# How the BCS superconductors violate the Gell-Mann-Low's Theorem?

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Some preliminary considerations:

1. $\mathcal{H} = \mathcal{H}_0+V$: the full Hamiltonian;
2. $V$: the perturbation part;
3. $\mathcal{H}_0$: the non-interacting Hamiltonian,whose eigenvalue problem can be solved exactly;
4. $|\eta_0\rangle$: the ground state of $\mathcal{H}_0$;
5. $U_{\alpha}^D$: the time evolution operator in Dirac picture;
6. $e^{-\alpha |t|} (\alpha>0)$ : 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 $(1)$.

asked Aug 31, 2017
recategorized Oct 7, 2017

I think that the potential must be relatively compact with respect to the kinetic part, in order that the theorem holds. For BCS in the thermodynamic limit, this is not the case.

## 1 Answer

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I have no definite answer, but I would guess that if the Gell-Mann-Low theorem "breaks down", then because the assumptions are not met, i.e. the limit does not exist. I mean, that's the only thing that can happen to mathematical theorems…

Also, a spontaneously broken symmetry is not necessarily an indicator that something is wrong with the theorem, either. Symmetry breaking usually means that there is a manifold of degenerate ground states, parametrized by the order parameter. The Gell-Mann low theorem could very well give you an eigenstate that is a symmetrized superposition of the eigenstates.

The best way to check what is going is probably to look at an interacting toy model with finitely many states.

answered Sep 1, 2017 by (705 points)

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