130_notes.dvi

∂

∂x′ 4

=

∂

∂x 4

(

−γj

∂

∂xj

+γ 4

∂

∂x 4

)

SPψ+

mc

̄h

SPψ = 0

SP−^1

(

−γj

∂

∂xj

+γ 4

∂

∂x 4

)

SPψ+

mc

̄h

ψ = 0

Sinceγ 4 commutes with itself but anticommutes with theγi, it works fine.

SP=γ 4

(We could multiply it by a phase factor if we want, but there is no point to it.)

Therefore, under aparity inversion operation

ψ′=SPψ=γ 4 ψ

Sinceγ 4 =







1 0 0 0

0 1 0 0

0 0 −1 0

0 0 0 − 1





, the third and fourth components of the spinor change sign while

the first two don’t. Since we could have chosen−γ 4 , all we know is thatcomponents 3 and 4
have the opposite parity of components 1 and 2.

36.11Bilinear Covariants

We have seen that the constantγmatrices can be used to make aconserved vector current

jμ=icψγ ̄μψ

that transforms correctly under Lorentz transformations. With 4 by 4 matrices, we should be able
to make up to 16 components. The vector above represents 4 of those.

TheDirac spinor is transformedby the matrixS.

ψ′=Sψ

This implies thatψ ̄=ψ†γ 4 transforms according to the equation.

ψ ̄′= (Sψ)†γ 4 =ψ†S†γ 4

Looking at the two transformations, we can write the inverse transformation.

Srot= cos

θ

2

+γiγjsin

θ

2

Sboost= cosh

χ

2

+iγiγ 4 sinh

χ

2

Srot−^1 = cos

θ

2

−γiγjsin

θ

2