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 $\epsilon$
Best Answer
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, \begin{equation} \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) \end{equation} 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 \begin{equation} \delta m ^2 = \lim _{ \epsilon \rightarrow 0 } \frac{ \lambda }{ 32 \pi ^2 } \left( \frac{ 2 }{ \epsilon } + \log 4 \pi - \log m ^2 - \gamma \right) \end{equation} Keeping only the divergent part: \begin{equation} \delta m ^2 = \lim _{ \epsilon \rightarrow 0 } \frac{ \lambda }{ 16 \pi ^2 } \frac{ 1 }{ \epsilon } \end{equation} 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).