Lagrangian of Yukawa Interaction and Quark Mixing

Question

I'm trying to understand the original of quark and lepton mass. Here's a paragraph from the book: "Massive neutrinos in physics and astrophysics" by R. N. Mohapatra and P. B. Pal :

The gauge invariance prevent adding bare masses fro them in the Lagrangian. They arise from the following Yukawa interactions allowed by gauge symmetry: $$ -\mathcal{L}_{Y}=\sum_{a,b}[h^{(u)}_{ab}\bar{q}_{aL}\hat{\phi}u_{bR}+h^{(d)}_{ab}\bar{q}_{aL}\phi d_{bR}+h^{(l)}_{ab}\bar{\psi}_{aL}\phi l_{bR}]+\mathrm{h.c.}\ .\tag{2.14} $$ Here, $a,b$ stand for generation indices and $$\hat{\phi}=i\tau_{2}\phi^{*}\ .\tag{2.15}$$ On substituting the non-zero vacuum expectation values for $\phi_{0}=\sqrt{\frac{\mu^{2}}{2\lambda}}=v/\sqrt{2}$, the following mass terms for up and down quarks as well as the charge leptons are generated: $$-\mathcal{L}_{mass}= \sum_{a,b}[\bar{u}_{aL}M^{(u)}_{ab}u_{bR}+\bar{d}_{aL}M^{(d)}_{ab} d_{bR}+\bar{l}_{Lb} M^{(l)}_{ab}l_{bR}]+\mathrm{h.c.}\ ,\tag{2.16}$$ where $$ M^{(f)}_{ab}=h^{(f)}_{ab}v/\sqrt{2} \tag{2.17}$$ with $f=u,d,l$. By an appropriate choice of the quark and lepton basis, the coupling matrices $h^{(u)}$ and $h^{(l)}$ can be chosen diagonal so that we have $u^{0}_{a}=u_{a}$ and $l^{0}_{a}=l_{a}$. The $M^{(d)}$ is however, a complex non-diagonal matrix in this basis and can be diagonalized by the following biunitary transformation: $$V_{L}M^{(d)}V_{R}^{\dagger}=D^{(d)}. \tag{2.18}$$

Here's what I don't understand: Why does the Yukawa interaction Lagrangian look like that? I first thought that it has something to do with $$\mathcal{L}_{Yukawa}=\mathcal{L}_{Dirac}+\mathcal{L}_{Klein-Gordon}-g\bar{\psi}\psi\phi\ ,$$ but I cannot see any obvious connection between these two Lagrangians mathematically. I also don't understand what those h-matrices $h^{(u)}_{ab},h^{(d)}_{ab},h^{(l)}_{ab} $in equation (2.14) are. In equation (2.15), why does $\hat{\phi}$ look like that? In equation (2.16), how does $\bar{q}_{aL}$ become $\bar{u}_{aL}$ and so on. Also, how can I know that the matrices $M^{(u)}_{ab}$ and $M^{(l)}_{ab}$ are diagonal and $M^{(d)}_{ab}$ are not? At last, why do I need to put subscripts for the matrix $V$?

I'd be very grateful for any one who finish reading and answer my lengthy questions.

Answer 1

In fact, the Yukawa Lagrangian is (more or less) only the term $\mathcal{L}_Y = -g \bar{\psi}\psi \phi$. The (massless) Dirac Lagrangian for fermions and Klein-Gordon Lagrangian (plus potential) for the Higgs are not shown in your formula. The main difference between the Yukawa Lagrangian and the simpler $-g \bar{\psi}\psi \phi$ is that the Standard Model has several fermionic fields, which are coupled in this Lagrangian.

The $h$ matrices are called Yukawa couplings, and are the equivalent to $g$ above. They are matrices instead of numbers because you will end up with terms like $-g \bar{s}d \phi$ (more to that later).

To understand the difference between $\bar{q}_{aL}$ and $\bar{u}_{aL}$ you have to remember that left-handed fermions come in doublets of isospin. The Higgs field is also a doublet of isospin $$\phi =\frac{1}{\sqrt{2}} \begin{pmatrix}\phi_1 + i \phi_2\\ \phi_0 +i\phi_3 \end{pmatrix}\ . $$ After electroweak spontaneous symmetry breaking, only the $\phi_0$ components has a non-zero value in the vacuum. If we replace in the Yukawa Lagrangian for $d$-type quarks or leptons, we get $$-\mathcal{L}_d = \sum_{a,b} h_{ab}^{(d)} \begin{pmatrix} \bar{u}_{aL} & \bar{d}_{aL} \end{pmatrix}\begin{pmatrix} 0 \\ \phi_0 \end{pmatrix} d_{bR} = \sum_{a,b} h_{ab}^{(d)} \phi_0 \bar{d}_{aL} d_{bR}\ . $$

So the fact that the Higgs field breaks the isospin symmetry selects only the lower component ($T_3 = +\frac{1}{2}$) of the isospin doublet. This explains the mass of $d$-type quarks and charged leptons.

But you need also a mass term for the $u$-type quarks. This is the purpose of the first term in the Yukawa Lagrangian, wich is a sort of "transposed version" of the other two terms. Here you need the charge-conjugated Higgs field $\hat{\phi}$. This Lagrangian reads $$-\mathcal{L}_u = \sum_{a,b} h_{ab}^{(u)} \begin{pmatrix} \bar{u}_{aL} & \bar{d}_{aL} \end{pmatrix}\begin{pmatrix} \phi_0 \\ 0 \end{pmatrix} u_{bR} = \sum_{a,b} h_{ab}^{(u)} \phi_0 \bar{u}_{aL} u_{bR}\ .$$

To understand the reason of the (non)-diagonal matrices, note that the Lagrangian is invariant under unitary transformations $$u_R^i \to V^{ij}_{uR} u_R^j \qquad \qquad d_R^i \to V^{ij}_{dR} d_R^j\\ u_L^i \to V^{ij}_{uL} u_L^j \qquad \qquad d_L^i \to V^{ij}_{dL} d_L^j$$ you can choose these matrices in order to diagonalize the masses $$D^{u,d} = \frac{v}{\sqrt{2}} V_{u,d L} h^{(u,d)} V_{u,d R}^\dagger\ . $$

But four different matrices is just overkill. You can't diagonalize $h^{(u)}$ and $h^{(d)}$ simultaneously, so it is chosen to diagonalize $u$-type quarks and rotate the $d$-type quarks using the CKM matrix. This election is just a convention, it could be done the other way.

The physical significance of this quark mixing is that charged can connect, with a small probability, quarks of different generations. You can describe it saying that the weak $d'$ quark is a superposition of $d$ and $s$ quark mass eigenstates, or saying that the weak $u'$ quark is a superposition of the $u$ and $c$ quark mass eigenstates; it makes no difference.