- #1
Luca_Mantani
- 33
- 1
Hi,
I am computing the anomalous dimension of a mass operator in the MSbar scheme, but i have a doubt. The following is the solution of an exercise given by a professor but i don't understand a passage. I have computed the counterterm ##\delta## and i have the formula
$$\gamma=-\mu \frac{d\delta}{d\mu}$$
Substituting my calculation and dropping all the constants (which i need but are not relevant for the question) i have
$$\gamma\propto \mu \frac{de^2(\mu)}{d\mu}\frac{1}{\epsilon}$$
Then I use the formula
$$\mu \frac{de(\mu)}{d\mu}=-\epsilon e + \beta(e)$$
The solution says that if I substitute and drop the ##\beta(e)## i get the finite result we need, since the ##\epsilon## in the numerator cancels the one in the denominator. But why we drop the ##\beta## term, which is finite? If we keep it, we are not able to cancel the ##1/\epsilon## and we get a divergent term, even if we know that the anomalous dimension is finite. What am i missing in this calculation?
Thanks in advance for the help!
I am computing the anomalous dimension of a mass operator in the MSbar scheme, but i have a doubt. The following is the solution of an exercise given by a professor but i don't understand a passage. I have computed the counterterm ##\delta## and i have the formula
$$\gamma=-\mu \frac{d\delta}{d\mu}$$
Substituting my calculation and dropping all the constants (which i need but are not relevant for the question) i have
$$\gamma\propto \mu \frac{de^2(\mu)}{d\mu}\frac{1}{\epsilon}$$
Then I use the formula
$$\mu \frac{de(\mu)}{d\mu}=-\epsilon e + \beta(e)$$
The solution says that if I substitute and drop the ##\beta(e)## i get the finite result we need, since the ##\epsilon## in the numerator cancels the one in the denominator. But why we drop the ##\beta## term, which is finite? If we keep it, we are not able to cancel the ##1/\epsilon## and we get a divergent term, even if we know that the anomalous dimension is finite. What am i missing in this calculation?
Thanks in advance for the help!