Scalar field divergent mass correction interpretation question (hierarchy problem) 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?
 A: 
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!
A: 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.
