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In the proof of the Gell-Mann and Low theorem (See equation (6.38) in Fetter and Walecka for an example), we assume that at time $T \rightarrow \infty$, we start with \begin{equation} \tag{1} |\psi (-T) \rangle_S = e^{i E_0 T} |E_0 \rangle \end{equation} in the Schrodinger picture where $|E_0 \rangle$ is the ground state of the free Hamiltonian. This helps us when we move to the interaction picture because \begin{equation} \tag{2} |\psi (-T) \rangle_I \equiv e^{-i H_0 T} |\psi (-T) \rangle_S =e^{-i H_0 T} e^{i E_0 T} |E_0 \rangle = |E_0 \rangle \end{equation} i.e. it becomes independent of any divergent phase factors as $T \rightarrow \infty$.

My question is: Why do we have the $e^{i E_0 T}$ phase factor in (1)? It feels more natural to me to set the initial state to just $|E_0 \rangle$ without the phase factor.

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I am not 100% sure what the question is, but I will try to say some things about the interaction picture in general hoping that they will prove to be useful to you. If there are any concerns left after you read this post, you can always comment.

First of all, a comment on Eq. (1) in your text: We know that in the Schroedinger picture, states are time evovled, whereas the operators are thought to be time-independent. On the other hand, in the Heisenberg picture, the operators carry the time-dependence and the states are thought to be independent of time. So, given two states $|\psi(T)\rangle_S$ and $|\psi_0\rangle$ in the Schroedinger and Heisenberg picture respectively, the following equation holds $$|\psi(T)\rangle_S=e^{iHt}|\psi_0\rangle$$ and furthermore it provides a way of associating states in the first picture with states in the second. Here, $H$ is the full Hamiltonian, which can always be written as the kinetic part, $H_0$ plus whatever interaction exists in the physical system under consideration, $V$. Assuming that the interaction is described in terms of a coupling constant, which we assume to be small, at the $T\rightarrow\infty$ limit, the interaction is thought to be neglected and hence $$|\psi(T)\rangle_S=e^{iHt}|\psi_0\rangle \xrightarrow[\text{}]{\lim_{T\to\infty}} |\psi(T)\rangle_S=e^{iH_0t}|\psi_0\rangle$$ Now, if instead of the mapping of an arbitrary state $|\psi_0\rangle$ from the Schroedinger picture to the Heisenberg one, I choose to study the mapping of the state that is eigenstate to the free Hamiltonian, from Schroedinger's picture to the Heisenberg one, then I get $$|\psi(T)\rangle_S=e^{iHt}|E_0\rangle \xrightarrow[\text{}]{\lim_{T\to\infty}} |\psi(T)\rangle_S=e^{iH_0t}|E_0\rangle= e^{iE_0t}|E_0\rangle$$ where I have labelled with $|\psi(T)\rangle_S$) the state in the Schroedinger picture (whose corresponding Heisenberg picture $|E_0\rangle$ is this time) again. This justifies some things about the form of Eq. (1) I think.

Then, we have the usual prescription that takes us from the Schroedinger picture to the Interaction picture (https://en.wikipedia.org/wiki/Interaction_picture). This is given by $$|\psi(T)\rangle_I=e^{-iH_0t}|\psi(T)\rangle_S$$ substituting the previous expression in the above-mentioned prescription, one obtains the state state $|E_0\rangle$ in the Interaction picture. $$|\psi(T)\rangle_I=e^{-iH_0t}e^{iE_0t}|E_0\rangle= e^{iE_0t}e^{-iH_0t}|E_0\rangle= e^{iE_0t}e^{-iE_0t}|E_0\rangle=|E_0\rangle$$ So, in other words, the interaction picture of a state that is eigenstate of the free Hamiltonian (in the Scrhoedinger picture) is given by that very same state itself. Or put simply, for eigenstates of the free Hamiltonian, it doesn't matter if one wors in the interaction picture or the Heisenberg one.

I hope this helps...

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  • $\begingroup$ Thank you for your answer. I undsrstand all the things that that you have said except one. Can we show that the full time-evolution operator $U_\epsilon (-T, 0)$ just becomes $e^{i H_0 T}$ as $T \rightarrow \infty$? $\endgroup$
    – Owl101010
    Commented Aug 4, 2022 at 13:37
  • $\begingroup$ We can not show that. Instead, we assume that the full time-evolution operator, in which the Hamiltonian is comprised by both the Kinetic and the Potential part, reduces to the Free time-evolution operator, as in the asymptotic regions of space (where colliding particles used to be in the far past and where colliding particles will be in the far future), colliding particles can be resembled to be free! Hence, the potential part in the Hamiltonian can be approximately neglected! $\endgroup$
    – schris38
    Commented Aug 4, 2022 at 13:45
  • $\begingroup$ It still seems a little ad hoc to me. It's as if we start with a state $|E_0 \rangle$ at $t = 0$, tacitly go to the interaction picture to give the $e^{i E_0 T}$ phase to the state and then explicitly go to the interaction picture to claim that the phase disappears. Does the end justify the means here? $\endgroup$
    – Owl101010
    Commented Aug 4, 2022 at 14:08
  • $\begingroup$ I am not sure I fully understand your concern. And by the way, states in the Heisenberg picture are time-independent. This means that the state is $|E_0\rangle$ at all times, not only at $t=0$ $\endgroup$
    – schris38
    Commented Aug 4, 2022 at 14:13
  • $\begingroup$ Sorry for not being able to convey my confusion. In shifting between the Schrodinger and the Heisenberg pictures, you assume there is a time $t = t_0$ at which both the pictures match. Here, in our discussion, we have set $t_0 = 0$. I guess I will start everything after setting $t_0 = -T$ and see whether things make more sense. $\endgroup$
    – Owl101010
    Commented Aug 4, 2022 at 14:22

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