For $\Lambda\to\infty$ we obtain $$\begin{aligned}
P_{\Phi,k}(p)-P_{\Phi,\Lambda}(p) &=&-V A_k^{-1} V.\end{aligned}$$ For $P_{\Phi,\Lambda}$ we can use $$P_{\Phi,\Lambda}(p) =  V.$$ The propagator $$G_k(p) = P_{\Phi,k}^{-1}(p) = \left(-V A_k^{-1} V + V\right)^{-1}
\label{eq:propagatorformal}$$ describes correlations of two particles with relative momentum $\vec q^\prime$ to two particles with relative momentum $\vec q$. We use here a matrix notation where $\vec q$ and $\vec q^\prime$ are indices and $P_{\Phi,k}(p,\vec q,\vec q^\prime)$ is inverted as a matrix for fixed value of the center of mass momentum $p$ which is conserved.

To understand the physical meaning of $G_k(p)$ we consider field configurations $\Phi$ for which $G_k$ has a pole, or for which $$\int_{\vec q^\prime} P_{\Phi,k}(p,\vec q, \vec q^\prime) \Phi(\vec q^\prime) = 0$$ holds. In the center of mass frame where $p=(p_0,0,0,0)$ and using the reduced mass $\mu=M_1 M_2/(M_1+M_2)$, this condition can be written as $$\int_{\vec q^\prime} V(\vec q,\vec q^\prime) \Phi(\vec q^\prime) = \left[p_0-\frac{1}{2\mu}(\vec q^2+R_k(\vec q^2))\right] \Phi(\vec q).$$ For $R_k=0$ this is, of course, just the Schrödinger equation for the two-body problem. This becomes clear in position space where one obtains $$\left[p_0-\frac{1}{2\mu}(-\vec \nabla^2+R_k(-\vec \nabla^2))-V(\vec x)\right] \Phi(\vec x)=0.$$

In principle, one could obtain the bound state propagator $G_k$ for arbitrary values of the center of mass momentum $p$ by inverting 