Why is Quark Mixing forbidden in the Lagrangian (pre CKM)

Question

The corresponding term in Lagrangian for the coupling of quarks to gauge fields reads $$ \sum_{i} \bar Q_i D_\mu \gamma^ \mu Q_i .$$

Considering the Yukawa terms it is generally stated, that no symmetry principle forbids generation mixing terms in the first place, therefore one writes in general

$$\sum_{i,j} \bar Y_{ij} Q_i \Phi Q_i .$$

After symmetry breaking the mass matrices are introduced, which aren't orthogonal. Global SU(2) invariance is then used to diagonalize them and this eventually leads to generation mixing through the $W$-Bosons, expressed through the CKM matrix.

I was wondering why generation mixing terms in the Lagrangian aren't allowed in the first place for the Quark-Gauge-Field coupling term, like it is for the Yukawa term. In other words: What prevents the Quark-Gauge part of the Lagrangian from being

$$ \sum_{i} \bar Q_i D_{\mu,ij} \gamma^ \mu Q_j,$$

which of course would mean that some gauge coupling matrices $g_{ij}$ appear in the $D_{\mu,ij}$ term, instead of the universal $g$.

This post imported from StackExchange Physics at 2014-10-11 09:49 (UTC), posted by SE-user JakobH

Comments

If by $D_{\mu\,,ij}$ you mean $D_{\mu\,,ij}=\delta_{ij}\partial_\mu-i g_{ij}W_\mu$, then is gauge invariance that enforces $g_{ij}=g\delta_{ij}$. It is pretty clear if you move the gauge coupling $g$ to the gauge boson kinetic term.

This post imported from StackExchange Physics at 2014-10-11 09:50 (UTC), posted by SE-user TwoBs

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Yes, thats what I meant by $D_{\mu\,,ij}$. The gauge boson kinetic term is $F_{\mu\nu}^a F^{\mu\nu a}= (\partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g' f^{abc} A_\mu^b A_\nu^c) ( ...) $. In my understanding allowing $g=g_{ij}$ would mean allowing the possibility that the weak force couples with difference strength to the different quarks, and with another strength, denoted $g'$ above to itself. This may be ugly, but I'm unfortunately not able to see why it should be forbidden by symmetry. Any further thoughts would be awesome!

This post imported from StackExchange Physics at 2014-10-11 09:50 (UTC), posted by SE-user JakobH

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As I've already said, it is gauge invariance that forbids it. If you don't see it, just try to do a gauge transformation (say abelian) to $\bar{Q}_{j}iD_{\mu\,,ji}Q_j$ which will not result to be invariant unless $g_{ij}=g\delta_{ij}$.

This post imported from StackExchange Physics at 2014-10-11 09:50 (UTC), posted by SE-user TwoBs