# QFT counter-terms

1. Jan 9, 2016

### anthony2005

Hello,

It's been a long time I am trying to accept renormalization in QFT but still I cannot be satisfied.
The usual pedagogical way one introduces renormalization is to cure infinities that arise from perturbative expansions.
Now, I can accept the statement that we are doing the perturbation expansion via the wrong coefficient, say the bare coupling $g_{0}$ . The physical quantity is instead $g=g_{0}-\delta g$ or we are dealing with the wrong mass, $m=m_{0}-\delta m$ .
One can then fix this, expressing our Lagrangian in terms of counterterms. But this means that the Lagrangian itself is infinite!! How can I believe in a theory based on a Lagrangian that makes no sense?? All considerations we can infer from the Lagrangian, say Noether's theorem, say the gauge symmetry itself, obviously is based on a finite Lagrangian.

In summary, what I see from a QFT book:
1) Define the Lagrangian for a specific theory (assuming of course it is a well-behaved mathematical structure)
2) Using the Lagrangian, extract the symmetries of the theory, impose gauge invariance.
3) Find out that the Lagrangian is indeed infinite. But still retain all considerations made.

What I am looking for is a textbook or notes that considers renormalization a priori. Starts from general assumptions about QFT (existence of a separable Hilbert space, Poincarè invariance, causality), justifies renormalization, obtain general theorems (e.g. spin-statistics), and then calculates amplitudes and cross-sections for specific theories, without the surprise "oh look it's divergent".

I know also there is the Wilson approach to QFT, which is satisfactory, but I cannot understand its connection to the usual approach. The RG in Wilson's approach is based on the variation of the cut-off, in the usual approach it's based on the variation of the renormalization point.

Thanks

2. Jan 9, 2016

### vanhees71

You don't describe the system by the bare Lagrangian but by the entire effective action, which is finite (order by order in perturbation theory). The classical action (integral over the Lagrangian) + the counter terms + the radiative corrections give a finite action functional (order by order in perturbation theory), and from this you calculate the observables (in terms of S-matrix elements), which are also finite and comparable to experiment.

An alternative treatment of quantum field theory, which doesn't involve infinities is the Epstein-Glaser approach. A textbook following this approach for QED is

G. Scharf, Finite Quantum Electrodynamics, The Causal Approach, Springer (1995)

3. Jan 10, 2016

### Staff: Mentor

I don't get the conventional approach either:
http://arxiv.org/pdf/1208.4700.pdf
'Roughly speaking, the program of removing the infinities from physically measurable quantities in relativistic field theory, the renormalization program, involves shuffling all the divergences into bare quantities. In other words, we can redefine the unmeasurable quantities to absorb the divergences so that the physically measurable quantities are finite'

This shuffling infinities around leaves me cold. Only Wilson's approach makes sense to me. I simply accept that and move on.

Thanks
Bill

Last edited: Jan 10, 2016
4. Jan 10, 2016

### atyy

As far as I understand, it's a miracle. Actually, I think even the Wilsonian approach has some miracles, like the ε-expansion. The strength of the Wilsonian approach is that the heuristics are so obviously right, we believe the mathematicians should be able to make it non-miraculous some day.

5. Jan 10, 2016

### vanhees71

Well, the conventional approach using appropriate regularizations and then subtracting infinities within a given renormalization scheme is much more convenient in perturbative calculations. The Wilson approach elucidates the physics behind the conventional renormalization procedure. There no contradiction between the two approaches.

6. Jan 10, 2016

### Staff: Mentor

I have read all over the place where they are equivalent, so I don't doubt it. Its just at my current level I haven't seen the detail of that claim.

Thanks
Bill

7. Jan 10, 2016

### A. Neumaier

The closest existing approximation to what you want is described in my insight article.

See also my paper Renormalization without infinities - a tutorial, which discusses renormalization on a much simpler level than quantum field theory.