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The contradiction between the Gell-Mann Low theorem and the identity of Møller operator $H\Omega_{+}=\Omega_{+}H_0$

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This question originates from reading the [proof of Gell-Mann Low thoerem](https://arxiv.org/abs/math-ph/0612030v1).

Let $|\psi_0\rangle$ be an eigenstate of $H_0$ with eigenvalue $E_0$, and consider the state vector
$$|\psi^{(-)}_\epsilon\rangle=\frac{U_{\epsilon,I}(0,-\infty)|\psi_0\rangle}{\langle \psi_0| U_{\epsilon,I}(0,-\infty)|\psi_0\rangle}$$

**Gell-Mann and Low's theorem:**
If the  $|\psi^{(-)} \rangle :=\lim_{\epsilon\rightarrow 0^{+}}|\psi^{(-)}_\epsilon\rangle$  exist, then $|\psi^{(-)} \rangle$ must be an eigenstate of $H$ with eigenvalue $E$. And the eigenvalue $E$ is decided by following equation:
$$\Delta E= E-E_0=-\lim_{\epsilon\rightarrow 0^+} i\epsilon  g\frac{\partial}{\partial g}\ln \langle\psi_0| U_{\epsilon,I}(0,-\infty)|\psi_0\rangle$$

However we learn in scattering theory,
$$U_I(0,-\infty) = \lim_{t\rightarrow -\infty} U_{full}(0,t)U_0(t,0) = \Omega_{+}$$
where $\Omega_{+}$ is the Møller operator. We can prove the identity $H\Omega_{+}=H_0 \Omega_{+}$  for the Møller operator. It says that the energy of a scattering state will not change when you turn on the interaction adiabatically.

My questions:

1.The only way to avoid these contradiction is to prove that $\Delta E$ for scattering state of $H_0$ must be zero. How to prove this? In general, it should be that for a scattering state there will be no energy shift, for discrete state there will be some energy shift. But the Gell-Mann Low theorem do not tell me the result.es

2.How to tackle this explicit case?
$$H_0 = \frac{\mathbf{p}^2}{2}- \frac{1}{|\mathbf{r}|}, \ H_I= \frac{1}{|\mathbf{r}|}$$
then $H=H_0+H_I=\frac{\mathbf{p}^2}{2}$. If we start with $|\psi_0\rangle$ is the ground state of $H_0$, i.e. the ground state of Hydrogen atom, and evolve by  $ U_{ I}(0,-\infty)|\psi_0\rangle $, what is the result of this state? Although in this case the system experience some level crossing, the theorem tells us if the state exists, then it must be some eigenstate of $H$. In this case the adiabatic theorem cannot be used, but the Gell-mann Low theorem still works.

3.The existence of $\lim_{\epsilon\rightarrow 0^{+}}|\psi^{(-)}_\epsilon\rangle$ is annoying. Is there some criterion of existence of $\lim_{\epsilon\rightarrow 0^{+}}|\psi^{(-)}_\epsilon\rangle$? Or give me an explicit example in which this does exixt.

4.It seems that the Gell-Mann Low theorem is a generalized adiabatic theorem, which can be used for discrete spectrum or continuous spectrum. How to prove Gell-Mann Low theorem reduces to the adiabatic theorem under the condition of the adiabatic theorem?

asked Oct 30 in Theoretical Physics by Alienware (185 points) [ revision history ]
edited Oct 30 by Arnold Neumaier

The correct identity for the Moeller operator is $H\Omega_+=\Omega_+H_0$, which doesn't have the implication you claim. 

@ArnoldNeumaier $\Omega_+=U_{I}(0,-\infty)=\lim_{\epsilon\rightarrow 0+ }U_{\epsilon,I}(0,-\infty)$ Using this identity, we require $\Delta E =0$

But your argument is meaningless since $\psi_0$ for scattering states are not normalizable. The eigenstates belong to the bound state spectrum, and this is indeed unchanged under scattering. 

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