I was reading Peskin&Schroeder's QFT book, and there was some discussion about how ##\left|0\right>##, the ground state of a free field and ##\left|\Omega \right>##, the ground state of an interacting field differ from each other, and they outlined how the latter can be obtained by propagating the former in imaginary time with the full interacting Hamiltonian... There was a mention that in this calculation it was assumed that there is some overlap between these two ground states: ##\left< 0 | \Omega \right> \neq 0##. Is it actually even possible for the ground states of two systems that have the same degrees of freedom to be orthogonal to each other? If you consider a simple 1D single particle system in nonrelativistic QM, it seems that the ground state wavefunction ##\psi_0 (x)## can always be chosen to be a nonnegative real-valued function. The only examples of nonoverlapping ground states of two such systems would be something like two particle-in-a-box systems where the infinite potential wells don't have any overlap between them (e.g. one is the interval ##[0,L]## and the other is the interval ##[2L,3L]##). Then the wavefunctions wouldn't differ from zero at any common point, but also the difference between the two box potentials couldn't be seen as a "perturbation" because it wouldn't even be a proper finite-valued function. However, if we have two hydrogenic atom problems with a particle moving in a ##1/r## potential, and the centers of the potentials in the two systems are separated by a large (macroscopic) distance, there would be practically no overlap between the ground states because the hydrogenic ground state wavefunction decreases exponentially with increasing distance from the attraction center. Is it possible for two systems of several fermions to have nonoverlapping ground states because of the antisymmetry requirement?