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This is a quite basic question but I confess it is something I didn't get up to this point.

When defining the Moller operators and hence the $\cal{S}$-matrix one usually considers "states $\Psi$ evolving with the full interacting theory" and "states $\Psi_0$ evolving with the free theory". This is alluded to in this answer.

This paper also makes this clear:

Typically, one is interested in the overlap of scattering state, i.e., true eigenstates of the full Hamiltonian. Since these are usually unavailable, one resorts to descriptor states and an operator in such states - the $\cal{S}$ matrix - that describes scattering and can be expanded in a perturbative series.

But I'm surelly missing something here. I mean, given any dynamics $W(t)$ be it either $U(t)$ or $U_0(t)$ or any other, we can evolve any state with it.

I mean, we can pretty much consider $U_0(t)\Psi$ or $U(t)\Psi_0$. This makes me wonder what people precisely mean with states evolving with the interacting/free theory.

The guess, from the quotation of the paper, is that they mean "eigenstates of the full hamiltonian" and "eigenstates of the free hamiltonian".

But then there's something wrong in my understanding. I mean, if $\Psi$ is eigenstate of the full Hamiltonian, then

$$H\Psi=E\Psi$$

and hence the evolution $U(t)\Psi$ is trivial, it is just $\Psi(t)=e^{-iE t}\Psi$ and the overlap with any other eigenstate is zero.

So what people mean with "states evolving with the interacting theory" or "states evolving with the free theory"? If it is about the corresponding eigenstates, why isn't the evolution trivial as it seems?

Is this because the full Hamiltonian is time-dependent? But then, for potential scattering with a Coulomb potential for instance the Hamiltonian is time-independent $V(\mathbf{R})=g/|\mathbf{R}|$ and it seems the evolution would really be trivial.

I'm clearly missing something really basic here. What is it? In summary:

  1. Why people talk about "states evolving with free/interacting theory"? Can't we use any state as initial condition for any evolution $W(t)$ be it free/interacting?

  2. How these states are characterized? Are they eigenstates of the free/interacting Hamiltonian? If so, why their evolution isn't trivial as outlined in the question?

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  • $\begingroup$ Can you be a little more clear about it? $\endgroup$ – GK A Feb 11 at 14:44
  • $\begingroup$ @GKA well I particularly think the question was already clear. Trying to make it even clearer I've added the two main points of doubt in simple sentences in the end. I hope it makes the question better. $\endgroup$ – user1620696 Feb 11 at 15:39
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In scattering theory, one would like to approximate, when time is large, a complicated interacting evolution with something simpler, i.e. the free evolution.

The basic physical justification goes as follows. If an interaction is suitably localized in space, and if one considers configuration that "escape at infinity", i.e. that as time goes by get very far from the region where the interaction is going on, then such configurations will asymptotically behave as a free theory but with an initial datum that is in general different from the original one.

Mathematically speaking, let $U(t)$ be the interacting evolution, and $U_0(t)$ the free one. The aim is to prove that for all (or almost all) $\psi$, there exists $\psi_{\pm}$ (the asymptotic state) such that $$\lim_{t\to\pm\infty}\lVert U(t)\psi - U_0(t)\psi_{\pm}\rVert=0\; .$$ If that is true, then the complicated system described by $U(t)\psi$ can be described, to a very good approximation if one waits enough time in the future or past, by the simpler evolution $U_0(t)\psi_{\pm}$. In this case we say that the state $\psi$ scatters, with corresponding asymptotic states $\psi_{\pm}$.

Since both $U_0(t)$ and $U(t)$ are unitary operators, the above is equivalent to saying that $$\lim_{t\to\pm\infty}\lVert U_0(t)^*U(t)\psi - \psi_{\pm}\rVert=0\; .$$ In other words, this amounts to studying the limit, in the strong topology, of $U_0(t)^*U(t)$ (and also, conversely, of $U(t)^*U_0(t)$).

There are, however, states for which one immediately sees that such convergence is impossible. Let $\psi_\lambda$ be an eigenvector of $U(t)$: $U(t)\psi_\lambda=e^{-it\lambda}\psi_\lambda$. Then $U_0(t)^*U(t)\psi_\lambda=e^{it(H_0-\lambda)}\psi_\lambda$, and such vector has strong limit as $t\to\pm\infty$ if and only if $\psi_\lambda$ is also an eigenvector of $H_0$ with eigenvalue $\lambda$. However, it is physically implausible that the interacting and free theories that are being compared share common eigenvalues and eigenvectors, and therefore one should get rid of such eigenvectors of $U(t)$, in order to prove the strong convergence of $U_0(t)^*U(t)$ (since these states clearly do not scatter). This is usually done projecting out of the pure point spectrum of $U(t)$. Physically, this also means that for Hamiltonians with pure point spectrum no state scatters, and this is because in this case the states cannot escape the region where there is interaction (Hamiltonians with pure point spectrum describe trapped systems). The converse is also true for $U^*(t)U_0(t)$, but usually the free reference dynamics $U_0(t)$ is chosen in such a way that it has purely continuous spectrum (e.g. the free particle Hamiltonian), and therefore it is not necessary to project out of the pure point spectrum, since the latter is empty.

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  • $\begingroup$ Thanks for the answer @yuggib. So we take one eigenstate $\Psi_0$ of the free Hamiltonian. Such a state describes by definition the system free of interaction. We then define the states evolving with the interacting dynamics exactly as those for which there are free states so that the limit identity holds. In other words, we define such states as those of the form $\Omega_- \Psi_0$ acting on eigenstates of the free Hamiltonian? $\endgroup$ – user1620696 Feb 11 at 15:59
  • $\begingroup$ @user1620696 Well, actually quite the opposite. The eigenstates (of either $H_0$ or $H$) are precisely states for which we know a priori that they will not scatter (i.e. for which the scattering approximation fails). The scattering theory is done precisely outside of the pure point spectrum of both $H_0$ and $H$. Usually however, $H_0$ has no eigenvalues and eigenvectors, and therefore one should be careful only to exclude the eigenvalues of the interacting Hamiltonian $H$. $\endgroup$ – yuggib Feb 11 at 16:25
  • $\begingroup$ The scattering theory is an approximation theory: the aim is to approximate a given (usually complicated) evolution, with a simpler one. This approximation is only valid asymptotically in time, and "it changes initial condition" (in the sense that the interacting evolution of one state is approximated by the free evolution of a different state). In addition, the approximation is valid only for states that "escape" for long time. And eigenvectors never escape. $\endgroup$ – yuggib Feb 11 at 16:33
  • $\begingroup$ sorry I used imprecise terminology, by eigenstate of the free Hamiltonian I had in mind the improper $|p\rangle$, but I shouldn't have used this terminology because this is not a true eigenstate (it isn't even an element of the Hilbert space). So your point is: the eigenstates (true ones, corresponding to the point spectrum), describe bound states and do not scatter - the limit can't hold, since they cannot look free at asymptotic times as they are bound. Thus they are unsuitable to use with the scattering approximation. Is that your point regarding the eigenstates? $\endgroup$ – user1620696 Feb 12 at 11:46
  • $\begingroup$ @user1620696 yes, this is precisely my point about eigenstates. $\endgroup$ – yuggib Feb 13 at 9:41

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