On-shell renormalization scheme

  • #1
16
0

Homework Statement


Show that, after considering all 1 particle irreducible diagrams, the bare scalar propagator becomes:

$$D_F (p)=\frac{i}{p^2-m^2-\Sigma (p^2)}$$

And that the residue of the pole is shifted to a new value, and beomes:

$$i\bigg(1-\frac{d\Sigma(p^2)}{dp^2}\bigg)^{-1}\bigg|_{p=\tilde{p}}$$

2. The attempt at a solution
Part a is no problem, I simply express the corrections as a geometric series in the standard way and find:

$$D_F(p)=D_0 (p) + D_0 (p) (-i\Sigma(p^2)D_0 (p)\ \dots\ = D_0 (p) \sum_{n=0}^\infty \big(-i\Sigma(p^2)D_0(p)\big)^n$$
$$=\frac{D_0 (p)}{1+i\Sigma(p^2)D_0 (p)}=\frac{i}{p^2-m^2-\Sigma(p^2)}$$

Where I have labelled the bare propagator as D0

Part b is also okay in terms of calculation. I find:

$$\tilde{p}^2=\Sigma(\tilde{p}^2)+m^2$$

As the location of the new pole. If I expand the denominator of the propagator around this new pole I find:

$$p^2-m^2-\Sigma(p^2)\approx 0+\frac{d}{dp^2}\bigg(p^2-m^2-\Sigma(p^2)\bigg)\bigg|_{p^2=\tilde{p}^2}\times (p^2-\tilde{p}^2) + \mathcal{O}\big((p^2-\tilde{p}^2)^2\big)$$
$$\implies D_F(p)\approx i\Bigg((p^2-\tilde{p}^2)\bigg(1-\frac{d\Sigma}{dp^2}\bigg|_{p^2=\tilde{p}^2}\bigg)\Bigg)^{-1}$$

The residue is therefore given by:

$$\mathrm{Res}(D_F(p),\tilde{p})=\lim_{p\rightarrow\tilde{p}}\bigg(D_F(p)\times(p^2-\tilde{p}^2\bigg)=i\bigg(1-\frac{d\Sigma(p^2)}{dp^2}\bigg)^{-1}\bigg|_{p=\tilde{p}}$$

As required.

Now my question is this: I have read that in the on-shell renormalization scheme we require:

$$\Sigma(p^2=m^2)=0,\ \ \ \ \ \ \frac{\partial \Sigma}{\partial p^2}=0$$

In order to ensure that the exact all-order two point function has a residue of 1 at the physical mass of the particle, m. However, quite simply, if I use this condition in my expression for the residue, I find a residue of i and not 1. This seems like a small, almost trivial issue, however it would lead to different counterterms if I were to try and demand that the residue by 1, as I would need to impose

$$\frac{\partial\Sigma}{\partial p^2}=1-i$$.

I can clearly see that if the propagator were defined without the factor of i in the numerator then this problem would disappear, is this "allowed"? Can we be so cavalier with the complex phase factor of the propagator?
 

Answers and Replies

  • #2
king vitamin
Science Advisor
Gold Member
478
230
I would guess that either you or your textbook mixed up two different very common notations. Some sources define the Green's function you've specified as ##i G## rather than ##G## itself, while some specify the Green's function as you say, but the renormalization condition is that the residue is fixed at ##i##. Finally, some sources like to work with Euclidean theories, which gets rid of the ##i##.

As a rule of thumb, the on-shell scheme simply tells you to keep the pole at the physical mass and to fix the residue at the tree-level value, whatever conventions you're using. There are way too many similar-yet-different conventions for QFT.
 

Related Threads on On-shell renormalization scheme

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
930
Replies
2
Views
898
Replies
1
Views
1K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
4
Views
935
Top