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I was pretty confident that things are simple, but unfortunately I must have missed something. We can always change between between the bases for Dirac spinors, using unitary transformation, because

$$ \partial_\mu \bar\Psi \gamma_\mu \Psi=\partial_\mu \bar\Psi \underbrace{U^\dagger U }_{=1}\gamma_\mu \underbrace{U^\dagger U }_{=1}\Psi= \partial_\mu\underbrace{ \bar\Psi U^\dagger}_{= \bar \Psi' } \underbrace{U \gamma_\mu U^\dagger}_{\gamma_\mu'} \underbrace{U \Psi }_{\Psi'}$$

The action of the projection operator $P_L = \frac{1-\gamma_5}{2}$ is obvious in the Weyl basis:

$$ \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} \begin{pmatrix} \Psi_L \\ \Psi_R \end{pmatrix} = \begin{pmatrix} \Psi_L \\ 0 \end{pmatrix}. $$

In a different basis:

$$ P_L^{{\mathrm{Weyl}}} \Psi^{{\mathrm{Weyl}}} = \Psi^{{\mathrm{Weyl}}}_{{\mathrm{left}}} \Rightarrow P_L' \Psi' = \frac{1-\gamma'_5}{2}\underbrace{ \Psi'}_{U\Psi^{{\mathrm{Weyl}}}}= \frac{1-U i \gamma_0 U^\dagger U \gamma_1 U^\dagger U\gamma_2U^\dagger U \gamma_3U^\dagger }{2}U\Psi^{{\mathrm{Weyl}}}= \frac{1-U i \gamma_0 \gamma_1 \gamma_2 \gamma_3 }{2}\Psi^{{\mathrm{Weyl}}} $$

and I was hoping that in the end $U P_L^{{\mathrm{Weyl}}} \Psi^{{\mathrm{Weyl}}}= \Psi'_{{\mathrm{left}}}$, which would show that the projection operator works in different bases, too. Unfortunately, I can't factor out $U$.

Any ideas about what I missed would be great!

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Either you incorrectly think that $1\cdot U=1$ or more probably have forgotten $U$ in the first term: $$P_L' \Psi' = \frac{1}{2}U\Psi^{{\mathrm{Weyl}}}-\frac{1}{2}\gamma_5'U\Psi^{{\mathrm{Weyl}}} = \frac{U-U i \gamma_0 \gamma_1 \gamma_2 \gamma_3 }{2}\Psi^{{\mathrm{Weyl}}}=U\Psi^{{\mathrm{Weyl}}}_{{\mathrm{left}}}=\Psi'_L$$

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