# Scalar Propagator form with width

1. Aug 1, 2014

### ChrisVer

I would like to ask when can someone add the width in a scalar particle's propagator. In general the scalar propagator can be:
$\frac{1}{k^{2}-m^{2}+i \epsilon} (\epsilon \rightarrow 0)$
However I read somewhere that if necessary one can include a width for the propagator:

$\frac{1}{k^{2}-m^{2}+i 0} \rightarrow \frac{1}{k^{2}-m^{2}+i m \Gamma}$

my question is when is it necessary and why?

2. Aug 1, 2014

### Einj

In principle every propagator should contain the width. However, for stable particles $\Gamma=0$, while for "long-living" particles $\Gamma$ is very small and hence one usually includes the width just for short living particles (like resonances).

The quantum mechanical reason for why the width should be there is that, in principle, the Hamiltonian (energy) of the system could always have an absorbitive part causing the decay of your particle.

A pretty good treatment of decays and widths can be found in - De Wit "Field theory in particle physics".

3. Aug 1, 2014

### ChrisVer

My problem is that in QFT I didn't see any reason to include the width anywhere.
For example take a theoretical $\Phi \phi^{2}$ theory ($L_{tri}= \lambda_{3}\Phi \phi^{2}$) which can give you an s-channel with the decay of $\phi,\phi$ into $\Phi$ and the last into $\phi,\phi$ again... the scheme is:
>---< $\phi (p_{1}) \phi(p_2) \rightarrow \Phi(k) \rightarrow \phi(p_3) \phi(p_4)$
In this case, the propagator comes out from the Feynman rules and is as I gave it in the first expression in my OP.
$\frac{1}{k^{2}-M_{\Phi}^{2} +i \epsilon}$
with $k$ the momentum of $\Phi$ intermediate state which you integrate.
Is there something wrong or missing in the above?

The same can also work for t or u channel

Last edited: Aug 1, 2014
4. Aug 1, 2014

### Einj

You propagator is the bare one. When you consider the full theory, i.e. including all the possible loop diagrams the propagator gets a correction of the kind:
$$\Delta =\frac{1}{p^2-m^2+i\Sigma(p)}.$$
If $\Sigma(p)$ only causes a shift on the real $p^0$-axis then the behavior of your propagator doesn't change qualitatively. However, if this correction induces also a shift in the imaginary axis then you have your width.

5. Aug 1, 2014

### ChrisVer

Thank you .
I hope the source you mentioned explains everything nicely, I'll look for it, because your last paragraph confused me a little.

6. Aug 1, 2014

### Einj

Yes, it does. I'm sorry it wasn't clear, you can find the same line of thought on the book I mentioned :D

7. Aug 1, 2014

### ChrisVer

No it wasn't unclear, I just didn't understand the "shift"-thing for the p0 Real or Im axis...

8. Aug 1, 2014

### Einj

Well, that's easy. Suppose that $\Sigma=Re\Sigma+iIm\Sigma$. Then your propagator becomes:
$$\frac{1}{p^2-m^2+iRe\Sigma-Im\Sigma}.$$
As you can see, the imaginary part of $\Sigma$ causes a shift on the real axis of your pole. However, now you can define $Re\Sigma = m\Gamma$ and you have your width.

9. Aug 1, 2014

### The_Duck

Another source you could look at is Srednicki Chapter 25, which works this out in an explicit calculation.