QuantumPhysics.dvi

Eliminating one or the other components, gives the standardenergy-momentum-mass relation,
(k^20 −~k^2 −m^2 )uL,R(~k) = 0. To solve the Dirac equation, it is useful to separate themassless case
from the massive one. Whenm= 0, we may choose the momentum as follows,kμ= (k 0 , 0 , 0 ,k)
wherek 0 =|k|. The equations foruL,uRthen read,

(k 0 −kσ^3 )uL = 0

(k 0 +kσ^3 )uR = 0 (22.49)

The solutions depend on the sign ofk, and we have,

k > 0

{

uL 2 = 0

uR 1 = 0

k < 0

{

uL 1 = 0

uR 2 = 0

(22.50)

For a left Weyl spinor, for example, withk > 0, there is a 1-dimensional space of solutions,
parametrized by uL 1 , corresponding to the definite helicity +1/2 ̄h. Form 6 = 0, we have two
linearly independent solutionsu(ℓ)forℓ= 1,2, which may be parametrized as follows,

u(Rℓ)=

i

m

(k 0 −~k·~σ)u(Lℓ) u(1)L =

(

1

0

)

u(2)L =

(

0

1

)

(22.51)

22.6 The conserved current of fermion number

The Dirac equation is invariant under phase rotations of thefieldψby anx-independent angleθ,

ψ(x)→ψ′(x) =eiθψ(x) (22.52)

and the associated conserved current is given by

jμ=ψ†γ^0 γμψ (22.53)

We begin by showing thatjμ indeed transforms as a 4-vector under Lorentz transformations.
Group theoretically, this works as follows. We decomposeψinto its Weyl components, and recast
the fermion number current in terms ofψL,R,

jμ=ψ†L ̄σμψL+ψ†RσμψR (22.54)

SinceψL ∼ (^12 ,0) andψR ∼ (0,^12 ), this combination does indeed give rise to a vector, since
ψ†L⊗ψL∼(^12 ,^12 ), as well asψ†R⊗ψR∼(^12 ,^12 ). More precisely, we need the transformation matrices
for the complex conjugate ofψ, which can be obtained as follows,

ψ(x) → ψ′(x′) =D(Λ)ψ(x)

ψ(x)† → (ψ′(x))†=ψ(x)†D(Λ)†

ψ ̄(x) → ψ ̄′(x′) =ψ(x)†D(Λ)†γ^0 =−ψ ̄(x)γ^0 D(Λ)†γ^0 (22.55)