To calculate the combination−γ^0 D(Λ)†γ^0 , we take Λ infinitesimally close to the identity, so that
D(Λ) = I+
1
2
ωμνSμν+O(ω^2 )= I+ω^0 iS 0 i+1
2
ωijSij+O(ω^2 )D(Λ)† = I+ω^0 iS 0 †i+1
2
ωijSij†+O(ω^2 ) (22.56)From the explicit representations ofSμν, we have that
S† 0 i = +S 0 i=γ^0 S 0 iγ^0
S†ij = −Sij=γ^0 Sijγ^0 (22.57)Thus, we have
D(Λ)† = I+ω^0 iγ^0 Si 0 γ^0 +1
2
ωijγ^0 Sijγ^0 +O(ω^2 )= −γ^0(
I−ω^0 iS 0 i−1
2
ωijSij+O(ω^2 ))
γ^0= −γ^0 D(Λ)−^1 γ^0 (22.58)Using this expression in the transformation law forψ ̄, we get
ψ ̄(x)→ψ ̄′(x′) =ψ ̄(x)D(Λ)−^1 (22.59)It is now straightforward to derive the transformation law of the fermion current,
jμ(x)→j′μ(x′) = ψ ̄′(x′)γμψ′(x′)
= ψ ̄(x)D(Λ)−^1 γμD(Λ)ψ(x)
= Λμνjν(x) (22.60)Finally, we check that the current is conserved,
∂μjμ=∂μ(ψγ ̄ μψ) =ψγ ̄ μ∂μψ+ (∂μψ ̄)γμψ (22.61)The first term on the rhs is simplified by using the Dirac equation,γμ∂μψ=mψ, while the second
term can be handled using the conjugate of the Dirac equation,
(∂μψ)†(γμ)†−mψ†= 0 (22.62)Using (γμ)†=γ^0 γμγ^0 , this becomes,
(∂μψ ̄)γμ+mψ ̄= 0 (22.63)Combining both equations gives∂μjμ= 0.