# Visual explanation for $|\psi(-\infty)\rangle^{\text{in}}= \lim_{t\rightarrow -\infty} e^{iH_0 t} e^{-iHt}|\psi\rangle$

I'm reading section 7.4, Scattering and the $$S$$-matrix of Quantum Field Theory: Lectures of Sidney Coleman. It says

For a scattering of particles in potential, we have a very simple formula for the S-matrix. We don't have to worry about in states and out states, because the in states are the states in the far past, and the out states are the states in the far future: $$|\psi(-\infty)\rangle^{\text{in}}= \lim_{t\rightarrow -\infty} e^{iH_0 t} e^{-iHt}|\psi\rangle=\lim_{t\rightarrow -\infty} U_I(0,t)|\psi\rangle. \tag{7.57}$$

Here $$|\psi(t)\rangle$$ is solution of the free Schrodinger equation $$|\psi(t)\rangle = e^{-iH_0(t-t')}|\psi(t')\rangle = U_0(t,t')|\psi(t')\rangle \tag{7.41}$$ where $$H_0=p^2/2m.\tag{7.42}$$ This ket $$|\psi(t)\rangle$$ represents the wave packet. In particular, $$|\psi\rangle = |\psi(0)\rangle$$. Furthermore, $$|\psi(t)\rangle^{\text{in}}$$ is the exact solution of the Hamiltonian so that $$|\psi(t)\rangle^\text{in} = e^{-iH(t-t')} |\psi(t')\rangle^\text{in}\tag{7.43}$$ The two states are connected by the requirement that if we look in the very far past, we can't make a difference.

$$\lim_{t\rightarrow -\infty} \lVert e^{-iH_0t}|\psi\rangle -e^{-iHt} |\psi\rangle^\text{in}\rVert =0.\tag{7.44}$$

I'm not able to make sense of the passage written above. We don't know what would $$e^{-iHt}|\psi\rangle$$ be since $$|\psi\rangle$$ is not an eigenstate of exact Hamiltonian.

In mathematical terms, we expect that for every quantum state $$\psi$$ in a Hilbert space $$\mathcal{H}$$, which is not an eigenstate of the Hamiltonian $$H=H_0+V(X)$$, we can find states $$\psi^{\mathrm{in/out}}$$ (called incoming/outgoing states) such that $$\mathrm{e}^{-\mathrm{i} tH} \psi$$ approaches $$\mathrm{e}^{-\mathrm{i} tH_0} \psi^{\mathrm{in/out}}$$ in the limit $$t\to \pm \infty$$, that is, $$\lim_{t\to\mp\infty} \|\mathrm{e}^{-\mathrm{i} tH} \psi - \mathrm{e}^{-\mathrm{i} tH_0} \psi^{\mathrm{in/out}}\| = 0.$$ Rewriting this formula using unitarity of $$\mathrm{e}^{\mathrm{i}t H_0}$$ we obtain a formula for the incoming/outgoing states: $$\psi^{\mathrm{in/out}} = \lim_{t\to\mp\infty}\mathrm{e}^{\mathrm{i}t H_0} \mathrm{e}^{-\mathrm{i} tH} \psi.$$ We don't expect to find such incoming/outgoing states for eigenstates of $$H$$ because $$\mathrm{e}^{-\mathrm{i}t H}$$ is just a complex number when applied to an eigenstate, that is, $$H\psi = E \psi$$, $$E\in\mathbb{R}$$, implies $$\mathrm{e}^{-\mathrm{i}t H} \psi = \mathrm{e}^{-\mathrm{i}t E} \psi$$.
Let's call the space of all incoming states $$\mathcal{H}^{\mathrm{in}}$$ and the space of all outgoing states $$\mathcal{H}^{\mathrm{out}}$$. If $$\mathcal{H}^{\mathrm{in}} = \mathcal{H}^{\mathrm{out}} = \mathcal{H}_c$$, where $$\mathcal{H}_c \subset \mathcal{H}$$ consists of all quantum states which are not eigenstates of $$H$$, then we say that the theory is asymptotically complete. That asymptotic completeness is satisfied in quantum mechanics has been verified for a very large class of potentials $$V(X)$$ (see e.g. Teschl: Mathematical Methods in Quantum Mechanics. Chapter 12).
• Unfortunately, your definition of $|\psi\rangle^{in}$ is completely different to Coleman's. See here. Oct 24, 2023 at 10:00