In the Ashok Das book "Lectures on quantum field theory" , it's written that in page 76 : therefore, the matrix $ S~ \gamma^0~ S^\dagger ~ \gamma^0$ must be proportional to the identity matrix (this can be easily checked by taking a linear combination of the sixteen basis matrices in (2.100) and calculating the commutator with $\gamma^\mu$ ). As a result, we can denote
$ S~ \gamma^0~ S^\dagger ~ \gamma^0 = b~ 1 $$$ S~ \gamma^0~ S^\dagger ~ \gamma^0 = b~ 1 $$
Do any one know how this can be made ?
Edit
Where Dirac equation
$iγ^ μ ∂ μ − m ψ(x) = 0 $.$$iγ^ μ ∂ μ − m ψ(x) = 0 .$$
Under a Lorentz transformation $x^ μ → x ^{′ μ} = Λ^μ_ν x^ ν $,$$x^ μ → x ^{′ μ} = Λ^μ_ν x^ ν ,$$
the transformed equation has the form
$iγ^ μ ∂_μ′ − m ψ ′ (x ′ ) = 0$,$$iγ^ μ ∂_μ′ − m ψ ′ (x ′ ) = 0,$$
where
$\psi'(x') = \psi'(\Lambda x)= S(\Lambda) \psi(x) $$$\psi'(x') = \psi'(\Lambda x)= S(\Lambda) \psi(x) $$
and
$\psi(x) = S^{-1}(\Lambda) \psi'(x') $$$\psi(x) = S^{-1}(\Lambda) \psi'(x') .$$
