Say your space-time lattice has 30 fermions on the vertices. (Would they have to form a path?) Swapping any two fermions (on the same row??) should make the amplitude of the lattice state negative. How does this work in lattice QCD? Let (X,T) be the lattice points. If I put a Fermions at a=(0,0), b=(10,0), c=(0,1) and d=(10,1) thus two fermion paths. By the time I made all possible swaps I get zero for the total amplitude.
I can see how it would work for Feynman graphs since you would calculate the probability as $|\Delta_t(a,b)\Delta_t(c,d)-\Delta_t(a,c)\Delta_t(b,d)|^2$ but that works because the path lengths are different. I'm not sure how it works for lattice QCD.
Do you have to make paths of fermions thorough the lattice and number the paths?