Ψ L 1 2 (1 γ 5)Ψ. Ψ R 1 2 (1 + γ 5)Ψ (2.3.1)

Transcription

1 .3 THE SPIN 1 FERMION FIELD We next desire to describe non-interacting (free) particles with spin 1.Aswe know these particles will obey Fermi- Dirac statistics due to the Pauli exclusion principle. Hence, we can expect the commutation rules to be changed for spin 1 fields since the canonical commutation rules implied symmetric wave functions under the interchange of the particles. For antisymmetric wave functions we will need anti-commutation relations for the fermion creation and annihilation operators. First, let s introduce the field that will be used to describe the spin 1 particles; the four component Dirac (bi-)spinor field, denoted by Ψ a (x), where a =1,, 3, 4. Recall that we can obtain the left-handed and right-handed components of Ψ by multiplying with respect to the γ 5 matrix Ψ L 1 (1 γ 5)Ψ Ψ R 1 (1 + γ 5)Ψ (.3.1) where γ 5 +iγ 0 γ 1 γ γ 3 with γ µ, µ =0, 1,, 3 are the four Dirac γ matrices. The Dirac matrices are four-by-four matrices that obey their defining Clifford algebra and have the conjugation properties γ µ γ ν + γ ν γ µ =g µν 1 (.3.) γ 0 = γ 0 γ i = γ i (.3.3) that is γ µ = γ 0 γ µ γ 0 (.3.4) (see section 1. for a review of the γ-matrices). The Dirac spinor field Ψ a (x) has the next simplest non-trivial transformation properties under the restricted Poincare group (recall that spin zero or scalar fields were invariant under L + i.e. they transformed according to the identity U 1 (a, Λ)Φ(Λx + a)u(a, Λ) = 1Φ(x) ). 154

2 The transformation law is given by the quantum mechanical operator transformation law analogous to equation (1..18) and (1.3.8), that is, for x =Λx + a, (ψ a(x )) < A Ψ a (x ) B > = L ab (Λ) <A Ψ b (x) B > (= L ab ψ b (x)) (.3.5) where ψ b (x) < A Ψ b (x) B >and we denote D ab (S) L ab. Since the transformation is implemented by unitary operators U(a, Λ) we have U 1 (a, Λ)Ψ a (x )U(a, Λ) = L ab (Λ)Ψ b (x) (.3.6) or multiplying on the left by U and on the right by U 1, we obtain where recall equation (1..131) U(a, Λ)Ψ a (x)u 1 (a, Λ) = L 1 ab (Λ)Ψ b(λx + a) (.3.7) Λ µν γ µ = Lγ ν L 1 (.3.8) that is Λ µν = 1 4 Tr [ γ µ Lγ ν L 1]. (.3.9) Since any finite P + transformation can be built up from the infintesimal, consider those x µ = x µ + ω µν x ν + a ν (.3.10) ω µν = ω νµ (.3.11) hence U(a, Λ) = e ia µp µ e i ω µνm µν 1+ia µ P µ i ωµν M µν (.3.1) while by equations (1..19), (1..98), (1..107) and (1..13) L ab = δ ab i 4 ω µνσ µν ab (.3.13) where now (σ µν ) ab i [γµ,γ ν ] ab (.3.14) 155

3

4 [ [M µν, Ψ a (x)] = i (x µ ν x ν µ )Ψ a (x) i ] σµν ab Ψ b(x) (.3.1) where D µν ab i σµν ab according to the notation of equations (1..115) and (1..116) and obeys the Lorentz algebra, equation (1..113). We can contract spinor fields and Dirac matrices in order to make scalars, vectors and tensors. Hence it can be shown that under Lorentz transformations ΨΨ is a Lorentz scalar, Ψγ 5 Ψ is a Lorentz pseudoscalar, Ψγ µ Ψ is a Lorentz vector, Ψγ 5 γ µ Ψ is a Lorentz pseudovector or axial vector, Ψσ µν Ψ is a Lorentz tensor, (.3.) where Ψ Ψ γ 0. For non-interacting spin 1 particles of mass m, Dirac discovered that the field equation is given by a first order partial differential equation known as the Dirac equation (i/ m)ψ(x) =0 (.3.3) where / γ µ µ. Writing out the indices we have i x µγ µ ab Ψ b(x) mψ a (x) =0, (.3.4) so we see that the Dirac equation is a matrix differential equation. Since the particles have mass m the relativistic relation p = m should be valid, that is, Ψ(x) should also obey the Klein-Gordon equation. To see this consider (i/ + m)(i/ m)ψ= ( / / m ) Ψ =0. (.3.5) However, and thus, / / = γ µ γ ν µ ν = 1 (γµ γ ν + γ ν γ µ ) µ ν = g µν 1 µ ν = (.3.6) ( + m ) Ψ a (x) =0, (.3.7) 157

5 each component of Ψ(x) obeys the Klein-Gordon equation. Furthermore, if (i/ m) Ψ = 0 then taking the hermitian conjugate we obtain Ψ ( i ) / m =0 (.3.8) that is, ( ) Ψ i leftarrow µ γ µ m =0. (.3.9) Multiplying by γ 0 and using the relation γ µ γ 0 = γ 0 γ µ yields ) Ψ γ ( i 0 µ γ µ m =0. (.3.30) Now we define the adjoint Dirac spinor as Ψ Ψ γ 0 (.3.31) so that the preceding equation takes the form ( ) Ψ(x) i +m =0 (.3.3) which is called the adjoint Dirac equation. Dirac showed that these field equations can be obtained as Euler-Lagrange equations from the Lorentz invariant Dirac Lagrangian ( i L = Ψ m ) Ψ. (.3.33) Notice that L = L, but often we write L = Ψ(i/ m) Ψ which is not hermitian but differs from the hermitian Lagrangian by a total divergence i µ ( Ψγ µ Ψ ),so the action is the same. Thus, either Lagrangian leads to the same Euler-Lagrange equations of motion, L Ψ L µ µ Ψ =0=(i/ m)ψ, (.3.34) that is the Dirac equation. 158

6 According to our canonical quantization procedure we define the momentum conjugate to Ψ by L 0 Ψ. However, typical of Lagrangians first order in t, the canonically conjugate momenta Π and Π are proportional to the field variables Ψ and Ψ, and hence, the fields and momenta are not independent quantities, i.e. the fields are conjugate to each other. We cannot naively apply our canonical quantization procedure. Dirac found that we must throw all the derivatives to one field when finding its conjugate momenta, choose it as the independent coordinate and then the other field will be the momentum. Either way we throw the derivative, the result is the same momentum-field relation. Thus, throwing the derivatives to Ψ by parts, L = i µ Ψγ µ Ψ m ΨΨ and ignoring the total divergence we define Π a (x) L 0 Ψ a (x) = iγ 0 Ψ(x). (.3.35) As we will see and as Jordan and Wigner discovered, we must further change the canonical commutation relations used for bosons to anticommutation relations for fermions. The quantization rules become δ(x 0 y 0 ){Π a (x), Ψ b (y)} = iδ ab δ 4 (x y) (.3.36) where {A, B} =[A, B] + = AB + BA, and δ(x 0 y 0 ){ Ψ(x), Ψ(y)} =0, (.3.37) and δ(x 0 y 0 ){ Π(x), Π(y)} =0=δ(x 0 y 0 ){Ψ(x), Ψ(y)}. (.3.38) Now the equal time anti-commutation relations (ETAR), (.3.36), are δ(x 0 y 0 ){ iγ 0 acψ c (x), Ψ d (y)γ0 db} = iδ ab δ 4 (x y), (.3.39) multiplying by γeaγ 0 bf 0, we obtain that is, in summary, δ(x 0 y 0 ){Ψ e (x), Ψ f (y)} = δ ef δ 4 (x y) (.3.40) δ(x 0 y 0 ){Ψ a (x), Ψ b (y)} = δ abδ 4 (x y) 159

7 δ(x 0 y 0 ){Ψ a (x), Ψ b (y)} =0. (.3.41) A few words are in order. The fact that these operators anticommute means that they are not like ordinary functions of x, rather they are elements of an infinite dimensional Grassmann algebra (they are anti-commuting numbers). Furthermore, we will see that these are the correct quantization rules for the Dirac fields to obey. They are correct in that they lead to a consistent definition of energy, momentum, and spin of Fermi-Dirac particles. This will become clearer as we proceed. Naturally we can find the Hamiltonian density given the momenta H = 0 Ψ Π L ( ) = i 0 Ψγ 0 Ψ+ Ψ i +m Ψ = Ψ iγ i x +m Ψ. (.3.4) i Thus, the Hamiltonian is H = ( d 3 xψ i ) i +m Ψ (.3.43) where we have again made H hermitian by adding total three-divergences. The Heisenberg equations are as before [H, Ψ(x)] = i Ψ(x). More generally we can apply Noether s Theorem to construct the energymomentum tensor and angular momentum tensor T µν ν Ψ L µ Ψ + ν Ψ L µ Ψ gµν L (.3.44) where now we must be careful taking derivatives with respect to anticommuting variables Ψ a Ψ b = δ ab Ψ b =0 Ψ a Ψ b = δ ab Ψ a 160

8 Ψ b =0 (.3.45) Ψ a and for Ψ b either Ψ b or Ψ b and A an arbitrary operator Ψ a Ψb A = Ψ b Ψ a A Ψ b A Ψ a. (.3.46) Similarly for. Thus we have the general chain rule for derivatives with respect Ψ a to anti-commuting quantities Ψ a AB = A Ψ a B +( 1) A A B Ψ a (.3.47) where A = 1 if A is an odd element of the Grassmann algebra i.e. if A is an anticommuting number or A = 0 if A is an even element of the Grassmann algebra i.e. if A is a commuting number. Thus the reason for the factors appearing to the left in the expression for T µν above. We shall also use the hermitian Lagrangian so that T µν is hermitian L = i Ψ Ψ mψψ. (.3.48) So L µ Ψ = i γµ Ψ L µ Ψ = i Ψγµ (.3.49) where the minus sign appears in the second term since µ Ψ Ψ= Ψ µ Ψ. (.3.50) Hence, T µν = i ν Ψγ µ Ψ+ i Ψγµ ν Ψ g µν L, (.3.51) the plus sign comes from the relation ν ΨΨ= Ψ ν Ψ. Thus we secure T µν = i [ Ψγ µ ν Ψ ν Ψγ µ Ψ ] g µν L = i. (.3.53) Ψγµ ν Ψ g µν L 161

9 Now we check directly that this is conserved µ T µν = i Ψ/ ν Ψ+ i Ψ ν Ψ i ν Ψ Ψ i ν Ψ/ Ψ ν L. (.3.53) ( ) Using the field equations (i/ m) Ψ = 0 and Ψ i +m = 0, we obtain ( ) ( ) i µ T µν = ν Ψ Ψ i ν i Ψ Ψ ν Ψ Ψ + i Ψ ν Ψ ν L ( ) i = ν Ψ Ψ m ν ΨΨ m Ψ ν Ψ ν L ( ) i = ν Ψ Ψ m ΨΨ ν L =0. (.3.54) Hence, the energy momentum tensor is conserved µ T µν =0. (.3.55) Notice however that T µν T νµ = i [ Ψγ µ ν Ψ Ψγ ν µ Ψ ν Ψγ µ Ψ+ µ Ψγ ν Ψ ] 0. (.3.56) We can put this difference into a more useful form by using the field equations i/ Ψ =mψ and iψ = mψ. By multiplying the Dirac equation by Ψγ µ γ ν and the adjoint equation by γ µ γ ν Ψ we obtain Hence, adding we find iψγ µ γ ν γ ρ ρ Ψ=mΨγ µ γ ν Ψ i ρ Ψγ ρ γ µ γ ν Ψ= mψγ µ γ ν Ψ. (.3.57) Ψγ µ γ ν γ ρ ρ Ψ+ ρ Ψγ ρ γ µ γ ν Ψ=0. (.3.58) 16

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11 where G ρµν = 1 [Hρµν + H µνρ + H νµρ ] = 1 ( 1 ) [ ɛ ρµνλ + ɛ µνρλ + ɛ νµρλ] Ψγ 5 γ λ Ψ (.3.68) = 1 Hρµν and ρ G ρµν = 1 ρh ρµν ρ G ρµν = i 4 = 1 4 ρɛ ρµνλ Ψγ 5 γ λ Ψ [ ] Ψγ µ ν Ψ Ψγ ν µ Ψ. (.3.69) Thus, according to (.3.5), (.3.67) and (.3.69), the symmetric Belinfante energy momentum tensor becomes Θ µν = i Ψγµ ν Ψ g µν L i 4 Ψγµ ν Ψ+ i 4 Ψγν µ Ψ Θ µν = i 4 Ψ [ γ µ ν +γ ν µ ] Ψ g µν L. (.3.70) It is conserved and symmetric µ Θ µν = µ T µν =0 (.3.71) Θ µν =Θ νµ. (.3.7) Note all this follows from Belinfante s general prescription, equations (.1.157), (.1.158) and (.1.159) H ρµν Π ρ rd rs µν Φ s (.3.73) which in our case yields H ρµν = D µν Ψ L ρ Ψ + ΨDµν L ρ Ψ = i Ψγρ D µν Ψ i ΨDµν γ ρ Ψ. (.3.74) 164

12 However, D µν = i σµν and D µν = γ 0 D µν γ 0 = i σµν, so that H ρµν = 1 4 Ψ{γρ,σ µν }Ψ = 1 ɛρµνλ Ψγ 5 γ λ Ψ. (.3.75) The generator of translations is given by P µ d 3 xθ 0µ = d 3 xt 0µ ( ) i = d 3 x Ψγ0 µ Ψ g 0µ L. (.3.76) Consequently, the Hamiltonian is as given by the Legendre transform of L, equation (.3.43) ( i H = P 0 = d 3 i x Ψγ0 Ψ 0Ψγ 0 Ψ i Ψγ0 Ψ+ i 0Ψγ 0 Ψ i ) Ψγi i Ψ+mΨΨ ( = d 3 xψ i ) (.3.77) γi i +m Ψ = d 3 x i ( ) Ψ Ψ Ψ Ψ since L = 0 upon use of the Dirac equations, and [ P 0, Ψ(x) ] = i Ψ(x). (.3.78) In addition, we have P = = = d 3 x i Ψγ0 Ψ ( ) d 3 x iψγ 0 Ψ ) d 3 x (iψ Ψ (.3.79) so that, since P is time independent, [ ] [ P, Ψ(x) = i d 3 y Ψ (y) ] y Ψ(y), Ψ(x) y 0 =x 0 = i d 3 y{ψ (y), Ψ(x)} y Ψ(y). y 0 =x 0 (.3.80) 165

13 Using the ETAR equation (.3.41) this yields [ P, Ψ(x)] = i Ψ(x). (.3.81) Thus, we have as we should for space-time translations [P µ, Ψ(x)] = i µ Ψ(x). (.3.8) According to Belinfante s procedure the angular momentum tensor is given by M µνρ = x ν Θ µρ x ρ Θ µν µ M µνρ =0 (.3.83) and M µν d 3 xm 0µν (.3.84) which yields [ [M µν, Ψ(x)] = i (x µ ν x ν µ )Ψ(x) i ] σµν Ψ(x). (.3.85) Finally, since Ψ is a complex field we also have a phase invariance because L is made from Ψ Ψ. As in the complex scalar case we introduce unitary operators U(α) =e iαq transforming the phase of Ψ and Ψ Since U(α) =e iαq, (.3.86) implies U (α)ψ(x)u(α) e iα Ψ(x) U (α)ψ(x)u(α) e iα Ψ(x). (.3.86) [Q, Ψ(x)] = Ψ(x) [ Q, Ψ(x) ] =+Ψ(x). (.3.87) Since L is invariant under this transformation δl = δl = U (α)lu(α) L=0 (.3.88) 166

14 we have by Noether s Theorem a conserved current in terms of the dynamical variables Ψ and Ψ whose zero component gives the charge Q ( J µ i Ψ L µ Ψ + L ) µ Ψ Ψ ( J µ = i i Ψγµ Ψ i ) Ψγµ Ψ J µ = Ψγ µ Ψ. (.3.89) Checking explicitly the conservation of this current we have using the Dirac equations µ J µ = iψi/ Ψ iψi Ψ+imΨΨ imψψ ( ) ( ) = iψ i m Ψ iψ i (.3.90) +m Ψ=0. Hence the U(1) phase symmetry current is conserved µ J µ =0. (.3.91) By Noether s theorem the U(1) charge is Q d 3 xj 0 (x) = d 3 xψγ 0 Ψ Q = d 3 xψ Ψ. (.3.9) Further Q = 0 since δl = 0, or more generally, the internal and space-time symmetry generators commute [Q, P µ ]=0 [Q, M µν ]=0. (.3.93) Finally we check explicitly that Q given by equation (.3.9) indeed induces phase transformations on the Dirac fields [Q, Ψ(x)] = d 3 y [ Ψ(y)γ 0 Ψ(y), Ψ(x) ] y 0 =x 0 = d 3 y{ψ a(y), Ψ(x)}Ψ a (y) y 0 =x 0 167

15 [Q, Ψ(x)] = Ψ(x) (.3.93) and similarly [ Q, Ψ(x) ] =+ Ψ(x) (.3.94) as required. Thus we have obtained the space-time properties of the Dirac field theory according to the canonical quantization procedure. To interpret this quantum field theory in terms of particle states and associated creation and annihilation operators, we Fourier transform Ψ(x) to momentum space, that is, expand it in terms of plane wave solutions of the Dirac equation Ψ(x) = d 4 k (π) 4 e ikx Ψ(k). (.3.95) As we have seen Ψ is a solution to the Klein-Gordon equation, ( + m ) Ψ=0, (.3.96) implying that Ψ a (k) =(π)δ ( k m ) w a ( k,k 0 ). (.3.97) [ Using δ(k m )= 1 δ(k 0 ω k )+δ(k 0 + ω k ) ], this yields Ψ(x) = [ e ikx w( (π) 3 ω k,+k 0 )+e +ikx w( ] k, k 0 ), (.3.98) k where k 0 = ω k = k + m, as usual, is understood. In addition to the Klein- Gordon equation, Ψ also obeys the Dirac equation, (i/ m)ψ(x) = 0, (i/ m)ψ(x) = Hence w a obeys the 4 4 matrix equation [ e ikx (/k m)w( (π) 3 ω k,+k 0 ) k e +ikx (/k + m)w( ] k, k 0 ) =0. (.3.99) (/k m)w( k,+k 0 )=0 (/k + m)w( k, k 0 )=0. (.3.100) 168

16 It is convenient to separate the c-number spinor solutions to the Dirac equation from the q-number operator coefficients. We define w a ( k,+k 0 ) b( k)u a ( k) w a ( k, k 0 ) d ( k)v a ( k) (.3.101) where b and d are q-number operators, i.e. creation and annihilation operators, while u a ( k) and v a ( k) are c-number spinor solutions to the momentum space Dirac equations (/k m)u( k)=0 (/k + m)v( k)=0. (.3.10) Since Ψ is complex, w a ( k,+k 0 ) and w a ( k, k 0 ) are independent, thus the coefficients b and d and the column vectors u and v are independent also. In order to solve (.3.10) for u and v we go to the rest frame of the particle, that is consider these equations at k = 0 and k 0 = m. So equations (.3.10) become m ( γ 0 1 ) u(0) = 0 m ( γ 0 +1 ) v(0) = 0. (.3.103) That is we obtain the γ 0 Dirac matrix eigenvalue equations γ 0 u(0) = +u(0) γ 0 v(0) = v(0). (.3.104) Hence, u and v are the +1 and 1 eigenvectors of γ 0 respectively. Recall that γ 0 = 1, so γ 0 has two pairs of eigenvalues ±1 as seen directly in the Dirac representation where γ 0 is diagonal. u(0) denotes the pair of +1 eigenvalue eigenvectors and v(0) the 1 pair. Since the γ 0 eigenvalues are degenerate we need another label for u( k) and v( k) to denote the independent eigenvectors. It is conventional as to how to choose the orthogonal pair of degenerate γ 0 eigenvectors. We choose them to be eigenvectors of σ 1 also, these are simultaneously diagonalizable since [ γ 0,σ ij] =0. As we will see σ ij will give the three components of the spin of the particle at rest, σ 1 being the projection of the spin onto the x 3 axis. So the four independent vectors in the k-frame are denoted by u (s) ( k) and v (s) ( k) where s =1, labels the 169

17 two u and the two v vectors. We take the rest frame vectors to obey the eigenvalue equations (.3.104) as stated γ 0 u (s) (0) = +u (s) (0) γ 0 v (s) (0) = v (s) (0) (.3.105) as well as the σ 1 eigenvalue equations, note σ 1 has eigenvalues ±1 also, σ 1 u (s) (0) = (3 s)u (s) (0) = ( 1) s+1 u (s) (0) σ 1 v (s) (0) = (3 s)v (s) (0) = ( 1) s+1 v (s) (0), (.3.106) that is for instance σ 1 u (1) (0) = +u (1) (0) σ 1 u () (0) = u () (0). (.3.107) Furthermore, we have that u (r) (0) and v (r) (0) are orthogonal and we normalize them, by convention, to m. So we take u (r) (0) u (s) (0) = mδ rs v (r) (0) v (s) (0) = mδ rs u (r) (0) v (s) (0) = v (r) (0) u (s) (0) = 0. (.3.108) Recall that in the Dirac representation while for i j γ 0 = ( ) 0 σ γ i i = σ i 0 σ ij = iγ i γ j ( ) σ = i i σ j 0 0 σ i σ j ( ) = ɛ ijk σ k 0 0 σ k. (.3.109) (.3.110) 170

18 Hence, σ 1 = ( ) σ σ 3 = Thus the simultaneous (γ 0,σ 1 ) eigenvectors are u (1) (0) = 1 0 m (.3.111) u () (0) = 0 1 m 0 0 v (1) (0) = 0 0 m 1 0 v () (0) = 0 0 m. (.3.11) 0 1 We also have creation and annihilation operators for each of these eigenmodes. Hence, the expressions (.3.101) for w a ( k, +k 0 ) and w a ( k, k 0 ) become a sum over the k-frame eigenvector basis u (s) ( k),v (s) ( k) w a ( k,+k 0 ) w a ( k, k 0 ) s=1 s=1 b s ( k)u (s) a ( k) The Dirac field Fourier transform, equation (.3.98), becomes d s( k)v (s) a ( k). (.3.113) Ψ a (x) = (π) 3 ω k [ b s ( k)u a (s) ( k)e ikx + d s( k)v a (s) ( ] k)e +ikx. (.3.114) s=1 171

19 (Compare this with the complex scalar field expansion, equation (..66) [ Φ(x) = (π) 3 a( ω k)e ikx + b ( ] k)e +kx.) (.3.115) k Now u (s) ( k) and v (s) ( k) still must be solutions of the Dirac equation (/k m)u (s) ( k)=0 (/k + m)v (s) ( k)=0. (.3.116) Since (/k + m)(/k m) =(/k m)(/k + m) =k m = 0 we can solve the above equations (alternatively we could just boost from the rest frame u (s) (0) and v (s) (0) to momentum k by means of (SL(,C)) Lorentz transformation S( k) to obtain the same answer) u (s) ( k)=a( k)(/k + m)u (s) (0) v (s) ( k)=b( k)(/k m)v (s) (0) (.3.117) with A and B arbitrary normalization constants chosen by convention. We choose u and v just to be the (SL(,C)) Lorentz transformation matrices to the moving frame k since the u(0) and v(0) are already normalized at rest as we desire. Hence, A( k) 1 m(ωk + m) B( k). (.3.118) Consequently, u (s) ( k)= (/k + m) m(ωk + m) u(s) (0) v (s) ( k)= ( /k + m) m(ωk + m) v(s) (0) (.3.119) where u (s) (0) and v (s) (0) are given above in the Dirac representation. To see this consider the Lorentz boost to go from the rest frame of the particle to the frame in which the particle has momentum k. That is with k µ rest =(m, 0, 0, 0). Hence Λ µν ( k) is given by k µ =Λ µν ( k)k rest ν, (.3.10) Λ 0 0( k) = cosh β 17

20 Λ i 0( k)=λ 0 i( k)= ki k sinh β k j Λ i j( k)=δj i + ki k [cosh β 1], (.3.11) k where tanh β = k ω k. Correspondingly the SL(,C) transformation, denoted S( k)= W ( k), that takes the spinor u (s) (0) from the rest frame to the k-frame, u (s) ( k), is W ( k)= (/k + m) m(ωk + m), (.3.1) such that Likewise, we have with W ( k)= u (s) ( k)=w ( k)u (s) (0). (.3.13) v (s) ( k)= W ( k)v (s) (0), (.3.14) ( /k + m) m(ωk + m) = γ 5W ( k)γ 5. (.3.15) Recalling the relation between SL(,C) transformations and Lorentz transformation matrices, equation (1..131), Λ µν ( k)γ µ = W ( k)γ ν W 1 ( k), or in a more direct form Λ µν ( k)= 1 4 Tr[γµ Wγ ν W 1 ]. (.3.16) Using equation (.3.15) we see that W ( k) yields the same Lorentz transformation matrix. Thus using the explicit form for W ( k), equation (.3.1), we verify that the Lorentz transformation obtained in equation (.3.16) is indeed the same as in equation (.3.10) and (.3.11). Thus, the Fourier transform breaks up into positve and negative frequency components Ψ + (x) = Ψ (x) = Ψ(x) =Ψ + (x) +Ψ (x) (π) 3 ω k (π) 3 ω k 173 s=1 s=1 b s ( k)u a (s) ( k)e ikx d s( k)v (s) a ( k)e +ikx (.3.17)

21 where the positive frequency plane wave solutions to the Dirac equation are and the negative frequency solutions are U (s) k (x) u (s) ( k)e ikx (.3.18) V (s) k (x) v (s) ( k)e +ikx (.3.19) with (i/ m)u (s) (x) =0=(i/ m)v (s) k (x). (.3.130) k Now we desire to invert these Fourier transforms and to do so we need the orthogonality properties of the plane wave solutions u and v. First consider the hermitian conjugate (in Dirac matrix space) u (r) ( k)= 1 m(ωk + m) u(r) (0)(/k + m), (.3.131) but γ µ = γ 0 γ µ γ 0 so that u (r) ( k)= 1 m(ωk + m) u(r) (0)(γ 0 /kγ 0 + m). (.3.13) However, hence and so γ 0 u (r) (0) = u (r) (0), (.3.133) u (r) (0)γ 0 = u (r) (0) (.3.134) u (r) ( (/k + m) k)=u (r) (0) m(ωk + m) γ0. (.3.135) Thus, the orthogonality properties are given by u (r) ( k)u (s) ( k)=u (r) (0) (/k + m)γ0 (/k + m) u (s) (0) m(ω k + m) = u (r) (0) ( /k + m +γ0 k 0 )(/k + m) u (s) (0) m(ω k + m) = u (r) (0) w kγ 0 (/k + m) m(ω k + m) u(s) (0) = u (r) (0) ω k(ω k + m) m(ω k + m) u(s) (0) + u (r) (0) ω kγ 0 γ i k i m(ω k + m) u(s) (0). (.3.136) 174

22 But γ 0 γ i γ 0 = γ i, so that u (r) (0)γ 0 γ i γ 0 u (s) (0) = u (r) (0)γ i u (s) (0) = u (r) (0)γ i u (s) (0), (.3.137) hence, u (r) (0)γ i u (s) (0) = 0. (.3.138) Since we normalized at rest to m, u (r) (0)u (s) (0) = mδ rs, (.3.139) we find u (r) ( k)u (s) ( k)=ω k δ rs. (.3.140) Similarly for the negative energy spinors v (s) ( k) and v (r) ( k)= v (r) ( /k + m) (0) m(ωk + m) γ0, (.3.141) which leads to v (r) ( k)v (s) ( k)=ω k δ rs. (.3.14) Furthermore, we have u (r) ( k)v (s) ( k)=u (r) (0) (/k + m)γ0 ( γ 0 /kγ 0 + m) v (s) (0) m(ω k + m) = u (r) (0) (/k + m)( /k + m)γ0 v (s) (0) = 0. m(ω k + m) (.3.143) Thus we secure u (r) ( k)v (s) ( k)=0=v (r) ( k)u (s) ( k). (.3.144) Equivalently we could introduce adjoint spinors u (r) ( k) u (r) ( k)γ 0 v (r) ( k) v (r) ( k)γ 0. (.3.145) Then, as usual, if (/k m)u (r) ( k)=0 (.3.146) 175

23 taking the hermitian conjugate yields u (r) ( k)(/k m) =0 = u (r) (. (.3.147) k)(γ 0 /kγ 0 m) Multiplying by γ 0 implies u (r) ( k)(/k m) =0, (.3.148) and similarly (/k + m)v (r) ( k) = 0 leads to v (r) ( k)(/k + m) =0. (.3.149) Hence, we can proceed as before, so just summarizing the results here for the spinor and adjoint spinor orthogonality relations in the k- frame u (r) ( k)u (s) ( k)=δ rs m v (r) ( k)v (s) ( k)= δ rs m u (r) ( k)v (s) ( k)=0=v (r) ( k)u (s) ( k). (.3.150) Besides orthogonality the spinors are complete; that is we have for positive energy spinors u (s) ( k)u (s) ( k)=(/k + m) (.3.151) s=1 where we used the fact that u (s) (0) are the +1 eigenvectors of γ 0 (see (.3.11)) and thus u (s) (0)u (s) (0) = 1 (1 + γ0 )m. (.3.15) s=1 176

24 It then follows that u (s) ( k)u (s) ( k)= s=1 = m (m)(ω k + m) (/k + m)(1 + γ0 ) (/k + m)γ 0 m (m)(ω k + m) (/k + m)(1 + γ0 ) (/k + m) (m) 1 [ = (/k + m)(/k + m)+/kk 0 (m)(ω k + m) k γ 0 + m γ 0 + m{/k,γ 0 } ] m = (4m)(ω k + m) [(/k + m)(/k + m)+ω k(/k + m)] m = (4m)(ω k + m) [k + m +m/k +ω k /k +ω k m] m = (4m)(ω k + m) [m(m +/k)+ω k(/k + m)] =/k + m. (.3.153) Similarly the negative energy spinors obey v (s) ( k)v (s) ( k)=(/k m) (.3.154) s=1 since v (s) (0) are the 1 eigenvalue eigenvectors of γ 0 (see equation (.3.11)) and thus v (s) (0)v (s) (0) = 1 (1 γ0 )m. (.3.155) s=1 Thus, we have completeness of the spinors 1 m [u(s) ( k)u (s) ( k) v (s) ( k)v (s) ( k)] = 1, (.3.156) s=1 where 1 is the 4 4 identity matrix. (Note that we have chosen our spinors to be eigenstates of γ 0 and σ 1 at rest. Alternatively, we could have chosen another combination of 4 4 γ matrices at rest or in another frame. Mandl and Shaw have chosen u and v to be eigenvectors of the helicity operator h k σ k (.3.157) 177

25 where σ (σ 3,σ 31,σ 1 ) and of course the Dirac equation /ku = mu and /kv = mv in the k-frame, not at rest as we have chosen. Also note that [/k,h] = 0. Each choice are linear combinations, boosts, and rotations of the other.) We can now invert the Fourier transforms of the Dirac field Ψ(x) = (π) 3 ω k s=1 [ b s ( k)u (s) ( k)e ikx + d s( k)v (s) ( k)e +ikx] (.3.158) and the adjoint Dirac field Ψ(x) = (π) 3 ω k [d s ( k)v (s) ( k)e ikx + b s( k)u (s) ( ] k)e +ikx s=1 (.3.159) to obtain b s ( k)= d s( k)= b s( k)= d s ( k)= d 3 xe +ikx u (s) ( k)γ 0 Ψ(x) d 3 xe ikx v (s) ( k)γ 0 Ψ(x) d 3 xψ(x)γ 0 u (s) ( k)e ikx d 3 xψ(x)γ 0 v (s) ( k)e +ikx. (.3.160) The canonical anti-commutation relations (CAR) can now be obtained for b and d and their hermitian conjugates by using the ETAR and the fact that they are time independent, ḃ =0= d, for example { b r ( p ),b s( } k) = d 3 xd 3 ye ipx e iky x 0 =y 0 = d 3 xd 3 ye ipx e iky x 0 =y 0 = d 3 xe i(p k)x u (r) ( p )γ 0 u (s) ( k) ( u (r) ( p )γ 0) ( u (s) ( ) } k) {Ψ a (x), Ψ b (y) a b ) ( u (r) ( p )γ 0) ( u (s) ( k) a b δ ab δ 3 ( x y ) =(π) 3 δ 3 ( p k)e i(ω p ω k )x 0 u (r) ( p )u (s) ( p ) (.3.161) but δ 3 ( p k)e i(ω p ω k )x 0 = δ 3 ( p k) and u (r) ( p )u (s) ( p )=ω p δ rs, so that {b r ( p ),b s( k)} =(π) 3 (ω k )δ rs δ 3 ( p k). (.3.16) 178

26 Similarly we find {d r ( p ),d s( k)} =(π) 3 (ω k )δ rs δ 3 ( p k). (.3.163) Since Ψ and Ψ anticommute as do Ψ and Ψ we have {b r ( p ),b s ( k)} =0={d r ( p ),d s ( k)} (.3.164) and {b r ( p ),d s ( k)} =0 (.3.165) and since u and v are orthogonal we also find {b r ( p ),d s ( k)} =0, (.3.166) the adjoints of the above expressions vanish as well. The expressions for the Hamiltonian, momentum, and charge operators can now be Fourier transformed. As usual we obtain these quantities by starting with P µ = d 3 x i Ψγ0 µ Ψ Q = d 3 xψγ 0 Ψ M µν = = d 3 x [ x µ Θ 0ν x ν Θ 0µ] d 3 x i 4 Ψ [ γ 0 x µ ν γ 0 x ν µ +(x µ γ ν x ν γ µ ) γ 0 0 (.3.167) + ( g 0µ γ ν g 0µ γ ν)] Ψ. 179

27 First we analyze the energy momentum operator P µ = d 3 x i d 3 l [ d (π) 3 ω k (π) 3 r ( ω k)v (r) ( k)e ikx l r=1 s=1 +b r( k)u (r) ( k)e +ikx] γ 0 x [b µ s ( l)u (s) ( l)e ilx + d s( l)v (s) ( l)e +ilx] = i d 3 d 3 l [ x d (π) 3 ω k (π) 3 r ( ω k)v (r) ( k)e ikx l r,s=1 +b r( k)u (r) ( k)e +ikx] [ γ 0 il µ b s ( l)u (s) ( l)e ilx + il µ d s( l)v (s) ( l)e +ilx] + [ik µ d r ( k)v (r) ( k)e ikx ik µ b r( k)u (r) ( ] k)e +ikx γ 0 [b s ( l)u (s) ( l)e ilx + d s ( l)v (s) ( ] l)e +ilx = i (π) 3 ω k r=1 [ d r ( k)d r ( k)ik µ ik µ b r ( k)b r ( ] k). (.3.168) So we find for the energy-momentum operator P µ = (π) 3 ω k s=1 k [b µ s( k)b s ( k) d s ( k)d s( ] k). (.3.169) The charge operator is found similarly to be Q = (π) 3 ω k s=1 [ b s( k)b s ( k)+d s ( k)d s( ] k). (.3.170) With an analogous expression for M µν which we will not bother to calculate here. We see that the Hamiltonian is given by H = P 0 = (π) 3 ω k s=1 ω k [b s ( k)b s ( k) d s ( ] k)d s ( k). (.3.171) First suppose that d and d obeyed the CCR rather than the CAR, then we can write H as H = ω (π) 3 k [b ω s( k)b s ( k) d s( k)d s ( ] k) k s=1 180

28 ω k (π) 3 ω k [d s ( k),d s( k)] (.3.17) s=1 where now we have that [d r ( k),d r]=(π) 3 ω k δ rr δ( k k), the CCR. Thus H would be given by H = ω (π) 3 k [b ω s( k)b s ( k) d s( k)d s ( ] k) E 0 (.3.173) k with E 0 s=1 (π) 3 ω k ω k [(π) 3 ω k δ ss δ( k k)], (.3.174) s=1 the infinite constant harmonic oscillator zero point energy for the four oscillators corresponding to the b (s) ( k) and d (s) ( k) degrees of freedom. The point is that if b and d obey CCR the Hamiltonian is unbounded below; it can become as negative as you like by adding more and more d particles to a state. Hence, there is no state with lowest energy, if E >is the lowest energy state d r( k) E >is ω k lower in energy! Hence, we have no stable ground state. The only consistent interpretation of the field theory is to demand that b and d obey the CAR. Thus, H = (π) 3 ω k s=1 ω k [b s( k)b s ( k)+d s( k)d s ( ] k) E 0. (.3.175) As in the scalar case the infinite constant E 0 is eliminated by a more careful definition of the Lagrangian and the energy-momentum tensor by means of the Wick normal product. As before, the Wick product of Fermi fields is defined so that the annihilation operators b and d are to the right of the creation operators b and d. Since these are Grassmann algebra valued we include a (+1) if the fields are ordered in an even permutation of the original field order and ( 1) if the fields are ordered in an odd permutation of the original field order. Thus, denoting the Wick product as N[ ] or : :, we have for instance N[Ψ(x)Ψ(y)] Ψ + (x)ψ + (y)+ψ (x)ψ (y)+ψ (x)ψ + (y) Ψ (y)ψ + (x). (.3.176) Recall the ordinary product is simply Ψ(x)Ψ(y) =Ψ + (x)ψ + (y)+ψ (x)ψ (y)+ψ (x)ψ + (y)+ψ + (x)ψ (y) { } = N[Ψ(x)Ψ(x)] + Ψ + (x), Ψ (y). (.3.177) 181

29 In general the Wick product of Fermi fields is defined by its interchange property N[ Ψ(x 1 ) Ψ(x n )] = ( 1) P N[ Ψ(x i1 ) Ψ(x in )] (.3.178a) where Ψ isψorψ and P is the parity of the permutation of (1,,,n) into (i 1,,i n ), its linearity N[( Ψ(x)+a Ψ (x)) Ψ(x 1 ) Ψ(x n )] = N[ Ψ(x) Ψ(x 1 ) Ψ(x n )] +an[ Ψ (x) Ψ(x 1 ) Ψ(x n )], (.3.178b) and its definition in the standard order N[ Ψ (x 1 ) Ψ (x m ) Ψ + (y 1 ) Ψ + (y n )] =+ Ψ (x 1 ) Ψ (x m ) Ψ + (y 1 ) Ψ + (y n ). (.3.179) Thus, the correctly defined coincident point composite operators are just defined by normal ordering the ordinary products of fields, for example ˆL = N[L] ˆT µν = N[T µν ] Ĵ µ N[J µ ] ˆM µνρ N[M µνρ ]. (.3.180) Since these differ from the naive expressions by infinite constants, none of our field equations, CAR, CCR of charges, and conservation equations change. So now P µ = d 3 xn[θ 0µ ] = (π) 3 ω k s=1 [ k µ b s( k)b s ( k)+d s( k)d s ( ] k). (.3.181) Since b and d obey CAR, Q can become negative instead of being positive definite, as is reasonable for a charge Q = d 3 xn[j 0 ] = (π) 3 ω k s=1 [ b s( k)b s ( k) d s( k)d s ( ] k). 18 (.3.18)

30 The operators b s( k) create particles with energy-momentum k µ and charge +1, while the d s( k) create particles with energy-momentum k µ and charge 1. The operators b s ( k) and d s ( k) annihilate these respective particles. But for a further clarification of the properties of the particles that b s( k) and d s( k) create let s consider the action of our charges, P µ, M µν and Q on b, b, d, and d by Fourier transforming Ψ and Ψ. Since [P µ, Ψ(x)] = i µ Ψ(x) [P µ, Ψ(x)] = i µ Ψ(x) [Q, Ψ(x)] = Ψ(x) [Q, Ψ(x)] = +Ψ(x) [M µν, Ψ(x)] = i[(x µ ν x ν µ )Ψ(x) i σµν Ψ(x)] [M µν, Ψ(x)] = i[(x µ ν x ν µ )Ψ(x) + i Ψ(x)σµν ], (.3.183) having used we find γ 0 σ µν γ 0 = σ µν, (.3.184) [P µ,b r ( k)] = k µ b r ( k) [P µ,b r ( k)] = +k µ b r ( k) [P µ,d r ( k)] = k µ d r ( k) [P µ,d r( k)] = +k µ d r( k) [Q, b r ( k)] = b r ( k) [Q, b r( k)] = +b r( k) [Q, d r ( k)] = +d r ( k) [M µν,b r ( k)] = [Q, d r ( k)] = d r ( k) d 3 xe ikx u (s) ( k)γ 0 [ i(x µ ν x ν µ ) 1 σµν ] Ψ(x) 183

31 [M µν,b r ( k)] = [M µν,d r ( k)] = [M µν,d r( k)] = d 3 x [ i(x µ ν x ν µ )Ψ+Ψ 1 ] σµν γ 0 u (s) ( k)e ikx d 3 x [ i(x µ ν x ν µ )Ψ+Ψ 1 ] σµν γ 0 v (s) ( k)e ikx d 3 xe ikx v (s) ( k)γ 0 [ i(x µ ν x ν µ ) 1 σµν ] Ψ(x). (.3.185) The angular momentum terms require further analysis. Rather than determining the above in complete generality let s discuss what operators we desire in order to label the states. As usual the states of the field theory will be labeled by the eigenvalues of a complete set of commuting observables. In general there are many choices for this complete set and we must choose one. We would like our states to be translationally invariant hence, the operators should commute with P µ. Furthermore, the states should have definite mass. Thus, P and P µ are two of the commuting observables. Actually only P is needed since once P and P are given, P 0 is determined due to the constraint that P 0 0 and P 0. In fact since P commutes with P µ, M µν, and Q it will label the irreducible representations of the algebra. Besides P there is one other invariant that one can make out of P µ and M µν by using the Pauli-Lubanski vector Note that W µ has the properties W µ 1 ɛ µνρσm νρ P σ. (.3.186) P µ W µ =0 [P µ,w ν ]=0 [M µν,w λ ]=i[g λν W µ g λµ W ν ]. (.3.187) Hence W = W µ W µ is a scalar and totally invariant i.e. commutes with P µ, M µν, and Q. Writing it out we have W = 1 Mµν M µν P M µρ M νρ P µ P ν. (.3.188) Recalling that J i = ɛ ijk M jk and K i = M 0i we can write the Pauli-Lubanski vector as W 0 = J P 184

32 W = K P + JP 0. (.3.189) Since [W, P µ ]=0=[W, M µν ]=[W,Q], P and W, when acting on each irreducible representation of the P µ, M µν, and Q algebra will be proportional to the identity (Schur s lemma: P and W commute with all operators P µ, M µν, and Q and are therefore proportional to the identity). Thus, the eigenvalues of P and W will label the different irreducible representation states. Since we desire translational invariance, P µ that is, P, will label the states within each irreducible representation. Also Q, since it commutes with P, will be needed to uniquely label the states. P and Q for each P and W eigenvalue will not be a complete set of commuting observables, there will still be spin up and down degeneracy of the states. Hence, we need an operator that commutes with ( P,Q) that specifies the spin. Now, [M 1, P 1 ]=ip is directly not an acceptible choice and consequently, the M µν operators are out since they are not translationally invariant. But recall the Pauli-Lubanski vector So W µ commutes with P µ and Q but [W µ, P ν ]=0=[W µ,q]. (.3.190) [W µ,w ν ]= iɛ µνρσ W ρ P σ. (.3.191) Thus, only one component of W µ can be chosen since the different components do not commute amongst themselves. Let s choose W 3. Our complete set of commuting observables will be (P,W, P,W 3,Q) and their eigenvalues will label the states. (Actually we should include Q with P and W as labeling the irreducible representation since Q commutes with all the other operators.) The way we have expanded the fields, in fact, has been in terms of the eigenfunctions of these operators so that the creation and annihilation operators will commute with these operators to give eigenvalues. More specifically, we have seen this for all the operators so far, for instance, [P µ,b r( k)] = k µ b r( k), (.3.19) except W 3 and W. Since W 3 is a vector operator and W is a scalar operator we can always transform them to the rest frame. Since [P µ,b r( k)] = k restb µ r( k) with k µ rest =(m, 0, 0, 0) in the rest frame we find W µ =(0,mJ ) in the rest frame. Thus, W J = W m is just the spin operator for the particle at rest. While m = J J in 185

33 the rest frame. As we can check explicitly the eigenvalue of J J = 1 (1 + 1 ) for spin 1 single particle states. Since W is a Lorentz invariant so are its eigenvalues and W m = s(s + 1) in all frames. In the rest frame the eigenvalue of W 3 m = J 3 are just the spin projections on the z axis, that is, ± 1 as we will see for a spin 1 particle. In general, for a spin s particle the projection of the spin on the z axis may take on the values s, s 1,, s +1, s. So we see that we need to consider the commutator of J with the operators b, d, b, and d at rest in order to find what states are created and annihilated. Hence, we desire J i = 1 ɛ ijkm jk. (.3.193) Integrating by parts we find [M jk,b r ( k)] = i(k j k k k k j k )b r( k) 1 d 3 xe ikx u (r) ( k)γ 0 σ jk Ψ(x) [M jk,b r( k)] = i(k j k k k k j k )b r( k)+ 1 d 3 xψ(x)σ jk γ 0 u (r) ( k)e ikx [M jk,d r ( k)] = i(k j k k kk j k )d r( k)+ 1 d 3 xψ(x)σ jk γ 0 v (r) ( k)e ikx [M jk,d r( k)] = i(k j k k k k j k )d r( k) 1 d 3 xe ikx v (r) ( k)γ 0 σ jk Ψ(x). (.3.194) Defining the 4 4 spin matrix Σ i 1 ɛ 1 ijk σjk (.3.195) we have, at zero momentum where the angular momentum terms L i = 1 ɛ ijki(k j k k k k j k ) vanish since k = 0 and recalling γ 0 u (r) (0) = +u (r) (0) and γ 0 v (r) (0) = v (r) (0), that [ J,b r (0)] = d 3 xu (r) (0) ΣΨ(x) [ J,b r(0)] = d 3 xψ(x) Σu (r) (0) 186

34 [ J,d r (0)] = d 3 xψ(x) Σv (r) (0) [ J,d r (0)] = d 3 xv (r) (0) ΣΨ(x). (.3.196) We see that due to our choice of spinors u and v we have Σ 3 u (r) (0) = ( 1) (r+1) 1 u(r) (0) Σ 3 v (r) (0) = ( 1) (r+1) 1 v(r) (0) (.3.197) and using we find Σ = Σ (.3.198) [J 3,b r (0)] = 1 ( 1)(r+1) b r (0) [J 3,b r (0)] = +1 ( 1)(r+1) b r (0) [J 3,d r (0)] = + 1 ( 1)(r+1) d r (0) [J 3,d r (0)] = 1 ( 1)(r+1) d r (0). (.3.199) We are now in a position to interpret the state content of our spin 1 field theory. Summarizing the commutators of the creation and annihilation operators with the CSCO we have [P µ,b r ( k)] = k µ b r ( k) [P µ,b r( k)] = +k µ b r( k) [P µ,d r ( k)] = k µ d r ( k) [P µ,d r( k)] = +k µ d r( k) [Q, b r ( k)] = b r ( k) [Q, b r( k)] = +b r( k) [Q, d r ( k)] = +d r ( k) [Q, d r( k)] = d r( k) 187

35 [J 3,b r (0)] = 1 ( 1)(r+1) b r (0) [J 3,b r(0)] = + 1 ( 1)(r+1) b r(0) [J 3,d r (0)] = + 1 ( 1)(r+1) d r (0) [J 3,d r(0)] = 1 ( 1)(r+1) d r(0). (.3.00) As in the scalar case, we assume that we are given an eigenstate of P µ, p >, where P µ p>= p µ p>. We see that P µ [b r ( k) p >]=(p µ + k µ )[b r ( k) p >] P µ [d r( k) p >]=(p µ + k µ )[d r( k) p >] (.3.01) while P µ [b r ( k) p >]=(p µ k µ )[b r ( k) p >] P µ [d r ( k) p >]=(p µ k µ )[d r ( k) p >]. (.3.0) Repeating the application of annihilation operators on the state leads eventually to a state for which H[b r1 ( k 1 ) d rn ( k n ) p >]=(p 0 ω k1 ω kn )[b r1 ( k 1 ) d rn ( k n ) p >](.3.03) goes negative. But H is a non-negative operator, thus, there must be a lowest energy state, 0 >, such that b r ( k) 0 >= 0 d r ( k) 0 >= 0 (.3.04) for all k and r =1,. This state is called the ground state, vacuum state or no particle state and we choose its normalization such that < 0 0 >= 1. Since P µ is defined to be normal ordered we have that the vacuum is the zero energy-momentum state, P µ 0 >= 0. The vacuum state has zero charge, Q 0 >= 0, and no intrinsic spin, M µν 0 >= 0, as well. The single particle states are given by the momentum value k, the spin ± 1, and the charge ±1 eigenvalues. Consequently, there are four states for each value of k. We have k, 1 ( 1)r+1, + >= b r( k) 0 > 188

36 k, 1 ( 1)r, >= d r( k) 0 >. (.3.05) Using the commutators in equation (.3.00), we have P k, 1 ( 1)r+1, ± >= k k, 1 ( 1)r+1, ± > H k, 1 ( 1)r+1, ± >= ω k k, 1 ( 1)r+1, ± > Q k, 1 ( 1)r+1, ± >= ± k, 1 ( 1)r+1, ± > J 3 0, 1 ( 1)r+1, + >= 1 ( 1)r+1 0, 1 ( 1)r+1, + > J 3 0, 1 ( 1)r, >= 1 ( 1)r 0, 1 ( 1)r, >. (.3.06) Thus, there are four states with momentum k, two having charge +1 which are conventionally called the particle state, and two having charge 1 which are conventionally refered to as the antiparticle state. The particle has two spin states, at rest the spin being ± 1 when projected on the z axis J 3 0, ± 1, + >= ±1 0, ± 1, + >. (.3.07) Correspondingly, the antiparticle has at rest two spin states ± 1 when projected on the z axis J 3 0, ± 1, >= ±1 0, ± 1, >. (.3.08) For every particle state there is a corresponding oppositely charged antiparticle state. We see that b 1 ( k) and b 1 ( k) create and annihilate, respectively, particles with momentum k, energy ω k, charge +1, and in the rest frame spin projected on the z axis of + 1. Similarly, b ( k) creates a particle with momentum k, energy ω k, charge +1, and in the rest frame spin projected on the z axis of 1 while b ( k) annihilates such a particle. From above we also see that d ( k) creates an antiparticle with momentum k, energy ω k, charge 1, and in the rest frame spin projected on the z axis of + 1 while d ( k) annihilates such an antiparticle. d 1 ( k) creates an antiparticle with momentum k, energy ω k, charge 1, and in the rest frame spin projected on the z axis of 1, and d 1( k) annihilates such an antiparticle. 189

37 The inner products of the one particle states are found by using the anticommutators of the creation and annihilation operators and < 0 0 >= 1 and < k, ( 1)r+1 < k, ( 1)r, + k +1 {, ( 1)r, + > =< 0 b r ( } k),b r ( k ) 0 > =(π) 3 (ω k )δ rr δ 3 ( k k ), k, ( 1)r {, > =< 0 d r ( k),d r ( } k ) 0 > =(π) 3 (ω k )δ rr δ 3 ( k k ) (.3.09) (.3.10) while < k, ( 1)r+1, + k, ( 1)r, >= 0. (.3.11) Hence, the resolution of the identity in the one particle subspace is 1 = (π) 3 ω k [ k, ( 1)r+1 r=1 + k, ( 1)r, >< k, ( 1)r, + >< k, ( 1)r+1, + ],. (.3.1) The multi-particle and multi-antiparticle states are made from multiple application of the creation operators ( k 1, + 1, +),, (k n, + 1, ) >= b 1 ( k 1 ) d ( k n ) 0 >. (.3.13) Recall that since b r and d r anticommute the fermion particle states are antisymmetric under the interchange of particles. The operators P µ, Q, and J 3 acting on the states give the sum of the individual particle eigenvalues ( H ( k 1, + 1, +),,( k n, + 1 n ), ) >= ω ki ( k 1, + 1, +),,( k n, + 1, ) > i=1 ( P ( k 1, + 1, +),,( k n, + 1 n ), ) >= ki ( k 1, + 1, +),,( k n, + 1, ) > i=1 190

38 Q ( k 1, + 1, +),, ( k n, + 1, ) >= (N + N ) ( k 1, + 1, +),, ( k n, + 1, ) > J 3 ( k 1, + 1, +),, ( k n, + 1, ) > = 1 (N + N + + N N ) ( k 1, + 1, +),,( k n, + 1, ) > (.3.14) where N ± = N ± + N ± with N ± and N ± the number of particles with charge ±1 and spin + 1 ( ) or 1 ( ), respectively. The number operator for the number of particles with charge +1 and spin + 1 is N + = (π) 3 ω k b 1 ( k)b 1 ( k) (.3.15) while for particles with charge +1 and spin 1 the number operator is N + = (π) 3 ω k b ( k)b ( k). (.3.16) The number operator for the number of antiparticles with charge 1 and spin + 1 is N = d (π) 3 ω ( k)d ( k) (.3.17) k and for antiparticles with charge 1 and spin 1 the number operator is N = (π) 3 ω k d 1 ( k)d 1 ( k). (.3.18) The total number operator being N = N + + N + + N + N. (.3.19) Finally, we can calculate the covariant anticommutator of the fields { } { } { } Ψ(x), Ψ(y) = Ψ + (x), Ψ (y) + Ψ (x), Ψ + (y). (.3.0) 191

39 Now as usual we Fourier transform the field operators and use the CAR { } Ψ + (x), Ψ (y) = d 3 l (π) 3 ω k (π) 3 ω l r,s=1 { b r ( k),b s( } l) u (r) ( k)e ikx u (s) ( l)e +ily = u (r) ( (π) 3 ω k)u (r) ( k)e ik(x y) k r=1 = (π) 3 (/k + m)e ik(x y) ω k =(i/ x + m) e ik(x y). (π) 3 ω k (.3.1) But recall i + (x y) (π) 3 ω k e ik(x y) (.3.) so { } Ψ + (x), Ψ (y) =(i/ x + m)i + (x y) is + (x y). (.3.3) Similarly { } Ψ (x), Ψ + (y) = v (r) ( (π) 3 ω k)v (r) ( k)e +ik(x y) k r=1 = (/k m)e +ik(x y) (π) 3 ω k = (i/ x + m) (π) 3 e +ik(x y), ω k (.3.4) but i (x y) (π) 3 ω k e +ik(x y), (.3.5) so { } Ψ (x), Ψ + (y) =(i/ x + m)i (x y) is (x y). (.3.6) 19

40

41 we find the contour integral for S d 4 k i(/k + m) is(x) = e ikx (π) 4 k m. (.3.31) Since k m =(/k + m)(/k m) we often symbolically write so that C 1 /k m is(x) = C (/k + m) k m (.3.3) d 4 k (π) 4 e ikx i /k m. (.3.33) Similarly is + d 4 k i (x) = C (π) 4 e ikx /k m + is d 4 k i (x) = C (π) 4 e ikx /k m. (.3.34) Also we have the trivial covariant anticommutators {Ψ(x), Ψ(y)} =0 { } Ψ(x), Ψ(y) =0. (.3.35) Besides the covariant anti-commutators, we can evaluate the vacuum expectation values of fermion fields and the vacuum expectation values of the time ordered products of the fermion fields. The vacuum expectation value of Ψ(x) and Ψ(y), < 0 Ψ(x)Ψ(y) 0 >, is simply is + (x y) < 0 Ψ(x)Ψ(y) 0 > =< 0 Ψ + (x)ψ (y) 0 > { } =< 0 Ψ + (x), Ψ (y) 0 > (.3.36) = is + (x y). The general field product vacuum expectation value, the Wightman function, is defined as W (n,n) (x 1,,x n,y 1,,y n ) < 0 Ψ(x 1 ) Ψ(x n )Ψ(y 1 ) Ψ(y n ) 0 > 194

42

43 This can be conveniently written in a form reminiscent of the Slater determinant < 0 Ψ(x 1 ) Ψ(x n ) Ψ(y 1 ) Ψ(y n ) 0 > is + (x 1 y 1 ) is + (x 1 y )... is + (x 1 y n ) is + (x y 1 ) is + (x y )... is + (x y n ) = δ n n ( 1) M... is + (x n y 1 ) is + (x n y )... is + (x n y n ) (.3.43) where { n M =, if n = even n 1, if n = odd. (.3.44) Besides the Wightman functions we will make use of the time ordered functions, that is, the vacuum expectation value of the time ordered products of the fermi fields. Since the fermi fields anticommute we will define the time ordered product to have a sign depending upon the permutation of the operators T Ψ(x) =Ψ(x) T Ψ(x) =Ψ(x) T Ψ(x)Ψ(x) =θ(x 0 y 0 )Ψ(x)Ψ(y) θ(y 0 x 0 )Ψ(y)Ψ(x), (.3.45) for instance. In order to make the notation concise, let Ψ stand for either Ψ or Ψ. Then we define in general T Ψ(x 1 ) Ψ(x n ) P ( 1) P θ(x 0 i 1 x 0 i )θ(x 0 i x 0 i 3 ) θ(x 0 i n 1 x 0 i n ) Ψ(x i1 ) Ψ(x in ) (.3.46) where P permutes (1,,,n) into (i 1,i,,i n ) and { ( 1) P +1, if P is even = 1 if P is odd. (.3.47) For instance, T Ψ(x)Ψ(y) = T Ψ(y)Ψ(x) (.3.48) and in general, T Ψ(x 1 ) Ψ(x n )=( 1) P T Ψ(x i1 ) Ψ(x in ). (.3.49) 196

44 Now we define the vacuum expectation value of these products as the Green functions, τ-functions, time ordered functions, or (n, n) point functions G (n,n) (x 1,,x n,y 1,,y n ) < 0 T Ψ(x 1 ) Ψ(x n )Ψ(y 1 ) Ψ(y n ) 0 > = P ( 1) P θ(x 0 i 1 x 0 i ) θ(x 0 i n+n 1 x 0 i n+n ) where here and < 0 Ψ(x i1 ) Ψ(x in+n ) 0 > Ψ(xi ) if i n Ψ(x i )={ Ψ(y i n ) ifi>n { xi if i n x i = if i>n. y i n (.3.50) (.3.51) (.3.5) W (n,n) vanishes unless n = n, we find the same for G (n,n), G (n,n) =0 if n n. (.3.53) We evaluate G (n, n) by using the results for W (n, n) and the properties of the θ functions. First, the two point function < 0 T Ψ(x)Ψ(y) 0 > = θ(x 0 y 0 ) < 0 Ψ(x)Ψ(y) 0 > θ(y 0 x 0 ) < 0 Ψ(y)Ψ(x) 0 > = θ(x 0 y 0 )is + (x y) θ(y 0 x 0 )is (x y). (.3.54) Since this combination occurs frequently we define it as and call it the fermion Feynman propagator, < 0 T Ψ(x) Ψ(y) 0 > S F (x y) (.3.55) S F (x) θ(x 0 )is + (x) θ( x 0 )is (x) = θ(x 0 )(i/ x + m)i + (x) θ( x 0 )(i/ + m)i (x). (.3.56) When we bring the 0 x through the θ(x 0 ) and θ( x 0 ) we pick up δ(x 0 )so S F (x) =(i/ x + m) [ θ(x 0 )i + (x) θ( x 0 )i (x) ] + γ 0 δ(x 0 ) [ + (x)+ (x) ], (.3.57) 197

45 but and by equation (..145) (x) + (x)+ (x) (.3.58) δ(x 0 ) (x) =0. (.3.59) Hence, S F (x) =(i/ x + m) [ θ(x 0 )i + (x) θ( x 0 )i (x) ]. (.3.60) Recall that the scalar Feynman propagator is given by F (x) [ θ(x 0 )i + (x) θ( x 0 )i (x) ] (.3.61) so that S F (x) =(i/ x + m) F (x). (.3.6) Thus, we finally secure < 0 T Ψ(x)Ψ(y) 0 > = S F (x y) =(i/ x + m) F (x y). (.3.63) Recalling the integral representation we had for F, equation (..01), F (x y) = d 4 k (π) 4 e ik(x y) i k m + iɛ we find S F (x y) = Again we symbolically write d 4 k (π) 4 e ik(x y) i(/k + m) k m + iɛ. (.3.64) i(/k + m) k m + iɛ i /k m (.3.65) so that < 0 T Ψ(x)Ψ(y) 0 > = S F (x y) d 4 k = (π) 4 e ik(x y) i (/k m). (.3.66) 198

46 Returning to the (n, n)-point function we can show that the functions factorize into products of propagators for instance < 0 T Ψ(x)Ψ(y) 0 >= S F (x y) < 0 T Ψ(x 1 )Ψ(x )Ψ(y 1 )Ψ(y ) 0 >=< 0 T Ψ(x 1 )Ψ(y ) 0 >< 0 T Ψ(x )Ψ(y 1 ) 0 > In general we find < 0 T Ψ(x 1 )Ψ(y 1 ) 0 >< 0 T Ψ(x )Ψ(y ) 0 >. (.3.67) < 0 T Ψ(x 1 ) Ψ(x n )Ψ(y 1 ) Ψ(y n ) 0 >= δ nn (1,,n) P (i 1,,i n ) ( 1) P n < 0 T Ψ(x a )Ψ(y ia ) 0 > (.3.68) a=1 where ( 1) P equals the parity of the fermion order due to the permutation P P x 1 y i1 x y i x n y in x 1 x n y 1 y n. (.3.69) Again we are able to write this as a determinant < 0 T Ψ(x 1 ) Ψ(x n )Ψ(y 1 ) Ψ(y n ) 0 > is F (x 1 y 1 ) is F (x 1 y )... is F (x 1 y n ) is F (x y 1 ) is F (x y )... is F (x y n ) = δ n n ( 1) M... is F (x n y 1 ) is F (x n y )... is F (x n y n ) (.3.70) where { n M = if n = even n 1 (.3.71) if n = odd. As before, the time ordered function of free fields is just the sum over all possible chronological pairings of the coordinates of the product of the associated two point functions. Since < 0 T Ψ(x 1 )Ψ(x ) 0 >=< 0 T Ψ(y 1 )Ψ(y ) 0 >= 0 chronological pairings occur only between {x i } and {y j }. As in the scalar case we can represent 199

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50 annihilation and creation operators we can apply the results immediately to our bosonized fields Φ(x) = η a (x)ψ a (x) Φ (x) = Ψ a (x)η a (x). (.3.85) Factoring the c-number Grassmann variables η and η from the products we obtain Wick s Theorem for free Fermi fields. Wick s Theorem: 1) Ψ(x 1 ) Ψ(x n ) Ψ(y 1 ) Ψ(y n )=N [ Ψ(x 1 ) Ψ(x n ) Ψ(y 1 ) Ψ(y n ) ] pairing pairings i 1 <i [ ] ( 1) P (i,j) < 0 Ψ(x i ) Ψ(y Ψ(x1 ) Ψ(y n ) j ) 0 >N Ψ(x i ) Ψ(y j ) P (i ( 1) 1,i,j 1,j ) < 0 Ψ(x i1 ) Ψ(y j1 ) 0 >< 0 Ψ(x i ) Ψ(y j ) 0 > [ ] Ψ(x 1 ) N Ψ(y n ) Ψ(x i1 )Ψ(x i ) Ψ(y j1 ) Ψ(y j ) + 3 pairings + ( 1) P < 0 Ψ(x 1 ) Ψ(y i1 ) 0 > < 0 Ψ(x n ) Ψ(y in ) 0 > (.3.86) P n= n ) T Ψ(x 1 ) Ψ(x n ) Ψ(y 1 ) Ψ(y n )=N [ Ψ(x 1 ) Ψ(x n ) Ψ(y 1 ) Ψ(y n ) ] contraction contractions i 1 <i [ ] ( 1) P (i,j) < 0 T Ψ(x i ) Ψ(y Ψ(x1 ) j ) 0 >N Ψ(y n ) Ψ(x i ) Ψ(y j ) P (i ( 1) 1,i,j 1,j ) < 0 T Ψ(x i1 ) Ψ(y j1 ) 0 >< 0 T Ψ(x i ) Ψ(y j ) 0 > [ ] Ψ(x 1 ) N Ψ(y n ) Ψ(x i1 )Ψ(x i ) Ψ(y j1 ) Ψ(y j ) + 3 contractions 03

51 + ( 1) P < 0 T Ψ(x 1 ) Ψ(y i1 ) 0 > < 0 T Ψ(x n ) Ψ(y in ) 0 >. (.3.87) P n= n Wick s Theorem also applies to the chronological product of normal products as discussed following equation (..17). Finally, let s check the Green function property of the fermion time ordered functions. Since S F (x) =(i/ x + m) F (x) (.3.88) and we find That is ( + m ) F (x) = iδ 4 (x) (.3.89) (i/ x m)s F (x y) =iδ 4 (x y). (.3.90) (i/ x m) < 0 T Ψ(x)Ψ(y) 0 >= iδ 4 (x y). (.3.91) As usual this follows directly from the ETAR, using T Ψ(x)Ψ(y) =θ(x 0 y 0 )Ψ(x)Ψ(y) θ(y 0 x 0 )Ψ(y)Ψ(x) (.3.9) we have 0 xt Ψ(x)Ψ(y) =θ(x 0 y 0 ) Ψ(x)Ψ(y) But from the ETAR θ(y 0 x 0 )Ψ(y) Ψ(x)+δ(x 0 y 0 ){Ψ(x), Ψ(y)}. (.3.93) δ(x 0 y 0 ){Ψ(x), Ψ(y)} = γ 0 δ 4 (x y). (.3.94) Thus, 0 xt Ψ(x)Ψ(y) =T Ψ(x)Ψ(y) +γ 0 δ 4 (x y) (.3.95) and (i/ x m)t Ψ(x)Ψ(y) =T (i/ x m)ψ(x)ψ(y)+iδ 4 (x y). (.3.96) Since (i/ x m)ψ(x) = 0 by the Dirac equation we have (i/ x m)t Ψ(x)Ψ(y) =+iδ 4 (x y). (.3.97) 04

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