"in" and "out" states, $\Psi^\pm$, with reference to Weinberg Vol. 1 pages 109 and 110 could be defined by $$\Psi_\alpha^\pm = \Omega(\mp \infty)\Phi_\alpha\tag{3.1.13}$$ where $$\Omega(\tau) = \exp(+iH\tau)\exp(-iH_0\tau).\tag{3.1.14}$$ This by itself defines $\Psi^\pm$s and so any great property this object couldn't be included in definition but it should be proven. but Weinberg includes in definition that $\Psi_\alpha^\pm$ are eigenvectors of total Hamiltonian $H$. something is missing! I'm graduated in math and tried over forty hours to find rigorous picture of this things and I couldn't!
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$\begingroup$ More on Weinberg eqs. (3.1.13) & (3.1.14): physics.stackexchange.com/q/207943/2451 $\endgroup$– Qmechanic ♦Commented Jan 29, 2022 at 4:29
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2 Answers
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We know that, for instace, $$\Psi - e^{itH}e^{-itH_0}\Phi \to 0\quad if \quad t\to +\infty.$$ If we heuristically reinforce this requirement also regarding the time derivative: $$\frac{d}{dt}\left[\Psi - e^{itH}e^{-itH_0}\Phi \right]\to 0\quad if \quad t\to +\infty,$$ we find (disergarding a number of issues concerning the domains of the operators) $$i\left[ He^{itH}e^{-itH_0}- e^{itH}e^{-itH_0}H_0 \right]\Phi\to 0\quad if \quad t\to +\infty.$$ In other words $$H \Psi = \Omega(+\infty) H_0 \Phi.$$ This implies (even more heuristically) the thesis. I expect that all this reasoning can be rigorously fixed under suitable hypoteses on $H-H_0$.
answered Apr 17 at 15:31
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This is an assumption of the Hamiltonian. Weinberg states
...suppose we can divide the time-translation generator $H$ into two terms, a free-particle Hamiltonian $H_0$ and an interaction $V$, $$H=H_0+V$$ in such a way that $H_0$ has eigenstates $\Phi_{\alpha}$ that have the same appearance as the eigenstates $\Psi^+_{\alpha}$ and $\Psi^-_{\alpha}$ of the complete Hamiltonian $$H_0\Phi_{\alpha}=E_{\alpha}\Phi_{\alpha}$$...
We are purely assuming that the full Hamiltonian has the same spectrum as the free Hamiltonian. This is generically the case in most physical circumstances in quantum field theory, where the spectrum of these two operators is continuous (it consists of unbounded particle momentum states). The only caveat though is within this assumption we are imposing that the particle states for the free Hamiltonian $\Phi_{\alpha}$ have the same masses as the ones for the full Hamiltonian. But this can be done trivially by redefining $V$.
