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Consider the Lippmann-Schwinger equation (LSE)

$$ |\psi\rangle = |\phi\rangle + \hat{G}_0(\epsilon) \hat{V} |\psi\rangle \tag{1}$$

where $\hat{G}_0(\epsilon) = \frac{1}{\epsilon - \hat{H}_0 + i\eta}$.

If $|\phi\rangle$ is an incoming state, how do I determine the corresponding time-dependent incoming state? Is it simply $\hat{U}_0(t)|\phi\rangle$? Here $\hat{U}_0(t) = e^{-it\hat{H}_0}$.

If that is the case, then what is the time-dependent scattered state? Is it $\hat{U}_0(t)|\psi\rangle$? Why?

I also know that $\hat{G}_0(\epsilon)$ is the Fourier image of $\hat{G}_0(t)$ (see e.g. this Phys.SE post). Maybe I should somehow apply inverse Fourier transform to LSE to get time-dependent functions?

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Minethlos
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  • The idea here is that the "incoming state" $|\phi\rangle$ is defined in an asymptotic region far away from the scattering potential $V$, or (better) far back in time, where/when the dynamics is described by the unperturbed Hamiltonian $H_0$ only. By definition $|\phi\rangle$ is also an eigenstate of $H_0$, so its asymptotic dynamics is trivial, although technically it is indeed ${\hat U}_0(t)|\phi\rangle$. Similarly, the scattered state $|\psi\rangle$ "resides" in the scattering region defined by $V$, where the dynamics is given by the perturbed Hamiltonian $H_0+V$. – udrv Apr 11 '16 at 22:36
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    The scattered $|\psi\rangle$ is also, by definition, an eigenstate of $H_0 +V$, so again its dynamics is trivial, but it is given by $e^{-it(H_0+V)t}|\psi\rangle$. True time-dependent dynamics arises only for wave packets constructed out of the $|\phi\rangle$-s and the corresponding $|\psi\rangle$-s, see https://en.wikipedia.org/wiki/Lippmann–Schwinger_equation#Creating_wavepackets on the Wikipedia page you cited. – udrv Apr 11 '16 at 22:36
  • @udrv Indeed, I hadn't read that paragraph of the wiki article. It answers many of my questions. So if I choose $\phi_g (t)$ as the incoming state, is $\psi_g (t)$ the scattered state? Why? Or in what sense is it the scattered state? – Minethlos Apr 11 '16 at 23:22
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    Yep, $\phi_g(t)$ generates $\psi_g(t)$. Note that the LSE actually generates a series expansion for $|\psi\rangle$ by iterative use of itself in the rhs: $$ |\psi\rangle = |\phi\rangle + {\hat G}_0(\epsilon) {\hat V}|\psi\rangle = |\phi\rangle + {\hat G}_0(\epsilon) {\hat V} |\phi\rangle + \left[{\hat G}_0(\epsilon) {\hat V} \right]^2|\psi\rangle = \dots = \ = \left[ {\hat I} + {\hat G}_0(\epsilon) {\hat V} + \left[{\hat G}_0(\epsilon) {\hat V} \right]^2 + \dots + \left[{\hat G}_0(\epsilon) {\hat V} \right]^n +\dots \right]|\phi\rangle = {\hat S};|\phi\rangle $$ – udrv Apr 12 '16 at 03:12
  • where the scattering operator $\hat S$ is linear, so indep. of $|\phi\rangle$. This makes it obvious that $\hat S$ maps any superposition $\phi_g$ into the corresponding superposition $\psi_g$. – udrv Apr 12 '16 at 03:12

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