Some preliminary considerations:

- $\mathcal{H} = \mathcal{H}_0+V$: the full Hamiltonian;
- $V$: the perturbation part;
- $\mathcal{H}_0$: the non-interacting Hamiltonian,whose eigenvalue problem can be solved exactly;
- $|\eta_0\rangle$: the ground state of $\mathcal{H}_0$;
- $U_{\alpha}^D$: the time evolution operator in Dirac picture;
- $e^{-\alpha |t|} (\alpha>0)$ : the adiabatic switching factor,by which we can relate $\mathcal{H}$ and $\mathcal{H}_0$:

$$\lim_{t\rightarrow\pm\infty} \mathcal{H}_\alpha = \mathcal{H}_0\quad ;\quad \lim_{t \rightarrow 0} \mathcal{H}_\alpha = \mathcal{H} \quad (\mathcal{H}_\alpha = \mathcal{H}_0 +e^{-\alpha |t|} V \quad \alpha>0 ) $$

Then the famous Gell-Mann-Low's theorem tell us if the statement

$$ \lim_{\alpha\rightarrow 0} \dfrac{U_\alpha^D(0,-\infty)|\eta_0\rangle}{\langle \eta_0|U^D_\alpha(0,-\infty)|\eta_0\rangle} = \lim_{\alpha\rightarrow 0} \dfrac{|\psi_\alpha^D(0)\rangle}{\langle\eta_0|\psi^D_\alpha(0)\rangle} \tag{1} $$

exists for every order of perturbation theory,then it is an exact eigenstate of $\mathcal{H}$.Very often we also assume that no crossings of the states occur during their evolution from the free states $|\eta_0\rangle$, namely the statement $(1)$ will be the ground state of $\mathcal{H}$.

My question is about how the BCS superconductors violates this theorem;in that case,the interaction $V$ leads to a new type of ground state with different symmetry and a lower energy than the adiabatic ground state $(1)$.