# Renormalization at different scales in $\phi^4$ theory

Starting from the bare Lagrangian-density $$\mathcal{L} = \frac{1}{2}(\partial^2 - m_0^2) \phi_0 - \frac{\lambda_0}{4!}\phi_0^4$$ one introduces the renormalized field and parameters as

$$\phi_0 = \sqrt{Z_\phi}\phi$$, $$m_0^2 = Z_m m^2$$ and $$\lambda_0 = Z_\lambda \lambda$$.

To fix the first two constants one can look at the two-point correlator in the vicinity of the one-particle pole at $$p^2 = m_p^2$$ (physical mass): $$\langle T\phi_0\phi_0\rangle(p)|_\text{pole} = \frac{iZ}{p^2 - m^2_p} = Z_\phi \langle T\phi\phi\rangle|_\text{pole}$$.

By choosing $$Z_\phi = Z$$ and $$Z_m = m_0^2/m_p^2$$ (which gives $$m = m_p$$) one gets the simple expression for the one-particle pole of the renormalized field,$$\langle T\phi\phi\rangle|_\text{pole} = \frac{i}{p^2-m^2}$$.

These conditions are equal to demanding $$(\langle T\phi\phi\rangle(p^2=m^2))^{-1} = 0$$ and $$\partial_{p^2}(\langle T\phi\phi\rangle (p^2=m^2))^{-1} = 1$$, with $$m=m_p$$. (*)

$$Z_\lambda$$ can be fixed similarly by looking at the four-point function (cross section, respectively) at vanishing 3-momenta and setting $$\lambda = \lambda_{p}|_{p^2=m^2}$$, where $$\lambda_p$$ is the physical value obtained from the according experiment conducted at the given momentum scale $$m^2$$.

Since the choice of the momentum scale $$\mu$$ we fix the constants at ("renormalization scale" $$\mu$$) is arbitrary, one could make a different choice. My professor expresses this as $$\lambda = \lambda_p|_{p^2=\mu^2}$$ and $$m^2 = m^2_p|_{p^2=\mu^2}$$.

While the first of the two expressions is easy understand (we conduct the four-point experiment at higher energy and set the coupling constant "equal" to the obtained value $$\to$$ running of the coupling), I'm quite puzzled by the latter one. What is $$m_p^2|_{p^2=\mu^2}$$ supposed to mean? Why should the physical mass be energy dependent? I guess you could change the condition $$m^2=m^2_p$$ (or $$\pi(m^2)=0$$) to $$m^2+\pi(m_p^2)=m_p^2$$, where $$\pi(p)$$ is the 1PI diagram for the renormalized field. But what would $$\mu$$ be here? How would (*) look like in this case?

Classical mechanics

Consider an analogy in a classical setting. Say you want to study some complex system, e.g., a bridge. It is very complicated to fully model all of its dynamics, so you first introduce a very rough approximation. For example, we decide that there is a single relevant degree of freedom. Of course, this d.o.f. depends on what exactly you want to study; for example, if you are focusing on, say, the thermal expansion of the bridge, then the most important degree of freedom could be its length, or volume. If you are focusing on its structural integrity, you would take its vertical curvature or something like that. Call this effective degree of freedom $$x(t)$$.

Again, describing the full dynamics of $$x(t)$$ is beyond reach, at least for now, so we content ourselves with more approximations. For example, we assume that $$x(t)$$ doesn't vary too much throughout our experiments. In that case, any Lagrangian describing its dynamics must take the form $$L=c_0 \dot x^2+c_1 x+c_2x^2+\mathcal O(x^3)$$ where $$c_0,c_1,c_2$$ are some arbitrary coefficients. (We don't take higher derivatives because of Ostrogradsky).

At this point, any prediction you compute using $$L$$ will depend on $$c_i$$. What is the value of these coefficients? Well, you need experiments. But you cannot measure the $$c_i$$ directly: coefficients in a Lagrangian are not measurable. What you can measure, for example, is the equilibrium position of $$x$$. Indeed, if $$x$$ is the length of the bridge, you just measure $$x(t)$$ for different values of $$t$$ and take the average. And the prediction for the equilibrium position, given $$L$$ above, is $$\langle x\rangle=-c_1/2c_2$$. So while you cannot measure $$c_1,c_2$$ directly, you can measure their quotient. Another object you can easily measure is the frequency of oscillations of $$x(t)$$ around $$\langle x\rangle$$; according to the Lagrangian above, this frequency is $$\omega=\sqrt{c_2/c_0}$$. So, again, even if you cannot measure $$c_0,c_2$$ directly, you can measure their quotient, by measuring the frequency $$\omega$$. We summarise this as follows: The coefficients of a Lagrangian are not directly measurable, but you can use the Lagrangian to compute predictions, which are measurable, and then use those predictions to fix the value of your coefficients. Once you've fixed all the free parameters, any new computation will be a true prediction of your model, something you can compare to experiments.

Needless to say, instead of measuring the average position and frequency, we could measure other observables, such as the energy or something else. These predictions will also fix the value of $$c_i$$, although the expression for $$c_i$$ will change. You are free to use whatever measurable you want. As long as you don't make an algebraic mistake, the model will make the exact same predictions no matter what choices you make. The algebraic form of the predictions will change -- it depends on what observables you used to fix the $$c_i$$ -- but their numerical value will not.

Another important point to stress is the following. We saw above that $$\omega=\sqrt{c_2/c_0}$$, so you may want to replace $$c_2\to\omega^2c_0$$ in the Lagrangian. You shouldn't really be doing this, for the following reason. Say you increase your precision and so the "small $$x$$'' approximation is not terribly good anymore. So you introduce a higher order term in the Lagrangian, $$c_3x^3$$. In this situation, the relation $$\omega=\sqrt{c_2/c_0}$$ is no longer true: the anharmonic oscillator has frequency $$\omega^2\sim \omega_0^2+c_3^2$$, where $$\omega_0=\sqrt{c_2/c_0}$$. Of course, $$\omega_0$$ is not measurable anymore: if you measure the frequency of the system, you'll get $$\omega$$, not $$\omega_0$$. The object $$\omega_0$$ is not useful anymore, I personally wouldn't even introduce a notation for it. It is irrelevant, I'd rather just stick to the arbitrary coefficients $$c_i$$, and true measurables like $$\omega$$.

Quantum mechanics

Now let's see how this works in the quantum case. Again, we want to describe a complex system. Unlike before, we no longer have a good mental picture of what the "microscopic dynamics" are. We don't have a useful concept of fundamental "quantum bridge". We don't really know what the "true" system is. We only have the effective, approximate picture: we assume that, whatever the right description is, an effective description should work, at least for small energies. So we introduce some "relevant" degree of freedom $$\phi(x)$$, and hope that it gives at least a rough approximation to the true dynamics, whatever that means.

Again, we hope that it makes sense to say that $$\phi(x)$$ stays "small" during our experiments so that an effective expansion $$\mathcal L=c_0(\partial\phi)^2+c_1\phi+c_2\phi^2+\mathcal O(\phi^3)$$ makes sense. Much like in the classical examples, the coefficients $$c_i$$ are not directly measurable.

Something you can measure, analogous to the frequency $$\omega$$ from before, is the ratio $$c_2/c_0$$. You measure this ratio as follows. First, define the function $$\Pi(p^2)$$ as the inverse of the expectation value $$\langle \phi^2\rangle$$ in Fourier space $$\langle \phi(p)^2\rangle=\frac{1}{\Pi(p^2)}$$ This function can be computed, from $$\mathcal L$$, by adding all one-particle-irreducible Feynman diagrams with two external legs. So you can express $$\Pi(p^2)$$ as some function of $$c_i$$. Next, you can also prove [ref.1] that if $$\Pi(p^2)$$ has some first-order root, $$\Pi(p^2)\propto(p^2-a)+\mathcal O((p^2-a)^2)$$ for some $$a$$, then in a laboratory you would observe a point-particle with mass $$\sqrt a$$, propagating through space. Moreover, the imaginary part of $$\Pi(a)$$ becomes the decay width of this particle. So, all in all, you can compute the mass and decay constant in terms of $$c_i$$, and also measure these parameters, which allows you to compute the value of $$c_i$$. Once you have the value of these constants, you can make any other prediction you want. In the example above, it turns out that $$a=c_2/c_0$$, and $$\Pi(p^2)$$ is purely real, so the particle is stable and has mass $$\sqrt{c_2/c_0}$$. (As in the classical case, you should not replace $$c_2\to m^2c_0$$ in the Lagrangian. The reason is pretty much the same: if you include higher order terms, the relation $$m^2=c_2/c_0$$ no longer holds, but rather $$m^2\sim c_2/c_0+c_3$$ or something like that. Again, you can define $$m_0^2=c_2/c_0$$, but this is of little use, because $$m_0$$ is not measurable anymore. I personally don't find the "bare mass" to be a useful concept at all. I prefer to work entirely in terms of the arbitrary coefficients $$c_i$$, and measurable things like $$m$$, and never introduce "bare", unmeasurable objects at all).

As in the classical case, you can choose other measurable quantities in order to fix $$c_i$$. (In practice, measuring mass is particularly convenient because it is the most relevant interaction, in a precise sense, and so it is the parameter that has the least inaccuracy). As $$\phi$$ is less physical than $$x$$, there is really no reason to stick to "physical" conditions. You can choose whatever prescription you want -- after all, the coefficients $$c_i$$ are not directly measurable, and $$\phi$$ has little meaning on its own. As long as you don't make algebraic mistakes, the model will make the exact same predictions for a given question.

For example, the physical mass, the one you measure in a laboratory (spectroscopy, or Breit-Wigner histograms), is defined by the condition of $$\Pi$$ having a first order root, i.e., $$\Pi(m^2)=0,\qquad \Pi'(m^2)=1$$ You could for example redefine $$\tilde\Pi(p^2)=\Pi(p^2+m^2-\mu^2)$$ such that the conditions become $$\tilde\Pi(\mu^2)=0,\qquad \tilde\Pi'(\mu^2)=1$$ This is just a change of notation, the value of $$\langle \phi(p)^2\rangle$$ stays the same. The only difference is that we now fix the value of $$c_i$$ in terms of $$\mu^2$$ instead of $$m^2$$. Of course, $$m^2$$ is directly measurable, while $$\mu^2$$ is just some arbitrary parameter, with no physical meaning, and not directly measurable.

The choice of how to fix the free parameters $$c_i$$ in terms of some condition is known as a choice of scheme. The "physical" choice in terms of measurable quantities like $$m$$ is known as the on-shell scheme. Other schemes are useful too, even if they do not involve parameters that are directly measurable. No prediction can depend on the choice of scheme; only the intermediate steps do.

A natural question is why would one want to express things in terms of $$\mu^2$$ instead of $$m^2$$. The answer is that, while this parameter is arbitrary, you can make a wise choice for it that simplifies things for you. For example, it turns out that the so-called leading logs [refs.2-5], that is, the largest power of a logarithm that appears to a given order in perturbation theory, have a form that is very constrained by consistency conditions. For example, by dimensional analysis and some other properties of healthy quantum theories, one can argue that they always take the form $$\sim\log^n(s/\mu^2)$$, with $$s$$ the center of mass energy. Therefore, if you choose $$\mu^2\sim s$$, i.e., if you take the free parameter $$\mu$$ to be around the energies of your experiments, then the leading logs vanish, and your lower order approximation becomes almost as accurate as having the leading logs to all orders in perturbation theory. This is why having an adjustable parameter like $$\mu$$ becomes useful. The physical mass, the one corresponding to the position of the pole of $$\langle \phi^2\rangle$$ is still $$m^2$$. This mass is measurable, and does not depend on any choices you can make. Its value is unique.

If we were able to compute all observables to all orders in perturbation theory, running couplings would be entirely useless. But we can't. So we do the following: if in a given low-order result we replace on-shell couplings for their running counterparts, such as $$m\to m(\mu)$$, then these low-order results become almost as accurate as having higher-order corrections: the large logs, to all orders in perturbation theory, become very small, and so their contribution is almost as already accounted for.

References

1. Sidney Coleman, lecture notes, section 19, https://arxiv.org/abs/1110.5013.

2. Bjorken & Drell - Relativistic quantum fields, section 19.15.

3. Schwartz - Quantum Field Theory and the Standard Model, section 23.1.

4. Srednicki - Quantum Field Theory, section 27.

5. Weinberg - Quantum theory of fields, Vol.2, chapter 18.

• There are a few things which I still don't understand. First of all you say you fix the $c_i$s by imposing conditions on $\tilde{\Pi}(\mu^2)$ (the derivative respectively). But all $c_i$ that fulfill the on-shell conditions also fulfill the off-shell ones for any $\mu$ automatically?? Sep 18, 2020 at 17:36
• Furthermore, what do you precisely mean with $m(\mu)$ in you last paragraph? Sep 18, 2020 at 17:38
• @user2224350 1) The value of the $c_i$ depend on the conditions you use. The formal statement is that $c_i$ is scheme-dependent. You can use whatever condition you want; observable objects are insensitive to such choices, while unobservable objects are typically not. 2) $m(\mu)$ is the running mass, see e.g. this PSE post. Sep 22, 2020 at 15:04