# General renormalization

1. Mar 26, 2012

### anthony2005

Hello everyone.
in standard approach to QFT, you study fields, S matrix, and you get a perturbative expression. You see that at higher terms you find infinities and so you renormalize.
Now, Weinberg states that the renormalization procedure of mass and fields must be done even with no infinities. I like this approach.
I want to have a general idea of renormalization, starting from the basic assumptions of QFT, without considering infinites or perturbation theory.
So we postulate that for a field theory (scalar for semplicity) it must be $<0|\phi\left(0\right)|1>=1$, where $|1>$ is a one-particle state (its existence is also postulated) and $|0>$ the vacuum. Then we also require that the mass operator $P^{2}$ has the physical mass $m^{2}$ as eigenvalue on $|1>$. Via the Källén–Lehmann decomposition we see that this $m^{2}$ is an isolated pole of the 2 point Green function.
In order to satisfy these two assumptions (not obvious in an interacting field), fields and mass must be rescaled, or renormalized, so we get a renormalized Lagrangian with the renormalized fields and masses, and containing counterterms. This is so done without having to deal with perturbation theory.
Will this automaticaly induce finite results in perturbation theory? A rescaling of the coupling constant of the theory is missing in this approach, and can this be done in a general non-perturbative way, as just did for mass and fields?

Any reference who treats QFT and renormalization in the most general possibile way are welcome.
Thank you

Last edited: Mar 26, 2012
2. Mar 26, 2012

### anthony2005

Sorry, I'm editing since I have one more question. When Weinberg at page 439 vol I writes down the renormalized scalar Lagrangian, what happens when the interaction $V\rightarrow0$? It should be that $Z\rightarrow1$ and $\delta m\rightarrow0$ so that $\phi \rightarrow\phi_{B}$ and $m\rightarrow m_{B}$, and so there is no difference in this case if using $m$ or $m_{0}$.
In the scattering theory, in the asymptotic states, what happens to the renormalized Lagrangian? I guess that you cannot put $V\rightarrow0$ for $t\rightarrow\pm \infty$, since $V$ contains the self-interaction term. What happens then, what limit is taken to the Lagrangian in order to get $\left(\square+m^{2}\right)\phi_{in/out}=0$ asymptotically?
Thank you.

3. Mar 28, 2012

### anthony2005

No answers? No need for detailed description, just a good reference is ok, i'll study from there..thanks

4. Mar 28, 2012

### Avodyne

Yes.
Yes.

See the text by Srednicki (draft version available free at his web page) for details.

5. Mar 29, 2012

### tom.stoer

I am not so happy to start a discussion regarding renormalization always with these infinities. Renormalization has a much broader (and mathematically more exact) meaning than that. The renormalization group describes how certain parameters and observables change when changing the (length or energy) scale of reference (in a calculation or an experiment).

6. Mar 29, 2012

### anthony2005

Ohhh thank you very much!!! I will read Srednicki, it seems pretty good.
I definitely agree with you, I always find the sentence "In order to make this integral finite we introduce.." and I hate this way, I think it's not the way one should deal with physics. And after, they introduce the μ stating that it must be introduced in order to cure the infinities, again, don't like that way. And at the end they talk about the renormalization group, which is how couplings (and Green functions etc..) change with this renormalization scale μ, but not how it changes with a coordinate scale $x \rightarrow xs$.
Nair talks about the chaning of the Green function with scale only after talking about this μ, and gets $G\left(e^{s}x_{1},...e^{s}x_{n},\lambda\right) = G\left(x_{1},...x_{n},\lambda'\right)\exp\left(\int_{\lambda}^{\lambda'}\frac{\Delta\left(u\right)}{\beta\left(u\right)}du\right)$
Sorry that $\in t$ is $\int$.
He then says (pag 174) "The effect of scaling the coordinates x is obtained by replacing the coupling constants in the Green’s function by the running coupling constant λ' and then there is an extra exponential factor. So I can make the constant coupling run only scaling the coordinates, without having to introduce the μ. But that expression of the Green function has the beta function inside, which was derived with the renormalization scaling. Is there a way to treat quantum field theory with the coordinate (or momenta) scale and not the renormalization one?
There is the paper "Eliminating infinities in the λϕ4 theory by simple scaling" by Cristiane M.L. de Aragãoa, C.E.I. Carneirob which seems answers what I am asking, but I can't read it, it's not for free.
Thank you

Last edited: Mar 29, 2012
7. Mar 29, 2012

### francesco85

Good afternoon to everybody! I would like to ask a question about what you have said: do you mean that once we find a wave-function renormalization such that $<0|\phi\left(0\right)|1>=1$ and that the two-point function spectral function has a pole at a particular $m^2$ then we have all the correlation functions finite?
Best,
Francesco

8. Mar 29, 2012

### anthony2005

That is exactly what I was asking. But I don't think all correlation functions would be finite, since there is still the coupling renormalization missing (no $\Delta\lambda$ counterterms). If they were finite and resolved all our problems I wouldn't know why Weinberg continued dealing with renormalization and fixing UV divergences in the following chapters. Anyway, lets hope an expert has a real answer :)

9. Mar 29, 2012

### francesco85

Ok, actually I agree :) .
So, maybe I have missed what Avodyne means when he answered yes to the question "Will this automaticaly induce finite results in perturbation theory?". Can you explain me please?
Best,
Francesco

10. Mar 29, 2012

### anthony2005

Hey, me again, maybe just those two conditions would give finite results at least in the UV, I've just realized that, for the $\phi^{4}$ theory, mass and field renormalization induces the coupling renormalization since $g_{b}\phi_{b}^{4} = g_{b}Z^{2}\phi^{4} = g\phi^{4}+\left(g_{b}Z^{2}-g^{4}\right)\phi^{4}$ and so there is the interaction term + the counterterm.

11. Mar 29, 2012

### francesco85

I think I don't understand: let us take the phi4 theory and let us call $\Sigma (p^2)$ the 1PI part of the two point correlation function which includes also the tree level term $m_b^2$. In the following the physical parameters are m_ph and \g_ph.
Then I think that we have that the two conditions we impose on the two point function (the physical mass as a pole and the Z chosen in the physical way) give only one relation among m_ph, m_b and g_b:
$\Sigma(m_{ph}^2)=m_{ph}^2$
The other condition yields a formula for Z in terms of m_ph, g_b and m_b (it does not add any information on g_b and m_b).
Basically we have g_b as a function of m_b and m_ph; the question is: is the 4-point function finite however we choose m_b and m_ph?

Last edited: Mar 29, 2012
12. Mar 30, 2012

### Avodyne

Sorry, I did not answer very precisely. It is also necessary to define the coupling constant in terms of a correlation function or scattering amplitude to get finite results. Again I recommend Srednicki as a place where all this is explained in great detail (for phi^3 theory in 6 dimensions!).