Dirac Matrices and Lorentz Spinors

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1 Dirac Matrices and Lorentz Spinors Background: In 3D, the spinor j = 1 representation of the Spin3) rotation group is constructed from the Pauli matrices σ x, σ y, and σ k, which obey both commutation and anticommutation relations σ i, σ j = iɛ ijk σ k and {σ i, σ j } = δ ij 1. 1) Consequently, the spin matrices S = i σ σ = 1 σ ) commute with each other as angular momenta, S i, S j = iɛ ijk S k, so they represent the generators of the rotation group. Moreover, under finite rotations Rφ, n) represented by MR) = exp iφn S ), 3) the spin matrices transform into each other as components of a 3 vector, M 1 R)S i MR) = R ij S j. 4) In this note, I shall generalize this construction to the Dirac spinor representation of the Lorentz symmetry Spin3, 1). Dirac Matrices are 4 anti-commuting 4 4 matrices γ µ, γ µ γ ν + γ ν γ µ = g µν ) The specific form of these matrices is not important as long as they obey the anticommutation relations 5) and different books use different conventions. In my class I shall follow the same convention as the Peskin & Schroeder textbook, namely the Weyl convention where in block notations ) ) 0 γ σ =, γ =. 6) 1 0 σ 0 Note that the γ 0 matrix is hermitian while the γ 1, γ, and γ 3 matrices are anti-hermitian. 1

2 Lorentz spin matrices. Given the Dirac matrices obeying the anticommutation relations 5), we may define the spin matrices as S µν = S νµ def = i 4 γµ, γ ν. 7) These matrices obey the same commutation relations as the generators Ĵµν = Ĵνµ of the continuous Lorentz group. Moreover, their commutation relations with the Dirac matrices γ µ are similar to the commutation relations of the Ĵµν with a Lorentz vector such as ˆP µ. Lemma: γ λ, S µν = ig λµ γ ν ig λν γ µ. 8) Proof: Combining the definition 7) of the spin matrices as commutators with the anticommutation relations 5), we have γ µ γ ν = 1 {γµ, γ ν } + 1 γµ, γ ν = g µν is µν. 9) Since the unit matrix commutes with everything, we have X, S µν = i X, γµ γ ν for any matrix X, 10) and the commutator on the RHS may often be obtained from the commutators or anticommutators: Leibniz rules for the A, BC = A, BC + BA, C = {A, B}C B{A, C}, {A, BC} = A, BC + B{A, C} = {A, B}C BA, C. 11) In particular, γ λ, γ µ γ ν = {γ λ, γ µ }γ ν γ µ {γ λ, γ ν } = g λµ γ ν g λν γ µ 1) and hence Quod erat demonstrandum. γ λ, S µν = i γλ, γ µ γ ν = ig λµ γ ν ig λν γ µ. 13)

3 Theorem: The S µν matrices commute with each other like Lorentz generators, S κλ, S µν = ig λµ S κν ig κν S µλ ig λν S κµ + ig κµ S νλ. 14) Proof: Again, we use the Leibniz rule and eq. 9): γ κ γ λ, S µν = γ κ γ λ, S µν + γ κ, S µν γ λ = γ κ ig λµ γ ν ig λν γ µ) + ig κµ γ ν ig κν γ µ) γ λ = ig λµ γ κ γ ν ig κν γ µ γ λ ig λν γ κ γ µ + ig κµ γ ν γ λ = ig λµ g κν is κν) ig κν g λµ + is λµ) ig λν g κµ is κµ) + ig κµ g λν + is λν) 15) = g λµ S κν g κν S λµ g λν S κµ + g κµ S λν, and hence S κλ, S µν = i γ κ γ λ, S µν = ig λµ S κν ig κν S µλ ig λν S κµ + ig κµ S νλ. 16) Quod erat demonstrandum. In light of this theorem, the S µν matrices represent the Lorentz generators Ĵµν in a 4-component spinor multiplet. Finite Lorentz transforms: Any continuous Lorentz transform a rotation, or a boost, or a product of a boost and a rotation obtains from exponentiating an infinitesimal symmetry X µ = X µ + ɛ µν X ν 17) where the infinitesimal ɛ µν matrix is antisymmetric when both indices are raised or both lowered), ɛ µν = ɛ νµ. Thus, the L µ ν matrix of any continuous Lorentz transform is a matrix exponential L µ ν = expθ) µ ν δ µ ν + Θ µ ν + 1 Θµ λ Θλ ν Θµ λ Θλ κθ κ ν + 18) of some matrix Θ that becomes antisymmetric when both of its indices are raised or lowered, Θ µν = Θ νµ. Note however that in the matrix exponential 18), the first index of Θ is raised 3

4 while the second index is lowered, so the antisymmetry condition becomes gθ) = gθ) instead of Θ = Θ. The Dirac spinor representation of the finite Lorentz transform 18) is the 4 4 matrix M D L) = exp i Θ αβs αβ). 19) The group law for such matrices L 1, L SO + 3, 1), M D L L 1 ) = M D L )M D L 1 ) 0) follows automatically from the S µν satisfying the commutation relations 14) of the Lorentz generators, so I am not going to prove it. Instead, let me show that when the Dirac matrices γ µ are sandwiched between the M D L) and its inverse, they transform into each other as components of a Lorentz 4 vector, M 1 D L)γµ M D L) = L µ νγ ν. 1) This formula makes the Dirac equation transform covariantly under the Lorentz transforms. Proof: In light of the exponential form 19) of the matrix M D L) representing a finite Lorentz transform in the Dirac spinor multiplet, let s use the multiple commutator formula AKA the Hadamard Lemma ): for any matrices F and H, exp F )H exp+f ) = H + H, F + 1 H, F, F H, F, F, F +. ) In particular, let H = γ µ while F = i Θ αβs αβ so that M D L) = exp+f ) and M 1 D L) = exp F ). Consequently, MD 1 L)γµ M D L) = γ µ + γ µ, F + 1 γ µ, F, F γ µ, F, F, F + 3) where all the multiple commutators turn out to be linear combinations of the Dirac matrices. 4

5 Indeed, the single commutator here is γ µ, F = i Θ αβ γ µ, S αβ = 1 Θ αβ g µα γ β g µβ γ α) = Θ αβ g µα γ β = Θ µ λ γλ, 4) while the multiple commutators follow by iterating this formula: γ µ, F, F = Θ µ λ γ λ, F = Θ µ λ Θλ νγ ν, γ µ, F, F, F = Θ µ λ Θλ ρθ ρ νγ ν,.... 5) Combining all these commutators as in eq. 3), we obtain MD 1 γµ M D = γ µ + γ µ, F + 1 γ µ, F, F γ µ, F, F, F + = γ µ + Θ µ νγ ν + 1 Θµ λ Θλ νγ ν Θµ λ Θλ ρθ ρ νγ ν + ) = δ ν µ + Θ µ ν + 1 Θµ λ Θλ ν + 6 1Θµ λ Θλ ρθ ρ ν + γ ν 6) L µ νγ ν. Quod erat demonstrandum. Dirac Equation The Dirac spinor field Ψx) has 4 complex components Ψ α x) arranged in a column vector Ψ 1 x) Ψ Ψx) = x) Ψ 3 x). 7) Ψ 4 x) Under continuous Lorentz symmetries x µ = L µ νx ν, the spinor field transforms as Ψ x ) = M D L)Ψx). 8) The classical field equation for the free spinor field is the Dirac equation a first-order differential equation iγ µ µ m ) Ψx) = 0. 9) The Dirac equation implies the Klein Gordon equation for each component Ψ α x). Indeed, 5

6 if Ψx) obey the Dirac equation, then iγ ν ν m ) iγ µ µ m ) Ψx) = 0 30) where the differential operator on the LHS is the Klein Gordon m + times a unit matrix. Indeed, iγ ν ν m ) iγ µ µ m ) = m + γ ν γ µ ν µ = m + 1 {γµ, γ ν } ν µ = m + g µν ν µ. 31) The Dirac equation transforms covariantly under the Lorentz symmetries its LHS transforms exactly like the spinor field itself. Proof: Note that since the Lorentz symmetries involve the x µ coordinates as well as the spinor field components, the LHS of the Dirac equation becomes iγ µ µ m ) Ψ x ) 3) where Consequently, µ xν = x µ x µ x ν = L 1) ν µ ν. 33) µψ x ) = L 1) ν µ M DL) ν Ψx) 34) and hence But according to eq. 3), γ µ µψ x ) = L 1) ν µ γµ M D L) ν Ψx). 35) so M 1 D L)γµ M D L) = L µ νγ ν = γ µ M D L) = L µ ν M D L)γ ν = L 1) ν µ γµ M D L) = M D L)γ ν, 36) γ µ µψ x ) = M D L) γ ν ν Ψx). 37) Altogether, iγ µ µ m ) Ψx) Lorentz iγ µ µ m ) Ψ x ) = M D L) iγ µ µ m ) Ψx), 38) which proves the covariance of the Dirac equation. Quod erat demonstrandum. 6

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