Custodial symmetry of the standard model symmetry group $SU(2)_L \times SU(2)_R$

Question

I am studying the standard model including the Higgs sector and electroweak interactions. Here, all of my terms have their usual meanings. Therefore my symmetry group is $SU(2)_L \times SU(2)_R \times U(1)_Y$. Here I have included $U(1)_Y$ within the larger group of $SU(2)_L \times SU(2)_R \approx O(4)$ as I want to describe electroweak interactions. Spontaneous symmetry breaking (SSB) and the Higgs mechanism reduce this group to $SU(2)_C \times U(1)_{EM}$. What I am finding confusing is understanding what happens to all of these generators when SSB occurs? In my understanding:

• $SU(2)_L$ has 3 generators, all of which are broken
• $SU(2)_R$ has 3 generators, but only $T^3_R$ is broken. The R subscript is just to note that this is a generator for the right-handed fermions
• $U(1)_Y$ has 1 generator which remains unbroken and is the gauge photon

In my understanding, the broken generators from $SU(2)_L \times SU(2)_R$ combine in some way to give us the three gauge bosons for weak interactions: $W^+, W^-, Z^0$, and the remaining two unbroken generators of $SU(2)_R$ are 'eaten' by the gauge bosons. So what are the generators for the remaining $SU(2)_C \times U(1)_{EM}$ symmetry group?

I understand the model when we just include $SU(2)_L \times U(1)_Y$ which results in just $U(1)_{EM}$ after SSB. What I am struggling to understand is the inclusion of the group $SU(2)_R$ and how all of this leads to the custodial symmetry? I also know that the different masses of the up and down quarks lead to an explicit custodial symmetry breaking, but it confuses me on how to incorporate this also.

Answer 1

I have included 5 links, including a "primary" one, which should help clear up the trail map for you; I think you've gone off it, but I can't be sure, as weird misconceptions have crept in.

The big picture, which it behooves you to work out the explicit transformation laws for, for all fields, "before SSB" is:

1. The gauge & Yukawa sectors of the SM have an $SU(2)_L \times U(1)_Y$ gauge symmetry.

2. The non-Yukawa fermion & Higgs potential sectors, however, have a larger global symmetry, with two more "R", generators scrambling the four Higgs d.o.f., beyond the hypercharge $U(1)_Y$, namely $SU(2)_L \times SU(2)_R\sim SO(4)$; you may think of these two new $SU(2)_R$ generators, explicitly broken by the Yukawa sector, as "custodial", as they are an approximate symmetry of the model, i.e., explicitly broken by the capricious array of Yukawa coupling constants y. These will enter as spoilers in radiative corrections, then, (unless the Yukawa couplings and hence masses of the fermions were equal among themselves).

• "After" SSB, these symmetries persist as symmetries, but three are realized nonlinearly ("SSBroken") and their goldstons Higgs-eaten by three of the four gauge bosons, with the Weinberg angle twist that reconfigures $T^3_L$ and $U(1)_Y$ into the cockeyed neutral current Z and the surviving vector $U(1)_{EM}$. The two (global) custodial generators and the charge one are still around, not SSBroken, and scrambling the (pseudo)-goldstons in the belly of the Ws and the Z among themselves. But for Weak mixing (set $\theta_W=0$ to keep track of them), these goldstons transform as a triplet of the surviving (imperfect) custodial SO(3).

I have accounted for all 6 generators―I'm not sure where you got the 7th from.

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Geeky

I also know that the different masses of the up and down quarks lead to an explicit custodial symmetry breaking, but it confuses me on how to incorporate this also.

The ideal $SU(2)_L \times SU(2)_R\sim SO(4)$ Yukawa sector would have only a common coupling y for both ups and downs, cavalierly in normalizations, $$ y(\overline{u_L}, \overline{d_L}) (\tilde\Phi, \Phi) \begin{pmatrix} u_R \\ d_R \end{pmatrix}=y(\overline{u_L}, \overline{d_L}) \begin{pmatrix} \phi^{0~~*} &\phi^+ \\ -\phi^- & \phi^0 \end{pmatrix}\begin{pmatrix} u_R \\ d_R \end{pmatrix}. $$

The v.e.v. of the middle Higgs matrix is $v{\mathbb I}$, yielding equal masses, $$ yv (\overline{u_L} u_R + \overline{d_L}d_R), $$ so you see that for different Yukawas you loose the full custodial symmetry, here (plain, strong) isospin symmetry.

Answer 2

If you're going to include an $SU(2)_R$, then your $U(1)$ would more likely be $U(1)_{B-L}$, for the baryon ($B$) minus lepton ($L$) quantum number, not $U(1)_Y$, since $I_{3R}$ is already present as a contribution to $Y$, so that the symmetry group would be $SU(2)_L × SU(2)_R × U(1)_G$, where I'm going to use $G = ½ (B - L)$ in the following.

Let $x_y$ denote the various sectors, where $x ∈ \{ν,e,u,d,\bar{ν},\bar{e},\bar{u},\bar{d}\}$ denote the neutrino, electron, up and down quarks (and their respective anti-particles), and $y ∈ \{L, R\}$ denote the left and right versions of each. Up to scaling, the spectra are given by $$\begin{matrix} · & \bar{e}_L & u_R & \bar{d}_L & ν_R & \bar{ν}_L & d_R & \bar{u}_L & e_R \\ I_{3L} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ I_{3R} & +\frac12 & +\frac12 & +\frac12 & +\frac12 & -\frac12 & -\frac12 & -\frac12 & -\frac12\\ G & +\frac12 & +\frac16 & -\frac16 & -\frac12 & +\frac12 & +\frac16 & -\frac16 & -\frac12\\ Y & +1 & +\frac23 & +\frac13 & 0 & 0 & -\frac13 & -\frac23 & -1\\ Q & +1 & +\frac23 & +\frac13 & 0 & 0 & -\frac13 & -\frac23 & -1\\ \\ · & \bar{e}_R & u_L & \bar{d}_R & ν_L & \bar{ν}_R & d_L & \bar{u}_R & e_L \\ I_{3L} & +\frac12 & +\frac12 & +\frac12 & +\frac12 & -\frac12 & -\frac12 & -\frac12 & -\frac12\\ I_{3R} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ G & +\frac12 & +\frac16 & -\frac16 & -\frac12 & +\frac12 & +\frac16 & -\frac16 & -\frac12\\ Y & +\frac12 & +\frac16 & -\frac16 & -\frac12 & +\frac12 & +\frac16 & -\frac16 & -\frac12\\ Q & +1 & +\frac23 & +\frac13 & 0 & 0 & -\frac13 & -\frac23 & -1 \end{matrix}$$ Similar spectra apply for each of the other two generations.

The symmetry breakings would be $SU(2)_R × U(1)_G → U(1)_Y$, where $Y = G + I_{3R}$, and $SU(2)_L × U(1)_Y → U(1)_Q$, where the electric charge is given by $Q = Y + I_{3L}$ or, in a more explictly parity-symmetric way, by $Q = G + I_{3R} + I_{3L}$.

The $B-L$ force could be called "brightness", in analogy with "color" for quarks. When combined with the $$ for quarks and $\bar{}$ for anti-quarks (and $$ for leptons and anti-leptons), the result is the "fermion cube". The cube would be oriented with opposite vertices at the top and bottom (the [anti-]lepton $$'s) and two pairs of three vertices just above and below the equator (the quark $$ and anti-quark $\bar{}$). It could arise, for instance by a symmetry breaking of $SU(4) → SU(3)_c × U(1)_G$, noting particularly that $SU(4)$ and $SO(6)$ have the same Lie algebras and cubic latices for spectra.

The existence of local $SU(3)_R$ and $U(1)_G$ symmetries is conjectural and is not part of the Standard Model. All you can say is that if $U(1)_G$ is a local symmetry, then there is an $I_{3R}$.

There's a whole cottage industry for $B-L$ symmetry-breaking. Most are playing it close to the collar and only tacking on $B-L$, not the whole $SU(2)_R$, such as:

• Asymmetric matter from $B − L$ symmetry breaking,
• Neutrinos with a linear seesaw mechanism in a scenario of gauged $B - L$ symmetry,
• Light, Long-Lived $B − L$ Gauge and Higgs Bosons at the DUNE Near Detector,
• Phenomenology of the minimal $B − L$ extension of the Standard Model at the LHC.

Fewer play it loose and throw in all of $SU(2)_R$, such as:

• Alternative formulation of left-right symmetry with $B − L$ conservation and purely Dirac neutrinos,
• Spontaneous $R$-Parity Breaking And Left-Right Symmetry.

Since $SU(2)_L × SU(2)_R$ has the same Lie algebra as $SO(4)$, then I could also include references on symmetry breaking for $SO(4) × SO(6)$, such as

• How to break $SO(10)$ via $SO(4) × SO(6)$ down to $SU(2)_L × SU(3)_c × U(1)$.