130_notes.dvi

Now take the sum and the difference of the two equations.

i ̄h~σ·∇~(φ(L)−φ(R))−i ̄h

∂

∂x 0

(φ(R)+φ(L)) +mc(φ(R)+φ(L)) = 0

i ̄h~σ·∇~(φ(R)+φ(L))−i ̄h

∂

∂x 0

(φ(L)−φ(R)) +mc(φ(R)−φ(L)) = 0

Now rewriting in terms ofψA=φ(R)+φ(L)andψB =φ(R)−φ(L)and ordering it as a matrix
equation, we get.

−i ̄h~σ·∇~(φ(R)−φ(L))−i ̄h

∂

∂x 0

(φ(R)+φ(L)) +mc(φ(R)+φ(L)) = 0

i ̄h~σ·∇~(φ(R)+φ(L)) +i ̄h

∂

∂x 0

(φ(R)−φ(L)) +mc(φ(R)−φ(L)) = 0

−i ̄h

∂

∂x 0

(φ(R)+φ(L))−i ̄h~σ·∇~(φ(R)−φ(L)) +mc(φ(R)+φ(L)) = 0

i ̄h~σ·∇~(φ(R)+φ(L)) +i ̄h

∂

∂x 0

(φ(R)−φ(L)) +mc(φ(R)−φ(L)) = 0

−i ̄h

∂

∂x 0

ψA−i ̄h~σ·∇~ψB+mcψA= 0

i ̄h~σ·∇~ψA+i ̄h

∂

∂x 0

ψB+mcψB= 0

(

−i ̄h∂x∂ 0 −i ̄h~σ·∇~

i ̄h~σ·∇~ i ̄h∂x∂ 0

)(

ψA

ψB

)

+mc

(

ψA

ψB

)

= 0

Remember thatψAandψBare two component spinors sothis is an equation in 4 components.

We can rewrite the matrix above as a dot product between 4-vectors. The matrix has a dot product
in 3 dimensions and a time component
(
−i ̄h∂x∂ 0 −i ̄h~σ·∇~
i ̄h~σ·∇~ i ̄h∂x∂ 0

)

= ̄h

[(

0 −i~σ·∇~

i~σ·∇~ 0

)

+

( ∂

∂x 4 0

0 −∂x∂ 4

)]

= ̄h

[(

0 −iσi

iσi 0

)

∂

∂xi

+

(

1 0

0 − 1

)

∂

∂x 4

]

= ̄h

[

γμ

∂

∂xμ

]

The4 by 4 matricesγμare given by.

γi =

(

0 −iσi

iσi 0

)

γ 4 =

(

1 0

0 − 1

)

With this definition, the relativistic equation can be simplified a great deal.
(
−i ̄h∂x∂ 0 −i ̄h~σ·∇~
i ̄h~σ·∇~ i ̄h∂x∂ 0

)(

ψA

ψB

)

+mc

(

ψA

ψB

)

= 0

ψ=

(

ψA

ψB

)

≡







ψ 1

ψ 2

ψ 3

ψ 4





