Simple power counting tells you that a scalar field coupled to some fermions at one-loop picks up a correction to the mass of the order $\Lambda^2$.

Based on this people say things like "it's natural to expect that the mass of the scalar is roughly the cut-off scale", which in this case is some GUT/Planck scale.

My question is this: is this really the right interpretation? If I'm doing perturbation theory and it's telling me that I have a correction as big as the largest scale in my problem (cut-off scale), it means I cannot trust the answer. It does not meant the answer is $m_\phi^2 \propto \Lambda^2$. The renormalized mass could still be far below $\Lambda$, but the current approach cannot see that. The correct and finite answer might emerge only after adding up all diagrams. There's no reason to try to fine-tune anything such that already at one-loop the mass is small. One must simply concede that the one-loop answer is not correct.

What is the correct interpretation?

  • What makes you expect higher loop corrections to cancel a large one-loop contribution? Higher loops will have answers suppressed by higher powers of the coupling and so, will be parametrically suppressed. So for small values of the coupling, how can higher order corrections save you? – Siva Apr 26 '14 at 4:22
  • "it's telling me that I have a correction as big as the largest scale in my problem (cut-off scale)" -- that's not quite true. Any corrections to the mass are suppressed by positive powers of the loop factor $\frac{g^2}{16 \pi^2}$. – Siva Oct 8 '14 at 7:38

If I'm doing perturbation theory and it's telling me that I have a correction as big as the largest scale in my problem (cut-off scale), it means I cannot trust the answer. It does not meant the answer is $m_\phi^2 \propto \Lambda^2$. The renormalized mass could still be far beyond $\Lambda$, but the current approach cannot see that.

I disagree with this about a small point, but for now let's assume it is absolutely correct. Then you still have a scalar field that you would like to be massless but your calculation says its mass is of the order of $\Lambda$ or higher. This means that the hierarchy problem is still there and we are only arguing about a detail on how it is formulated.

Now the small point: It is actually very useful to know how the mass scales with the cutoff and there is a lot of information in knowing that $m_\phi^2 \propto \Lambda^2$ as opposed to for example $m_\phi^2 \propto \log\frac{\Lambda^2}{\mu^2}$ or anything else.

The way to think about it is this: Imagine another "fictitious" cutoff $\Lambda_f$ with $\Lambda_f\ll\Lambda$. Then your previous calculation will give $m_\phi^2 \propto \Lambda_f^2$, but now you are in a region where you can trust perturbation theory! Your calculation says that if you use to different fictitious cutoffs with $\Lambda_{f1}=2\Lambda_{f2}$ then the mass correction for the second theory will be 4 times bigger than the mass correction for the first theory.

Hope this helps!

  • My original post was supposed to say "far below $\Lambda$", not "beyond". I meant to say that the true correction could be smaller, but we do not know. – user129 Apr 28 '14 at 22:27
  • No, this cannot happen. If the true correction was much smaller than $\Lambda$, perturbation theory would work fine and give you the correct result. The fact that perturbation theory breaks down means you are guaranteed that the corrections are big. The only thing you can argue about is how big and if we can trust that they scale like $\Lambda^2$, but they will be big for sure. And this makes the hierarchy problem unavoidable. – Heterotic Apr 29 '14 at 9:09
  • 1
    Hope that you will be notified. But there are functions like $\frac{\mu^4}{\mu^2+\Lambda^2}$ that can give small value even while every term of series in $\Lambda$ can be very large. Are there any arguments that rule out such a possibility? – OON Nov 15 '14 at 9:13

It has to do with the scale to which you would like your physics to be valid. In general, we would like it to be correct up to a GUT scale. But in doing so, you would introduce a very large correction to mass of the scalar particle.

A simple dimensional analysis will show that if you go to higher order using the $\phi^4$ interaction then your diagrams will still be quadratically divergent, and there may be a cancellation depending upon the sign of the terms which will come next.

But there is no priori that there should be a cancellation of this very higher order and hence we have to say that it must be fine tuned to give sensible result.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.