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 $$\Lambda^\dagger_\frac{1}{2}=exp(\frac{i}{2}(S^{\mu\nu})^\dagger\omega_{\mu\nu})$$?
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$\begingroup$ That would be the same. $\endgroup$– Qmechanic ♦Commented Feb 3, 2020 at 7:27
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$\begingroup$ @Qmechanic Could you explain, why? $\endgroup$– v_talCommented Feb 3, 2020 at 7:34
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1$\begingroup$ Because $\omega_{\mu\nu}$ does not carry Dirac indices. $\endgroup$– Qmechanic ♦Commented Feb 3, 2020 at 7:38
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$\begingroup$ @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? $\endgroup$– v_talCommented Feb 3, 2020 at 7:54
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1$\begingroup$ 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. $\endgroup$– GoldCommented Feb 4, 2020 at 1:23
2 Answers
$\omega_{\mu\nu}$ 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 $\psi_\alpha \psi_\beta^\dagger = -\psi_\beta^\dagger \psi_\alpha$
Expanding the exponential:
\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.