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A lagrangian is a scalar under any relevant symmetry group (at least in standard theory). This would make me think that we can take any single term within any lagrangian and transpose it without repercussions -- doing any subsequent algebra correctly (taking fermionic anti-commutation into account, for instance), of course. Is this correct?

I'm not sure how the fact that the degrees of freedom here are operator-valued fields can mess with this conclusion -- even if the transposition regards only Lorentz spinor indices.

For instance, consider an interaction term of a doubly-charged vector boson with a pair of same-sign charged leptons. Can we 'impose' the first equality below

$$ U^{++}\bar{\ell_b^c}\gamma_\mu P_L \ell_a=(U^{++}\bar{\ell_b^c}\gamma_\mu P_L \ell_a)^T=-U^{++}\bar{\ell_a^c}\gamma_\mu P_R\ell_b $$

(where $\ell^c$ is the charge conjugate of $\ell$ and the $P$ are chirality projectors)?

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Generally people do not work with operator-valued Lagrangians. In fact, the only source I can think of off the top of my head that does is Weinberg's QFT volume 1, and if I recall correctly he only uses it to work out some things for QED. Working with operator-valued Lagrangians will generically get you into a lot of trouble with ordering issues.

However, let me also say that, indeed, the Lagrangian is a scalar in the sense that all indices, whether they be Lorentz or internal, must be contracted in a sensible way. So necessarily, this object is invariant under the operator of "transpose" where "transpose" in defined to mean exchanging the orders of these indices.

If your Lagrangian is operator-valued, it will not necessarily be invariant under transposing terms as operators. In general, if you are going to work with objects like the ones you're describing, I recommend you switch to using index notation as using position-based notation for indicating matrix multiplications tends to conflict with operator multiplication ordering (essentially, you're trying to use adjacency of symbols to indicate both operator multiplication and matrix multiplications). It's generally a good idea to not overload notation in this way, and it would seem that this is precisely what's causing your confusion.

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  • $\begingroup$ In a general sense (regarding the lagrangian in the way people usually do), then, the consequences of the first equality in my example can be carried on? Thank you! $\endgroup$
    – GaloisFan
    Feb 18 at 22:31

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