Some preliminary considerations:
- H=H0+V: the full Hamiltonian;
- V: the perturbation part;
- H0: the non-interacting Hamiltonian,whose eigenvalue problem can be solved exactly;
- |η0⟩: the ground state of H0;
- UDα: the time evolution operator in Dirac picture;
- e−α|t|(α>0) : the adiabatic switching factor,by which we can relate H and H0:
limt→±∞Hα=H0;limt→0Hα=H(Hα=H0+e−α|t|Vα>0)
Then the famous Gell-Mann-Low's theorem tell us if the statement
limα→0UDα(0,−∞)|η0⟩⟨η0|UDα(0,−∞)|η0⟩=limα→0|ψDα(0)⟩⟨η0|ψDα(0)⟩
exists for every order of perturbation theory,then it is an exact eigenstate of H.Very often we also assume that no crossings of the states occur during their evolution from the free states |η0⟩, namely the statement (1) will be the ground state of 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).