the free Dirac equation reads,
(
−m σμ∂μ
̄σμ∂μ −m
)(
ψL
ψR)
= 0 (22.41)Separating the two components,
σμ∂μψR−mψL = 0
σ ̄μ∂μψL−mψR = 0 (22.42)we see that the massmproduces a coupling or a mixing between the left and right Weyl spinors.
When the massmvanishes, the equations for left and right Weyl spinors decouple, and we are
left with the Dirac equation for a massless spinor,
σμ∂μψR = 0
σ ̄μ∂μψL = 0 (22.43)It is now consistent to retain only one of these 2-component spinors, and set the other to zero. For
example, settingψR= 0 produces the left Weyl spinor equation,
σμ∂μψL= 0 (22.44)Note that, by construction, this equation is also Lorentz-invariant, just as the full Dirac equation
was. Whenm 6 = 0, it is of course inconsistent to setψR = 0, since the Dirac equation then
automatically implies that alsoψL= 0.
22.5 Elementary solutions to the free Dirac equation
The free Dirac equation, (γμ∂μ−m)ψ(x) = 0 is a linear partial differential equation withx-
independent coefficients, and may thus be solved by Fourrier analysis. The elementary solutions
are labeled by the 4-momentumk,
ψk(x) =u(k)e−ik·x (22.45)whereu(k) is a 4-component spinor which satisfies the algebraic equation
(iγμkμ+m)u(k) = 0 (22.46)In the Weyl basis, we decomposeu(k) intouL(k) anduR(k), for which the equation reads,
(
m iσμkμ
iσ ̄μkμ m
)(
uL
uR)
= 0 (22.47)or separating out space from time components,
i(−k 0 +~k·~σ)uL+muR = 0
i(+k 0 +~k·~σ)uR+muL = 0 (22.48)