Fermions in the Standard Model of elementary particles

In Chapter 7 we need some more details about the matrices $\Sigma_a$ and $M$ that appear in the Lagrangian (5.36). They are completely specified in the Standard Model of elementary particles[41], that explains with great precision a very large amount of experimental data.

In its detailed formulation it is convenient to use complex Dirac fields that can be decomposed into their real and imaginary parts:

$$\Psi_{\mathcal C} = \Psi_{\mathcal R} + i \Psi_{\mathcal I},... ...{c} \Psi_{\mathcal R} \\ \Psi_{\mathcal I} \end{array}\right).$$ (5.52)

In the rest of this Section we drop the subscript $\mathcal C$ and we assume that all the Dirac fields are complex. We indicate by $\Sigma_a$ and $M$ complex matrices that operate on complex fields. They have half the dimension of the matrices indicated by the same symbols in the preceding Section. Note that not all the real matrices can be considered as complex matrices with half dimension: this possibility is a relevant physical assumption. We also adopt the standard notation

$$\overline\Psi = - i \Psi^{\dagger} C \qquad \Psi^{\dagger} = \Psi^{*T}.$$ (5.53)

It is easy to show that the Lagrangian

$$\lambda^D = 2^{-1} d \overline \Psi \gamma^i \Psi \wedge \eta_i - 2^{-1} \overline \Psi \gamma^i d \Psi \wedge \eta_i$$

$$- 2^{-2} \phi \overline \Psi (\gamma^i \Sigma_{[jk]} + \Sigma_{[jk]} \gamma^i) \Psi \omega^{[jk]} \wedge \eta_i$$

$$- \overline \Psi \gamma^i \Sigma_a \Psi \omega^a \wedge \eta_i - \overline \Psi M \Psi \eta,$$ (5.54)

when written in terms of real fields takes the form (5.36). In a similar way we get

$$T^i_k = - 2^{-1} A_k \overline \Psi \gamma^i \Psi + 2^{-1} \overline \Psi \gamma^i A_k \Psi,$$ (5.55)

$$W^l = - 2^{-1} \overline \Psi \gamma^l \gamma^5 \Psi, \quad ... ...\Psi \Psi, \quad W^5 = - 2^{-1} \overline \Psi \gamma^5 \Psi.$$ (5.56)

A Dirac field can also be decomposed into the sum of two (necessarily complex) left and right-handed Weyl fields

$$\Psi^L = 2^{-1} (1 + i \gamma_5) \Psi, \qquad \Psi^R = 2^{-1} (1 - i \gamma_5) \Psi.$$ (5.57)

They transform separately under proper orthochronous Lorentz transformations (namely under $SL(2, \mathbf{C})$) and they satisfy separate Dirac equations only for vanishing mass. A massless neutrino is described in the original version of the Standard Model by a left-handed Weyl field that has no right-handed partner. We, however, assume that neutrinos have a small mass.

In order to explain the absence of the parity symmetry in the weak interactions, one assumes that the internal gauge symmetries have a $\it chiral$ character, namely they act in a different way on the left and right-handed Weyl fields. This means that the matrices that represent the infinitesimal transformations of the gauge group $\mathcal{G} = SU(2)_W \times SU(3)_C \times U(1)_Y$ have the form

$$\Sigma_a = 2^{-1} (1 + i \gamma_5) \Sigma^L_a + 2^{-1} (1 - i \gamma_5) \Sigma^R_a,$$ (5.58)

where the anti-Hermitian matrices $\Sigma^L_a$ and $\Sigma^R_a$ represent two different representations of the Lie algebra of $\mathcal{G}$. Note that $C \gamma^i \Sigma_a$ is anti-Hermitian, as it must be in order to have a real Lagrangian.

In the following few formulas we write explicitly the weak isotopic spin indices $t_3 = \pm 1/2$ on which the 2-dimensional complex representation of $SU(2)_W$ operates and the hypercharge indices $y$. We remember that the electric charge is given by $y/2 + t_3$. The two components of the Higgs field have $y = 1$ and weak isotopic spin $\pm 1/2$. Note that both the spinors

$$\left(\begin{array}{c} \Xi_{1/2} \\ \Xi_{-1/2} \end{array}\r... ...in{array}{c} \Xi^*_{-1/2} \\ - \Xi^*_{1/2} \end{array}\right).$$ (5.59)

transform according to the same representation of $SU(2)_W$.

The same representation acts on the left-handed Fermion fields, while the right-handed field are invariant under $SU(2)_W$. It follows that the new fields

$$\Psi'_{y+1} = v^{-1} (\Psi^L_{y, 1/2} \Xi_{-1/2} - \Psi^L_{y, -1/2} \Xi_{1/2}) + \Psi^R_{y+1},$$

$$\Psi'_{y-1} = v^{-1} (\Psi^L_{y, 1/2} \Xi^*_{1/2} + \Psi^L_{y, -1/2} \Xi^*_{-1/2}) + \Psi^R_{y-1}$$ (5.60)

are both invariant under $SU(2)_W$. The hypercharge index $y$ takes the value $-1$ for leptons and $1/3$ for quarks.

For the potential of the Higgs field we adopt the expression

$$V(\Xi) = \lambda (\Xi^{\dagger} \Xi - v^2)^2,$$ (5.61)

that takes its minimum value for $\Xi^{\dagger} \Xi = v^2$ After the spontaneous symmetry breaking, the Higgs field takes (for instance) the vacuum expectation value

$$<\Xi_{1/2}> = 0, \qquad <\Xi_{-1/2}> = v = v^*$$ (5.62)

and we have, disregarding fluctuations of the Higgs field around its vacuum expectation value, $\Psi \approx \Psi'$. It follows that the Fermions acquire their physical masses if we put

$$\hbar \overline \Psi M \Psi = \overline \Psi' K m K^{\dagger} \Psi',$$ (5.63)

where $m$ is a diagonal mass matrix and $K$ a unitary matrix that contains the Cabibbo-Kobayashi-Maskawa (CKM) matrix that mixes the $d, s, b$ quark fields and possibly another matrix that mixes the neutrino flavours. By means of this equation and eq. (5.60), it is easy to find an explicit expression in terms of the Higgs field $\Xi$ for the matrix $M$ that appears in the Lagrangian. Note that the expression (5.63) is invariant with respect to the internal symmetry group and that the matrix $C M$ is anti-Hermitian, as it is required in order to obtain a real Lagrangian.

It is important to remark that in the Standard Model there is no natural correspondence between the left handed and right handed Weyl fields that permits a natural introduction of Dirac fields. This correspondence is generated by the mass term only after the Higgs field has acquired a vacuum expectation value.

However, in order to introduce higher geometric symmetries, we have to specify the action of $SL(4, \mathbf{R})$ on the Dirac fields and this group contains elements that do not commute with $\gamma^5$ and mix Weyl fields with different helicities. In other words, the $SL(4, \mathbf{R})$ transformations do not commute with the internal gauge transformations given by eq. (5.58). In Chapter 7 we solve the problem by modifying this formula.