On the naturalness problem

Question

I know that there are several questions about the naturalness (or hierarchy or fine-tunning) problem of scalars masses in physics.stackexcange.com, but I have not found answers to any of the following questions. Suppose that we add to the SM Lagrangian the following piece:

$(\partial b)^2-M^2 \, b^2-g\, b^2 \, h^2+ \, ....$

where $b$ is a real scalar field (that is not contained in the SM) and $h$ is the Higgs real field. Then the physical mass $m_P$ of the Higgs is given by the pole of its propagator (I am omitting numerical factors):

$m^2_P=m^2_R (\mu)+I_{SM}(\mu)-g\, M^2\, ln(M/\mu)$

where $m_R(\mu)$ is the renormalized Higgs mass, $I_{SM}(\mu)$ (which also depends on the SM couplings and masses) is the radiative contribution of the SM fields (with the Higgs included) to the two point function of the Higgs fields (note that is cut-off independent because we have subtracted an unphysical "divergent" part) and the last term is the one-loop contribution of the new field $b$ (where we have also subtracted the divergent part).

I have two independent questions:

1. The contribution of the $b$ particle (the last term) is cut-off independent (as it has to be) so the correction to Higgs mass is independent of the limit of validity of the theory, contrary to what is usually claimed. However, it does depend on the mass of the new particle. Therefore, if there were no new particles with masses much higher than the Higgs mass, the naturalness problem would not arise. It could be new physics at higher energies (let's say beyond 126 GeV) as long as the new particles were not much heavier than the Higgs (note that I'm not discussing the plausibility of this scenario). Since this is not what people usually claim, I must be wrong. Can you tell me why?

2. Let's set aside the previous point. The naturalness problem is usually stated as the fine-tunning required to have a Higgs mass much lighter than the highest energy scale of the theory $\Lambda$, which is often taken as GUT scale or the Planck scale. And people write formulas like this: $\delta m^2 \sim \Lambda^2$ that I would write like that: $m^2_P=m^2 (\Lambda) + g\, \Lambda^2$. People think it is a problem to have to fine-tune $m^2 (\Lambda)$ with $\Lambda^2$ in order to get a value for $m^2_P$ much lower than $\Lambda^2$. And I would also think that it is a problem if $m^2 (\Lambda)$ were an observable quantity. But it is not, the observable quantity is $m^2_P$ (the pole of the propagator). I think that the misunderstanding can come from the fact that "interacting couplings" (coefficients of interacting terms instead of quadratic terms) are observables at different energies, but this is not the case, in my opinion, of masses. For example, one talks about the value of the fine structure constant at different energies, but the mass of the electron is energy independent. In other words, the renormalized mass is only observable at the energy at which it coincides with the physical mass (the specific value of the energy depends on the renormalization procedure but it is usually of the order of the very physical mass), while one can measure (i.e. observe) interacting couplings at different energies and thus many different renormalized couplings (one for every energy) are observables. Do you agree?

*(Footnote: since free quarks cannot be observed the definition of their masses is different and one has to give the value of their renormalized mass at some energy and renormalization scheme.)

Thank you in advance.

Answer 1

1. You are right.

• The naturalness problem has been formulated as that the loop correction to the scalar mass squared is quadratically divergent, using the momentum cutoff $\Lambda$, $$ \delta m^2 \propto \Lambda^2. \quad \text{(wrong)} $$ The most misleading consequence is that a light field, e.g. in the Standard Model, would give divergent correction as above. This cannot be.
• After renormalization, the cutoff must disappear in the observable quantity. As you said the loop correction by the $b$ field is $$ \delta m^2 \propto M^2 \log M. $$ If the $b$ field is light so that $M$ is comparable to the original Higgs mass, this correction is sizably small.
• However, this also shows the UV sensitivity of the scalar field. The larger the mass in the loop, the more correction to the Higgs mass. We may expect a bunch of fields at the GUT or the Plack scale $M \sim M_{\text{Pl}}$, which defines the naturalness problem.

2. Also, I agree with you. The observable Higgs mass is the entire combination. However, with some care.

• The pole mass is defined as the location of the pole of the propagator with loop correction by the self-energy $$ \Gamma^{(2)}(m_p^2) = [ m_B^2 - p^2 - \tilde \Sigma(p^2) ]_{p^2=m_p^2} = 0. $$ The pole mass itself has no predection because we defined it as above. For instance, if we take this as the renormalization condition, which is the on-shell renormalization scheme, any loop correction does not shift the location of the pole by construction.
• What is important is the departure of the scalar mass from the pole mass, as a function of scale or the external momentum $$ m^2 (p^2) = m_p^2 + \tilde \Sigma (p^2) - \tilde \Sigma (m_p^2) - (p^2-m_p^2) \frac{d \tilde \Sigma_p }{dp^2} (m_p^2). $$ The additional terms are from the field-strength renormalization. Recall that we can only measure the electric charge $e^2(p^2)$ with respect to the reference charge $e^2(0)$ defined at a scale $p^2=0$. Likewise, we may measure the relative scalar mass squared from the reference mass, here, the above pole mass.
• In the language or RG flow, the pole mass (or the renormalization condition in general) is like of the boundary condition that we freely set and the solution is the rest of the terms. The boundary condition is not a prediction, but the profile of the solution is the prediction.
• You may verify that the cutoff $\Lambda$ disappears. In fact, this is the formula giving us the loop correction of the $b$-field as above.
• Of course, we may take another reference mass which can always be transformed by a finite RG running.
• Lastly, in the effective field theory, the (bare) mass is not what is given but is a parameterization of our ignorance fitted by the experiment. How? We may observe the physical mass $m^2(p^2)$ at a scale and invert it. The most convenient one is again the pole mass $m_p^2$ and from which we can obtain the bare mass $$ m_B^2 = m_p^2 + \tilde \Sigma(m_p^2), $$ which you meant by $m(\Lambda^2)$.