# Renormalization and counterterms

1. Oct 23, 2013

### metroplex021

Hi all,

Can anyone tell me whether the counterterms introduced in renormalized perturbation theory (see e.g. Chapter 10 of Peskin and Schroeder) have any physical interpretation? In particular, are they taken to model the self-interactions that take us from a 'bare' to a 'dressed' particle?

Thanks a lot!

2. Oct 23, 2013

### dauto

The counterterms have no physical interpretation. They are a book keeping device introduced to separate the physically significant renormalization effects from the mess of infinities that arise ultimately from our ignorance of the physics at short scales.

3. Oct 24, 2013

### DarMM

In the Lagrangian we have an constant $\lambda^i_0$ multiplying the $i$-th term of the Langrangian, let's say.

It turns out that, unlike classical physics, this constant is not directly interpretable as a physical coupling or mass, but simply as just a number in the Lagrangian. Rather some related constant $\lambda^i$ would be physically meaningful.

To make calculations easier we split $\lambda^i_0$ into $\lambda^i_0 = \lambda^i + \delta\lambda$, this allows the physical coupling to be visible from the beginning, rather than being buried in the poles of some correlation function and involving a complicated calculation to extract.

It's also more convenient as $\lambda^i$ is an input to your theory, a value you fix to experiment, so again it is better to have it visible from the beginning. The counterterms are simply the $\delta\lambda$.

Now all of this would have to be done anyway, even if QFT had no divergences. The interesting thing though, is that it turns out that in many theories, like QED or QCD, the $\delta\lambda$ are essentially the negatives of any divergence in the theory and so cancel them out.

This has nothing to do with ignorance of physics at short scales. The infinities of QFT do not originate from our ignorance of physics at low scales.

4. Oct 24, 2013

### dauto

I agree with all of the above except the last part. The counter terms are formally infinite in the calculation because the calculation extrapolates QFT to infinitesimally short scales. That extrapolation is not experimentally warranted. It is conceivable (some would say extremely likely) that at some short scale some new physics must be taken into consideration. It is in that sense that the counterterms are sweeping under the rug our ignorance of the short scale physics. In essence we're declaring that this short scale physics has no other observable consequence at the low energy phenomena aside from a few counterterms. Note that this declaration is akin to the requirement that the theory at hand be renormalizable.

5. Oct 24, 2013

### DarMM

True, in fact the counterterms are rigorously infinite, not just formally. This was proven by Kallén years ago for QED for example.

True.

True.

This where I disagree. The counterterms are simply the difference between the constants of the Lagrangian and the physical constants that directly parameterise the theory. Pure Yang-Mills for example is well-defined on all scales. It might be physically incorrect on very short scales, but it still makes sense and makes well-defined (but probably false) predictions on those scales.
I can write down pure Yang-Mills on $\mathbb{R}^4$ and compute its counterterms. There is nothing being swept into them, as I'm only working with pure Yang-Mills, they are simply the computable difference between the bare and physical constants.

6. Oct 24, 2013

### Avodyne

Yes, here we have to distinguish asymptotically free theories like QCD (which are believed to exist at arbitrarily small scales) and non-asymptotically free theories (like QED and the full Standard Model) which are believed to require an explicit modification (often referred to as an "ultraviolet completion") at some sufficiently small scale.

7. Oct 24, 2013

### metroplex021

Thanks people. So just to be clear: the counterterms do not in themselves model e.g. the vacuum polarization effects that are taken to 'screen' the electron at finite distances and make the charge well-defined?

8. Oct 26, 2013

### Avodyne

As explained above, counterterms just correspond to a renaming of parameters in the lagrangian, and this can have no physical, measurable effect.