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Dirac equation is $i \hbar \gamma^\mu \partial_\mu \psi - m c \psi = 0 $

To show its Lorentz invariance, we convert spacetime into $x'$ and $t'$ from $x$ and $t$ and then

$( iU^\dagger \gamma^\mu U\partial_\mu^\prime - m)\psi(x^\prime,t^\prime) = 0$

The question is, how does one show from the above equation the following equation follows?:

$U^\dagger(i\gamma^\mu\partial_\mu^\prime - m)U \psi(x^\prime,t^\prime) = 0$

where $U$ is some unitary matrix for lorentz transformation for $\psi$.

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    $\begingroup$ $U$ cannot be unitary: There are no nontrivial unitary finite dimensional representations of Lorentz group. $U$ has to be replaced by a generic $U^{-1}$. Next notice that $m$ should be interpreted as $mI$ where $I$ is the identity matrix and $I= U^{-1}U$, so... $\endgroup$ – Valter Moretti Jul 25 '14 at 9:24
  • $\begingroup$ what about en.wikipedia.org/wiki/… ? It says that $U$ is unitary... $\endgroup$ – user50374 Jul 25 '14 at 9:34
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    $\begingroup$ It is wrong or at least quite confused! It is true that if you transform the $\gamma^\mu$s with a unitary matrix you preserve the structure of Clifford algebra, but there is *no* unitary representation $SL(2,\mathbb C) \ni \Lambda \mapsto U(\Lambda) \in SU(4)$ as the one you intend to use. $\endgroup$ – Valter Moretti Jul 25 '14 at 9:42
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Actually, "unitary representation" is meant with respect to the spinors, which do not form a finite-dimensional space and therefore allow a unitary representation of the proper Lorentz group. The action is defined by $D(\Lambda)\psi(x)=U(\Lambda)\psi(\Lambda^{-1}x)$, and you can simply calculate that this is unitary on your spinor space. However, this does not(!) mean that the matrix $U$ is actually unitary. Therefore I also assume, that you mean $U^{-1}$ instead of $U^\dagger$.

To your problem: Just notice that $U$ is actually a matrix constant with respect to $x'$, therefore it commutes with $\partial'_\mu$, and therefore you have $(iU^{-1}\gamma^\mu U\partial'_\mu-m)=(iU^{-1}\gamma^\mu \partial'_\mu U-U^{-1} mU)=U^{-1}(i\gamma^\mu \partial'_\mu-m)U$.

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