- #1
Aleolomorfo
- 73
- 4
Hello everybody!
I have a big question about the renormalization: I do not understand why the "renormalization condition" is to impose the tree level result. Now I will explain it better.
Let's take, for example, the electron self energy. The tree-level contribution is the simple fermionic propagator ##\frac{i}{\displaystyle{\not}p-m}##.
If I calculate the complete propagator (using the sum of all 1-particle-irreducible), the result is
$$\frac{i}{\displaystyle{\not}p-m-\Sigma(\displaystyle{\not}p,m)}$$
which is like the tree level result but the pole is shifted.
Renormalizing the electron self energy implies to redefine the electron mass. I define the "renormalized mass" as the position of the pole in the propagator. Consequently, to define the "physical measurable mass" I need to find the pole:
$$\displaystyle{\not}p-m-\Sigma(\displaystyle{\not}p,m)|_{\displaystyle{\not}p=m_R}=0$$
After finding the renormalized mass I Taylor-expand the full propagator arounf ##m_R## and to find exactly the tree level result (mutatis mutandis with the renormalized mass) I will need also to redefine the electronic field with ##Z_2##.
If I am not wrong this is the idea behind the renormalization of the electron self-energy. However, I do not understand why I want the tree level result and I redefine things to obtaint it. For example, why the "physical mass" , the one that I measure in an experiment, is the shifted pole of the propagator? I think I understand how to do the things but I do not understand why.
Thanks in advance!
I have a big question about the renormalization: I do not understand why the "renormalization condition" is to impose the tree level result. Now I will explain it better.
Let's take, for example, the electron self energy. The tree-level contribution is the simple fermionic propagator ##\frac{i}{\displaystyle{\not}p-m}##.
If I calculate the complete propagator (using the sum of all 1-particle-irreducible), the result is
$$\frac{i}{\displaystyle{\not}p-m-\Sigma(\displaystyle{\not}p,m)}$$
which is like the tree level result but the pole is shifted.
Renormalizing the electron self energy implies to redefine the electron mass. I define the "renormalized mass" as the position of the pole in the propagator. Consequently, to define the "physical measurable mass" I need to find the pole:
$$\displaystyle{\not}p-m-\Sigma(\displaystyle{\not}p,m)|_{\displaystyle{\not}p=m_R}=0$$
After finding the renormalized mass I Taylor-expand the full propagator arounf ##m_R## and to find exactly the tree level result (mutatis mutandis with the renormalized mass) I will need also to redefine the electronic field with ##Z_2##.
If I am not wrong this is the idea behind the renormalization of the electron self-energy. However, I do not understand why I want the tree level result and I redefine things to obtaint it. For example, why the "physical mass" , the one that I measure in an experiment, is the shifted pole of the propagator? I think I understand how to do the things but I do not understand why.
Thanks in advance!