Renormalization and the Hierarchy Problem

Question

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.)

Answer 1

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.

Answer 2

First, as you said, $\Lambda$ is not physical. Thus, it disappears in the physical quantity and is taken to be infinity.

The most straightforward example is the one-loop vacuum polarization, defining the electric charge. In Peskin and Schroder, Eq. (7.91) shows \begin{align*} \widehat{\Pi}_2\left(q^2\right) &\equiv \Pi_2\left(q^2\right)-\Pi_2(0)\\ &=-\frac{2 \alpha}{\pi} \int_0^1 d x x(1-x) \log \left(\frac{m^2}{m^2-x(1-x) q^2}\right). \end{align*}

Each individual $\Pi_2(q^2)$ contains the regularization-dependent $\Lambda$ or $1/\epsilon$, the above combination does not. It is because the above is expressed as the relative quantity from the reference value $\Pi_2(0)$, at which the electric charge is defined. The high momentum modes do not see the details and are canceled to accuracy ${\cal O}(q^2/\Lambda^2).$ This is the renormalization and in reality we measure the electric charge in terms of the reference one.

Next, you are right: the bare mass is not an observable. It is one component of the loop-corrected two-point vertex by the self-energy $\tilde \Sigma(p^2)$ $$ \langle \phi(p) \phi(-p) \rangle_{1PI} = p^2 - m_B^2 - \tilde \Sigma(p^2), $$ where $p$ is the external momentum. As before, let us fix the reference mass as e.g. the pole mass $m$ such that $$ [p^2 - m_B^2 - \tilde \Sigma(p^2)]_{p^2=m^2} =0. $$ We can replace the bare mass with the pole mass $$ \langle \phi(p) \phi(-p) \rangle_{1PI} = p^2 - m^2 - \tilde \Sigma(p^2) + \tilde \Sigma(m^2). $$ If we observe this, we indirectly observe the bare mass. Not separately, but the whole combination is the only observable.

In fact, from the LSZ reduction formula, etc., the field-strength renormalized one \begin{align*} \langle \phi(p)_r \phi(-p)_r \rangle_{1PI} &\equiv \Gamma^{(2),ren} \\ &= p^2 - m^2 - \tilde \Sigma(p^2) + \tilde \Sigma(m^2) + (p^2-m^2) \frac{d \tilde \Sigma}{d p^2} (m^2). \end{align*} is the only observable. We observe the mass only indirectly via the propagator $$ \frac{i}{\Gamma^{(2),ren}(p^2)}. $$

You may check here that the quadratic $\Lambda^2$ inside $\tilde \Sigma(p^2)$ disappears here.