Can quarks of different generations enter a $Z$ boson vertex?

Reading through my particle physics book (Thomson's "Modern Particle Physics"), it appears that when calculating the possible decays of the $Z$ boson, we do not consider decays such as $Z\rightarrow u \overline{c}$. Why is that? As far as I understand, we can have decays such as $W^+\rightarrow c\overline{b}$ where different generations mix (with the according CKM coefficients), and I don't get how is the $Z$ different.

I have an idea as to why, and I will illustrate it here, but I'm not sure my reasoning is correct. Let's say the weak eigenstates (denoted with $'$) of the quarks are related to the mass eigeinstates as such : $$\begin{pmatrix} u'\\ c'\\ t'\end{pmatrix} = U\begin{pmatrix}u\\ c\\ t\end{pmatrix}$$ with U an unitary matrix. Thus, the general Z annihilation vertex can be written as:

$$\frac{g_z}{2}\begin{pmatrix}\overline u &\overline c & \overline t\end{pmatrix}U^{\dagger}\gamma^{\mu}(c_v-c_a\gamma^5)U\begin{pmatrix}u\\ c\\ t\end{pmatrix}$$

And with $UU^{\dagger} = I$, we would get that indeed, the coupling of $u$ with $\overline c$ (for example), would be 0. However I'm not convinced with this argument since the compact way of writing the vertex for all possible interactions is just a "notation". Is there any other way(maybe more physical, or at least more rigorous) to explain the absence of coupling of quarks from different generations, with the $Z$ boson? Or could anyone give insight into my explanation, if it is right?

Edit, extra info : I recently had a similar question which was why can't a electron appear in a annihilation vertex with a neutrino. The answer was straightforward : lepton number violation. This is of course just an experimental fact, and we came up with lepton number to explain it. Could it be that there is something similar here ? I searched but couldn't find a similar concept for this case.

The topic - the general form of neutral current interactions

Suppose the $SU_{L}(2)\times U_{Y}(1)$ electroweak interactions. With corresponding gauge couplings $g_{1}, g_{2}$ and fixed quarks doublets $Q_{i}$, $$\tag 0 Q_{1} = \begin{pmatrix} u\\ d \end{pmatrix}, \quad Q_{2} = \begin{pmatrix}c \\ s\end{pmatrix}, \quad Q_{3} = \begin{pmatrix}t \\ b \end{pmatrix}$$ and the quarks column $q_{a}$, $$\tag 1 q_{a} = (u,c,t,d,s,b)^{T},$$ The quarks ''names'' are fixed by the diagonalization of the Higgs interaction term (i.e., the mass term).

You may write down the most general neutral currents interaction lagrangian $$\tag 2 L_{\text{int}} = g_{1}\bar{Q}_{i}\gamma^{\mu}P_{L}\sigma_{3}a_{ij}Q_{j}(c(\theta)Z_{\mu} - s(\theta)A_{\mu}) +g_{2}\bar{q}_{a}\gamma^{\mu}c_{ab}q_{b}(s(\theta)Z_{\mu}+c(\theta)A_{\mu}) + h.c.$$ Here

1. $a_{ij}$ is the $3\times 3$ unitary matrix acting in a 3-dimensional space of quarks doublets $Q_{i}$ $(0)$,
2. $c_{ab}$ is the $6\times 6$ unitary matrix acting in a 6-dimensional space of quarks $q_{a}$ $(1)$,
3. $\sigma_{3} = \text{diag}(1,-1)$ is the isospin zero generator of the $SU_{L}(2)$ group,
4. $\theta$ denotes the Weinberg angle relating the couplings $g_{1},g_{2}$ to the EM interaction constant $\alpha$. $c(\theta)$ denotes $\cos(\theta)$, and so on.

In general, the forms of $a_{ij}$ and $c_{ab}$ are restricted. Since You already know the electric charges of quarks (precisely, the charges of an upper quarks are $\frac{2}{3}$, while the charges of the lower quarks are $-\frac{1}{3}$), the matrix $c_{ab}$ has the form $$\tag 3 c_{ab} = \begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix},$$ with $A,B$ being the unitary $3\times 3$ matrices. ALso, with precise values of electric charges, You have to impose some constrains on diagonal elements of $c_{ab}, a_{ij}$, but we're not interested in this at the moment.

From $(2)$ You can extract the $Z$-boson interaction part: $$\tag 4 L_{\text{int}} = \bar{q}_{a}\gamma^{\mu}(c^{ab}_{V} - c^{ab}_{A}\gamma_{5})q_{b}Z_{\mu} + h.c.,$$ with $c^{ab}_{V,A}$ having the general form of $(3)$. For example, in the Standard model we have $c^{ab}_{V,A} = \delta^{ab}c^{b}_{V,A}$ (see below), which means that there is assumed that the neutral current interactions are quark species diagonal.

Experimental origin - no so much oscillations

So why we choose $$\tag 5 c^{ab}_{V,A} = \delta^{ab}c^{b}_{V,A}?$$ The answer is the experiment.

Before I'll discuss the case of quarks, let's establish the difference between the quarks and leptons cases. In the leptons case, You also may write down the matrices $c_{A,V}^{ab}$ in the form $(3)$. However, there is experimental fact to write them in the form of $(4)$ - the conservation of the so-called lepton numbers. There are 3 different lepton numbers, and this immediately imposes $(5)$. In the case of quarks such argument isn't valid. Unlike the case of leptons, there is only one conserved quark global number - the so-called baryon number (all of quarks are massive, unlike the leptons, and this kills 2 "missing" numbers). All quarks have the same baryon number, and the restriction to conserve it doesn't reduce the general form $(3)$ of $c^{ab}_{V,A}$ to a simpler one.

Let's now go to the reason why $c^{ab}$s are reduced to $(5)$ in the case of quarks. Suppose the neutral mesons $M^{0}$ - the bounded states of two quarks with zero total electric charge. They are $$M^{0} = \{ B^{0} = d\bar{b}, \quad \bar{B}^{0} = \bar{d}b, \quad D^{0} = c\bar{u}, \quad \bar{D}^{0} = \bar{c}u, \quad K^{0} = d\bar{s}, \quad \bar{K}^{0} = \bar{d}s \}$$ With the general form $(3)$ of the matrices $c^{ab}_{V,A}$ there are tree-level oscillations $$M^{0} \to M^{0}$$ There are also tree-level decays on a lepton-antilepton pair $l\bar{l}$, $$M^{0} \to l \bar{l},$$ if we take into account the lepton part of $(4)$. Such processes violate some quantum numbers like strangeness.

Apart from these tree-level processes, there are also corresponding loop-mediated processes. In compare to the formers, the latters are suppressed by the factor of $\frac{m_{q}^{2}}{m_{Z}^{2}}<<1$, where $m_{q}$ is the mass of given quark, while $m_{z}$ is the mass of $Z$-boson. However, they are possible even in the case of $(5)$.

The experiment says that the above processes are strongly suppressed. This requires to set $c^{ab}_{V,A}$ to the form $(5)$.

• I agree with you that because of the electroweak gauge symmetry, the neutral current is associated with a diagonal operator. But why does this ensure that different flavors cannot interact ? I mean, we could have a $(\overline{u} \overline{d})$ double on one side, and $(c s)$ doublet on the other side. The fact that the neutral current is diagonal would not prevent $\overline{u}$ from interacting with $c$, right ? (Though I must admit, I'm not sure as how antiparticle $SU(2)$ doublets work, but I hope you get my point. – Frotaur Jan 2 '17 at 21:48
• @Frotaur : sorry, I misunderstood Your question. I've edited my answer; it also includes the information about Your "extra question" about the comparison with leptons. – Name YYY Jan 2 '17 at 23:48