Why is an imaginary part necessary in the Lippmann-Schwinger equation?

[Sparticle99]

I have a question on the interpretation of the LS-equation.

$$\psi^{\pm} = \phi + \frac{1}{E- \mathcal{H} \pm \epsilon} V\psi^{\pm}$$ (1)

Deriving the equation was simply because the Hamiltonian can be written as H=H0 + V where H and H0 have the same eigenvalues and H0 is a free Hamiltonian. The equation can be written as

$$H_0|\phi> = E|\phi>$$

So an interaction in the equation makes

$$(H_0+V)|\psi> = E|\psi>$$

Continuity states that as $$\psi rightarrow \phi$$ and $$V \rightarrow 0$$ then the solution has (E − H0) as singular. You can avoid the singularity when you assume a small part of the denominator in

$$\psi^{\pm} = \phi + \frac{1}{E- \mathcal{H} \pm \epsilon} V\psi^{\pm}$$

was imaginary. My question is as follows: This would clearly make the Hamiltonian as negative. The solution to the equation almost derives

$$\psi^{\pm} = \phi + \frac{1}{E- \mathcal{H} \pm \epsilon} V\psi^{\pm}$$

as an automatic solution when considering it was obtained to avoid nasty singularities. Is there any physical reason to assume the Hamiltonian has an imaginary part?

 

[vanhees71]

It is easier to understand scattering theory in the time-dependent wave-packet formalism than going directly to the Lippmann-Schwinger equation in momentum space by the usual hand-waving arguments. A quite good discussion of this issue is given in the textbook by Messiah.

 

[Bill_K]

The purpose of the iε is to insure that the solution contains only outgoing waves. If you look at the matrix elements of the kernel in the coordinate representation, it is just the Green's function:

<x'|(E - H₀ ± iε)^-1|x"> ~ e^{±ip·|x'-x"|/ħ} / |x'-x"|

 

[Sparticle99]

Before obtaining the solution, it would be fair to say your Hamilatonian is by definition, complex no? I will elaborate: brb

 

[Sparticle99]

$$\psi^{\pm} = \phi + \frac{1}{E- \mathcal{H} \pm i\epsilon} e\phi'\psi^{\pm}$$

As has been mentioned, the imaginary part is there so that the solution will have out-going waves only. But that takes manipulatin of the L-S equation. Before a solution is sought, the integral expression the wave function expands to the infinite future and infinite past.

since $$\hat{\mathcal{H}}_D = \mathcal{H} + i\chi$$ from breaking the complex part into two descriptions as shown
before, will allow us to treat the Lippmann-Schwinger as having two solutions as well for the imaginary part of the hat-Hamiltonian.

Knowing that $$\chi = \chi_R + \chi_D$$

Then

$$\hat{\mathcal{H}} = \mathcal{H} + i(\chi_R + \chi_D)$$

which is making the relaxational part and the diffusive part.

$$\psi^{\pm} = \phi + \frac{1}{E- {\mathcal{H}}+i(\chi_R + \chi_D)} e\phi'\psi^{\pm}$$

Since in the most mathematical formal definition of quantum waves desribed under a time-symmetric analysis of the wave function
in the Transactional Interpretation, the deformed wave function $$\psi^{\pm}$$ when under integration has the solution
of a wave function for the in-phase and out-phase both extending to the infinite past and infinite future. These are analgous
to understanding wave functions where one half retarded and one half advanced wave propagate into the past and the other into
the future. This means that the equation

$$\psi^{\pm} = \phi + \frac{1}{E- {\mathcal{H}} + i(\chi_R + \chi_D)} e\phi'\psi^{\pm}$$

can be decoupled to make to solution for the advanced and retarded wave solutions. This means before messing with the Lippmann Schwinger equation, it can be used to satisfy an imaginary Hamiltonian with a wave function under the mathematical representation of the Transactional Interpretation.