Green's function in Hamiltonian vs. Path Integral QFT

Question

For a spacetime $M$, the Green's function for the operator $\Delta+m^2$ is the following distribution on $M\times M$: $$G(x,y):=\langle \phi(x)\phi(y)\rangle=\int_{C^\infty(M)}\mathcal D\phi\,\phi(x)\phi(y) e^{-S(\phi)/\hbar},~~~~S(\phi)=\int_{x\in M}\phi(\Delta+m^2)\phi$$ If we wanted to compute this correlator in the Hamiltonian approach to QFT, we would interpret $\phi(x),\phi(y)$ as the appropriate time-dependent operators, but would have to specify the state of the field, i.e. we would have to write $$G(x,y):=\langle \phi(x)\phi(y)\rangle=\left<\psi\right|\phi(x)\phi(y)\left|\psi\right>$$ For some state $\left|\psi\right>$. What is that state?

Answer 1

In general, $|\psi\rangle$ is the ground state of your theory, i.e., the state with the lowest energy.

In free theories, $|\psi\rangle$ is written $|0\rangle$, and is defined through $H_0|0\rangle=0$, or equivalently, $a_{\boldsymbol k}|0\rangle=0$, with $a_{\boldsymbol k}$ the annihilation operator.

In interacting theories, $|\psi\rangle$ is written $|\Omega\rangle$, and is defined as the eigenket of $H$ with the lowest possible eigenvalue (energy). Note that we assume (it is an axiom of any QFT) that $H$ is bounded below, i.e., $|\Omega\rangle$ always exists.

In general, we define $H=H_0+H_I$, with $H_0$ the free (quadratic) part of the hamiltonian, and $H_I$ as the interaction. Note that both $|0\rangle$ and $|\Omega\rangle$ are well-defined, but in general they are different $|0\rangle\neq|\Omega\rangle$ (unless $H_I=0$).

With this, we define the propagator of any theory as $$ \Delta(x,y)\equiv \langle 0|\mathcal T\ \phi_0^\dagger(x)\phi_0(y)|0\rangle $$ and the correlator (or $n$-point function) as $$ G(x,y)\equiv \langle \Omega|\mathcal T\ \phi^\dagger(x)\phi(y)|\Omega\rangle $$

Finally, we can use Wick's theorem, together with the Dyson series and the Gell-Mann and Low's formula, to write $G(x,y)$ in terms of $\Delta(x,y)$. Note that the propagator is independent of interactions (see e.g. this answer of mine) and $G(x,y)$ is not, but if $H_I=0$, then $\Delta(x,y)=G(x,y)$.