# 6.1. Dirac Equation. Hamiltonian. Dirac Eq.

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1 6.1. Dirac Equation Ref: M.Kaku, Quantum Field Theory, Oxford Univ Press (1993) η μν = η μν = diag(1, -1, -1, -1) p 0 = p 0 p = p i = -p i p μ p μ = p 0 p 0 + p i p i = E c 2 - p 2 = (m c) 2 H = c p 2 + (m c) 2 Hamiltonian Set H = α p c + β m c 2 = -α i p i c + β m c 2 H + = H α i + = α i β + = β p μ i ħ μ = i ħ c t, x i = i ħ i ħ t = ħ c α +β m c 2 = ħ c α i i i c t, H 2 = (p c) 2 + m c 2 2 = -p i p i c 2 + m c 2 2 α p α p = p 2 x i +β m c 2 = α p α p c 2 + ( α p β + β α p ) m c 3 + β 2 m c 2 2 α p β + β α p = 0 β 2 = 1 In component form: α i p i α j p j = -p i p i = 1 2 αi α j + α j α i p i p j = p i p i α i α j + α j α i = α i, α j = 2 δ i j where the anti-commutator is defined by { A, B } = [ A, B ] + = A B + B A α i β + β α i p i = 0 α i β + β α i = α i, β = 0 Dirac Eq. β 2 = 1 i ħ β t = ħ c β α + m c 2 i or i ħ β c t + β α - m c = 0 where i ħ γ μ μ -m c = 0 = γ μ p μ - m c γ μ = ( β, β α ) γ μ + = ( β, α β ) = β ( β, β α ) β = γ 0 γ μ γ 0 Dirac Equation:

2 2 6.1._DiracEquation.nb i ħ γ μ μ -m c ψ = 0 = γ μ p μ - m c ψ Since p μ p μ = (m c) 2, this is equivalent to writing p μ p μ = γ μ p μ = +m c γ 0, γ 0 = { β, β } = 2 β 2 = 2 γ 0, γ i = β, β α i = β 2 α i + β α i β = β β, α i = 0 γ i, γ j = β α i, β α j = β α i β α j + β α j β α i = -β 2 α i α j + α j α i = - α i, α j = -2 δ i j { γ μ, γ ν } = 2 η μν γ μ, γ ν = 2 η μν i.e., { γ μ } forms a Clifford algebra so that ψ is a spinor ( spin 1 2 representation of the Lorentz group ). Note: { γ μ } is just the set of linear combination coefficients for p μ that gives they take the same values in all Lorentz frames, i.e., γ μ is not a 4-vector. However, as shown later ψ γ μ ψ is a 4-vector. (See J.J.Sakurai, Advanced Quantum Mechanics, 3.4 ) p μ p μ. Therefore, Lorentz transformation Under a Lorentz transformation, x μ x μ ' μ = Λ ' ν x ν μ' = Λ ν μ' ν γ μ γ μ ' = γ μ ψ ψ ' with ψ ' (x') = S(Λ) ψ(x) where S(Λ) is some representation of the Lorentz group. Since Λ ν μ' notation & write is the inverse of Λ ν μ ', some author (e.g., Kaku) prefers to dispense with the primed Λ ν μ Λ ν μ ' & Λ -1 μ μ ν Λν' Thus, Λ -1 μ ν ν = μ' i ħ γ μ μ -m c ψ(x) = 0 i ħ γ μ ' μ ' -m c ψ ' (x') = 0 = i ħ γ μ ' Λ ν μ' ν -m c S(Λ) ψ(x) i.e., S -1 i ħ γ μ' Λ ν μ' ν -m c S ψ = 0 i ħ S -1 γ μ' Λ ν μ' S ν -m c ψ = 0 Dirac eq. is covariant under the Lorentz transformation (c.f. Kaku, 3.5). S -1 γ μ' Λ ν μ' S = γ ν or S γ ν S -1 = γ μ' ν Λ μ' = Λ -1 μ ν γ μ i.e., S -1 γ μ S = Λ ν μ γ ν Generator of the Lorentz group is M μν = L μν σ μν

3 6.1._DiracEquation.nb 3 where L μν deals with the space-time part and σ μν = i 2 γ μ, γ ν where γ μ = η μν γ ν Hence, S(Λ) = exp - i 4 σ μν ω μν where ω μν are the antisymmetric parameters of Λ. Using γ μ + = γ 0 γ μ γ 0 & γ 0 γ 0 = 1 (from Dirac Eq.) we have σ + μν = - i 2 γ + ν, γ + μ = i 2 γ0 γ μ γ 0, γ 0 γ ν γ 0 = γ 0 σ μν γ 0 so that S + = exp i 4 γ0 σ μν γ 0 ω μν = γ 0 exp i 4 σ μν ω μν γ 0 (see below) = γ 0 S -1 γ 0 i.e., S is not unitary & S -1 = γ 0 S + γ 0 Note : We can write e γ0 σ γ 0 = γ 0 e σ γ 0 because γ 0 γ 0 = 1 γ 0 σ γ 0 n = γ 0 σ n γ 0 e γ0 σ γ 0 = 1 + γ 0 σ γ γ0 σ γ = γ σ σ γ 0 = γ 0 e σ γ 0 Conjugate Field ψ i ħ γ μ μ -m c ψ = 0 ψ + i ħ γ μ + μ +m c = 0 ( f f ) The appearance of γ μ + makes the pair of eqs unsymmetrical. More importantly, under a Lorentz transformation Λ, ψ ' (x') = S(Λ) ψ(x) ψ + ' (x') = ψ + (x) S + (Λ) so that ψ + ' (x') ψ ' (x)' = ψ + x S + S ψ(x) ψ + x ψ(x) i.e., ψ + ψ is not a Lorentz scalar. In order to construct tensor quantities, we introduce the conjugate field ψ ψ + γ 0 ψ + = ψ γ 0 Under a Lorentz transformation Λ, ψ ' (x') = ψ + ' (x') γ 0 = ψ + x S + γ 0 = ψ(x) γ 0 S + γ 0 = ψ(x) S -1 ψ ' (x') ψ ' (x') = ψ(x) S -1 S ψ(x) = ψ(x) ψ(x) is a Lorentz scalar. The conjugate eq. ψ + i ħ γ μ + μ +m c = 0

4 4 6.1._DiracEquation.nb can also be rewrite as ψ γ 0 i ħ γ μ + μ +m c = 0 ψ i ħ γ 0 γ μ + γ 0 μ +m c = 0 ψ i ħ γ μ μ +m c = 0 which is preferred form. γ 5 It s useful to introduce one more γ matrix γ 5 γ 5 = i γ 0 γ 1 γ 2 γ 3 = -i γ 0 γ 1 γ 2 γ 3 ( Index 5 can t be raised or lowered using η μν since μ, ν = 0, 1, 2, 3 ) γ μ γ ν = -γ ν γ μ μ ν γ μ γ 5 = -γ 5 γ μ (changes sign 3 times) & γ 5 = - i 4! ε μνστ γ μ γ ν γ σ γ τ where ε μνστ = -ε μνστ & ε 0123 = 1. ( Using ε μνστ instead of ε μνστ honors the Einstein summation rule but introduces a minus sign ). { γ μ, γ ν } = 2 η μν γ 5 2 = -γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 = γ 1 γ 2 γ 3 γ 1 γ 2 γ 3 = -γ 2 γ 3 γ 2 γ 3 = -γ 3 γ 3 = 1 γ μ + = γ 0 γ μ γ 0 γ 0 γ 0 = 1 γ 5 + = -i γ 3 + γ 2 + γ 1 + γ 0 + = -i γ 0 γ 3 γ 2 γ 1 γ 0 γ 0 = i γ 3 γ 2 γ 1 γ 0 = γ 5 Tensors Similarly, one can construct other types of tensors as follows type form number scalar ψ(x) ψ(x) 1 vector ψ(x) γ μ ψ(x) 4 tensor ψ(x) σ μ ν ψ(x) 6 pseudovector ψ(x) γ 5 γ μ ψ(x) 4 pseudoscalar ψ(x) γ 5 ψ(x) 1 For example, using S -1 γ μ S = Λ μ ν γ ν, we have ψ ' (x') γ μ ψ ' (x') = ψ(x) S -1 γ μ Sψ (x) = Λ μ ν ψ(x) γ ν ψ(x) so ψ γ μ ψ indeed transforms as a vector. γ 5 is a pseudoscalar because γ μ γ ν = -γ ν γ μ μ ν γ 5 = i γ 0 γ 1 γ 2 γ 3 = - i 4! ε μνστ γ μ γ ν γ σ γ τ S -1 γ 5 S = - i 4! ε μνστ S γ μ γ ν γ σ γ τ S -1

5 6.1._DiracEquation.nb 5 = - i 4! ε μ μνστ Λ ν μ' Λν' Λ σ σ' Λ τ τ' γ μ' γ ν' γ σ' γ τ' = - i 4! (det Λ) γμ' γ ν' γ σ' γ τ' = (det Λ) γ 5 Spin Spin is given by s i = ħ 4 ϵ i j k σ j k = i ħ 8 ϵ i j k γ j, γ k = s i i.e., s 1 = 1 2 ħ σ 23 = i ħ 4 [γ 2, γ 3 ] = i ħ 4 (γ 2 γ 3 -γ 3 γ 2 ) = i ħ 2 γ 2 γ 3 Similarly, [s 1, s 2 ] = - ħ2 4 [ γ 2 γ 3, γ 3 γ 1 ] = - ħ 2 4 ( γ 2 γ 3 γ 3 γ 1 - γ 3 γ 1 γ 2 γ 3 ) = - ħ2 4 ( -γ 2 γ 1 + γ 1 γ 2 ) = - ħ 2 γ 1 γ 2 = i ħ s 3 s i, s j = i ħ ε i j k s k 2 Standard Representation 4-D (standard) representation of { γ μ } : γ 0 = = γ 0 γ i = where σ i are the Pauli matrices: which satisfy σ 1 = σ i, σ j = 2 i ε i j k σ k σ i σ j = i ε i j k σ k + δ i j I with I being the 2 2 identity matrix. σ 2 = 0 -i i 0 s 3 = i ħ 2 γ 1 γ 2 = i ħ 2 γ1 γ 2 = i ħ 2 = -i ħ 2 σ 1 σ 2 0 γ 5 = i γ 0 γ 1 γ 2 γ 3 = i = i = i = i 0 σ 1 σ 2 = ħ 2 0 σ 1 -σ σ 1 -σ 1 0 σ σ 3 0 σ 1 -σ 1 0 -σ 2 σ σ 1 σ 2 σ 3 σ 1 σ 2 σ σ 1 σ 2 σ 3 -σ 1 σ 2 σ 3 0 σ 1 σ 2 σ 3 = i σ 3 σ 3 = i 0 σ i -σ i 0 0 -σ 2 σ 3 = -γ i σ 3 = σ i, σ j = 2 δ i j I 0 σ 2 -σ σ 2 -σ σ 3 -σ 3 0

6 6 6.1._DiracEquation.nb γ 5 = 0 I H = α p c + β m c 2 & γ μ = ( β, β α ) = γ 0, γ 0 α H = cγ 0 ( γ p + m c ) γ p = H = c 0 σ p -σ p 0 = c m c σ p σ p -m c m c σ p -σ p m c Lagrangian Since L must be a Lorentz scalar, we write L = c ψ i ħ γ μ μ -m c ψ which gives the Dirac & its conjugate eqs trivially. ψ = ψ + γ 0 ψ + = γ 0 ψ since γ 0 + = γ 0 ( ψ ψ ) + = ψ + ψ + = ψ + γ 0 ψ = ψ ψ i.e., ψ ψ is hermitian. Also, ψ γ μ ψ + = ψ + γ μ + ψ + = ψ + γ 0 γ μ γ 0 ψ + = ψ γ μ ψ where γ 0 ψ + = γ 0 γ 0 ψ = ψ was used. Hence, L is also hermitian. Noether Currents See 3.5._NoetherCurrents.pdf. L is invariant under a global phase transformation ψ(x) e -i ϵ ψ(x) ϵ = real const ψ + (x) e i ϵ ψ + x ψ (x) e i ϵ ψ (x) For infinitesimal transformation ψ(x) (1 - i ϵ ) ψ(x) δ s ψ(x) = -i ϵ ψ(x) ϵ Δ δ s ψ (x) = i ϵ ψ (x) ϵ Δ Conserved Noether current is j μ = L μ ψ Δ + L μ ψ Δ = -i L μ ψ ψ + i L μ ψ ψ = ħ c ψ γ μ ψ Chiral transformation is defined as ψ(x) e -i ϵ γ 5 ψ(x) ϵ = real const

7 6.1._DiracEquation.nb 7 ψ + (x) ψ + (x) e i ϵ γ 5 ( γ 5 + = γ 5 ) since Hence ψ (x) ψ + (x) e -i ϵ γ 5 γ 0 = ψ (x) e -i ϵ γ 5 ( γ 5 γ 0 = -γ 0 γ 5 ) ψ(x) ψ(x) ψ(x) e -2 i ϵ γ 5 ψ(x) ψ(x) γ μ ψ(x) ψ(x) e -i ϵ γ 5 γ μ e -i ϵ γ 5 ψ(x) = ψ(x) γ μ ψ(x) γ 5 γ μ = -γ μ γ 5 L = c ψ i ħ γ μ μ -m c ψ γ μ e -i ϵ γ 5 = e i ϵ γ 5 γ μ L' = c ψ i ħ γ μ μ -e -2 i ϵ γ 5 m c ψ which means L is chiral invariant only if m = 0. In which case, conserved Noether current is j μ = L μ ψ Δ + L μ ψ Δ L = -i μ ψ γ L 5 ψ -i μ ψ ψ γ 5 = ħ c ψ γ μ γ 5 ψ ( for massless particles only ) Since γ 5 is a pseudo-scalar, this is a pseudo-current.

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