# Scalar pseudo-Goldstones from hypothetical $SU(3)_V$ symmetry breaking in QCD

The eight light pseudoscalar mesons of QCD are the pseudo-Goldstone bosons of the spontaneously broken chiral (axial) $SU(3)_A$ quark flavor symmetry.

If we consider the hypothetical case of also breaking the vector part $SU(3)_V$ of this flavor symmetry, would the resulting pseudo-Goldstone bosons be scalar instead of pseudoscalar? Would they mediate a similar interaction as their pseudoscalar cousins do, with the only difference that their interaction is not spin-dependent? And in the chiral limit of massless quarks, would these massless scalar Goldstone bosons induce a $V(r)\propto 1/r$ Coulomb potential instead of a $V(r)\propto \exp(-mr)/r$ Yukawa potential and thus mediate a long-range force?

I think you may be asking a few related questions, so here's my attempt to answer some of them:

## Are there examples of scalar Goldstone bosons?

Yes, at least at a hypothetical level. One simple example in particle physics is the dilaton, the Goldstone boson of spontaneous breaking of scale symmetry.

Remarks:

1. When scale symmetry is broken by strong dynamics, the dilaton is often related to the radion, the radial excitation in the size of an extra dimension, which is related to the dilaton through the holographic principle.

2. Scale symmetry (and the larger conformal symmetry that contains it) is anomalous in our universe. Renormalization violates scale invariance, which is why we have things like anomalous dimensions---literally the scaling dimensions of our fields and operators appear to change at different length scales. This means that the dilaton is at best a pseudo-Goldstone boson.

I believe that in this case you do get a Coulomb potential.

Remark: in your question you ask if you get a Coulomb potential, which is a very interesting question. But part of the reason why it is interesting is that a pseudoscalar particle does not mediate a Coulomb potential, but rather a spin-dependent $1/r^3$ potential.

## Why are our favorite Goldstone bosons pseudoscalars?

Our "favorite Goldstone bosons" in particle physics are the longitudinal modes of the $W$ and $Z$ (pseudoscalars) and the neutral pions/kaons (pseudoscalar mesons!). Nothing in Goldstone's theorem says that the Goldstone mode has to be pseudoscalar, so what's going on?

This is a little subtle and I'm not sure if I have the whole story, so I invite more knowledgeable readers to correct me!

1. The parity of a spin-0 particle---that is, whether it is a scalar or pseudoscalar---is not really well defined in a field theory until you have a sense of what (charge--)parity means in that theory. In most particle theories we play with, the sense of parity comes from how the particle interacts with fermions. This boils down to: (in 4-component fermion notation) whether there's a $i\gamma^5$ in the interaction or (in 2-component fermion notation) whether the coupling is imaginary. (These two are equivalent because the hermitian conjugate of the latter term gives the negative term in the $\gamma^5$.)

2. You can parameterize the Goldstone $\pi(x)$ in the usual Coleman--Callan--Wess--Zumino way as $$U(x) = \exp(i\pi^a(x) T^a/v)\, v \ ,$$ where $v$ is the vev (the order parameter of spontaneous symmetry breaking), and $T^a$ are the generators of the broken symmetry. $U(x)$ is a linear representation under the whole symmetry group. For the Higgs this is like saying $$H(x) = \exp(i\pi^a T^a/v)\, \begin{pmatrix}0\\v+h(x)\end{pmatrix}$$ where we've added in the radial mode $h(x)$ for clarity. The $\pi^a$ here are the Goldstones that are eaten by the $W$ and $Z$. $H(x)$ transforms as an SU(2) doublet, but we've pushed the Goldstone degrees of freedom into the exponential. What this boils down to is the following rule:

To identify the Goldstones: transform the vev by a general broken generator, parameterized by some direction $\epsilon^a$. Promote this parameter to a dynamical field, $\epsilon^a \to \pi^a(x)/v$. That is the Goldstone.

1. When you have a compact symmetry, the generators are always Hermitian. This is simply the generalization of $e^{i\theta}$ which is compact and lies on the unit circle versus $e^{x}$ which may take any value in (0,$\infty$). This, in turn, tells you something about the factors of $i$ floating around when you couple the Goldstones to fermions. As we mentioned in (1), the factors of $i$ when coupling to fermions have to do with whether a spin-0 is scalar of pseudoscalar.

2. In simple cases, we assume that the fermions are also linear representations of the global symmetry. Then you may use the global symmetry to write down the allowed interaction terms with respect to the order parameter of symmetry breaking (e.g. writing the Yukawa interactions with respect to the Higgs doublet in the Standard Model).

When you turn the crank, I believe this gives you that the Goldstone of a spontaneously broken internal symmetry interacts with fermions as a pseudoscalar. Now there are lots of caveats. Here are two:

• We've used one definition of CP: how the field interacts with fermions. There are other sense of CP that pop out purely from group theory. See, for example this recent thesis.

• If you're not too attached to the Goldstone being a true Goldstone, but are allowing for explicit breaking terms, then even the Goldstones of a spontaneously broken compact approximate symmetry may have scalar interactions. This is the case in models of a pseudo-Goldstone Higgs. The standard theory breaks $SO(5)\to SO(4)$, where $SO(4)$ contains the four degrees of freedom of a Higgs doublet. These are all identified as Goldstones of the breaking, but the global $SO(5)$ is broken explicitly by small terms. At loop level, these generate the usual Higgs properties (electroweak symmetry breaking potential, scalar interactions to Standard Model fermions).

## Can $SU(3)_L\times SU(3)_R$ chiral symmetry break to $SU(3)_V$?

I do not think this is possible. What you're proposing is that $$SU(3)_L\times SU(3)_R\to SU(3)_A ,$$ however, the axial part of chiral symmetry simply is not a group. If you compute the commutation relations of $SU(3)_A$ generators, you get elements of $SU(3)_V$. So the algebra of $SU(3)_A$ does not close.

I hope that helps!