# Addition of Complex Term in Lippmann Schwinger Equation

1. Jan 31, 2012

### richard wakefield

Hello PF People,

This is probably a very simple question but I don't really get it.

In Lippmann Schwinger Equation we add an infinitesimal term to the denominator in order to avoid singularity for when E is an eigenvalue of $\hat{H}_0$. This is fine, but why it has to be a complex number? Adding a real constant will also save denominator from becoming singular. That confuses me.

2. Feb 1, 2012

### Tarantinism

Nobody? It seemed to be a good question.

My point of view: I guess the reason is to avoid more singularities, as could be the case of continuous spectra. Knowing that the limit $\epsilon \rightarrow 0$ would be the same in all directions (imaginary, real or "mixed complex")

There are no complex energies in the spectrum, so, that should be "safer".

3. Feb 1, 2012

### Bill_K

The eigenvalue spectrum of H0 is continuous, all real numbers E0 > 0. In order to give meaning to the inverse operator (E - H0)-1 it is necessary to avoid the entire positive real axis. The fact that we use +iε rather than -iε has to do with the outgoing boundary conditions.

4. Feb 1, 2012

### Tarantinism

Ahh, i forgot that was indeed the free hamiltonian, then seems obvious. I have my scattering theory rusty, damn it.