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To show Lorentz invariance of Dirac equation P&S section 3.2 swap $\Lambda$ and $S(\Lambda)$ as both matrices commute. But why is it true? For example taking $${\cal J}^{01}=\left(\begin{matrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{matrix}\right)\,,$$ and $$S^{03}=\left(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1\\ \end{matrix}\right)\,,$$ $ [{\cal J}^{01},S^{03}]\neq0 $. What am I missing?

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$({\cal J}^{\mu\nu})^{\rho}{}_{\sigma}$ acts in a 4-vector representation of the Lorentz group while $(S^{\mu\nu})^a{}_b$ acts in the Dirac spinor representation, i.e. they don't live in the same representation despite both happen to be given by $4\times4$ matrices.

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  • $\begingroup$ I understand that logically, but when I have an expression in matrix representations of both, I have just two matrices that obviously do not commute. What's the mathematical excuse for swapping them? $\endgroup$ – v_tal Feb 18 at 12:41
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    $\begingroup$ They live in different spaces. $\endgroup$ – Qmechanic Feb 18 at 12:43
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    $\begingroup$ @v_tal Qmechanic means to say that they act on different Hilbert spaces. For example, you can swap momentum operator of particle 1 with that of particle 2 for a composite system, because they live in different spaces. Am I correct, @Qmechanic? $\endgroup$ – Abhay Hegde Feb 18 at 13:27
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Feb 18 at 13:37

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