# Physical intuition for bare quantities to be infinite

I don't know much about interacting field theory, as far as I understand, in the interacting field theories, when doing renormalization, one usually assume the bare quantities are infinite to cancel the infinities.

But what is bare quantity? What is the physical intuition of bare quantity? Are the quantities we measured are not "true" in some sense? What do we mean by bare quantities are infinite?

• My way to understand this (could be wrong!) is that you can choose the parameters in the Lagrangian as you please. The mass and coupling for example do not need to be the measured mass and coupling. One then realizes that one could even choose these to have a finite part plus a part depending on a regulating parameter that diverges when the regulator is removed. This is a good thing to do because one can choose this formally divergent part to cancel the infinite contributions from 1PI diagrams. The "bare" quantity is the unphysical parameter with a formally divergent part.
– Gold
Commented Oct 26, 2021 at 13:54
• The quantities we measure are true; they are properties of the physical world which are finite and make sense. It is the bare quantities that are a fiction introduced for mathematical convenience (or inconvenience!) The problem is that the bare quantities differ by an infinite amount from the quantities which actually characterise the physical behaviour. Commented Oct 26, 2021 at 15:43
• This is not just about QFT, but physics in general - the quantities/parameters/variables (or whatever you may want to call them) used in a theory need not be observable themselves. They're often crutches required to compute an observable quantity. Observables are what are measured or observed (e.g. cross sections for particle physics), and a fundamental tenet of physics is that all measurements must produce finite numbers. So as long as observables are finite, it is fine if intermediate variables are not. Commented Nov 8, 2021 at 18:56

Let me explain the way in which I understand this (correctios are highly appreciated!). Suppose you have the $$\phi^4$$ Lagrangian for example $${\cal L}=\frac{1}{2}\partial^\mu \phi\partial_\mu \phi-m^2\phi^2-\frac{\lambda}{4!}\phi^4\tag{1}.$$

Then you derive the Feynman rules and start deriving Green's functions $$\langle\Omega|T\{\phi(x_1)\cdots \phi(x_n)\}|\Omega\rangle$$ in terms of Feynman diagrams. The thing is that as soon as you add loops you encounter divergences.

Now it is possible to nail down the divergences to a finite set of diagrams, the one-particle irreducible (1PI) diagrams. When you study the loop 1PI diagrams you find that they diverge. Being more precise one chooses a regularization, so that with a finite regulator the diagram is finite, but in the limit in which the regulator is removed these diagrams diverge.

The computation you did used the Lagrangian density in (1) and it goes on with whatever parameters you choose there. The parameters $$m$$ and $$\lambda$$ do not have to be the actual mass and coupling one would measure. Moreover there is one extra parameter hidden there which is the normalization of the field. One could rescale $$\phi\mapsto N\phi$$ at will. This would be one extra parameter in that Lagrangian.

Now, the diagram you computed with a finite regulator is also a function of these parameters. Since they are quite arbitrary you can fine tune them in such a way that they have a finite part plus one part which depends on the regulator which cancels the part of the diagram which diverges as the regulator is removed.

This means that now, when you remove the regulator, the diagram is finite. One finally relates the actual measured quantities with those diagrams, and they come out finite.

The arbitrary parameters in the Lagrangian, which you have chosen to have a finite part plus one part depending on a regulator which diverges as the regulator is removed are called bare. The finite part of these parameters are called renormalized.

In practice one starts with the normalization of the field. To make this parameter explicitly write $$\phi = \sqrt{Z}\phi_R$$. Then (1) becomes

$${\cal L} = \frac{1}{2}Z \partial^\mu \phi_R \partial_\mu \phi_R - Z m^2\phi^2_R-\frac{\lambda}{4!}Z^2\phi^4_R\tag{2}.$$

Observe that the "arbitrary parameters of the Lagrangian" can now be taken to be $$Z$$, $$Zm^2$$ and $$Z^2\lambda$$. Now write $$Z = Z_R+\delta_Z$$, $$Zm^2=m_R^2+\delta_m$$ and $$\lambda Z^2=\lambda_R+\delta_\lambda$$. This is the splitting of the bare parameters into a finite renormalized part plus a part which will depend on the regulator and diverge as it is removed. These are called the counterterms. One now expands out $${\cal L}$$ and one gets a Lagrangian density as

$${\cal L} = \frac{1}{2}Z_R\partial^\mu \phi_R\partial_\mu \phi_R -m_R^2\phi^2_R-\frac{\lambda_R}{4!}\phi^4_R+\frac{1}{2}\delta_Z \partial^\mu \phi_R\partial_\mu \phi_R -\delta_m \phi^2_R - \frac{\delta_\lambda}{4!}\phi^4_R\tag{3}$$

If we set $$Z_R = 1$$, which is just a choice really, the first three terms is a Lagrangian density with the same functional form as (1) but where the parameters are the renormalized $$m_R$$ and $$\lambda_R$$. One can view the remaining terms as "interaction terms" which just add some new Feynman rules to the theory.

So the intuition you seek for is this: bare quantities are the arbitrary parameters of the Lagrangian which we have splitted into a finite renormalized part plus a formally divergent, regulator dependent, counterterm, so as to use the later to cancel the divergences in the 1PI diagrams. They are infinite by design so that you may cancel said divergences.

A final comment is that the renormalized parameters need not what you actually measure. In fact the measured mass is to be defined as the pole mass: the pole in the field propagator. You can set $$m_R = m_P$$ but not necessarily and then you have a relation between the two.