1
$\begingroup$

I am wondering whether the formation of a hierarchical vacuum condensate yields an instantaneous or sequential symmetry breaking in a cosmological phase transition. Let me illustrate this question with the example of QCD:

The cosmological phase transition of chiral symmetry breaking in QCD happens instantaneously at a temperature $T\sim \Lambda_{\rm QCD}$. In this transition, the quark condensate $|\langle \bar{q}q\rangle|=v~{\rm diag}(1,1,1)$ forms, where $v\sim\Lambda_{\rm QCD}$. This condensate spontaneously breaks the approximate $SU(3)_L\times SU(3)_R$ flavor symmetry of the three lightest quarks $(u,d,s)$ down to $SU(3)_V$ and gives an equal mass contribution to these three quarks.

Now let us assume a hypothetical world in which a more general quark condensate, e.g., $|\langle \bar{q}q\rangle|=v~{\rm diag}(a,b,c)$ with $a>b>c$, gives different contributions to the quark masses. (Such a condensate is of course not allowed in QCD since it would break the $SU(3)_V$ symmetry.)

Would such a hierarchical condensate still instantaneously break the original flavor symmetry $SU(3)\times SU(3)$ at a temperature $T\sim v\sim\Lambda_{\rm QCD}$? Or would it sequentially break the symmetry, e.g., down to $SU(2)^2\times U(1)$ at temperatures $T\sim a v$, and down to $U(1)^3$ at $T\sim b v$?

$\endgroup$
  • $\begingroup$ Could you perhaps give a reference where the idea of instantaneous symmetry breaking for the chiral symmetry comes from? It does not sound feasible. $\endgroup$ – flippiefanus Feb 12 '18 at 12:35
  • $\begingroup$ So you are envisioning a mystery strong dynamics controlled by a theory dramatically different than a vector gauge theory like QCD, right? But then, why are you calling your freak condensate $v \sim \Lambda_{QCD}$? If this freak condensate emerged at two scales, then the chiral symmetry would break in two steps... but I hope you are not asking about theories that would actually do this... $\endgroup$ – Cosmas Zachos Feb 13 '18 at 15:04
  • $\begingroup$ @CosmasZachos: you are right that I do not consider the standard quark condensate of QCD. I still call it $v\sim\Lambda_{\rm QCD}$ for illustrative purposes. So you agree with the last sentence of my question above, i.e., that a non-unit condensate, such as $|\langle \bar{q}q\rangle|=v~{\rm diag}(1,2,3)$, would actually break the symmetry in two steps? And it would not be possible that such a condensate instantaneously yields three different mass contributions? $\endgroup$ – Thomas Feb 15 '18 at 18:20
  • 1
    $\begingroup$ Sure. Highly unconventional, but two dramatically different condensation scales would lead to mass disparities at different stages. I'm just reading off the premise for the condensate. $\endgroup$ – Cosmas Zachos Feb 15 '18 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.