This question originates from reading the proof of Gell-mann Low thoerem.

$H=H_0+H_I$, let $|\psi_0\rangle$ be an eigenstate of $H_0$ with eigenvalue $E_0$, and consider the state vector defined as
$$|\psi^{(-)}_\epsilon\rangle=\frac{U_{\epsilon,I}(0,-\infty)|\psi_0\rangle}{\langle \psi_0| U_{\epsilon,I}(0,-\infty)|\psi_0\rangle}$$ where the definition of $U_{\epsilon,I}(0,-\infty)$ can be found in the above paper.

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

My question:

  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? In general, it should be that for scattering state there will be no energy shift, for discrete state there will be some energy shift. But Gell-Mann Low theorem do not tell me the result.

  2. It seems that the Gell-Mann Low theorem is more powerful than adiabatic theorem which requires that there must exist gap around the evolving eigenstate. And Gell-Mann Low theorem can be applied to any eigenstate of $H_0$ no matter whether the state is discrete, continuous or degenerate and no matter whether there is level crossing during evolution. However the existence of $\lim_{\epsilon\rightarrow 0^{+}}|\psi^{(-)}_\epsilon\rangle$ is annoying, which heavily restrict the application of this theorem. Is there some criterion of existence of $\lim_{\epsilon\rightarrow 0^{+}}|\psi^{(-)}_\epsilon\rangle$? Or give me an explicit example in which this doesn't exixt.

  3. It seems that Gell-Mann Low theorem is a generalized adiabatic theorem, which can be used in discrete spectrum or contiunous spectrum. How to prove Gell-Mann Low theorem can return to adiabatic theorem in condition of adiabatic theorem. Need to prove that the $\lim_{\epsilon\rightarrow 0^{+}}|\psi^{(-)}_\epsilon\rangle$ exist given the requirement of the adiabatic theorem.

  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/110018/2451 $\endgroup$
    – Qmechanic
    Commented Jan 11, 2017 at 21:27
  • $\begingroup$ @346699 Reading your question I have the impression that the two theorems together provide the proof that the the $g\partial/\partial g \ln \langle\ldots\rangle$ can't diverge as $1/\epsilon$, so that indeed $E=E_0$. Now, you seem to reject this: what is the reason for rejecting it? Do you have a better or independent source of intuition, or any reason to believe, what $g\partial/\partial g \ln \langle\ldots\rangle$ should be instead for scattering states? If not, I would accept the consequence implied by the two theorems for scattering states. $\endgroup$
    – TwoBs
    Commented Nov 18, 2017 at 15:35
  • $\begingroup$ You stumbled on a very tricky and technical point. Go through chapter 9 in : books.google.fr/… and you shall find the answer. In general (as it was mentioned above) the GML formula shifts the energies of only discrete states, and thus consistency is recovered. $\endgroup$
    Commented Jan 17, 2019 at 16:30

3 Answers 3


The Gell-Mann Low theorem applies only to eigenvectors, i.e. to the discrete part of the spectrum. Hence it does not apply to scattering states. The latter are not eigenvectors since they are not normalizable. Your formula for $\Delta E$ is meaningless for them since the inner product on the right hand side is generally undefined unless $\psi_0$ is normalizable.

[The equation for the Moeller operator] ''says the energy of scattering state will not change when you turn on the interaction adiabatically.'' No. It only says that $H$ and $H_+$ must have the same total spectrum; it says nothing about energies of individual scattering states.

Moreover, a more rigorous treatment (e.g. in the math physics treatise by Thirring) shows that your equation holds at best on the subspace orthogonal to the discrete spectrum (which almost always exhibits energy shifts), and that certain assumptions (relative compact perturbations) must be satisfied that it holds on this projection. These assumptions are not satisfied when the continuous spectrum of $H$ and $H_0$ is not identical, e.g., when $H_0$ is for a free particle and $H$ for a harmonic oscillator or a Morse oscillator, or vice versa.


The second and third statement seemingly are not necessarily true without any further assumptions: If one takes the trivial example $H_0 = H_i$, then the eigenstates don't change, there are neither more nor less eigenstates, and even the continous energy spectrum is changed: All energies are multiplied by $1+e^{-\epsilon |t|}$.

  • $\begingroup$ It says $H_i$ can be viewed as perturbation, compared with $H_0$. $\endgroup$
    – 346699
    Commented Aug 6, 2016 at 7:20
  • $\begingroup$ What exactly is that supposed to mean? If it just means "small", one might as well take $H_i = \epsilon_2 H_0$ with $\epsilon_2$ a very small number. Then the same argument applies, except that the energies are rescaled by $1+\epsilon_2 e^{-\epsilon |t|}$. Perhaps the paper means a particular form of interaction - like a combination of at least 3 annihilation or creation operators? Perhaps the statements are just valid for first order perturbation theory? $\endgroup$
    – QuantumAI
    Commented Aug 6, 2016 at 7:39
  • $\begingroup$ In QM, interaction term will not be proportional to $H_0$ $\endgroup$
    – 346699
    Commented Aug 6, 2016 at 8:20
  • $\begingroup$ The above saying may not be stated rigorously, but I think it's right. Because 1 is what Gell-Mann-Low theorem says and 2 is what Lippmann-Schwinger theorem says. However the problem is that when we prove Lippmann-Schwinger theorem, we've assumed the energy of continuous spectrum doesn't shift. And when we prove Gell-Mann-Low theorem, we've assumed there will be an energy shift for discrete spectrum. $\endgroup$
    – 346699
    Commented Aug 6, 2016 at 8:31
  • $\begingroup$ I've corrected my question, do you have any idea? Thanks. $\endgroup$
    – 346699
    Commented Feb 19, 2017 at 2:46

Actually, the original Gell-Mann-Low Theorem(GLT) can be deduced by the Adiabatic Theorem(AT), which requires no degeneracy and discrete spectrum. However, in more general cases, say, degenerate states, AT is failed while GLT survives by some technical tricks, for instance, well-re-desired initial state.


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