Rigorous derivation of the ground state projector using euclidean time evolution

Usually one argues that the euclidean path integral is able to recover the ground state of a system along the following lines:

Take the time evolution operator $$U(t,t_0)=e^{-iH(t-t_0)}$$. Transform to euclidean time by $$\tau = it$$ to obtain the euclidean time evolution operator $$U_E(\tau,\tau_0)=e^{-H(\tau-\tau_0)}$$. Suppose $$H$$ has a discrete non-degenerate basis $$|E_n\rangle$$ so that

$$U_E(0,T)=e^{HT}=\sum_{n} e^{E_n T} |E_n\rangle\langle E_n|.$$

One then usually says: well in the $$T\to -\infty$$ only the ground state projector should contribute and hence $$U_E(0,-\infty)\sim |\Omega\rangle\langle \Omega|$$ where $$|\Omega\rangle$$ is the ground state. One finally uses a path integral representation for $$U_E(0,T)$$ in the coordinate/field configuration basis.

This is more or less what I've seem done, but it is very far from rigorous.

1. In what sense the limit is taken? Is it pointwise with respect to the action of $$U_E(0,T)$$ on vectors in the Hilbert space? Is it a strong limit or a componentwise limit?

2. Why "only the ground state projector survives"? If the limit were componentwise, we would evaluate $$\lim_{T\to -\infty}e^{E_nT}$$. But this is zero for all $$E_n > 0$$, so if the ground state has $$E_0 > 0$$, the limit would annihilate it as well.

3. What if the spectrum is continuous? Things seem much worse. It seems that the existence of an energy gap is a requirement for this to work. Is it the case?

Is there some way to sharp this argument so as to make it rigorous?

• I think you are right about point 3., indeed for massless particles the $S$ matrix is not well defined, at least with the usual perturbative approach. – MannyC Jul 9 at 2:00
• I've been wondering the same, especially point $2.$, did you end up finding anything? – user2723984 Aug 1 at 13:06