# A What is a Mass Independent Renormalization Scheme?

1. Aug 4, 2016

### Luca_Mantani

Hi everyone,
i'm not able to find the exact definition of mass independent renormalization scheme. I often read that the MS-bar scheme is mass independent, but why? And why this feature help us to compute the beta function?

Luca

2. Aug 4, 2016

### vanhees71

The $\overline{\mathrm{MS}}$ scheme is defined within dimensional regularization, which is very elegant from the calculational point of view, but it's not very intuitive.

As the most simple example take simple $\phi^4$ theory with the Lagrangian
$$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m^2}{2} \phi^2 -\frac{\lambda}{4!} \phi^4,$$
which is renormalizable. The divergent parts are (besides the here not so interesting vacuum diagrams) the self-energy (leading to wave-function and mass renormalization), and the four-point function (leading to coupling-constant renormalization).

The counterterms and the renormalization scheme are thus determined by these divergent pieces. One can write the counter-term Lagrangian in the form
$$\delta \mathcal{L} = \delta Z \frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi) - \frac{1}{2} (\delta m^2 + m^2 \delta Z_m) \phi^2 - \frac{\delta \lambda}{4!} \phi^4.$$
Here the $\delta Z$, $\delta Z_m$ and $\delta \lambda$ are all dimensionless (i.e., of energy dimension 0). Only $\delta m^2$ has dimension 2.

The most intuitive MIR scheme is defined by introducing a mass scale $M$ as the renormalization scale and define the renormalized quantities by the renormalization conditions
$$\Sigma(p^2=0,m^2=0)=0 \; \Rightarrow \; \delta m^2=0.$$
This is allowed, because that's a quadratically divergent quantity and thus IR safe. So you can define it at $m^2=0$.

All other divergences are logarithmic and thus cannot be defined at $m^2=0$ and all external momenta at 0. Thus one defines them at $m^2=M^2$, i.e.,
$$[\partial_{p^2} \Sigma(p^2,m^2)]_{p^2=0,m^2=M^2}=0 \; \Rightarrow \; \delta Z,$$
$$[\partial_{m^2} \Sigma(p^2,m^2)]_{p^2=0,m^2=M^2}=0 \; \Rightarrow \; \delta Z_m,$$
$$\Gamma^{(4)}(s=t=u=0,m^2=M^2)=-\lambda \; \Rightarrow \; \delta \lambda.$$
All the counter terms are thus only dependent on the renormalization scale $M^2$ via the dimensionless renormalized coupling constant $\lambda$, but not on $m^2$. That's why this scheme is a mass-independent renormalization scheme, and you have $\delta Z=\delta Z(\lambda)$, $\delta Z_{m}=\delta Z_m(\lambda)$. The RG parameters like $\beta$ don't depend on $m/M$ but only on $M$ through $\lambda$, and you get a homogeneous RG equation that is not so difficult to solve.

For details, see my QFT manuscript

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

Chpt. 5.

The derivation of the $\beta$ function and other RG equation coefficients within the MIR, MS (or $\overline{\text{MS}}$) scheme in the context of dim. reg. see Sect. 5.11. In Sect. 5.12 also other ("non-MIR") schemes are treated. The corresponding RG equations are more difficult since the RG coefficients do not only depend on the renormalization scale $M$ via the dimensionless renormalized coupling $\lambda$ but also explicitly via $m/M$.

3. Aug 4, 2016

### A. Neumaier

Some misprints:

p.129: /2: positron, /8: Coulomb, /middle: hints->it hints, neglects->neglect

p.130: nonuniform capitalizations of 'quantum field theory' - I'd use no capitals at all. /mid: principally->in principle. You should mention that the scheme independence is only after summing all contributions; at low order there may be big differences and the mass scale matters. /-2: nowadays->today's

p.131: /11: loose->lose, /13: especially->in particular

Last edited: Aug 4, 2016
4. Aug 4, 2016

### Luca_Mantani

Thank you very much for the explanation! I'll get a look at the lectures but i think i grasped the basic idea behind it, thank you again.

5. Aug 5, 2016

### vanhees71

Thanks a lot. I corrected all typos. Where would you put the comment on scheme independence? It's of course the point of the RG equations to resum to leading logarithmic order to achieve this (approximate) independence in perturbation theory where it is applicable, i.e., for small coupling.

6. Aug 5, 2016

### A. Neumaier

On p.130 where you say ''this dependence should change nothing with respect to S-matrix elements'', or perhaps one paragraph later. If necessary with a forward reference to the resumming.

There is still a ''quantum Field theory" and a "quantum Field Theory" alongside the preferable ''quantum field theory". Probably also on other pages.

7. Aug 5, 2016

### vanhees71

Thanks again. I hope, now it's better.