Expansion of $S(\Lambda)$ for infinitesimal Lorentz transformation of Dirac spinors

Question

When dealing with deriving transformation properties of Dirac spinors under Lorentz transformations, we use $S(\Lambda)$ matrix, such that: $$\psi'(x') = S(\Lambda)\psi(x).$$ One of the steps is expanding $S$ for infinitesimal Lorentz transformations, and it is usually written as: $$S(\Lambda) = I - \frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}.$$ I'm having trouble understanding this form and I need confirmation of my suspicions. Textbooks usually take this sort of as an ansatz, but I want to understand why. First, I understand the identity matrix part intuitively.

Second, in my understanding $\omega_{\mu\nu}M^{\mu\nu}$ is just a nice notation to represent the fact that we want $S$ to linearly depend on all Lorentz parameters gathered in $\omega$ and all Lorentz group generators gathered in $M$. But since these are antisymmetric tensors when taken like this (only 6 free parameters), the summation convention gives us 2 times the linear combination of all of them. That explains also the $1/2$ factor.

Third, for fixed $\mu$ and $\nu$, $M^{\mu\nu}$ is just a 4x4 matrix, and the antihermitian one. Imaginary unit factor makes sure that the whole $\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}$ part is hermitian, which ensures the $S$ to be unitary.

And finally, I don't have an idea why the minus, but it's consistently showing up in textbooks.

This is my intuitive approach to understanding it, but if you have a more formal way, or several of them, please include them.

Edit: $S(\Lambda)$ is not unitary so that part can be ignored, which leaves open question - why $i$? As @Prahar and @MadMax said both $i$ and $-$ are just a convention.

Answer 1

Echoing @Prahar's comments, for $$ S(\Lambda) = exp(-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}) $$ to be Unitary, the exponent $$ -\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu} $$ has to be anti-Hermitian, which translated to $M^{\mu\nu}$ should be Hermitian, because of the extra $i$ in the exponent. (The minus sign does not matter, it's just a convention. You can re-define $\omega_{\mu\nu}$ as $\omega_{\mu\nu}' = -\omega_{\mu\nu}$ to absorb the minus sign.)

Is $M^{\mu\nu}$ Hermitian? Well, only the 3D space rotations are Hermitian, while the Lorentz boosts $M^{0\nu}$ are NOT Hermitian.

Therefore, $S(\Lambda)$ in general is NOT Unitary.

That said, there is the dubious claim of existence of Wightman's unitary $U(\Lambda)$ for Lorentz transformation: $$ \begin{equation} U(\Lambda)^\dagger \psi(x) \space U(\Lambda) = S(\Lambda) \psi(\Lambda^{-1}x) \tag{1}\end{equation} $$ where $\psi$ is a quantum spinor field transforming in the $(\frac{1}{2},0)$ irrep of $SL(2,\mathbb{C})$.

See another post here for my challenge to the proof of existence of unitary $U(\Lambda)$ for the spinor field.