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 $

Do any one know how this can be made ?

Edit

Where Dirac equation

$iγ^ μ ∂ μ − m ψ(x) = 0 $.

Under a Lorentz transformation $x^ μ → x ^{′ μ} = Λ^μ_ν x^ ν $,

the transformed equation has the form

$iγ^ μ ∂_μ′ − m ψ ′ (x ′ ) = 0$,

where

$\psi'(x') = \psi'(\Lambda x)= S(\Lambda) \psi(x) $

and

$\psi(x) = S^{-1}(\Lambda) \psi'(x') $