Dirac Matrices and Lorentz Spinors

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1 Dirac Matrices and Lorentz Spinors Background: In 3D, the spinor j = 2 1 representation of the Spin3) rotation group is constructed from the Pauli matrices σ x, σ y, and σ z, which obey both commutation and anticommutation relations Consequently, the spin matrices [σ i, σ j ] = 2iɛ ijk σ k and {σ i, σ j } = 2δ ij ) S = i 4 σ σ = 1 2 σ 2) commute with each other like 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 γ µ generalize the anti-commutation properties of the Pauli matrices σ i to Minkowski dimensions: γ µ γ ν + γ ν γ µ = 2g µν ) The γ µ are 4 4 matrices, but there are several different conventions for their specific form In my class I shall follow the same convention as the Peskin & Schroeder textbook, namely the Weyl convention where in 2 2 block notations ) ) 0 γ σ =, γ =. 6) σ 0 Note that the γ 0 matrix is hermitian while the γ 1, γ 2, and γ 3 matrices are anti-hermitian. Apart from that, the specific forms of the matrices are not important, the Physics follows from the anti-commutation relations 5). 1

2 Lorentz spin matrices generalize S = 4 i σ σ rather than S = 1 2σ. In 4D, the vector product becomes the antisymmetric tensor product, so we define S µν = S νµ def = i 4 [γµ, γ ν ]. 7) Thanks to the anti-commutation relations 5) for the γ µ matrices, the S µν obey the commutation relations of the Lorentz generators Ĵµν = Ĵνµ. Moreover, the commutation relations of the spin matrices S µν 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 2 {γµ, γ ν } [γµ, γ ν ] = g µν iS µν. 9) Since the unit matrix commutes with everything, we have [X, S µν ] = i 2 [X, γµ γ ν ] for any matrix X, 10) and the commutator on the RHS may often be obtained from the Leibniz rules for the commutators or anticommutators: [A, BC] = [A, B]C + B[A, C] = {A, B}C B{A, C}, {A, BC} = [A, B]C + B{A, C} = {A, B}C B[A, C]. 11) In particular, [γ λ, γ µ γ ν ] = {γ λ, γ µ }γ ν γ µ {γ λ, γ ν } = 2g λµ γ ν 2g λν γ µ 12) and hence Quod erat demonstrandum. [γ λ, S µν ] = i 2 [γλ, γ µ γ ν ] = ig λµ γ ν ig λν γ µ. 13) 2

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 2iS κν) ig κν g λµ + 2iS λµ) ig λν g κµ 2iS κµ) + ig κµ g λν + 2iS λν) 15) = 2g λµ S κν 2g κν S λµ 2g λν S κµ + 2g κµ S λν, and hence [ S κλ, S µν] = 2 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 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 2 Θ αβs αβ). 19) The group law for such matrices L 1, L 2 SO + 3, 1), M D L 2 L 1 ) = M D L 2 )M D L 1 ) 20) 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. 21) 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 2 matrices F and H, exp F )H exp+f ) = H + [ H, F ] [[ H, F ], F ] [[[ H, F ], F ], F ] +. 22) In particular, let H = γ µ while F = i 2 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 [[ 2 γ µ, F ], F ] + 1 [[[ 6 γ µ, F ], F ], F ] + 23) 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 2 Θ αβ [ γ µ, S αβ] = 1 2 Θ αβ g µα γ β g µβ γ α) = Θ αβ g µα γ β = Θ µ λ γλ, 24) 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, ) Combining all these commutators as in eq. 23), we obtain MD 1 γµ M D = γ µ + [ γ µ, F ] [[ γ µ, F ], F ] + 1 [[[ 6 γ µ, F ], F ], F ] + = γ µ + Θ µ νγ ν Θµ λ Θλ νγ ν Θµ λ Θλ ρθ ρ νγ ν + ) = δ ν µ + Θ µ ν Θµ λ Θλ ν + 6 1Θµ λ Θλ ρθ ρ ν + γ ν 26) L µ νγ ν. Quod erat demonstrandum. Dirac Equation and Dirac Spinor Fields History: Originally, the Klein Gordon equation was thought to be the relativistic version of the Schrödinger equation that is, an equation for the wave function ψx, t) for one relativistic particle. But pretty soon this interpretation run into trouble with bad probabilities negative, or > 1) when a particle travels through high potential barriers or deep potential wells. There were also troubles with relativistic causality, and a few other things. Paul Adrien Maurice Dirac had thought that the source of all those troubles was the ugly form of relativistic Hamiltonian Ĥ = ˆp 2 + m 2 in the coordinate basis, and that he could solve all the problems with the Klein-Gordon equation by rewriting the Hamiltonian as a first-order differential operator Ĥ = ˆp α + mβ = Dirac equation i ψ t = i α ψ + mβψ 27) where α 1, α 2, α 3, β are matrices acting on a multi-component wave function. Specifically, all 5

6 four of these matrices are Hermitian, square to 1, and anticommute with each other, {α i, α j } = 2δ ij, {α i, β} = 0, β 2 = 1. 28) Consequently α ˆp ) 2 = αi α j ˆp iˆp j = 1 2 {α i, α j } ˆp iˆp j = δ ij ˆp iˆp j = ˆp 2, 29) and therefore Ĥ 2 Dirac = α ˆp + βm) 2 = α ˆp ) 2 + {αi, β} ˆp i m + β 2 m 2 = ˆp m 2. 30) This, the Dirac Hamiltonian squares to ˆp 2 + m 2, as it should for the relativistic particle. The Dirac equation 27) turned out to be a much better description of a relativistic electron which has spin = 1 2 ) than the Klein Gordon equation. However, it did not resolve the troubles with relativistic causality or bad probabilities for electrons going through big potential differences e Φ > 2m e c 2. Those problems are not solvable in the context of a relativistic single-particle quantum mechanics but only in quantum field theory. Modern point of view: Today, we interpret the Dirac equation as the equation of motion for a Dirac spinor field Ψx), comprising 4 complex component fields Ψ α x) arranged in a column vector Ψ 1 x) Ψ Ψx) = 2 x) Ψ 3 x), 31) Ψ 4 x) and transforming under the continuous Lorentz symmetries x µ = L µ νx ν according to Ψ x ) = M D L)Ψx). 32) The classical Euler Lagrange equation of motion for the spinor field is the Dirac equation i Ψ + i α Ψ mβψ = 0. 33) t 6

7 To recast this equation in a Lorentz-covariant form, let β = γ 0, α i = γ 0 γ i ; 34) it is easy to see that if the γ µ matrices obey the anticommutation relations 5) then the α and β matrices obey the relations 28) and vice verse. Now let s multiply the whole LHS of the Dirac equation 33) by the β = γ 0 : 0 = γ 0 i 0 + iγ 0 γ mγ 0) Ψx) = iγ iγ i i m)ψx), 35) and hence iγ µ µ m ) Ψx) = 0. 36) As expected from Ĥ2 Dirac = ˆp2 + m 2, the Dirac equation for the spinor field implies the Klein Gordon equation for each component Ψ α x). Indeed, if Ψx) obey the Dirac equation, then obviously iγ ν ν m ) iγ µ µ m ) Ψx) = 0, 37) but the differential operator on the LHS is equal to the Klein Gordon m times a unit matrix: iγ ν ν m ) iγ µ µ m ) = m 2 + γ ν γ µ ν µ = m {γµ, γ ν } ν µ = m 2 + g µν ν µ. 38) The Dirac equation 36) 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 ) 39) where µ xν = x µ x µ x ν = L 1) ν µ ν. 40) 7

8 Consequently, µψ x ) = L 1) ν µ M DL) ν Ψx) 41) and hence But according to eq. 23), γ µ µψ x ) = L 1) ν µ γµ M D L) ν Ψx). 42) 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, 43) so γ µ µψ x ) = M D L) γ ν ν Ψx). 44) Altogether, iγ µ µ m ) Ψx) Lorentz iγ µ µ m ) Ψ x ) = M D L) iγ µ µ m ) Ψx), 45) which proves the covariance of the Dirac equation. Quod erat demonstrandum. Dirac Lagrangian The Dirac equation is a first-order differential equation, so to obtain it as an Euler Lagrange equation, we need a Lagrangian which is linear rather than quadratic in the spinor field s derivatives. Thus, we want L = Ψ iγ µ µ m ) Ψ 46) where Ψx) is some kind of a conjugate field to the Ψx). Since Ψ is a complex field, we treat it as a linearly-independent from the Ψ, so the Euler Lagrange equation for the Ψ 0 = L Ψ L µ µ Ψ = iγ ν ν m ) ) Ψ µ 0 47) immediately gives us the Dirac equation for the Ψx) field. 8

9 To keep the action S = d 4 xl Lorentz-invariant, the Lagrangian 46) should transform as a Lorentz scalar, L x ) = Lx). In light of eq. 19) for the Ψx) field and covariance 45) of the Dirac equation, the conjugate field Ψx) should transform according to Ψ x ) = Ψx) M 1 D L) = L x ) = Lx). 48) Note that the M D L) matrix is generally not unitary, so the inverse matrix MD 1 L) in eq. 48) is different from the hermitian conjugate M D L). Consequently, the conjugate field Ψx) cannot be identified with the hermitian conjugate field Ψ x), since the latter transforms to Ψ x ) = Ψ x) M D L) Ψ x) MD 1 L). 49) Instead of the hermitian conjugate, we are going to use the Dirac conjugate spinor, see below. Dirac conjugates: Let Ψ be a 4-component Dirac spinor and Γ be any 4 4 matrix; we define their Dirac conjugates according to Ψ = Ψ γ 0, Γ = γ 0 Γ γ 0. 50) Thanks to γ 0 γ 0 = 1, the Dirac conjugates behave similarly to hermitian conjugates or transposed matrices: For a a product of 2 matrices, Γ 1 Γ 2 ) = Γ 2 Γ 1. Likewise, for a matrix and a spinor, Γ Ψ) = Ψ Γ. The Dirac conjugate of a complex number is its complex conjugate, c 1) = c 1. For any two spinors Ψ 1 and Ψ 2 and any matrix Γ, Ψ 1 ΓΨ 2 = Ψ 2 ΓΨ 1 ). The Dirac spinor fields are fermionic, so they anticommute with each other, even in the classical limit. Nevertheless, Ψ αψ β ) = +Ψ β Ψ α, and therefore for any matrix Γ, Ψ 1 ΓΨ 2 = + Ψ 2 ΓΨ 1 ). 9

10 The point of the Dirac conjugation 50) is that it works similarly for all four Dirac matrices γ µ, γ µ = +γ µ. 51) Proof: For µ = 0, the γ 0 is hermitian, hence γ 0 = γ 0 γ 0 ) γ 0 = +γ 0 γ 0 γ 0 = +γ 0. 52) For µ = i = 1, 2, 3, the γ i are anti-hermitian and also anticommute with the γ 0, hence γ i = γ 0 γ i ) γ 0 = γ 0 γ i γ 0 = +γ 0 γ 0 γ i = +γ i. 53) Corollary: The Dirac conjugate of the matrix M D L) = exp i 2 Θ µνs µν) 19) representing any continuous Lorentz symmetry L = expθ) is the inverse matrix M D L) = M 1 D L) = exp + i 2 Θ µνs µν). 54) Proof: Let X = i 2 Θ µνs µν = Θ µν[γ µ, γ ν ] = Θ µνγ µ γ ν 55) for some real antisymmetric Lorentz parameters Θ µν = Θ νµ. The Dirac conjugate of the X matrix is X = 1 4 Θ µνγ µ γ ν = 1 4 Θ µνγ ν γ µ = 1 4 Θ µνγ ν γ µ = 1 4 Θ νµγ µ γ ν = 1 4 Θ µνγ µ γ ν = X. 56) Consequently, X 2 = X X = +X 2, X 3 = X X 2 = X 2 X = X 3,..., X n = X) n, 57) 10

11 and hence expx) = n 1 n! Xn = n 1 n! X)n = exp X). 58) In light of eq. 55), this means exp i 2 Θ µνs µν) = exp + i 2 Θ µνs µν), 59) that is, M D L) = MD 1 L). 60) Quod erat demonstrandum. Back to the Dirac Lagrangian: Thanks to the theorem 60), the conjugate field Ψx) in the Lagrangian 46) is simply the Dirac conjugate of the Dirac spinor field Ψx), Ψx) = Ψ x) γ 0, 61) which transforms under Lorentz symmetries as Ψ x ) = Ψ x ) = M D L) Ψx) = Ψx) M D x) = Ψx) MD 1 L). 62) Consequently, the Dirac Lagrangian L = Ψ iγ µ µ m ) Ψ = Ψ γ 0 iγ µ µ m ) Ψ 46) is a Lorentz scalar and the action is Lorentz invariant. 11

12 Hamiltonian for the Dirac Field Canonical quantization of the Dirac spinor field Ψx) just like any other field begins with classical Hamiltonian formalism. Let s start with the canonical conjugate fields, Π α = L 0 Ψ α ) = iψγ 0) α = iψ α 63) the canonical conjugate to the Dirac spinor field Ψx) is simply its hermitian conjugate Ψ x). This is similar to what we had for the non-relativistic field, and it happens for the same reason the Lagrangian which is linear in the time derivative. In the non-relativistic field theory, the conjugacy relation 63) in the classical theory lead to the equal-time commutation relations in the quantum theory, [ ˆψx, t), ˆψy, t) ] = 0, [ ˆψ x, t), ˆψ y, t) ] = 0, [ ˆψx, t), ˆψ y, t) ] = δ 3) x y). 64) However, the Dirac spinor field describes spin = 2 1 particles like electrons, protons, or neutrons which are fermions rather than bosons. So instead of the commutations relations 64), the spinor fields obey equal-time anti-commutation relations { ˆΨα x, t), ˆΨ β y, t) } = 0, { ˆΨ α x, t), ˆΨ β y, t)} = 0, { ˆΨα x, t), ˆΨ β y, t)} = δ αβ δ 3) x y). 65) Next, the classical Hamiltonian obtains as H = d 3 x Hx), H = iψ 0 Ψ L = iψ 0 Ψ Ψ γ 0 iγ i γ m ) Ψ 66) = Ψ iγ 0 γ + γ 0 m ) Ψ where the terms involving the time derivative 0 cancel out. Consequently, the Hamiltonian 12

13 operator of the quantum field theory is Ĥ = d 3 x ˆΨ x) iγ 0 γ + γ 0 m ) ˆΨx). 67) Note that the derivative operator iγ 0 γ +γ 0 m) in this formula is precisely the 1-particle Dirac Hamiltonian 27). This is very similar to what we had for the quantum non-relativistic fields, Ĥ = ) d 3 x ˆψ 1 x) 2M 2 + V x) ˆψx), 68) except for a different differential operator, Schrödinger instead of Dirac. In the Heisenberg picture, the quantum Dirac field obeys the Dirac equation. To see how this works, we start with the Heisenberg equation i t ˆΨ α x, t) = [ ˆΨα x, t), Ĥ] = d 3 y [ ˆΨα x, t), Ĥy, t)], 69) and then evaluate the last commutator using the anti-commutation relations 65) and the Leibniz rules 11). Indeed, let s use the Leibniz rule [A, BC] = {A, B}C B{A, C} 70) for A = ˆΨ α x, t), B = ˆΨ β y, t), C = iγ 0 γ + γ 0 m ) βγ ˆΨ γ y, t), 71) so that BC = Ĥy, t). For the A, B, C at hand, {A, B} = δ αβ δ 3) x y) 72) while {A, C} = iγ 0 γ y + γ 0 m ) βγ { ˆΨ α x, t), ˆΨ γ y, t)} = diff.op.) 0 = 0. 73) 13

14 Consequently [ ˆΨα x, t), Ĥy, t)] [A, BC] = {A, B} C B {A, C} = δ αβ δ 3) x y) iγ 0 γ + γ 0 m ) βγ ˆΨ γ y, t) 74) 0, hence [ ˆΨα x, t), Ĥ] = d 3 y δ 3) x y) iγ 0 γ + γ 0 m ) ˆΨ αγ γ y, t) = iγ 0 γ + γ 0 m ) ˆΨ αγ γ x, t), 75) and therefore i 0 ˆΨx, t) = iγ 0 γ + γ 0 m ) ˆΨx, t). 76) Or if you prefer, iγ µ µ m ) ˆΨx) = 0. 77) 14