Our problem to find a similarprobability and flux for Dirac theory is similarbut a little more
difficult. Start with the Dirac equation.
(
γμ∂
∂xμ+
mc
̄h)
ψ= 0Since the wave function is a 4 component spinor, we will use theHermitian conjugate of the
Dirac equationinstead of the complex conjugate. Theγmatrices are Hermitian.
γμ∂ψ
∂xμ+
mc
̄hψ= 0∂ψ†
∂(xμ)∗
γμ+mc
̄h
ψ†= 0The complex conjugate does nothing to the spatial component ofxμbut does change the sign of
the fourth component. To turn this back into a 4-vector expression, we can change the sign back by
multiplying the equation byγ 4 (from the right).
∂ψ†
∂xkγk+∂ψ†
∂(x 4 )∗γ 4 +mc
̄hψ†= 0∂ψ†
∂xkγkγ 4 −
∂ψ†
∂x 4γ 4 γ 4 +
mc
̄hψ†γ 4 = 0−
∂ψ†γ 4
∂xkγk−∂ψ†γ 4
∂x 4γ 4 +mc
̄hψ†γ 4 = 0Definingψ ̄=ψ†γ 4 , theadjoint spinor, we can rewrite the Hermitian conjugate equation.
−
∂ψ ̄
∂xkγk−
∂ψ ̄
∂x 4γ 4 +
mc
̄hψ ̄= 0−
∂ψ ̄
∂xμγμ+mc
̄hψ ̄= 0This is the adjoint equation. We now multiply the Dirac equation byψ ̄from the left and multiply
the adjoint equation byψfrom the right, and subtract.
ψγ ̄μ∂ψ
∂xμ+
mc
̄hψψ ̄ +∂ψ ̄
∂xμγμψ−mc
̄hψψ ̄ = 0ψγ ̄μ∂ψ
∂xμ+
∂ψ ̄
∂xμγμψ= 0∂
∂xμ(
ψγ ̄μψ)
= 0
jμ=ψγ ̄μψ∂
∂xμjμ= 0