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A universe with no Higgs; what would it look like?

+ 3 like - 0 dislike
681 views
More than 2 years have passed since the discovery of the Higgs boson. Although many things have said on the need of the Higgs particle the most common answer is that "we need the Higgs in order the particles, in specific the fermions, to acquire mass". I understand this as I do understand that most of the mass comes from the strong interactions between quarks and gluons in the nuclei. Despite this, I still have not found a qualitative picture of a universe without a Higgs particle. Any thoughts?
asked Jan 9, 2015 in Theoretical Physics by Outlander (95 points) [ no revision ]
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Sure, but the Higgs exists. Ergo the other Higgs-less theories are not that interesting after that.
It does not matter. "A" Higgs exists. In this answer I am not explaining the consequences of various Higgs models but of the simple SM-like model with no Higgs. As for the "similar observable effect" of the Higgs particle this is just null. I am not a phenomenologist but I really thought that people doubting the Higgs were convinced by now.
I lack the technical skills and computing facilities to make a precise analysis in the complete set of data on the subject. After a lot of discussions being an active member of the academic community, and examining the data presented by the two collaborations, the papers and the presentations, I have come into the conclusion that my fellow experimentalists have indeed found some Higgs boson. Not only this is the universally accepted view on the Higgs boson, but frankly speaking, the data speaks for itself. Certainly one can argue for various Higgs-like models but one must be quite paranoic to believe that these data are due to some other "effect".
I will retract the characterization "paranoic" because in no way I suggest that people supporting that stuff suffer from something like these. I would change it to "uneducated" and possibly augment it to "ignorant". I do not accept sceptic because it means there is an actual alternative which is favored by a significant amount of academics. This is not true.

I.e. there is no doubt.
No, I have proof. If something falsifies it I will change my view. I have "no doubt" always within the range I am eligible to. That means I can only follow papers and trust in the academic credibility of around 3,000 scientists working in the LHC and ATLAS.

In any case, basic knowledge of particle physics and (especially) Lie algebras should be enough. In the same way I "know" string theory is correct. In the same way people knew they had to search for the Higgs boson 40 years ago. I like critisism a lot but only when it is well based and with good arguments.

Vladimir, with all due respect, I think it is a bit curious that you re "sceptic", as you say, in both standard renormalization but also the discovery of Higgs (and from what I understand the Higgs mechanism itself). One might wonder what else you might be "sceptic" about too.
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"Paranoic" is a wrong word. "Sceptic" is better.
If you do not have any doubt, aren't you a "fanatic"?

1 Answer

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A Higgs-less Universe would have no Higgs field and no Higgs mechanism (surprise surprise). There is one huge consequence, namely the fact that in that scenario no electroweak symmetry breaking happens.

Let us concider the most simplistic scenario where the Standard Model (SM) Lagrangian contains no Higgs sector. We will take the gauge group to be the SM one, that is
\begin{equation}
SU(3)_{c} \times SU(2)_L \times U(1)_Y,
\end{equation}
and we will assume that gauge couplings are the same and likewise is the fermion sector of the theory. We stress again that only the Higgs sector is completely absent. Then, such an elimination would have absolutely no effect to the QCD part of the theory. QCD would still be a confining theory for color-triplet quarks and color-octet gluons into color-singlet hadrons, i.e. mesons and baryons. Now, if electroweak symmetry remains unbrocken, the what in the SM is asymptotically free $SU(2)_L$ force, would become confining too. This imples the existence of confined weak isospin particles into weak isospin singlets. Remember though that QCD spontaneously breaks the chiral symmetry of the massless $u$ and $d$ quarks, at some scale $\Lambda_{\text{QCD}}$,  as
\begin{equation}
SU(2)_L \times SU(2)_R \to SU(2)_V,
\end{equation}
resulting in a condensate capable of not making the electroweak symmetry apparent even when the Higgs mechanism is absent. Three Goldstone bosons would appear. Then, as mentioned, the $\langle \bar{q}q \rangle = \langle \bar{q}_Lq_R + \bar{q}_Rq_L \rangle$ would make the left- and right-handed quarks transform differently under $SU(2)_L \times U(1)_Y$. Then, because of $I_3$ and $Y_W$ numbers of the left- and right-handed quarks the condensate would transform as a weak isodoublet with weak hypercharge $Y_W = |1|$ and this would end up breaking $SU(2)_L \times U(1)_Y \to U(1)_{EM}$. Then the massless pions of this theory would ``dissapear" from the spectrum as they would become the longitudinal components of the weak gauge bosons. The spectrum would also contain a gauge field $A_{\mu}$, the photon. Since the weak bosons acquired mass the $SU(2)_L$ would not be confining. Still, the quarks and the leptons, would remain massless since in QCD-like electroweak symmetry breaking there does not exist a Yukawa couplings generating mechanism and since there is no Higgs boson the interactions of this theory would become strongly coupled. 

Now, note that altough the pions would be absent, the spectrum of this Higg-less SM would consist of light hadrons and despite the fact that the spectrum would would be very similar to the one of the SM there would be very significant, as we will see, differences. In the Higgs-less SM, as we mentioned before, the quark masses would be zero, so, for example, there would be no mass hierarchy or mass difference between the $u$ and $d$ quarks able to overcome the electromagnetic self-energy of the proton which leads to the latter decaying to the neutron in the SM. Then the bets would be on the $\gamma$ and $Z^0$ mass modulations and it would turn out that the difference in the masses of the $u$ and $d$ quarks would be $\mathcal{O}(1)$ MeV. This would have dramatic effects! The $\beta$-decay would go "crazy". In the SM a free neutron decays through $n \to pe^{-}\bar{\nu}_e$ with $\tau \approx$ 15 minutes. If the sign in the mass difference changed we could have $p \to ne^{+}\nu_e$ with $\tau \approx \mathcal{O}(10^{-12})$s for weak bosons near $\Lambda_{\text{QCD}}$. This implies that no hydrogen atoms could be formed since the protons would decay! No hydrogen also would mean that in this Higgs-less SM that the lightest nucleus would be a neutron. Overall the whole theory of Big Bang nucleosynthesis would be different since atoms would be impossible to form. Going a step further though let us assume that some elements could somehow form. They would be completely unfamiliar and different to the ones we know and of course unstable.

Finally, another remark is that the electron would also be massless as we have already realized. This would mean that the Bohr radius, that in the SM is of $\mathcal{O}(10^{-9})$m, would be infinite making matter not well defined in the sense that it would be without integrity. Then we realize that there could not be compact atoms and as a result bonds would also make no sense. There would be no chemistry, no biology and as a result no life and all that in an unstable vacuum that would continuously produce $e^{+}e^{-}$ pairs. 


We are glad that the Higgs field exists and the electroweak symmetry breaking happened to allow us to be here although we cannot say that there are not alternatives that would create a Universe like ours without the nessecity of a Higgs mechanism. I think the above are quite accurate but maybe a phenomenologist (unlike me) can provide a better explanation.

answered Jan 9, 2015 by conformal_gk (3,535 points) [ no revision ]
Thanks for this great answer - and indeed, there's no reason for quark confinement to be impossible/unlikely in case of a Higgs-less universe; I've deleted my answer (and I can't believe I forgot to mention the absence of electroweak symmetry breaking!).

Below the Z mass, there are Fermi current-current dimension six interactions, like \(\sim G_F ~\bar{u} \gamma_\mu u ~\bar{e} \gamma^\mu e\). If the quark condensate is non-zero, wont these give rise aftrer Fierzing to a tiny fermion mass, of order \(G_F \Lambda_{QCD}^3\) ?

I do not think this will have any affect in what I am saying. I am not really sure about these interactions, which seem non-renormalizable (correct me if wrong) but they have nothing to do with the Higgs. Now, I am not sure about your question.

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