Mathematical Principles of Theoretical Physics

4.6. WEAK INTERACTION THEORY 237

Thus we obtain

∂μJwμ=i

gw

hc ̄

ωaWμbψ γμ[σa,σb]ψ=− 2

gw

̄hc

(4.6.20) εabcωaWμbJμc.

Here we used[σa,σb] =i 2 εabcσcandJcμ=ψ γμσcψ.Note that

εabcωaWμb=

∣ ∣ ∣ ∣ ∣ ∣

~i ~j ~k

ω 1 ω 2 ω 3

Wμ^1 Wμ^2 Wμ^3

∣ ∣ ∣ ∣ ∣ ∣

=~ω×W~μ,

where~ω= (ω 1 ,ω 2 ,ω 3 ),W~μ= (Wμ^1 ,Wμ^2 ,Wμ^3 ). Hence, (4.6.20) can be rewritten as

(4.6.21) ∂μJμw=−

2 gw

hc ̄

(~ω×~Wμ)·J~μ,

andJ~μ= (J 1 μ,J 2 μ,J 3 μ). The weak current densityJaμis as

Jaμ=θaμδ(r), θaμ the constant tensor,

andW~μin (4.6.21) is replaced by the average value

(4.6.22)

1

|Bρw|

∫

Bρw

W~μdx,

whereBρw⊂Rnis the ball with radiusρw. Similar to the case (4.5.14) for the strong interac-
tion, the average value (4.6.22) is

W~μ=~ζμ/ρw, ~ζμ= (ζμ^1 ,ζμ^2 ,ζμ^3 ).

Thus, (4.6.21) can be expressed as

(4.6.23) ∂μJwμ=−κ δ(r)/ρw,

andκis a parameter, written as

(4.6.24) κ=

2 gw

hc ̄

~θμ·(~ω×~ζμ).

Putting (4.6.23) in (4.6.14) we deduce that

−∆φw+k 02 φw=−

gwκ

ρw

δ(r),

whose solution is given by

(4.6.25) φw=−

gwκ

ρw

1

r

e−k^0 r.

Therefore we obtain the solution of (4.6.14) in the form (4.6.25).