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In Paul Langacker's The Standard Model and Beyond, equation 3.80 says the following

$$ \mathcal{L}' = \overline{\psi} \mathrm{i} \partial ^{\mu} \gamma _{\mu} e^{- \mathrm{i} \beta ^i L^i} e^{\mathrm{i} \beta ^i L^i} \psi $$

While what I would assume is

$$ \mathcal{L}' = \psi ^{\dagger} e^{- \mathrm{i} \beta ^i L^i} \gamma ^0 \mathrm{i} \partial ^{\mu} \gamma _{\mu} e^{\mathrm{i} \beta ^i L^i} \psi $$

The reasons are, first, the transformation matrix $e^{- \mathrm{i} \beta ^i L^i}$ may not commute with $\gamma$ matrices, second, $\beta ^i$ might be local functions and therefore should not be moved crossing the derivative.

Thank you in advance!

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  • $\begingroup$ Why do you think that $e^{-i\beta^{i}L^{i}}$ does not commute with $\gamma$'s? $\endgroup$ – Greg.Paul Jul 18 '18 at 18:49
  • $\begingroup$ @Greg.Paul I think they are both matrices therefore they do not necessarily commute each other. Did I misunderstand? $\endgroup$ – zyy Jul 18 '18 at 19:03
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I think I find out why!

Field $\psi$ represents $n$ fermions, from $\psi _1$ to $\psi _n$. Each $\psi _i$, however, is a one by four vector.

Matrices $L^i$ are $n$ by $n$ matrices, $\gamma ^{\mu}$ are four by four matrices.

With above information, it is not hard to realize that each components in transformation matrices is seeing each individual components of $\psi$, though being a vector, as a whole.

Since there is no mixing in kinetic terms, $\overline{\psi}_i$ should match $\psi _i$, and for each $\overline{\psi}'_i \mathrm{i} \gamma ^{\mu} \partial _{\mu} \psi '_i$ term, only components of transformation present, which are scalars. Therefore they could perfectly commute through $\gamma$ matrices, they are not acting in the same space!

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