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I can understand the quadratic divergent corrections to Higgs bare mass which is referred to as the hierarchy problem.

But I don't understand how there won't be any hierarchy problem if we do not introduce a UV cutoff in the loop integrals, why won't we have such a problem when we integrate up to infinity...how can see it mathematically?

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The hierarchy problem has to be framed in the context of beyond standard model physics. You have to distinguish between 5 mass scales, namely

  1. $m$: the mass of the particle in concern, e.g. Higgs mass $m_H$.
  2. $\Lambda$: the UV cutoff scale of the regularization scheme (in dimensional regularization (DR), $\frac{1}{\epsilon}$ plays the role of $\Lambda$, where $\epsilon = d -4$). At the end of the renormalization procedure, $\Lambda$ can be safely sent to infinity, thanks to the painstakingly crafted counter terms.
  3. $Q$: the energy scale of the incoming/outgoing particles involved in a scattering process.
  4. $\mu$: the renormalization scale, which is an arbitrary scale to anchor the scattering amplitude (or coupling 'constant') as a function of $\frac{Q}{\mu}$ (or $ln(\frac{Q}{\mu}$)). The renormalization scale $\mu$ is a fiat scale that is set forth by human convention/convenience. Usually $\mu$ is set to the typical the energy scale $Q_0$ of a scattering process.
  5. $M$:the mass scale where beyond standard model (BSM) physics effect comes into the picture. $M$ could be either the grand unification scale $M_{GUT}$ or Planck scale $M_P$. In the effective field theory framework, the BSM Langrangian terms are suppressed by a factor of $\frac{Q}{M}$.

Assuming that there are BSM Langrangian terms, the hierarchy problem concerns the uncanny fine-tuning to arrive at the tiny value of ${m}$ compared with ${M}$, unless there is a spontaneously broken symmetry (technical naturalness) constraining the otherwise large BSM quantum loop corrections (of order $M$) to $m$. The specific regularization scheme or its concomitant cutoff scale $\Lambda$ ($\frac{1}{\epsilon}$ in DR) has nothing to do with the hierarchy problem. I don't blame you, actually even some professional physics papers got confused (for instance, some might erroneously argue that there is no hierarchy problem with dimensional regularization).

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  • $\begingroup$ As you've been told multiple times before, please try to minimise the number of times you edit your posts. Thanks. $\endgroup$ – AccidentalFourierTransform Aug 27 '19 at 21:24
  • $\begingroup$ thank you for your answer, I have a question concerning the same issue, the hierarchy problem is often stated as the hierarchy of the electromagnetic, strong and weak force over the gravitational force, does does SUSY explains this problem... ?? $\endgroup$ – 23rduser Aug 29 '19 at 8:50
  • $\begingroup$ @AkshanshSingh, SUSY folks hitch their hope to the purported convergence of different coupling constants at the GUT scale (via running Q or $\mu$ all the way up to $M_{GUT}$)... however, "why weak gravity" is a wrong question. The correct question is why the masses of elementary particles (e.g. electron mass $m_e$) are so tiny, so that the gravity between them ($\sim m_e^2/r^2$ in the Newton limit) is weak compared to the other forces. If the masses of the elementary particles were of the order of Planck mass ($m_e \sim M_P$), gravity would have been as strong as the other forces! $\endgroup$ – MadMax Aug 29 '19 at 14:31
  • $\begingroup$ @MadMax ok, and how does SUSY propose to explain this? $\endgroup$ – 23rduser Aug 29 '19 at 17:12
  • $\begingroup$ @AkshanshSingh, given the null result from LHC, the SUSY enterprise seems to be a dead end, a dashed hope, a miserable failure, a red herring, a wild goose chase, a collective gloom haunting physics community... I can go on for ever. So if you are still interested in knowing the details, go check similar posts here at PSE or post a new question which would surely be marked as redundant by our diligent and respectful moderators. $\endgroup$ – MadMax Aug 29 '19 at 19:34
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The hierarchy problem in its basic form is a statement about large differences in physical scales. If a theory has no UV cut-off (effectively sending it to infinity), and just a single physical scale (in this case the electro-weak scale), there are no scales to compare, and there can be no hierarchy. The single physical scale is just a parameter of the theory.

Now we have good reason to believe the Standard Model is only valid up to a certain cut-off scale $\Lambda$, say at most the Planck scale. Now you can talk about a hierarchy, and look for an explanation as why these scales are separated, even though from an effective field theory point of view it would naively seem that the electro-weak scale should be $\mathcal{O}(\Lambda)$.

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  • $\begingroup$ This is an oversimplification. Not every difference between scales creates a hierarchy problem. $\endgroup$ – knzhou Aug 27 '19 at 21:54
  • $\begingroup$ True, but it address the main point of the question. You are welcome to add more elaborate explanations :) $\endgroup$ – Sparticle Aug 28 '19 at 8:55
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You simply have to introduce a UV cutoff to define the theory. The Standard Model Lagrangian is not UV-finite - there are divergent diagrams in perturbation theory, so it's impossible to "integrate up to infinity" unless you introduce a regulator.

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  • $\begingroup$ the book and other notes say that there will be no inconsistencies if we integrate up to infinity because SM is renormalizable theory. The UV cutoff is introduced because we have to treat SM as an EFT because it does not explain all physics especially gravity. $\endgroup$ – 23rduser Aug 27 '19 at 13:32
  • $\begingroup$ Book: SUSY in particle physics by IAN J. R. AITCHISON $\endgroup$ – 23rduser Aug 27 '19 at 13:42
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    $\begingroup$ That's not quite correct. You can integrate up to infinity after you introduce a regulator. You introduce a UV cutoff $\Lambda$, relate the low-energy observables (masses, couplings) to their bare counterparts at the UV scale and then you can take the limit $\Lambda \to \infty$ in a sane way. In a truly UV-finite theory, you never have to introduce a regulator. $\endgroup$ – Hans Moleman Aug 27 '19 at 13:42

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