I have a question on the interpretation of the LS-equation.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\psi^{\pm} = \phi + \frac{1}{E- \mathcal{H} \pm \epsilon} V\psi^{\pm}[/tex] (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

[tex]H_0|\phi> = E|\phi>[/tex]

So an interaction in the equation makes

[tex](H_0+V)|\psi> = E|\psi>[/tex]

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

[tex]\psi^{\pm} = \phi + \frac{1}{E- \mathcal{H} \pm \epsilon} V\psi^{\pm}[/tex]

was imaginary.

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

[tex]\psi^{\pm} = \phi + \frac{1}{E- \mathcal{H} \pm \epsilon} V\psi^{\pm}[/tex]

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?

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# Lippmann-Schwinger equation

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