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Trying to understand some about Lorentz invariance and representation theory, I thought that the best way is with an example of application: Show the Lorentz invariance of the Dirac Equation $$(i \hbar \gamma^\mu \partial_\mu - m c )\psi(x) = 0 \tag1$$ where the $x$ denotes the four-position.

So I use the transformation rules: $$\psi^\prime(x^\prime) = S[\Lambda] \psi(x) \implies \psi(x) = S^{-1}[\Lambda] \psi^\prime(x^\prime) \tag2$$ $$ \partial_\mu = \frac{\partial}{\partial x^\mu} = \frac{\partial x^{\mu^\prime}}{\partial x^\mu} \frac{\partial}{\partial x^{\mu^\prime}} = \Lambda_\mu^{\mu^\prime} \frac{\partial}{\partial x^{\mu^\prime}} = \Lambda_\mu^{\mu^\prime} \partial_{\mu^\prime} \tag3 $$

Replacing $(2)$ and $(3)$ on $(1)$, and premultiplying by $S[\Lambda]$, we get: $$S[\Lambda](i \hbar \gamma^\mu \Lambda_\mu^{\mu^\prime} \partial_{\mu^\prime} - m c )S^{-1}[\Lambda] \psi^\prime(x^\prime) = 0 \tag4$$ And then distributing I get: $$(i \hbar X^{\mu^\prime} \partial_{\mu^\prime} - m c )\psi^\prime(x^\prime) = 0 \tag5$$ where $X^{\mu^\prime} = S[\Lambda] \Lambda_\mu^{\mu^\prime} \gamma^\mu S^{-1}[\Lambda]$. And I showed that $$[X^{\mu^\prime},X^{\nu^\prime}] = 2\eta^{\mu^\prime \nu^\prime} \tag6$$ So my first question. Is $(6)$ enough for saying that $X^{\mu^\prime}$ are gamma matrices and with this say that the Dirac equation is Lorentz Invariant?

How can I get the Transformation relation $(2)$? I know that cames from the action of the Lorentz group, but I saw that the action takes the form: $\psi^a(x) \rightarrow S[\Lambda]_b^a \psi^b(\Lambda^{-1}x)$ wich I don't understand very well and just confuses me. So how I procced with knowing $x' = \Lambda x$?

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  • $\begingroup$ You mean how do you invert the matrix S in (2) to go from the 1st eqn to the 2nd? $\endgroup$ – Cosmas Zachos Apr 12 '18 at 15:31
  • $\begingroup$ Basically one of my doubts is how to get $\psi^\prime(x^\prime)$ starting from $x' = \Lambda x$, mostly the reasoning behind that. $\endgroup$ – Julian Ar. Apr 12 '18 at 17:42
  • $\begingroup$ S is the spinor space transformation matrix. For a constant spinor, that would be the entire Lorentz transformation. For x-dependence, you also transform the argument. $\endgroup$ – Cosmas Zachos Apr 12 '18 at 18:42

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