# Does path intergral formula only works in perturbative situation?

I'm learning quantum field theory. In Peskin & Schroeder, when they derive $$\int {D\phi(x)\phi ({x_1})\phi ({x_2})\exp [i\int {{d^4}x\mathcal{L(x)}] = \left\langle {{\phi _b}|{e^{ - iHT}}T\{ \phi ({x_1})\phi ({x_2})\} {e^{ - iHT}}|{\phi _a}} \right\rangle } }$$ and what to make it nearly equal to two-point correlation function, they says:

...Just as in Section 4.2, this trick projects out the vacuum state $$|\Omega\rangle$$ from $$|\phi_a\rangle$$and $$|\phi_b\rangle$$ (provided that these states have some overlap with $$|\Omega\rangle$$, which we assume). For example, decomposing $$|\phi_a\rangle$$ into eigenstates $$|n\rangle$$ of $$H$$, we have $${e^{ - iHT}}|{\phi _a}\rangle = \sum\limits_n {{e^{ - i{E_n}T}}|n\rangle \left\langle {n|{\phi _a}} \right\rangle } \rightarrow\left\langle {\Omega |{\phi _a}} \right\rangle {e^{ - i{E_0}\infty (1 - i\epsilon )}}|\left. \Omega \right\rangle,$$ cf. eqs. (4.27)-(4.28). Thus we obtain a simple formula $$\left\langle {\Omega |T\{ \phi ({x_1})\phi ({x_2})\} |\Omega } \right\rangle = \lim_{T \to \infty (1 - i\epsilon )} \frac{{\int {D\phi (x)\phi ({x_1})\phi ({x_2})} \exp [i\int {{d^4}x{{\mathcal L}}} {\rm{ }}]}}{{\int {D\phi (x)} \exp [i\int {{d^4}x\mathcal L]}}}.$$

In Section 4.2 p.86 they illustrate this track as below:

We must assume that $$|\Omega \rangle$$ has some overlap with $$|0\rangle$$, that is, $$\langle \Omega | 0 \rangle\ne0$$(if this were not the case, $$H_I$$ would in no sense be a small perturbation).

Actually I haven't get what that means.

1. Does it mean if we are working in non-perturbative situation, the $$|\Omega\rangle$$ is orthogonal to $$| 0 \rangle$$ in the Hilbert space?

2. Why when it comes to a non-pertubative situation, the formula between path integral and correlation function still works? Do we require a stricter proof?

Haag's theorem is the precise mathematical statement that non-perturbatively, $$\langle \Omega | 0 \rangle = 0$$. People sometimes take this to mean that there is something fundamentally wrong with the interaction picture, but from my point of view