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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 quarkthe quarks column $q$$q_{a}$, $$ \tag 1 q = (u,c,t,d,s,b)^{T}, $$$$ \tag 1 q_{a} = (u,c,t,d,s,b)^{T}, $$ 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 $$ \tag 2 L_{\text{int}} = \bar{Q}_{i}\gamma^{\mu}P_{L}\sigma_{3}a_{ij}Q_{j}g_{1}(c(\theta)Z_{\mu} - s(\theta)A_{\mu}) +g_{2}\bar{q}_{a}\gamma^{\mu}c_{ab}q_{b}(s(\theta)Z_{\mu}+c(\theta)A_{\mu}) + h.c. $$$$ \tag 2 L_{\text{int}} = g_{1}\bar{Q}_{i}\gamma^{\mu}P_{L}\sigma_{3}a_{ij}Q_{j}(c(\theta)Z_{\mu} - s(\theta)A_{\mu}) +g_{2}\bar{q}_{a}\gamma^{\mu}c_{ab}q_{b}(s(\theta)Z_{\mu}+c(\theta)A_{\mu}) + h.c. $$ Here $a_{ij}$ is the $3\times 3$ unitary matrix acting in a 3-dimensional space of quarks doublets $Q_{i}$ $(0)$

  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, while $c_{ab}$ is the $6\times 6$ unitary matrix acting in a 6-dimensional spaceforms of quarks $q_{a}$$a_{ij}$ and $(1)$$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. Since theALso, with precise values of electric charges are known, You have to impose some constrains on diagonal elements of $c_{ab}, a_{ij}$, but we're not interested in this at the moment.

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 includetake into account the lepton part of $(4)$. Such processes violate some quantum numbers like strangeness.

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 quark column $q$, $$ \tag 1 q = (u,c,t,d,s,b)^{T}, $$ 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 $$ \tag 2 L_{\text{int}} = \bar{Q}_{i}\gamma^{\mu}P_{L}\sigma_{3}a_{ij}Q_{j}g_{1}(c(\theta)Z_{\mu} - s(\theta)A_{\mu}) +g_{2}\bar{q}_{a}\gamma^{\mu}c_{ab}q_{b}(s(\theta)Z_{\mu}+c(\theta)A_{\mu}) + h.c. $$ Here $a_{ij}$ is the $3\times 3$ unitary matrix acting in a 3-dimensional space of quarks doublets $Q_{i}$ $(0)$, while $c_{ab}$ is the $6\times 6$ unitary matrix acting in a 6-dimensional space of quarks $q_{a}$ $(1)$. 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. Since the electric charges are known, You have to impose some constrains on diagonal elements of $c_{ab}, a_{ij}$, but we're not interested in this at the moment.

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 $$ M^{0} \to l \bar{l}, $$ if we include the lepton part of $(4)$. Such processes violate some quantum numbers like strangeness.

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.

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.

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This is just the experimental fact thatThe topic - the general form of neutral current interactions

Suppose the neutral current operator is diagonal in fermion fields space; this serves as$SU_{L}(2)\times U_{Y}(1)$ 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 quark column $q$, $$ \tag 1 q = (u,c,t,d,s,b)^{T}, $$ The quarks ''names'' are fixed by the physical origindiagonalization of Your statement.the Higgs interaction term (i.e., the mass term).

The formal way to describe this fact was realized in the Standard model. In this theory,You may write down the electroweak gauge symmetrymost general neutral currents interaction lagrangian $$ \tag 2 L_{\text{int}} = \bar{Q}_{i}\gamma^{\mu}P_{L}\sigma_{3}a_{ij}Q_{j}g_{1}(c(\theta)Z_{\mu} - s(\theta)A_{\mu}) +g_{2}\bar{q}_{a}\gamma^{\mu}c_{ab}q_{b}(s(\theta)Z_{\mu}+c(\theta)A_{\mu}) + h.c. $$ Here $a_{ij}$ is based onthe $SU_{L}(2)\times U_{Y}(1)$ group. It acts$3\times 3$ unitary matrix acting in thea 3-dimensional space of fermion fields of left fermionquarks doublets (having non-zero weak isospin and fixed hypercharge) and left$Q_{i}$ $(0)$, while $c_{ab}$ is the $6\times 6$ unitary matrix acting in a 6-right singletsdimensional space of quarks $q_{a}$ $(1)$. Since You already know the electric charges of quarks (having onlyprecisely, the hypercharge) correspondingly. Thecharges of an upper quarks are $SU(2)$ group interactions in general mix fermions$\frac{2}{3}$, while the charges of the lower quarks are $U(1)$ remains them unchanged. Let's clarify$-\frac{1}{3}$), the situation aboutmatrix $c_{ab}$ has the form $$ \tag 3 c_{ab} = \begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix}, $$ with $SU(2)$$A,B$ being the unitary $3\times 3$ matrices. Since the electric charges are known, You have to impose some constrains on diagonal elements of $c_{ab}, a_{ij}$, but we're not interested in this at the moment.

There are three generators ofFrom $(2)$ You can extract the $SU(2)$ group, being Pauli matrices$Z$-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 $\sigma$. Two$c^{ab}_{V,A}$ having the general form of them ($\sigma_{x}, \sigma_{y}$) are anti-diagonal$(3)$. For example, whilein the otherStandard model we have $c^{ab}_{V,A} = \delta^{ab}c^{b}_{V,A}$ ($\sigma_{z}$see below), which means that there is assumed that the neutral current interactions are quark species diagonal one. With

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 linear combinationexperiment.

Before I'll discuss the case of quarks, let's establish the first twodifference between the quarks and leptons cases. In the leptons case, You also may write down the matrices electrically charged vector bosons (You know$c_{A,V}^{ab}$ in the form $(3)$. However, there is experimental fact to write them asin the form of $W^{\pm}$)$(4)$ - the conservation of the so-called lepton numbers. There are associated3 different lepton numbers, and we're dealing withthis immediately imposes $(5)$. In the charged currents interactioncase of quarks such argument isn't valid. WithUnlike the latter a linear combinationcase of electrically neutral photon andleptons, there is only one conserved quark global number $Z$-boson is associated the so-called baryon number (all of quarks are massive, unlike the leptons, and we're dealing withthis kills 2 "missing" numbers). All quarks have the neutral currentssame baryon number, and the restriction to conserve it doesn't reduce the general form $(3)$ of $c^{ab}_{V,A}$ to a simpler one.

Now, sinceLet's now go to the reason why $\sigma_{3}$ is diagonal$c^{ab}$s are reduced to $(5)$ in the fermion fields space, it's obvious thatcase of quarks. Suppose the neutral currents interaction is diagonal in both flavor and mass basesmesons $M^{0}$ - 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 $$ M^{0} \to l \bar{l}, $$ if we include the lepton part of $(4)$. Such processes violate some quantum numbers like strangeness.

P.SApart from these tree-level processes, there are also corresponding loop-mediated processes. Once You fixIn compare to the set of quark doublets and all hyperchargesformers, the only generality whichlatters are suppressed by the factor of $\frac{m_{q}^{2}}{m_{Z}^{2}}<<1$, where $m_{q}$ is remainedthe mass of given quark, while $m_{z}$ is the formmass of $Z$-boson. However, they are possible even in the Higgs interaction termcase of (i$(5)$.e

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

This is just the experimental fact that the neutral current operator is diagonal in fermion fields space; this serves as the physical origin of Your statement.

The formal way to describe this fact was realized in the Standard model. In this theory, the electroweak gauge symmetry is based on $SU_{L}(2)\times U_{Y}(1)$ group. It acts in the space of fermion fields of left fermion doublets (having non-zero weak isospin and fixed hypercharge) and left-right singlets (having only the hypercharge) correspondingly. The $SU(2)$ group interactions in general mix fermions, while the $U(1)$ remains them unchanged. Let's clarify the situation about the $SU(2)$.

There are three generators of the $SU(2)$ group, being Pauli matrices $\sigma$. Two of them ($\sigma_{x}, \sigma_{y}$) are anti-diagonal, while the other ($\sigma_{z}$) is the diagonal one. With the linear combination of the first two matrices electrically charged vector bosons (You know them as $W^{\pm}$) are associated, and we're dealing with the charged currents interaction. With the latter a linear combination of electrically neutral photon and $Z$-boson is associated, and we're dealing with the neutral currents.

Now, since the $\sigma_{3}$ is diagonal in the fermion fields space, it's obvious that the neutral currents interaction is diagonal in both flavor and mass bases.

P.S. Once You fix the set of quark doublets and all hypercharges, the only generality which is remained is the form of the Higgs interaction term (i.e., Yukawa interaction matrices).  

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 quark column $q$, $$ \tag 1 q = (u,c,t,d,s,b)^{T}, $$ 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 $$ \tag 2 L_{\text{int}} = \bar{Q}_{i}\gamma^{\mu}P_{L}\sigma_{3}a_{ij}Q_{j}g_{1}(c(\theta)Z_{\mu} - s(\theta)A_{\mu}) +g_{2}\bar{q}_{a}\gamma^{\mu}c_{ab}q_{b}(s(\theta)Z_{\mu}+c(\theta)A_{\mu}) + h.c. $$ Here $a_{ij}$ is the $3\times 3$ unitary matrix acting in a 3-dimensional space of quarks doublets $Q_{i}$ $(0)$, while $c_{ab}$ is the $6\times 6$ unitary matrix acting in a 6-dimensional space of quarks $q_{a}$ $(1)$. 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. Since the electric charges are known, 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 $$ M^{0} \to l \bar{l}, $$ if we include 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 $(5)$.

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This is just the experimental fact that the neutral current operator is diagonal in fermion fields space; this serves as the physical origin of Your statement.

The formal way to describe this fact was realized in the Standard model. In this theory, the electroweak gauge symmetry is based on $SU_{L}(2)\times U_{Y}(1)$ group. It acts in the space of fermion fields of left fermion doublets (having non-zero weak isospin and fixed hypercharge) and left-right singlets (having only the hypercharge) correspondingly. The $SU(2)$ group interactions in general mix fermions, while the $U(1)$ remains them unchanged. Let's clarify the situation about the $SU(2)$.

There are three generators of the $SU(2)$ group, being Pauli matrices $\sigma$. Two of them ($\sigma_{x}, \sigma_{y}$) are anti-diagonal, while the other ($\sigma_{z}$) is the diagonal one. With the linear combination of the first two matrices electrically charged vector bosons (You know them as $W^{\pm}$) are associated, and we're dealing with the charged currents interaction. With the latter a linear combination of electrically neutral photon and $Z$-boson is associated, and we're dealing with the neutral currents.

Now, since the $\sigma_{3}$ is diagonal in the fermion fields space, it's obvious that the neutral currents interaction is diagonal in both flavor and mass bases.

P.S. Once You fix the set of quark doublets and all hypercharges, the only generality which is remained is the form of the Higgs interaction term (i.e., Yukawa interaction matrices).