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  On the naturalness problem

+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user drake
asked Jul 18, 2012 in Theoretical Physics by drake (885 points) [ no revision ]
Most voted comments show all comments
All you've noticed is yet another runaway instability in the Higgs mass, from adding heavy particles. But you don't need heavy particles. This has been asked many times here already--- the bare couplings and bare masses are observable you extract them from very high energy scattering experiments. They have to be tuned just so to make the Higgs nearly massless, and this is ridiculous. That's heirarchy.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user Ron Maimon
Sure, Kostya. That expresion has been regulated with dimensional regularization. The meaning of $\mu$ depends on the subtraction scheme one chooses. In Minimal Subtraction (or its sister MS bar), $\mu$ is the parameter with mass dimension one has to introduce to keep the couplings dimensionless. In non-minimal subtraction schemes, $\mu$ is the energy scale at which one subtracts the cut-off ($1/\epsilon$) dependent part. Of course, the argument does not change if one uses another regularization procedure like Pauli-Villars or a sharp cut-off. Thank you.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user drake
@drake: Generally, what Kostya is saying is that $\mu$ is a stand-in for a cutoff in dim-reg, and it serves the same role. The $\mu$ dependent stuff is cutoff dependent in the usual way.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user Ron Maimon
$\mu$ is not a high-energy or ultraviolet parameter. I think we all know what is $\mu$ in dimensional regularization. It does not cut-off any integral as a sharp cut-off or an arbitrary big mass in Pauli-Villars.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user drake
@drake: Yes, that's one of the intuition problems in dim-reg--- it superficially looks like $\mu$ is not cutting things off. It's a substraction scale, and the subtraction scale is a stand-in for the cutoff.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user Ron Maimon
Most recent comments show all comments
@RonMaimon For a given coupling, one needs a log small enough in order to make perturbation theory. The "effective" expansion parameter is the coupling times the log.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user drake
@drake: Yeah, I agree, but if the coupling is small enough, it's consistent to a reasonably high order no matter what the log is. It's not a clear mistake to use a wrong $\mu$, it just means you have to go to higher order to get the same accuracy.

This post imported from StackExchange Physics at 2014-04-03 18:31 (UCT), posted by SE-user Ron Maimon

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