# Spontaneously broken linear sigma model in Peskin & Schroeder: where is the miracle?

P&S spend almost 12 pages discussing the renormalisability of the spontaneously broken linear sigma model and give a detailed calculation of the cancellation of divergences at one-loop level and call this a miracle. Now I think the only thing they have done is shifted one field $\phi_N (x)\longrightarrow v+\sigma (x)$ where $v$ is a constant. If looked at from the pov of $\phi^4$ theory in four dimension we know the theory is renormalisable, so why is this not obvious for the theory $(\phi_1, \cdots , \phi_{N-1}, v+ \sigma$)? I.e. Why do they claim this is a miracle? I understand that the symmetry of the ground state is broken as one specific direction is chosen, but why should this impact the renormalisability of the theory? I.e. Why do they claim this is a miracle?

• You are spectacularly misreading the book. In particular, you don't seem to understand the contrapositive of the miracle, the actual point they are making in the last paragraph of p 353. The miracle is not renormalizability, which they assure, but the fact the corrections regiment themselves to preserve the symmetry which is still there, despite the apparent disappearance of the visible manifestations of the symmetry. They are telling you there are no quantum anomalies in this system. – Cosmas Zachos Oct 13 '17 at 14:57
• And just before that paragraph they say "It would be a miracle if these three parameters were able to absorb all the infinities arising in the divergent amplitudes [...]" And the paragraph you mention says "If this miracle did not occur [...] we could still make this theory renormalisable by introducing new, symmetry breaking, terms in the Lagrangian, [etc]". I really don't think I am misreading the book at all. I see the point your are making reg anomalies, but anomalies appear 300 pages later only. So I think there was no need to be such an arrogant twat in your answer. – Oбжорoв Oct 13 '17 at 19:29
• Right. The issue is not renormalizability as you seem to agree. It is the absence of a finite number of specific undesirable terms that checks the gluttony of the renormalization procedure. Your question distinctly misrepresents that and puts unwarranted words in the mouths of the authors, words they never uttered. They are not cancelling anomalies, there simply aren't any. – Cosmas Zachos Oct 13 '17 at 19:36
• I'd willingly withdraw my question if I understood your point about it, but I really don't. I'm quoting the authors verbatim and am simply asking why they are saying this. Maybe you can let me know if the following is correct. It is by no means clear that a symmetry of the classical Lagrangian remains a symmetry in the quantum theory. This is especially the case here where the fields are "shifted" and the Lagrangian becomes more complicated. The miracle is that the symmetry is still there. I have recently found that some people prefer to call it hidden than than broken symmetry. Makes sense? – Oбжорoв Oct 14 '17 at 7:28
• Yes, that's their point. – Cosmas Zachos Oct 14 '17 at 10:26

I believe they are making a pedagogical point here that for you may be intuition/obvious.

The point of this section is that there may be more distinct terms in a Lagrangian than there are free parameters, and these distinct terms will give rise to different divergent diagrams, but if those distinct terms are related by a symmetry then so will their divergent diagrams. Therefore, you will need less renormalizing counterterms than you may expect. The authors hint at this in the last paragraph of their introduction to chapter 11.

For example, the linear sigma model Lagrangian has only $$3$$ parameters ($$\mu^2$$, $$\lambda$$, and an assumed common field strength) but technically $$N+N+\frac{N(N+1)}{2}$$ distinct terms. In such a case one might naively wonder "to renormalize this theory, do I need $$3$$ counterterms, or $$N+N+\frac{N(N+1)}{2}$$ counterterms?" Of course, there is an $$O(N)$$ symmetry relating the terms which brings the number of distinct terms down to $$3$$, so at this point it's nice but not too surprising to know that we only need $$3$$ counterterms to completely renormalize the theory.

But then the situation gets hairier: What if that symmetry of the original Lagrangian is not preserved by the (true) vacuum? In that case, in order to do standard perturbation theory we will need to expand our fields/Lagrangian about the vacuum state, hiding the symmetry of the original Lagrangian. Upon doing a perturbative expansion in our new fields (to calculate, say, 1PI diagrams), we will now get different diagrams with (potentially) different divergences. This is shown to occur in the linear sigma model. Now that we no longer have the explicit symmetry of the original Lagrangian, will we need more renormalizing counterterms? The authors show that, to one-loop order in the linear sigma model, only $$3$$ counterterms are needed, no more.

Another point of that chapter which Dr. Zachos pointed out in his comments is the notion of anomalous spontaneous symmetry breaking, which actually (I believe) is not relevant for the "miracle" of needing only 3 counterterms, but which is nonetheless very interesting and important. The linear sigma model with $$\mu^2=0$$ does not have any anomalous $$O(N)$$ symmetry breaking, as the authors argue (the final argument for this is given in section 13.2), but there do exist other QFT's which exhibit anomalous symmetry breaking - e.g. the Coleman-Weinberg model, which you can explore in the "final project" of Part 2 of Peskin & Schroeder.

• Thanks for clarifying this. I have now had much time to understand this issue and it is much clearer to me. I find that P&S often make statements in their book that you can only really understand much later. Certainly I find myself often going back and thinking "ah that is why they have written that there..." – Oбжорoв Oct 10 '18 at 13:20

Renormalization means splitting parameters of theory (field strength, charge) into physical- and counter- terms. Number and type of this parameters is constrained by the symmetries of original Lagrangian.

In broken symmetry it is no longer obvious what form original Lagrangian had, even less obvious is that one would have to introduce finite number of counterterms to renormalize it since there are no obvious symmetries to limit them.

• I am not sure that your answer does anything else than restating the problem. In fact if you look at the classical Lagrangian in terms of $\pi$ and $\sigma$Eq 11.9 in P&S and would give each interaction a separate coupling, then you can introduce 6 counterterms and 2 field strength renormalisation factors. There are also 8 superficially divergent diagrams. Hence, where is the miracle? – Oбжорoв Oct 13 '17 at 15:17