In Chapter 1 of his book String Theory in a Nutshell, Kiritsis states the following.

The [Standard M]odel is unstable as we increase the energy (hierarchy problem of mass scales) and the theory loses predictivity as one starts moving far from current accelerator energies and closer to the Planck scale. Gauge bosons are protected from destabilizing corrections because of gauge invariance. The fermions are equally protected due to chiral symmetries. The real culprit is the Higgs boson.

Let $\phi$ be a quantum field and define $G(p,q):=\langle \widehat{\phi}(p)\widehat{\phi}(q)\rangle$ (this is of course a vacuum expectation value). Then, it turns out that, $G(p,q)=\delta (p+q)H(p^2)$. Under the assumption $H$ has a unique pole which is a non-negative real number (anybody have a proof of why this would be true in general?), define the mass of $\phi$ to be the square-root of this pole.

My understanding was that this mass is given to us by experiment; instead, in calculations, we use this condition to calculate the counter-terms, not the mass. Is my understanding of this in-correct?

For some reason I was interpreting his words to mean something along the lines of "In quantum field theory, you can calculate particle masses via perturbation theory, and sometimes certain symmetries tells us that anomalies cannot arise, but in the case of the Higgs boson, there is no such symmetry, and so for the result to come out correct, some crazy cancellation has to occur." (now that I think about it, my assumption regarding his meaning here must have been heavily influenced by what I have been told about the Hierarchy Problem from other sources). This of course, doesn't jive with my aforementioned understanding of masses in quantum field theory (i.e. that you don't calculate them).

Can somebody explain to me the Hierarchy Problem given my understanding of mass in quantum field theory? What precisely are the theorems that "protect" gauge bosons and fermions? I've seen time and again people talk about some how finely-tuned cancellation needs to happen regarding the Higgs. Could somebody tell me very explicitly just what this cancellation is, and where and how it comes up? How does this imply the instability of the Standard Model?

Thanks much in advance.

  • $\begingroup$ Did you mean to also include the next paragraph in the blockquote? $\endgroup$
    – David Z
    Commented Aug 29, 2013 at 2:56
  • 1
    $\begingroup$ @DavidZ You mean the next paragraph in my question? No, that was written by me, included for the purposes of clarifying just what I mean by mass (and what I mean by "my understanding of mass") in this context (as opposed to, for example, just some coefficient in the Lagrangian). $\endgroup$ Commented Aug 29, 2013 at 3:21

1 Answer 1


The gauge bosons such as the photon and chiral fermions may be massless due to the symmetry – gauge symmetry and chiral symmetry, respectively. We may consistently require the theory to respect these symmetries at the exact level, including all quantum corrections.

For the gauge symmetry, it means that $1+1=2$ polarizations of the photon or another particle become unphysical. Only 2 transverse polarizations are left which is only possible if the little group is just $SO(2)$ i.e. if the particle stays massless. So the gauge symmetry protects the masslessness of the photon to all orders. The degree of freedom for the longitudinal photon is just not there – so the mass can't arise.

The same holds for gluons i.e. confined non-Abelian gauge fields although the mass is only meaningful at short distances.

For W-bosons and Z-bosons, the mass is nonzero but at energies much higher than the Higgs scale from the Higgs that gives them the mass, all these masses may be neglected and the longitudinal W-boson or Z-boson may be interpreted as a field from the Higgs multiplet again while the field from the gauge field decouples just like for the photon again. So for spontaneously broken gauge bosons, the mass is protected against large corrections and divergences that would go beyond the tree-level Higgs mass which is however nonzero.

For fermions, the mass term mixes two 2-component spinors, $m\chi_\alpha \psi^\alpha$ etc. However, one may impose a $U(1)$ symmetry rotating the phase of $\chi$ separately (or just $\psi$ separately) which bans the mass term. It is OK for gauge fields and fermions to postulate that Nature likes the symmetries above, at least in an approximate form, which simply explains that the mass is zero (or small if the symmetries are approximate).

For a scalar field, there is no symmetry that could ban the mass terms. And indeed, all corrections to the mass that could occur – finite or divergent ones – do occur. So the loop corrections to $m_H^2$ are of order $c\Lambda^2$, here $\Lambda$ is a cutoff, and they have to cancel with the accuracy $10^{30}$ (Planck scale cutoff is 15 orders of magnitude above the Higgs mass and it's being squared). Cancellation of such terms with the relative accuracy $10^{-30}$ which is needed is "unlikely" for a generic value of all the parameters – the probability is of order $10^{-30}$ by itself. If you reject the anthropic selection etc. and assume that the calculation of the probability above is basically right, then the light Higgs mass is extremely unlikely and an explanation – one that modifies the calculation of the probability that the cancellation is this precise – is needed.

SUSY, for example, may cancel the quadratic corrections (one can see it in two ways: either the Higgs has an approximately equally heavy higgsino that may be protected by the chiral symmetry as a fermion; or from SUSY cancellation to the Higgs scalar mass between particles and their superpartners). The expected order-of-magnitude Higgs mass is of order the superpartners' masses (mass differences) and not $\Lambda$. One may still ask why the tree-level Higgs mass is much smaller than the Planck mass but this is an easier problem to explain by various mechanisms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.