@Slereah if you could see this transformation in details:
$$
\begin{aligned}\left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) & \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\ &=\Lambda_{\frac{1}{2}} \Lambda_{\frac{1}{2}}^{-1}\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \end{aligned}
$$

I disappointed why he factored (lambda's_1/2) instead of

\begin{aligned}\left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) & \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\ &=\Lambda_{\frac{1}{2}} \Lambda_{\frac{1}{2}}^{-1}\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\
&= \Lambda_{\frac{1}{2}} \left[i (\Lambda_{\frac{1}{2}}^{-1} \gamma^{\mu} \Lambda_{\frac{1}{2}} )\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \psi\left(\Lambda^{-1} x\rig…

@bolbteppa Actually, I followed P&S till this point but I am saying that when performing any transformation you transform EACH element in the equation, hence, gamma should be transformed as well as I write it

$\begin{aligned}\left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) & \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) \\ &=\left[i \Lambda_{\frac{1}{2}} \gamma^{\mu}\Lambda_{\frac{1}{2}}^{-1}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\L…