Consider the scattering of a particle by a central time-independent potential $V$, that is limited to a finite region in the space. the hamiltonian is $H = H_{0} + V$. Using the Moller operators, we can express the state of the particle at time $t = 0$: $$|\psi \rangle = \Omega_{+} |\psi_{in} \rangle = \lim_{t \rightarrow -\infty} U^{\dagger}(t) U_{0}(t) |\psi_{in} \rangle,$$ where $|\psi_{in} \rangle$ is an asymptotic state and $U(t) = e^{-iHt}$, $\ U_{0}(t) = e^{-iH_{0}t}$ are time evolution operators. Deriving and integrating $U^{\dagger}(t) U_{0}(t)$, it can be shown that you can write $$|\psi \rangle = |\psi_{in} \rangle + i \int_{0}^{\infty} d\tau \ U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle.$$ The last integral is convergent and absolutely convergent, so it can be repleced by $$\lim_{\epsilon \rightarrow 0^{+}}\int_{0}^{\infty} d\tau \ e^{-\epsilon \tau} U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle.$$ My questions are:

1) How we know that this integral converge? it is not obvious to me.

2) What is the need to use this $\epsilon$? It always appear in scattering theory, why is it important?


1 Answer 1


Your potential function $V$ is time independent, and is limited to a finite region of space. That means that $V\in[a,b]$ where $a\geq0$, $b\geq0$. The result is that

$$|\psi \rangle = |\psi_{in} \rangle + i \int_{0}^{\infty} d\tau \ U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle$$

can be written as

$$|\psi \rangle = |\psi_{in} \rangle + i \int_{a}^{b} d\tau \ U^{\dagger}(\tau)V U_{0}(\tau) |\psi_{in} \rangle$$

because for all $\tau\notin[a,b]$ $V=0$. I make the assumption that none of the functions in your integrand are singular anywhere on the interval $[a,b]$. If this is all true, then the integrand is zero.


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