# What is the difference between additive and multiplicative renormalisations?

1. Mar 4, 2009

### Bobhawke

Some papers I am looking over at the moment mention both additive and multiplicative renormalisations. What is the difference between them?

2. Mar 5, 2009

### blechman

i guess you should provide a specific reference, but here's a guess.

multiplicative renormalization of a coupling $g$ has a counterterm of the form:

$$\delta g= g^af(\lambda_i)$$

for some power $a$, while additive renormalization would be

$$\delta g= h(\lambda_i)$$

Here $\lambda_i$ are all the other couplings in the theory. To be general, I include masses as couplings.

An example of the first kind is a fermion mass, while the second kind would be a scalar mass (assuming no SUSY).

3. Mar 6, 2009

### Naty1

In additive renormalization, physicsts wave their hands horizontally; in multiplicative renormalizations, physicsts wave their hands vertically.

Actually Richard Feynman had the funniest description, relating to "hokus pocus" .... Roger Penrose mentions in his massive THE ROAD TO REALITY on page 674:

Wikipedia has quite a discussion at http://en.wikipedia.org/wiki/Renormalization#Renormalizability

4. Mar 9, 2009

### Bobhawke

The context in which I came across the terms is the following:

It is a paper on lattice QCD. It first discusses how the Wilson term can be added to the fermion part of the Lagrangian to avoid the fermion doubling problem. It then says that this breaks chiral symmetry since the extra term looks like a mass term. It then says

"The quark mass has both multiplicative and additive renormalisations due to the explicit breaking of chiral symmetry by the wilson term"

5. Mar 9, 2009

### blechman

This should be interpreted precisely as I said in my last post.

Normally fermions are "multiplicatively renormalized" since any fermion mass corrections must be proportional to the mass itself by chiral symmetry. However, since chiral symmetry is explicitly broken on the lattice by removing the fermion tastes, you can now get a term that is NOT proportional to the mass, but proportional to $a^{-1}$. If we were to take this seriously, we would say that there is a "hierarchy problem" here, just like the Higgs in the standard model; but there really isn't since the whole fermion taste/Wilson term is an artifact of our lattice description and not really there.

6. Mar 9, 2009

### Bobhawke

Thanks Blechman!