# Why does transformation matrix commute with $\gamma$ matrices?

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!

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

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!