Can dimensional regularization solve the fine-tuning problem?

I have recently read that the dimensional regularization scheme is "special" because power law divergences are absent. It was argued that power law divergences were unphysical and that there was no fine-tuning problem. I was immediately suspicious.

Let us take $\lambda\phi^4$ theory. For the renormalized mass $m$ (not a physical mass) with dimensional regularization, $$m^2_\text{phys} = m^2(\mu) + m^2_\text{phys}\frac{\lambda}{16\pi^2}\ln({\mu^2}/{m^2_\text{phys}})$$ This looks promising, but $m$ is a renormalized mass, not a true parameter of a Lagrangian that will be set by some new physics, like string theory or whatever it is.

For the Lagrangian mass with a cut-off regulator, $$m^2_\text{phys} = m_0^2(\Lambda) + \frac{\lambda}{16\pi^2} \Lambda^2$$ which is basically what I understand to be the fine-tuning problem. We would need incredible cancellations for $m_0\ll \Lambda$ at the low scale. Here I understand $m_0$ to be a "real" parameter determining the theory, whereas $m$ in dimensional regularisation was just an intermediate parameter that scheme.

I suspect that these two equations are related by the wave-function renormalization, $$m_0^2=Z m^2 = (1+\text{const}\Lambda^2/m^2 + \ldots) m^2$$ If I am correct, not much has improved with dimensional regularization. We've sort of hidden the fine-tuning in the wave-function renormalization.

You don't see the fine-tuning in dimensional regularization because you are working with a renormalized mass. The bare Lagrangian mass $m_0$ is the one being set at the high-scale by some physics we don't know about. So it's $m_0$ that we need to worry about being fine-tuned. With dimensional regularization, we see that $m$ isn't fine-tuned, but that isn't a big deal.

Have I misunderstood something? I feel like I am missing something. Can dimensional regularization solve the fine-tuning problem? Is dimensional regularization really special?

EDIT

I am not necessarily associating $\Lambda$ with a massive particle, just a massive scale at which $m_0$ is set to a finite value.

It seems to me that the dimensional regularisation cannot help me understand how $m_0$ runs, or the tuning associated with setting it at the high scale, especially as it obliterates information about the divergences. I have no idea how quickly to take the $\epsilon\to0$ limit.

I can do something like,

$$m_0^2 = Z m^2 = ( 1 +\lambda/\epsilon) m^2\\ m_0^2(\epsilon_1) - m_0^2(\epsilon_2) = m^2 \lambda (1/\epsilon_1 - 1/\epsilon_2)$$ Now if I take $\epsilon_1$ somehow so that it corresponds to a low scale, and $\epsilon_2$ somehow corresponds to a high scale. $m_0^2(\epsilon_1)$ needs to be small for a light scalar. But then I need to fine tune the massive number on the right hand side with the bare mass at a high scale. The fine tuning is still there. Admittedly this is very informal because I have no idea how to really interpret $\epsilon$

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Honestly if you find a way to interpret $\frac{1}{\epsilon}$ divergences physically purely based on dim reg let me know. Any arguments I have seen are ultimately grounded in the intuition one gets from more 'physical' regulators like a hard cutoff or Pauli- Villars. – DJBunk Feb 17 '14 at 0:38
Every regulator carries an opinion about the UV physics. Intuitively, I think dimensional regularization assumes that there is no new UV physics. Further, one consistent way to compute divergent integrals is to just drop the power-law divergences, which is what dim-reg does. So it is unphysical inasmuch as you expect to see new UV physics. – Siva Apr 12 '14 at 19:39

Dimensional regularization (i.e., dim-reg) is a method to regulate divergent integrals. Instead of working in $4$ dimensions where loop integrals are divergent you can work in $4-\epsilon$ dimensions. This trick enables you to pick out the divergent part of the integral, as using a cutoff does. However, it treats all divergences equally so you can't differentiate between a quadratic and logarithmic divergence using dim-reg. All it really does is hide the fine-tuning, not fix the problem.

As an example lets do the mass renormalization of $\phi^4$ theory. The diagram gives, $$\int \frac{ - i \lambda }{ 2} \frac{ i }{ \ell ^2 - m ^2 + i \epsilon } \frac{ d ^4 \ell }{ (2\pi)^4 } = \lim _{ \epsilon \rightarrow 0 }\frac{ - i \lambda }{ 2} \frac{ - i }{ 16 \pi ^2 } \left( \frac{ 2 }{ \epsilon } + \log 4 \pi - \log m ^2 - \gamma \right)$$ where I have used the master formula'' in the back of Peskin and Schoeder, pg. A.44 (note that this $\epsilon$ doesn't have anything to do with the $\epsilon$ in the propagator). This gives a mass renormalization of $$\delta m ^2 = \lim _{ \epsilon \rightarrow 0 } \frac{ \lambda }{ 32 \pi ^2 } \left( \frac{ 2 }{ \epsilon } + \log 4 \pi - \log m ^2 - \gamma \right)$$ Keeping only the divergent part: $$\delta m ^2 = \lim _{ \epsilon \rightarrow 0 } \frac{ \lambda }{ 16 \pi ^2 } \frac{ 1 }{ \epsilon }$$ This is the same result as the one you arrived at above, but uses a different regulator. You regulated your integral using a cut-off, I did using dim-reg. The mass correction diverges as $\sim \frac{1}{ \epsilon }$. This is where the sensitivity to the UV physics is stored.

A cutoff, which is a dimensionful number, tells you something very physical, the scale of new physics. The $\epsilon$ is unphysical, just a useful parameter.

With a cutoff, depending on how badly your divergence is, you will get different scaling with the cutoff; it will be either logarithmic, quadratic, or quartic (which has real physical significance, namely, how sensitive the result is tothe high energy physics). However, dim-reg regulated integrals always diverge the same way, like $\frac{1}{ \epsilon }$. Dim-reg doesn't care how your integral diverges. It can be a logarithmically divergent integral but using dim-reg you will still get a $\frac{1}{ \epsilon }$ dependence. The reason for this is that $\epsilon$ is not a physical quantity here. Its just a useful trick to regulate the integrals.

Since dim-reg hides the type of divergences that you have, people like to say that dim-reg solves the fine-tuning problem, because by using it you don't get to see how badly your divergence is. This viewpoint is clearly flawed since the quadratic divergences are still there, they just appear to be on the same footing as logarithmic divergences when you use dim-reg.

In short the fine-tuning problem isn't really fixed using dim-reg but if you use it then you can pretend the problem isn't here. This is by no means a solution to the fine-tuning, unless someone develops an intuition for why dim-reg is the correct'' way to regulate your integrals, i.e., a physical meaning for $\epsilon$ (which its safe to say there isn't one).

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Thanks. Where is the fine-tuning hidden in dim. reg.? Can you be explicit? Using dim. reg., what's the easiest way to see that the sensitivity to the UV for scalar fields is still there? – innisfree Feb 14 '14 at 13:46
I updated the answer a bit. The sensitivity to new physics is in the fact that $1/\epsilon$ diverges. The mass correction is infinite and hence the theory is fine-tuned. The point is that using dim-reg this would occur whether you have a log divergence (good) or a quadratic (bad). But the type of divergence is the key differentiator between natural and unnatural theories. – JeffDror Feb 14 '14 at 14:42
I really disagree with this answer but I am not going downvote because I find it a bit antagonistic in this situation. I posted my answer below, please feel free to comment and we can discuss. – DJBunk Feb 14 '14 at 19:17
@DJBunk: Seems to me like we have similar answers... I'm not sure what you disagree with. The fact that dim-reg hides the fine-tuning? If you use dim-reg at all times, you will never know when you have log divergences and when you have quadratic divergences. So in this sense it hides it. This is what people mean when they say (and people do say it) to use dim-reg and not worry about fine tuning. Though I clearly emphasized I do not at all agree with this viewpoint. – JeffDror Feb 14 '14 at 19:47
@innisfree, by writing a cutoff what we mean is that there are some massive particles in our theory. The cutoff is the new physics at some scale. – JeffDror Feb 14 '14 at 20:08

The fine tuning you describe isn't present in the $\phi^4$ model. You need some some other heavy fields around to see it. For example, couple your $\phi$ to a heavy fermion of mass M, Then when you will have a shift in your mass $\delta m^2 \sim M^2$. There will be other factors of $\pi$ and coupling constants and such, but the problem remains - for large M you will have to tune the bare parameter $m_0$ to get the correct value for $m_{\text{physical} }$. Stated this way, the fine tuning problem appears for a hard cutoff or dim reg.

This relates to to your original question in the following way. When you use a hard cutoff you integrate up to $\Lambda$. But what is $\Lambda$? It is the scale of new physics, which is in this case is the scale of the heavy fermion we integrated out, namely M.

So to finally answer your question - no dim reg absolutely does not solve the fine tuning problem. Moreover, it does not even obscure it, all the same pathologies are still there.

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Not sure I follow. Why do I need to associate $\Lambda$ with a new super massive particle? Why can't I just say, $\Lambda$ is some high scale at which my Lagrangian parameter $m_0^2$ is set by a more fundamental theory? Then i see a hellish fine tuning in the RG running. But where is that in dim reg? – innisfree Feb 14 '14 at 19:53
Btw yes I agree that if there is a massive particle dim reg shows the fine tuning problem just like cut off regularisation – innisfree Feb 14 '14 at 20:10
@innisfree : It is like JeffDror said above - $\Lambda$, the cutoff, is the energy scale at which our model fails and new physics comes in. The energy scale at which our model fails is that one at which there are new particles of mass $\Lambda$ that we were otherwise ignoring. – DJBunk Feb 15 '14 at 19:45
Massive particles aren't necessary for the argument, only a massive scale is strictly neccessary. It's plausible that the massive scale is there because it's associated with a new massive particle. But on the other hand, the properties of the new physics could be very special. – innisfree Feb 16 '14 at 14:21
So $\Lambda$ need not correspond to a massive particle. That said, there isn't any other way to "tell" dim reg about some new physics at the scale $\Lambda$, so maybe it is reasonable to put in a particle of mass $\Lambda$ to represent the new physics, regardless of whether there is truly a massive particle – innisfree Feb 16 '14 at 14:22