# Why do we have freedom to choose a renormalization scale in massless QFT theories?

I thought that the freedom to choose renormalization conditions arises from the freedom to choose the arbitrary renormalization parameters. Let me exemplify this in a

### Massive $$\phi^4$$ scalar theory

Let's denote with a subscript $$_0$$ bare quantities and let $$\phi$$ be the renormalized field defined by $$\phi_0 =: \sqrt{A} \, \phi$$ for some arbitrary constant $$A$$, so that $$$$\begin{split} \mathcal{L} &= \frac{1}{2}(\partial \phi_0)^2 - \frac{1}{2}m_0^2\phi_0^2-\frac{1}{4!}\lambda_0\phi_0^4 \\ &\equiv \frac{1}{2} A(\partial \phi)^2 - \frac{1}{2}m_0^2A\phi^2-\frac{1}{4!}\lambda_0A^2\phi^4 \\ &\equiv \frac{1}{2}(\partial \phi)^2 - \frac{1}{2}\tilde{m}^2\phi^2-\frac{1}{4!}\tilde{\lambda}\phi^4 + \frac{\delta_Z}{2}(\partial \phi)^2 - \frac{\delta_m}{2}\phi^2-\frac{\delta_{\lambda}}{4!}\phi^4 \\ \end{split}$$$$

This is just a rewriting, identically true for any choice of the renormalization parameters $$A, \tilde{m}, \tilde{\lambda}$$ upon $$A = 1 + \delta_Z , \quad A m_0^2 = \tilde{m}^2 + \delta_m , \quad A^2 \lambda_0 = \tilde{\lambda} + \delta_{\lambda}$$

Let's focus on $$A$$ and $$\tilde{m}$$. From Dyson resummation and Kallen-Lehmann representation the Fourier bare propagator is $$\mathcal{F}\left[ \langle\Omega|T\phi_0(x)\phi_0(y)|\Omega\rangle \right](p) \equiv G_0(p) = \frac{i}{p^2-m_0^2 - M_0^2(p^2)} = \frac{iZ}{p^2-m^2} + \text{regular at } p^2 = m^2$$ where $$-iM_0^2(p^2)$$ is the sum of Fourier amputated 1PI bare diagrams and $$m$$ is the physical mass.

Then the Fourier renormalized propagator has, on one side, the same Dyson structure of the bare one upon $$m_0 \rightarrow \tilde{m}$$ and $$M_0^2 \rightarrow M^2$$, on the other it's simply $$\frac{G_0(p)}{A}$$ per definition of renormalized field:

$$\mathcal{F}\left[ \langle\Omega|T\phi(x)\phi(y)|\Omega\rangle \right](p) \equiv G(p) = \frac{i}{p^2-\tilde{m}^2 - M^2(p^2)} = \frac{i \, Z/A}{p^2-m^2} + \text{regular at } p^2 = m^2$$ where $$-iM^2(p^2)$$ is the sum of Fourier amputated 1PI renormalized diagrams and $$m$$ is again the physical mass.

In particular the renormalized propagator still has a pole at the physical mass and it's residue there it's $$i \, Z/A$$:

$$p^2 - \tilde{m}^2 - M^2(p^2) |_{p^2 = m^2} = 0 \Rightarrow M^2(m^2) = m^2 - \tilde{m}^2 \quad \quad \quad (*)$$

$$i \, Z/A = \frac{i}{1-\frac{d}{dp^2} M^2(p^2)|_{p^2 = m^2}} \Rightarrow \dot{M}^2(m^2) = 1-(Z/A)^{-1} \quad \quad (**)$$

These are all facts. But we still have the freedom to choose $$A$$ and $$\tilde{m}$$: an obvious choice is

$$A \overset{!}{=} Z , \quad \tilde{m} \overset{!}{=}m \Rightarrow M^2(m^2) = 0, \quad \dot{M}^2(m^2) = 0$$

Another possibile choice is to introduce some renormalization scale $$\mu$$ so that

$$M^2(-\mu^2) = 0, \quad \dot{M}^2(-\mu^2) = 0$$

We can choose 2 conditions because we have 2 free parameters, $$A$$ and $$\tilde{m}$$. This choice implicitly constraints these parameters in function of $$\mu$$ via equations $$(*), (**)$$; this dependence is described by Callan-Symanzik equation.

Question 1: is my understanding correct so far?

### Massless theory

In the massless theory all goes the same way, but since no $$\tilde{m}$$ is introduced we have only one degree of freedom, $$A$$, to choose. Following the same reasoning as before without $$m_0, m$$ and $$\tilde{m}$$ around the renormalized propagator must still have a pole at $$p^2 = 0$$, so equation (*) seems constrained to $$M^2(0)=0$$.

Question 2: Why do we have the freedom to choose again $$M^2(-\mu^2)=0$$ ?

• You do introduce $\bar{m}$ in the massless model. You also introduce $m_0$. This is the same number of parameters as before. What makes it massless is that the pole of the two point function $m=0$. – octonion Oct 16 '18 at 0:45
• If we introduced $\tilde{m}$ and $m_0$ and just set $m=0$ this would mean for the Fourier renormalized propagator $G(p^2 = -\mu^2) = \frac{i}{-\mu^2 - \tilde{m}^2}$, where $\tilde{m}$ depends on $\mu$ via the CS equation. But Peskin&Schroeder, pag. 408, reads (in this notation) $G(p^2 = -\mu^2) = \frac{i}{-\mu^2}$, as if no $\tilde{m}$ was ever introduced. – DavideL Oct 18 '18 at 17:12
• Okay fair enough, so they absorbed $\tilde{m}^2$ into $M^2$. There's no problem with that, $\tilde{m}^2$ was only introduced for convenience. The important thing is there is a bare mass and thus a mass counterterm $\delta_m$. You can see how they explicitly calculate in this scheme on page 412. – octonion Oct 18 '18 at 18:30
• The counterterms $\delta_m$ etc. are parameters too. Maybe that is the key to your question. – octonion Oct 18 '18 at 18:33
• Thanks, this makes sense. Let me see if I get it. Being massless means that the pole of the two point function is at $p^2 = 0$. There still is a bare mass and thus mass counterterm. This answers my original question on the number of degrees of freedom necessary to choose the ren. cond. As for my previous comment, you're saying that the ren. condition is actually chosen to be $-\tilde{m}^2 - M^2(-\mu^2) = 0$. Fair enough. Does this mean that $\tilde{m}$ is the running coupling, depending on $\mu$ by this constraint? – DavideL Oct 18 '18 at 18:43

In the massless theory all goes the same way, but since no $$\tilde{m}$$ is introduced we have only one degree of freedom, $$A$$, to choose... Question 2: Why do we have the freedom to choose again $$M^2(-\mu^2)=0$$ ?
For massless $$\phi^4$$ scalar theory, you still have the freedom to choose the mass renormalization condition. The reason is that nothing prevents the scalar field from gaining a mass from quantum corrections. So there is practically no difference between massive and massless $$\phi^4$$ scalar theories. This is actually all the fuss about surrounding the "naturalness" of the Higgs model: Higgs (as a $$\phi^4$$ scalar theory) can have mass quantum corrections all the way up to the Planck scale, regardless of the initial mass the Higgs field has.