Name: V. Shtabovenko Date: 03/25/20-11:13:13 PM Z
Hi,
I agree that the phi phi -> phi phi example is not consistent with
the
renormalization example. Moreover, it is even incorrect, since the
using
\delta_g from Peskin the singularity doesn’t cancel (due to a typo in
a
code this issue is not detected). This should be now corrected,
thanks.
The reason why one should use $\delta_g=\frac{3\Delta
g}{32\pi^2}$
(from the renormalization example) and not
$\frac{3\Delta g^2}{32\pi^2}$ (as in Peskin) has to do with the way
you
set up your counter-term Lagrangian. This is IMHO often not written in
a
very clear in the text books.
Eq 2.1 is arXiv:1606.09210 provides a much better starting point. If
you
solve Eqs. 2.2 for Z_2, Z_1 and Z_4 and substitute that back into Eq
2.1 you
will get the renormalized Lagrangian that I used in my FeynRules model
https://github.com/FeynCalc/feyncalc/blob/master/FeynCalc/Examples/FeynRules/Phi4/Phi4.fr
The count-term piece for the vertex has an explicit power of the
coupling, therefore $\delta_g$ will contain only g and not g^2.
What is usually done in most text books or pedagogical lecture notes
is
to introduce a wave function renormalization constant Z but keep m and
g
bare (at first).
Then you have
L = Z (del_mu phi)^2 - 1/2 Z m_0^2 phi^2 - Z^2 g_0 / 4! phi^4
Now instead of writing m_0^2 = Z_m m^2, g_0 = Z_g g (as one
normally
does when renormalizing a new model from scratch) they prefer to have
Z m_0 = m^2 + delta m^2
Z^2 g_0 = g + delta g
where delta m^2 and delta g must cancel your poles.
Obviously, if you calculate things this way, your delta g will have
two
powers of g, since the CT vertex contains only delta g and no explicit
powers of g.
There is of course nothing wrong with the way it is done in the text
books,
but you have to be careful when comparing the renormalization
constants
from different sources. When in doubt, check how they explicitly define
the
renormalized Lagrangian.
Cheers,
Vladyslav
Am 25.03.20 um 21:07 schrieb Huan Souza:
> Hi,
>
> I was studying the $phi^4$ theory renormalisation (MS,MSbar)
> example, kindly made available by you. However, the result for the
g
> counterterm is a bit strange to me ($\delta_g=\frac{3\Delta
> g}{32\pi^2}). In the other example you made available,
> $\phi\phi$->$\phi\phi$, the value used was
$\delta_g=\frac{3\Delta
> g^2}{32\pi^2}$ (which is the value showed in Peskin and
Schroeder,
> eq. 12.45). In this way, could you enlighten me about this
> divergence between the results?
>
> Since now, thank you for your time and patience.
>
> Best regards,
>
> Huan Souza.
>