ μ ) αα ɛ αβ ɛ α β(σ μ ) β β = (iσ 2 ) β(σ μt α ) ββ (iσ 2 ) βα. (II.7) We can write the trace condition as (σ μ ) α α ( σ ν ) αα =+2g μν (II.8) 204

20 and equation (II.5) has the simple form x μ =+ 1 2 ( x) α α( σ μ ) αα =+ 1 2 Tr[ x σμ ]. (II.9) The σ μ matrices are given by ( ) 1 0 ( σ 0 ) αα =+(σ 0 ) 0 1 αα αα ( ) 0 1 ( σ 1 ) αα = (σ 1 ) 1 0 αα αα ( ) 0 +i ( σ 2 ) αα = (σ 2 ) αα ( σ 3 ) αα i 0 ( αα ) αα = (σ 3 ) αα. (II.10) We can readily derive the completeness properties of the Pauli matrices (σ μ ) α α ( σ ν ) αα =+2g μν (II.11) (σ μ ) α α ( σ μ ) ββ =+2δα β δ β α. (II.12) Further products of two yield (σ μ ) α α ( σ ν ) αβ +(σ ν ) α α ( σ μ ) αβ =2g μν δ β α ( σ μ ) αα (σ ν ) α β +( σ ν ) αα (σ μ ) α β =2g μν δ α β. (II.13) If S is an element of SL(2,C) (that is 2 2 complex matrices with determinant equal to one) with matrix elements Sα β,whereα labels the rows and β labels the columns, and x is a Hermitian matrix, then we can define the transformed matrix x as ( x ) α α = Sα β ( x) β β β S α (II.14) with S the complex conjugate of S, again with α labelling the rows and β labelling the columns, or taking the transpose we have (S ) β α =(S ) β α, with β labelling the rows and α labeling the columns of S. Since dets = S1 1 S2 2 S1 2 S2 1 =1wehave det x =det x. (II.15) 205

21 Calculating the determinant we find det x =(x 0 +x 3 )(x 0 x 3 ) (x 1 ix 2 )(x 1 +ix 2 )=(x 0 ) 2 (x 1 ) 2 (x 2 ) 2 (x 3 ) 2 = x μ x μ =det x = x μx μ ; (II.16) the determinant is the Minkowski interval and is invariant. Thus, the transformation x = S xs (II.17) corresponds to a Lorentz transformation, Λ μν, of the coordinates. In order to determine it in terms of the SL(2,C) matrix S consider x μ = 1 2 ( x ) α α ( σ μ ) αα whereweidentify = 1 2 S α β ( x) β β β S α ( σ μ ) αα = 1 2 S β Λ μν x ν α Λ μν 1 2 Tr[Sσν S σ μ ] S β α (σ ν ) β β( σ μ ) αα x ν (II.18) (II.19) that is, Λ μν (σ μ ) α α = 1 2 S γ γ β S or more simply written β (σν ) γ γ ( σ μ ) ββ (σ μ ) α α = S γ = S γ α S γ α (σ ν ) γ γ, Λ μν σ μ = Sσ ν S. β γ S β (σν ) γ γ δα β δ β α (II.20) So for every element ±S of SL(2,C) there is an element Λ of the Lorentz group, the mapping of SL(2,C) into L + is 2 to 1 since ±S Λ. We can use the SL(2,C) matrices to define the spinor representations of the Lorentz group. The spinor transformation laws are given by ψ α(x ) S β α ψ β (x) ψ α (x ) ψ β (x)(s 1 ) α β (II.21) 206

22 where Sα β (S 1 ) γ β = δα γ and ψ α and ψ α are two different two component complex spinors transforming (as we will see) as the ( 1, 0) representation of 2 the Lorentz group, the ψ are called Weyl spinors. Similarly, we can use S and (S ) 1 to define two more different Weyl spinors, the complex conjugates of ψ, denoted ψ, which transform as (0, 1 ) representations of the Lorentz group 2 ψ α(x ) ψ β (x)(s ) β α ψ α (x ) (S 1 ) α β ψ β(x) (II.22) where, for the adjoint matrices (S ) α β, α labels the rows and β labels the columns. As with tensors, higher rank spinors transform just like products of the basic rank 1 spinors, for example, ψ α 1 α n α 1 α m (x )=S β 1 α 1 ψ α 1 α n (x )=S β 1 α 1 Since S is special, i.e. det S = 1, we have Sα βn n ψ β1 β n (x) S βm α 1 ψ β1 β n β1 β m (x)(s ) β 1 α1 (S ) β m αm. (II.23) ( S (S 1 ) α β = 2 S 1 2 S1 2 ) 2 S1 1 αβ or in matrix notation = ɛ αγ ɛ βδ S γ δ S 1 = ɛs T ɛ. Further ɛ is an anti-symmetric invariant second rank spinor, that is (II.24) (II.25) ɛ αβ = S γ α S δ β ɛ γδ or again in matrix form ɛ = SɛS T = SS 1 ɛ = ɛ. Since the indices can be confusing, let s write this out explicitly ɛ 12 = S 1 1 S 2 2 ɛ 12 + S 2 1 S 1 2 ɛ 21 = det S = 1 =ɛ

23 So indeed ɛ αβ = ɛ αβ, ɛ is an invariant second rank spinor. Hence, we can use ɛ to lower and raise indices of the spinors analogous to the invariant metric tensor g μν which lowers and raises vector indices ψ α = ɛ αβ ψ β ψ α = ɛ αβ ψ β ψ α = ɛ α β ψ β ψ α = ɛ α β ψ β. (II.26) As a consequence of (II.26) the transformation law for ψ α, for instance, follows from that of ψ α ψ α (x )=ɛ αβ ψ β(x )=ɛ αβ S γ β ψ γ (x) = ɛ αβ S γ β ɛ γδψ δ (x) = ɛ δγ S γ β ɛ βαψ δ (x) =ψ δ (x)(s 1 ) δ α. (II.27) Thus, we can contract similar spinor indices to make Lorentz scalars ψ α (x )ψ α(x )=(S 1 ) α β S γ α ψ β (x)ψ γ (x) = δ γ β ψ β (x)ψ γ (x) =ψ α (x)ψ α (x) (II.28) and similarly for ψ α ψ α. Also using the properties of the Pauli matrices we can make a four vector object whose vector index then contracts with another four vector index in order to make a scalar, for example ψ α (x )(σ μ ) α α μ ψ α (x )=(S 1 ) β α (S 1 ) α β(λ 1 ) ν μ ψβ (x)(σ μ ) α α ν ψ β(x). But (S 1 ) β α (σ μ ) α α (S 1 ) α βλ ( 1 μν =(S 1 ) β α (σ μ ) α α (S 1 ) α β 2 Tr[Sσν S σ ) μ ] = 1 2 (S 1 ) α β (S 1 ) α β(sσ ν S ) δ δ (σ μ) α α ( σ μ ) δδ hence =(S 1 ) β α (S 1 ) α β(sσ ν S ) δ δδ α δ δ δ α =(S 1 ) β α (S 1 ) α β(sσ ν S ) α α =(σ ν ) β β, ψ α (x )(σ μ ) α α μ ψ α (x )=ψ α (x)(σ μ ) α α μ ψ α (x). (II.29) (II.30) 208

24 As stated, ψ ψ is a Lorentz invariant. Finally, let s consider infinitesimal Lorentz transformations x μ = x μ + ω μν x ν (II.31) where now, for infinitesimal transformations, S differs from the identity by an infinitesimal matrix Σ Sα β = δα β +Σα β S β α Since ɛ is invariant we have that = δ β α +Σ β α. ɛ αβ = Sα γ Sβ δ ɛ γδ =(δα γ +Σα γ )(δβ δ +Σβ δ )ɛ γδ (II.32) = [ δα γ δ β δ +Σ α γ δ β δ +Σ β δ δ ] α γ ɛγδ = ɛ αβ + ɛ γβ Σα γ + ɛ αγσ γ β which implies that Σ is symmetric. With lowered indices we have Σ βα Σ αβ =0. (II.33) (II.34) Now given ω μν we desire Σ αβ ;usingλ μν = 1 2 Tr[Sσν S σ μ ] we find g μν + ω μν = 1 2 ( δ β α ) ( +Σα β (σ ν ) β β δ β α ) +Σ β α ( σ μ ) αα = 1 2 (σν ) α α ( σ μ ) αα Σ β α (σ ν ) β α ( σ μ ) αα (σν ) α βσ β α ( σ μ ) αα = g μν Σ α β (σ ν ) β α ( σ μ ) αα + 1 β Σ α ( σ μ ) αα (σ ν ) 2 α β. (II.35) Thus, we must find a solution for ω μν = 1 2 Σ β α (σ ν ) β α ( σ μ ) αα Σ β α ( σ μ ) αα (σ ν ) α β = 1 2 Tr[ Σσ ν σ μ +Σ σ μ σ ν]. (II.36) Multiplying by σ μ and σ ν we have (σ μ ) γ γ ( σ ν ) δδ ω μν = 1 [ (σμ ) γ γ ( σ ν ) δδ (σ ν ) γ γ ( σ μ ) δδ ] ω μν 2 209

25 Using Σ α α =0=Σ α α, we find =2Σγ δ δ δ γ +2Σ δ γ δγ δ. (II.37) Σγ δ = 1 [ (σμ ) γ γ ( σ ν ) γδ (σ ν ) γ γ ( σ ν ) γδ] ω μν (II.38) 8 and similarly Σ δ γ = 1 [ ( σν ) δγ (σ μ ) γ γ ( σ μ ) δγ ] (σ ν ) γ γ ω μν. (II.39) 8 The above commutators of the Pauli matrices arise frequently and so we define the matrices (σ μν ) α β i [ (σ μ ) α α ( σ ν ) αβ (σ ν ) α α ( σ μ ) αβ] = i 2 2 (σμ σ ν σ ν σ μ ) α β ( σ μν ) α β i 2 Thus we secure [ ( σ μ ) αα (σ ν ) α β ( σ ν ) αα (σ μ ) α β] = i 2 ( σμ σ ν σ ν σ μ ) α β. (II.40) Σ β α = i 4 ω μν(σ μν ) α β (Σ ) β α = +i 4 ω μν( σ μν ) β From our definitions of (σ μ ) α α we see that (σ μ ) α α ( σ ν ) αβ = g μν δ β α α. i(σ μν ) β α (II.41) ( σ μ ) αα (σ ν ) α β = g μν δ α β i( σ μν ) α β. The infinitesimal spinor transformations can now be obtained (II.42) ψ α(x )=S β α ψ β (x) =ψ α (x) i 4 ω μν(σ μν ) β α ψ β (x) ψ α (x) 1 2 ω μν(d μν ) β α ψ β (x) (II.43) and ψ α(x )= ψ β(x)(s ) β α = ψ α (x)+ i 4 ω μν ψ β(x)( σ μν ) β α ψ α (x) 1 2 ω μν( D μν ) β α. (II.44) 210

26 Hence, the spinor representations are given by (D μν ) α β +i 2 (σμν ) α β ( D μν ) β α i 2 ( σμν ) β α. (II.45) We must check that these matrices indeed obey the Lorentz algebra as did the tensor representation matrices. After some tedious Pauli matrix algebra we find [σ μν,σ ρσ ] α β = 1 4 [(σμ σ ν σ ν σ μ ), (σ ρ σ σ σ σ σ ρ )] β α = 2i (g μρ σ νσ g μσ σ νρ + g νσ σ μρ g νρ σ μσ ) β α. (II.46) Thus the spinor representation obeys the Lorentz algebra [D μν,d ρσ ] β α =(g μρ D νσ g μσ D νρ + g νσ D μρ g νρ D μσ ) β α, (II.47) and ψ α is the ( 1, 0) spinor representation of the Lorentz group. Similarly the 2 commutation relation for σ μν can be worked out and we find that the complex conjugate dotted spinors ψ α are the (0, 1 ) representation of the Lorentz group 2 with the D μν obeying the Lorentz algebra. As with tensors, we find the intrinsic variations of the spinor fields are given by δψ α = ψ α (x) ψ α(x) =δψ α δx μ μ ψ α For Poincare transformations δ ψ α = ψ α (x) ψ α (x) =δ ψ α δx μ μ ψ α. (II.48) x μ = x μ + ω μν x ν + ɛ μ we find that δψ α = 1 2 ω μν [ (x μ ν x ν μ )δ β α ] (D μν ) α β ψβ (x) ɛ μ μ ψ α (x) and δ ψ α = 1 2 ω μν i 2 ω μν(m μν ) β α ψ β (x)+iɛ μ P μ ψ α (x) [ (x μ ν x ν μ )δ β α ( D ] μν ) β α ψ β (x) ɛμ μ ψ α (x) (II.49) 211

27 i 2 ω μν(m μν ) β α ψ β (x)+iɛμ P μ ψ α (x). (II.50) As with tensors, the P μ and M μν obey the defining commutation relations of the Poincare group. Note that in the rest frame for ψ(x), assuming it describes a massive particle with rest frame four momentum p μ =(m, 0, 0, 0), J i 1 2 ɛ ijkm jk = i 2 ɛ ijkd jk = 1 4 ɛ ijkσ jk =+ 1 2 σi, (II.51) hence the third component of the intrinsic angular momentum J 3 has eigenvalues ± 1 and the particle has spin 1. Similarly finding K we have that 2 2 N N = 1 ( 1 +1) and N N =0,soψ 2 2 α is the ( 1, 0) representation of the 2 Lorentz group. Similarly we find that ψ α is the (0, 1)representationofthe 2 Lorentz group. For finite Poincare transformations x μ =Λ μν x ν + a μ (II.52) we again exponentiate the generators to obtain ψ α(x) =S β α ψ β (Λ 1 (x a)) and with = [ e +iaμ P μ e i 2 ωμν(λ)m μν] β ψ β(x) (II.53) α ψ α(x) = ψ β(λ 1 (x a))(s ) β α = [ ] e +iaμ P μ e i ωμν(λ)m μν β 2 α Λ μν = 1 2 Tr[Sσν S σ μ ] ψ β (x) (II.54) (II.55) and ω μν (Λ) as given earlier e 1 2 ωμν(λ)(xμ ν x ν μ) x ρ =(Λ 1 ) ρ σx σ. (II.56) Thus we have found all representations of the Poincare group. Finally, let s relate our two-component Weyl spinors to the usual Dirac four-component spinors. We can realize the Clifford algebra defining the 212

28 4 4 Dirac matrices γ μa b by using the Pauli matrices, this representation being referred to as the Weyl basis (or representation) or the chiral basis (or representation). Defining the Dirac matrices as ( 0 σ γ μ μ ) σ μ, (II.57) 0 that is, γ 0a b = ab γ 1a b = ab i γ 2a 0 0 +i 0 b = 0 +i 0 0 i ab γ 3a b =. (II.58) Thus, the γ μ obey the defining Dirac anti-commutation relations ab γ μ γ ν + γ ν γ μ =2g μν 1. (II.59) Also, we can define an additional matrix γ 5 = ( σ 0 ) σ 0 = γ 5 +iγ 0 γ 1 γ 2 γ (II.60) 213

29 In this basis the 4 component complex Dirac spinor, denoted ψ a D, is given in terms of two Weyl spinors ψ α and χ α ( ) ψd a ψα χ α = ψ 1 ψ 2 χ 1. (II.61) χ 2 a Under a Lorentz transformation the Dirac spinor transforms as ψ a D(x )=L a bψ b D(x) (II.62) where ( ) S L a β b = α 0 0 (S 1 ) α β. (II.63) ab Further since Λ μν σ μ = Sσ ν S and Λ μν σ μ = S 1 σ ν S 1 we have Λ μν γ a μb = La c γνc d (L 1 ) d b. (II.64) For left, ψ DL, and right, ψ DR, handed spinors we have ψ DL 1 2 (1 γ 5)ψ D and ψ ψ = χ 1 = χ 2 ψ 1 ψ (II.65) ψ DR 1 2 (1 + γ 5)ψ D 0 0 = χ 1. χ 2 (II.66) Thus, we have that ψ DL corresponds to our ( 1 2, 0) spinor ψ α while ψ DR corresponds to our (0, 1 2 ) spinor χ α. If the Dirac spinor is a Majorana spinor, ψ M,thatisψ D is self charge conjugate, then ψ C M = C ψ T M ψ M (II.67) 214

30 with the charge conjugation matrix C given by C = iγ 2 γ ( iσ 2 ) = 0 i σ 2 = with Cγ μ C 1 = γ μt and the conjugate Dirac spinor is defined as ψ D ψ Dγ 0. (II.68) (II.69) The Majorana condition (II.67) implies that C ψ M T = iγ 2 γ 0 γ 0 ψm ( 0 iσ 2 )( ) ψ = i σ 2 0 χ ( ) ( ) χα ψα = ψ α ψ M = χ α, (II.70) or ψ = χ, ψ = χ. Hence we find that a 4 component Majorana spinor is made up of a 2 component Weyl spinor and its complex conjugate ( ) ψα ψ M = ψ α. (II.71) Needless to say the Weyl representation for the Dirac γ matrices is not the only way we could have reralized the Dirac algebra γ μ γ ν + γ ν γ μ =2g μν 1. (II.72) After all, this remains invariant under unitary transformations U, U = U 1 and so (II.72) becomes ˆγ μ U γ μ U, ˆγ μˆγ ν +ˆγ νˆγ μ =2g μν 1. (II.73) 215

31 Further we can use these unitary transformations to define linear combinations of the Weyl spinor components to form a four-component complex spinor ˆψ D U ψ D. (II.74) Under a Poincare transformation we have ˆψ D (x )=U ψ D (x )=U Lψ D (x) =U LUU ψ D (x) = U LUψ D (x) = ˆL ˆψ D (x) (II.75) where ˆL U LU. As before we have for the hatted transformations Λ μν ˆγ μ =Λ μν U γ μ U = U Lγ ν L 1 U = U LUU γ ν UU L 1 U = ˆLˆγ ν ˆL 1. (II.76) Thus, all relations go through as before with all quantities replaced by their hatted values. There are several common choices for the four- component Dirac quantities. We have first defined the Weyl (or chiral) representation, in brief review in obvious notation ( γ μ 0 σ μ ) Weyl σ μ. (II.77) 0 That is ( ) 0 1 γweyl 0 = 1 0 ( o σ γweyl i i ) ( 0 σ i ) = σ i = 0 σ i. (II.78) 0 The Weyl basis Dirac spinor, now denoted ψ Weyl,isgivenas ψ Weyl Under Poincare transformations ( ) ψα χ α = ψ 1 ψ 2 χ 1. χ 2 (II.79) ψ Weyl(x )=Lψ Weyl (x) (II.80) with L = ( ) S 0 0 S 1. (II.81) 216

32 Left handed and right handed chiral spinors are defined by ψ Weyl L 1 2 (1 γ 5 Weyl)ψ Weyl (x) = ( ) ψα 0 ψ Weyl R 1 2 (1 + γ 5 Weyl)ψ Weyl (x) =( 0 χ α ) (II.82) with ( ) 1 0 γ 5 Weyl = Another common representation is that of Dirac γ μ Dirac U γ μ Weyl U (II.83) with U 1 ( ) U = 1 ( ) 1 1. (II.84) Hence γ μ Dirac = 1 ( (σ + σ) μ (σ σ) μ ) 2 ( σ σ) μ (σ + σ) μ (II.85) that is ( ) 1 0 γdirac 0 = 0 1 ( 0 σ γdirac i i ) ( 0 σ i ) = σ i = 0 σ i. (II.86) 0 Writing out all the components in order to be explicit, we have γdirac = γdirac 1 =

33 The γ 5 matrix becomes i γdirac i 0 = 0 i 0 0 i γdirac 3 = γ 5 Dirac U γ 5 Weyl U = (II.87) ( ) 0 1. (II.88) 1 0 Note the γ matrices in all representations obey γ 0 = γ 0,γ i = γ i,γ 5 = γ 5. The Dirac four component spinors (or bi-spinors as they are sometimes called) in the Dirac representation are ψ Dirac U ψ Weyl = 1 ( ) )( ψ χ = 1 ( ) (ψ + χ). (II.89) 2 ( ψ + χ) Hence, the chiral spinors are given by ψ Dirac L = 1 2 (1 γ 5 Dirac)ψ Dirac = U ψ Weyl L = 1 ( ) ψ 2 ψ ψ Dirac R = 1 2 (1 + γ 5 Dirac)ψ Dirac = U ψ Weyl R = 1 ( ). (II.90) 2 χ χ Another common representation is the Majorana representation in which all the γ matrices have pure imaginary matrix elements. In this basis we have γ μ Majorana U γdiracu μ (II.91) with the unitary transformation matrix also being hermitian and given as U 1 ( 1 σ 2 ) 2 σ 2 = U. (II.92) 1 218

34 Let us connect this classical discussion with the treatment of the fields as quantum mechanical operators. Theorem: The Quantum Mechanical Poincare Group. Every continuous unitary representation up to a phase of P+ can be brought, by an appropriate choice of phase factor, into the form of a continuous representation (a,s) U(a,S) of the inhomogeneous SL(2,C). The multiplication law becoming U(a, S) =U(a 1,S 1 )U(a 2,S 2 ) (II.93) for (a, S) =(a 1 + S 1 a 2 S 1,S 1 S 2 ). (II.94) Recall that the inhomogeneous SL(2,C) transformations are defined by x = S xs + a (II.95) where a is a two-by-two Hermitian matrix corresponding to the space-time translation by the four vector a μ. It is understood that the SL(2,C) transformations are performed before the translation. So x 2 = S 2 xs 2+ a 2 (II.96) and x 1 = S 1 x 2 S 1+ a 1 = S 1 S 2 xs 2S 1+ a 1 + S 1 a 2 S 1 = S 1 S 2 x(s 1 S 2 ) + a 1 + S 1 a 2 S 1 S xs + a. (1) Hence, (a, S) =(a 1,S 1 )(a 2,S 2 )=(a 1 + S 1 a 2 S 1,S 1 S 2 ) gives the composition law for ISL(2,C). Finally, let s just point out that if we have an operator, perhaps depending on space-time, A(x), an observer in another frame describes the operator in the same way, it is only translated, rotated or boosted as compared to the original frame. That is, an observer S uses A(x ) to study states Ψ while an observer S uses A(x) to study states Ψ. Since the theory is to be 219

35 relativistically invariant the corresponding matrix elements, the experimental observations, should transform covariantly like a tensor or a spinor, hence, <φ (a,s) A (α) (x ) ψ (a,s) >= D (α) (β) (S) <φ A(β) (x) ψ > (II.97) where x μ =Λ μ ν(s)x ν + a μ and ψ (a,s) >= U(a, S) ψ >so that U 1 (a, S)A (α) (x )U(a, S) =D (α) (β) (S)A(β) (x) (II.98) is the corresponding transformation law for operators; in particular our field operators will transform thusly. Note that ϕ (α) (x ) < φ (a,s) A (α) (x ) ψ (a,s) > is like the classical field ϕ (α) (x ) with the original field ϕ (α) (x) < φ A (α) (x) ψ > so that they transform as ϕ (α) (x )=D (α) (β) (S)ϕ(β) (x). (II.99) Thus quantum mechanical operators transform as U(a, S)A (α) (x)u 1 (a, S) =D 1(α) (β) (S)A(β) (x ). (II.100) We are now ready to study the quantum mechanical representations of the inhomogeneous SL(2,C). Since U(a,S) is unitary we can always write it as the exponential map. In addition, we have that where U(a, S) =U(a, 1)U(0,S) U(a, 1) e iaμp μ U(0,S) e i 2 ωμν(s)mμν (II.101) where the Hermitian (since U is unitary) operators are P μ the space-time translation generators identified with the energy and momentum operators and M μν are the Lorentz transformation (rotation) generators identified with the angular momentum operators. a μ is just the translation vector a μ = 1 Tr[ 2 a σμ ] while ω μν (S) are just the angles of rotation in the x μ x ν plane parameterizing the finite SL(2,C) transformation S, that is S = e i 4 ωμν(s)σμν, (II.102) this is related to Λ μν by Λ μν =Λ μν (S) = 1 2 Tr[ Sσ ν S σ μ]. (II.103) 220

36 For infinitesimal transformations x μ = x μ + ɛ μ + ω μ νx ν, (II.104) with ɛ μ and ω μν infinitesimal parameters, we can expand the unitary operators to first order U(a, S) =1 + iɛ μ P μ i 2 ω μνm μν (II.105) where now ω μν (S) =ω μν, the infinitesimal rotation angles. Recall that since g μν = (g αμ + ω αμ )g αβ (g βν + ω βν ) = g νμ + ω νμ + ω μν, (2) we find ω μν = ω νμ is antisymmetric and hence so is the generator M μν = M νμ. To make more explicit the identification of M μν with rotations consider the transformation U = 1 iω 12 M 12 describing the change in the state vector. The corresponding infinitesimal coordinate change is x 0 = x 0 x 1 = x 1 ω 12 x 2 x 2 = x 2 + w 12 x 1 x 2 = x 3. (II.106) This is a rotation in the x 1 x 2 plane. Hence, im ij is the generator of rotations in the x i x j plane for the state vectors and corresponds to the total angular momentum operator J i 1 2 ɛ ijkm jk =(M 23, M 31, M 12 ). (II.107) For an infinitesimal Lorentz boost along the x 1 direction x 0 = x 0 x 1 ω 01 x 1 = x 1 x 0 ω 01 x 2 = x 2 221

37 x 3 = x 3, the state vector is transformed by U =1 iω 01 M 01. (II.108) (II.109) Hence, im 0i generates Lorentz boosts along the i th axis for the state vectors. We write K i for the three-vector K i M 0i. Since P μ, M μν are the genrators of the Poincare or SL(2,C) group they obey commutation relations which characterize their group multiplication law (the commutation relations for P + and SL(2,C) generators are the same) U(a 1,S 1 )U(a 2,S 2 )=U(a 1 + S 1 a 2 S 1,S 1 S 2 ) U(a, S) 1 = U( S 1 as 1,S 1 ). Using the above laws we find U(a 1, 1)U(a 2, 1) = U(a 1 + a 2, 1) (II.110) (II.111) which implies [P μ, P ν ] = 0. Further, we have U(0,S 1 )U(a, 1)U(0,S)=U(S 1 as 1, 1) that is U(0,S 1 )e iaμp μ U(0,S)=e (Λ 1 (S)a) μp μ. For infinitesimal a μ this yields (II.112) (II.113) U 1 (0,S)a μ P μ U(0,S)=Λ 1ν μ (S)a ν P μ (II.114) or e i 2 ωμν(s)mμν a λ P λ e i 2 ωμν(s)mμν =Λ 1ν μ (S)a ν P μ. (II.115) For infinitesimal S we have a λ P λ i + a λ 2 ω μν[m μν, P λ ] = (δμ ν ωμ ν )a ν P μ Thus we obtain the commutator = a λ P λ aλ 2 (ω μλ ω λμ )P μ. (3) [M μν, P λ ]=i[g λν P μ g λμ P ν ]. (II.116) 222

38 Finally we obtain the angular momentum commutation relations by considering the infinitesimal S transformations U(0,S 1 )U(0,S )U(0,S)=U(0,S 1 S S) (II.117) or e i 2 ωμν(s)mμν e i 2 ω ρλ Mρλ e i 2 ω αβ(s)m αβ = e i 2 ωμν(s 1 S S)M μν. (II.118) Now the parameter describing the product of Lorentz transformations S 1 S S is found by considering the action of the three successive transformations on x μ. First we transform to then to and finally back by x μ 1 =Λ μν (S)x ν, x α 2 =Λ αβ (S )Λ βν (S)x ν, x μ 3 =Λ 1μα (S)Λ αβ (S )Λ βν (S)x ν. (II.119) (II.120) (II.121) For S infinitesimal we have Λ αβ (S )=g αβ + ω αβ (II.122) so Hence, we have that x μ 3 = Λ 1μα (S)g αβ Λ βν (S)x ν +Λ 1μα (S)ω αβλ βν (S)x ν = (g μν +Λ 1μα (S)ω αβλ βν (S))x ν. (4) ω μν (S 1 S S)=Λ 1 μα (S)ω αβ Λ βν (S) (II.123) and thus, U(0,S 1 )M μν U(0,S) = Λ 1αμ (S)Λ νβ (S)M αβ = Λ μα (S)Λ νβ (S)M αβ. (5) Taking S to be infinitesimal also, we find ω ρλ i 2 [Mρλ, M μν ] = (ω μα g νβ + g μα ω νβ )M αβ 223

39 = 1 2 ω ρλ[g ρμ g λα g νβ g λμ g ρα g νβ + g μα g ρν g λβ g μα g λν g ρβ ]M αβ = 1 2 ω ρλ[g ρμ M λν g λμ M ρν + g ρν M μλ g λν M μρ ]. (6) We finally secure the angular momentum commutation relations [M μν, M ρλ ]=i(g μλ M νρ g μρ M νλ + g νρ M μλ g νλ M μρ ). (II.124) As before with the space-time differential operators we define J i 1 2 ɛ ijkm jk and see that they obey the algebra K i M 0i (II.125) [J i, J j ]=+iɛ ijk J k [K i, K j ]= iɛ ijk J k [J i, K j ]=+iɛ ijk K k. (II.126) Hence J are the angular momentum operators, K the boost operators and P μ the translation operators. In particular let s consider the action of the space-time translations further. In the Heisenberg representation the states are independent of time while the operators depend on time. Thus the space-time translation of our operators is determined by the action of P μ. Recall Poincare invariance implies <φ (a,s) A (α) (x ) ψ (a,s) >= D (α) (β) (S) <φ A(β) (x) ψ > (II.127) or equivalently U(a, S)A (α) (x)u 1 (a, S) =D 1(α) (β) (S)A(β) (x ). (II.128) For x = x+ a we find e iaμp μ A (α) (x)e iaμp μ = A (α) (x + a). (II.129) 224

40 For a μ infinitesimal we expand the exponentials and Taylor expand the operator (1 + ia μ P μ )A (α) (x)(1 ia μ P μ )=A (α) (x)+a μ μ A (α) (x), (II.130) which implies ia μ [P μ,a (α) (x)] = a μ μ A (α) (x). Thus for the translation operator we find (II.131) [P μ,a (α) (x)] = i μ A (α) (x) = P μ A (α) (x) (II.132) with P μ i μ, as earlier. Likewise, we can consider Lorentz transformations x μ =Λ μν (S)x ν e i 2 ωμν(s)mμν A (α) (x)e i 2 ωμν(s)mμν = D 1(α) (β) (S)A(β) (x ). (7) For infinitesimal Lorentz transformations x μ = x μ +ω μν x ν and D 1(α) (β) (S) = δ (α) (β) ωμν [D μν ] (α) (β) this becomes (1 i 2 ωμν M μν )A (α) (x)(1 + i 2 ωμν M μν ) = A (α) (x) Hence 1 [ 2 ωμν (x μ ν x ν μ )δ (α) (β) [D μν] (α) (β)] A (β) (x). (8) [M μν,a (α) (x)] = i [ (x μ ν x ν μ )δ (α) (β) [D μν] (α) (β)] A (β) (x) = M μν A (α) (x), (9) with M μν as we found earlier for the classical fields. 225