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The hierarchy problem is roughly: A scalar particle such as the Higgs receives quadratically divergent corrections, that have to cancel out delicately with the bare mass to give the observed Higgs mass. I have a couple of related questions about that:

Why is this a problem, isn't this just ordinary renormalization? Other particles receive similar divergent corrections - not quadratically divergent, but still. The regulator gives you a parameter $\Lambda$ that you'd like to take to infinity, but can't, because the corrections would blow up. Renormalization adds a counterterm that cancels the divergent term, and stuffs it into the bare mass. Now $\Lambda$ is gone from the expression for your measurable parameter, and you can let it go to infinity. I know you can choose a finite value of $\Lambda$, and regard the theory as an effective field theory, valid up to that scale. But that doesn't seem necessary if the divergence vanishes into the bare parameter.

Framed differently: Why is the quadratic divergence in case of the Higgs a problem, but not the logarithmic one in QED? If you insert a value for $\Lambda$, say $m_\mathrm{Pl.}$, OK, then $\Lambda^2 \gg \log \Lambda$. But if we don't and keep $\lim_{\Lambda\rightarrow\infty}$ in mind, then... infinity is infinity... and haven't we got rid of $\Lambda$ by renormalizing anyway?

The second part was touched in another question: Why worry about what value the bare mass has, isn't it unphysical and unobservable anyway? I always thought that it is just a symbol, like $m_0 = \lim_{x\rightarrow\infty} x^2$, and it is meaningless to ask how many GeV it is. Just like it's meaningless to ask about the value of a delta function at zero (while it is well-defined if you integrate over it with a test function). But according to this comment by Ron Maimon, the bare mass is experimentally accessible. Is it? I thought you can keep pushing and pushing to higher energies, but will principally never observe the bare mass, just as you cannot observe a bare electron charge (you'll hit the Planck scale or the Landau pole first).

(Apologies that I put two questions in one, but I have a strong feeling that they might share the same answer.)

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jdm
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Let us suppose that that the Standard Model is an effective field theory, valid below a scale $\Lambda$, and that its bare parameters are set at the scale $\Lambda$ by a fundamental, UV-complete theory, maybe string theory.

The logarithmic corrections to bare fermion masses if $\Lambda\sim M_P$ is a few percent of their masses. The quadratic correction to the bare Higgs mass squared is $\sim M_P^2$. A disaster! - Phenomenologically we know that the dressed mass ought to be $\sim -(100 \,\text{GeV})^2$.

You are right that the SM is in any case renormalisable: our calculations are finite regardless of our choice of $\Lambda\to\infty$. But we have many reasons to believe that we should pick $\sim M_P$.

Also, if there are new massive particles, their contributions to the RG cannot be absorbed into the bare mass; they will affect the RG for the renormalised running mass.

PS apologies if I've repeated things you know and have written in the question.

innisfree
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    "we have many reasons to believe that we should pick ~ M_P" Someone should state these reasons. – Mitchell Porter Feb 28 '14 at 19:58
  • Sounds like you are volunteering @MitchellPorter ;) or are you suggesting I am mistaken and that there are no such reasons? – innisfree Feb 28 '14 at 20:01
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    Thanks, the repetition is OK, I'm trying to reaffirm what I basically should know :-). But some confusion remains: I understand, if you insert a finite value for Λ, you have to adjust $m_0$, and get the hierarchy $|m_0| \ll |\delta m|$. But if you leave Λ open, with the intent to let it go to infinity, then $m_0$ must cancel the divergence (not just a large number!). Then $m_0$ is just symbolic, and has no well-defined value of its own. I thought the main purpose of renormalization was eliminating the divergent Λ depencency from the equation! ... – jdm Feb 28 '14 at 20:43
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    So... I don't understand why people are surprized that choosing a value for Λ implies a hierarchy. I mean you explicitly put in the hierarchy by saying Λ = Mpl... I also don't understand why the bare mass should have a meaningful (measurable) value, since it is not observable... or is it? – jdm Feb 28 '14 at 20:43
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    @MitchellPorter: I think the argument goes: We expect the SM to break down at the latest at $M_P$, since that's where quantum gravity comes into play. Choose $\Lambda = M_P$ $\Rightarrow$ hierarchy problem $\Rightarrow$ BSM physics (e.g. SUSY). But if you don't go up that far, and say new physics comes already at $M_{BSM} = 1$ TeV, then there is no hierarchy problem (or you solved it, depending on how you look at it), but in any case $\Rightarrow$ BSM physics. Am I right? I've always found this argument oddly circular. (Nevertheless, I'd like to see the reasoning for why exactly M_P too) – jdm Feb 28 '14 at 20:57