Why does the $Z$ boson not change quark flavour? Is it just an observation from Particle Accelerators data?
I was trying to understand what is it that mathematically stops the $Z$ boson from changing quark flavour, if it is known?
 A: It does not mediate flavor changing because the couplings in the gauge basis are of the form $ c_L Z_\mu \bar{\psi}^i_L \gamma^\mu \psi^i_L$ and $c_R Z_\mu \bar{\psi}^i_R \gamma^\mu \psi^i_R$ with $i$ flavor index and $c_{L,R}$ flavor independent by gauge invariance (that is, $c_L$ and $c_R$ are proportional to the identity in flavor space). In this basis the mass terms is schematically of the form $m_{ij}\bar{\psi}^i_{L}\psi^j_{R}+h.c.$. Therefore, diagonalizing $m_{ij}=[U^\dagger m^{diag}V]_{ij}$ with two unitary matrix $U$ and $V$ in flavor space, we leave the gauge interactions invariant, that is flavor blind:
$$
(m_{ij}\bar{\psi}^i_{L}\psi^j_{R}+h.c.)+c_L Z_\mu \bar{\psi}^i_L \gamma^\mu \psi^i_L+\ldots=(m^{diag}_{ii}\bar{\Psi}^i_{L}\Psi^i_{R}+h.c.)+c_L Z_\mu \bar{\Psi}^i_L \gamma^\mu \Psi^i_L+\dots
$$
where $\psi_L=U \Psi_L$ and $\psi_R= V \Psi_R$.
The situation is different instead for charged currents because they couple fields $\psi$ and $f$ of different type (i.e. charge) $W_\mu \psi^i_L f^i_R$ that rotate differently under the transformation that diagonalizes the mass terms $m^{(\psi)}_{ij}\psi^i_L \psi^j_R$ and $m^{(f)}_{ij}f^i_L f^j_R$. Said differently, the charged currents are sensitive to two mass terms since they involve two different charges, whereas the neutral charges deal with one type of mass term at the time.
