In what sense, renormalization conditions are arbitrary?

Question

Let $F$ and $x$ be measureable quantity (e.g. scattering amplitude and in-going momentum of particles)

Suppose we have $F(x)=g+\delta+g^2M(x)$ where $g$ is coupling constant, $\delta $ the counterterm and $M$ the divergence part. Then we need renormalization condition to determine $\delta$ and then cancel infinity in $M$.

However, my teacher told me that the renormalization conditions are "arbitrary". I don't understand what does he mean, but in general, arbitrary seems to mean we can find some renormalization conditions such that $\delta=-g^2M(x)$.

Or even more aggressive, we can just directly pick $\delta$ for different orders to create any expression of $F$ we want, which is ridiculous!

Or can we prove we can't find such a set of renormalization conditions s.t. $\delta=-g^2M(x)$, or in other words, renormalization conditions are naturally controlled somehow?

So in what sense, renormalization conditions are arbitrary?

EDIT:

Let me offer an example. In the standard renormalization steps of $\phi^3$ theory, we will have a expression like $$M(Q_0)=g_R$$ to define the parameter $g_R$ and then derive our counterterm.

However, we can also use some werid expression like $$M(Q_0)=g_R+lng_R$$ to define this parameter $g_R$.

The point here is we "define" $g_R$ this way. So can we in principle construct some definition so that the counterterm agree with what I mentioned above?

EDIT2:

In $\phi^4$ theory, with cut-off $\Lambda$, we have $$iM=\lambda + C\lambda^2[ln(\frac{\Lambda^2}{s})+(s \rightarrow t)+(s \rightarrow u)]$$

If we define $$\lambda _R=M(s_0,t_0,u_0)$$

and suppose $$\lambda=\lambda_R+\delta_{\lambda(2)}$$ where 2 implies second order. Then we will have a finite solution.

$$M(s,u,t,s_0,u_0,t_0)$$

But if we define $$2\lambda _R=M(s_0,t_0,u_0)$$

We can also have another expression $$M(s,u,t,s_0,u_0,t_0)$$,which is not the same as the previous one.

Answer 1

The theory is specified by some parameters, which are numbers. You can think of them as terms in a Lagrangian, given letters like $m$ and $\lambda$. Experimentally, we measure values for various processes like cross sections.

Within the theory, we can calculate cross sections (or decay rates, or other observables) in terms of the parameters of the theory. Then, matching to the values determined by experiment, fixes the values of those parameters.

Conceptually, that's all that is going on. In detail, however, there are a few places where there are ambiguities in what I said, which let you carry out the process in different ways.

First, you can choose which experimental values to match against.

Second, you can choose to parameterize the theory in different ways. For example, in a theory with a mass term $m^2 \phi^2$, you can choose the parameter $m$ to represent the physical mass -- meaning that when you match against experiment, you fix the value of $m$ to be the observed mass (Or, more formally, you set $m$ to be the pole of the propagator). But you can also choose to parameterize your theory in a way where $m$ only cancels certain divergent terms, and the observable mass is a combination of what you call $m$ as well as other terms.

To summarize, you have freedom to choose how to define the parameters in the Lagrangian, and which experiments to compare to. This freedom leads to a degree of arbitrariness in how the parameters are related to experimental values. However, so long as you are consistent with the scheme you choose, and if you could compute quantities exactly, once you have fixed the values of the parameters within a given scheme, the predictions you make for observable quantities with the theory will be independent of the scheme you used to make those predictions.

An important caveat is that we never actually compute anything exactly in an interesting quantum field theory. There is usually some kind of perturbative expansion involved. And different ways of parameterizing the theory can lead to the expansion converging faster or more slowly. So in practice there may be a reason to prefer one scheme over another, even though in principle they are all ultimately equivalent.

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To say a little bit more about why there's maybe more freedom in quantum field theory than you'd expect from classical mechanics...

When we write down the Lagrangian and say we are going to quantize it, that is somewhat of a white lie. There are additional parameters that can be introduced through the process of quantization. In particular, there is a cutoff scale, or else some regularization parameter that is introduced in order to make the loop integrals finite. (We can understand this regularization parameter physically through Wilsonian renormalization as being the scale at which the effective field theory is defined.) No physical observables depend on this cutoff parameter. However, the parameters in the Lagrangian are chosen to cancel this cutoff dependence. The specific way you implement this cancellation is a free choice -- for example, you can choose the parameters to only cancel the divergence, or you can add extra finite pieces to the parameters as well.

Ultimately the freedom in how to define the Lagrangian parameters in terms of the cutoff and observable quantities boils down to our choice of how to parameterize the theory. Different parameterizations are ultimately equivalent, although in practice some parameterizations might be more useful. But it can be confusing that there are in some sense extra parameters in quantum field theory not included in the Lagrangian that you need to account for to make the theory well defined.

Answer 2

Let's say you want to calculate the scattering cross section of an electron and a positron annihilating into a pair of photons. You use Feynman diagrams and conclude that $$ \sigma=g+\delta+g^2\biggl(2\pi+37\frac{p^2}{p^2+m^2}\biggr)+\mathcal O(g^3) $$ or some other complicated expression. As in the OP, $\delta$ is a counterterm and the $g^2$ term comes from some Feynman diagram. $p^2$ is the center-of-mass momenta of your experiment.

Now, the $2\pi$ term inside the parentheses can be shifted by choosing $\delta$ appropriately. But no matter what you do to $\delta$, the $37$ term will not shift: $\delta$ cannot depend on the momenta $p^2$ of your particles. So the $37$ term is a universal, scheme-independent prediction that you can match to experiments.

More generally, some parts of observables such as $\sigma$ can be shifted by redefining couterterms, and some other parts cannot. The former are arbitrary, unphysical, and unobservable, while the former are the true predictions of your theory, the things you can actually measure. The general rule is that couterterms can depend on coupling constants, but not on position or momenta (unless you are willing to give up locality, which we almost never are).

So no: you cannot make $F(x)$ be whatever you want. There are some rules for what $\delta$ can be. Anything that satisfies these rules is scheme-dependent and arbitrary. But once you mod out by this arbitrariness, the result is unique.

(In the theories you will often find in particle physics, the counterterms in the Lagrangian are usually of the form $\delta_1(\partial\phi)^2+\delta_2\phi^2$. These lead to Feynman rules $-\delta_1p^2+\delta_2$. Therefore, it is usually the case that, if you expand observables in powers of $p^2$, then the first two terms are unobservable, while the rest are all observable. The details of course depend on the specific system you are dealing with.)