# Lorentz transformation of Dirac spinor

I'm wondering again what I'm missing in my understanding. In Peskin and Schroeder, as well as in other sources, the spinor representation of Lorentz transformation is given by $$\Lambda_\frac{1}{2}=exp(-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu})$$. Where $$\omega_{\mu\nu}$$ is antisymmetric tensor representing the Lorentz transformation and $$S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$$. In the same sources they write that $$\Lambda^\dagger_\frac{1}{2}=exp(\frac{i}{2}\omega_{\mu\nu}(S^{\mu\nu})^\dagger)$$. The question, using the matrix identity $$(AB)^T=B^TA^T$$ shouldn't we have $$\Lambda^\dagger_\frac{1}{2}=exp(\frac{i}{2}(S^{\mu\nu})^\dagger\omega_{\mu\nu})$$?

• That would be the same. Feb 3, 2020 at 7:27
• @Qmechanic Could you explain, why? Feb 3, 2020 at 7:34
• Because $\omega_{\mu\nu}$ does not carry Dirac indices. Feb 3, 2020 at 7:38
• @Qmecanic Probably the fact that I just started the Dirac field chapter is the reason that I'm not familiar with this term (Dirac indices). Is there a mathematical explanation for that? Feb 3, 2020 at 7:54
• The point is that for each $\mu,\nu$ the object $S^{\mu\nu}$ is a matrix, while $\omega_{\mu\nu}$ is a real number. So $\omega_{\mu\nu}S^{\mu\nu}$ is just a linear combination of matrices with real coefficients. The dagger only affects the matrices.
– Gold
Feb 4, 2020 at 1:23

$$\omega_{\mu\nu}$$ are simply real numbers, while $$S^{\mu\nu}$$ are matrices that act on the Dirac spinors. The components of the matrix $$S^{\mu\nu}$$ would be $$(S^{\mu\nu})_{\alpha\beta}$$ where $$\alpha$$ and $$\beta$$ are the "Dirac indices". I.e. the components of the spinor $$S^{\mu\nu} \psi(x)$$ are $$(S^{\mu\nu} \psi(x))_\alpha = (S^{\mu\nu})_{\alpha\beta} \psi_\beta(x)$$.
In general the components $$(S^{\mu\nu})_{\alpha\beta}$$, $$(\gamma^\mu)_{\alpha\beta}$$, $$\eta_{\mu\nu}$$ etc. are all just real numbers and therefore commute with each other. The exception is the components $$\psi_\alpha$$, which anticommute because they are fermion fields: $$\psi_\alpha \psi_\beta = -\psi_\beta \psi_\alpha$$ and $$\{\psi_\alpha(\vec{x}), \psi_\beta^\dagger(\vec{y})\} = \delta_{\alpha\beta} \delta^{(3)}(\vec{x} - \vec{y})$$. Even in the classical limit, they are anticommutative Grassmann numbers: $$\psi_\alpha \psi_\beta = -\psi_\beta \psi_\alpha$$ and $$\psi_\alpha \psi_\beta^\dagger = -\psi_\beta^\dagger \psi_\alpha$$
\begin{align} Λ = e^{−\frac{i}{2}ω_{μν}S^{μν}} = e^{−\frac{i}{2}(ω_{01}S^{01} + ω_{02}S^{02}...+ ω_{32}S^{32})} \end{align}
$$ω_{0i}$$ and $$ω_{i0}$$ are infinitesimal angles for boost (so that $$cosh(ω_{0i}) = \gamma$$, and $$sinh(ω_{0i}) = v_i\gamma$$ for finite boosts) and $$ω_{ij}$$ are infinitesimal angles for rotations. Just real numbers.