# Lorentz transformation of the Spinor Field

The mistake you are making is in "daggering" the object $\omega_{\mu\nu}$. For each $\mu, \nu = 0,\dots 3$, the symbol $\omega_{\mu\nu}$ is a real number, so its dagger (which is really just complex conjugation in this case) does nothing; $(\omega_{\mu\nu})^\dagger = \omega_{\mu\nu}$.

When we say that $\omega_{\mu\nu}$ is an antisymmetric real matrix, we really mean that the matrix with these numbers as components is such a matrix, not that $\omega_{\mu\nu}$ is a matrix for each $\mu$ and $\nu$.

The last step you performed is incorrect. \begin{aligned} \psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\& = \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\& = \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right) \end{aligned} $\omega$ is real, simply means that $$$$(\omega_{\mu \nu})^\dagger = \omega_{\mu \nu}$$$$ So we have \begin{aligned} \psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\& = \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\& = \psi^\dagger \left(1+\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right) \end{aligned} To find the Lorentz invariance of $\bar\psi\psi$ you are missing $\bar\psi = \psi^\dagger\gamma^0$. When this $\gamma^0$ passes through $S^{\mu\nu}$ it fixes the problem. Solve it and share with us again.