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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?

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The $-1$ factor for closeed fermion loops, together with the combinatorics of the way paths connect, makes any lattice worldline configuration with more that one fermion line on a link add to zero. The figure below gives an example

enter image description here

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  • $\begingroup$ Interesting. It seems like if you just state that fermions must form paths and only one fermion can go through a vertex would this be enough to give the right results? Is there a ink somewhere for rules for fermions on a lattice? $\endgroup$ – zooby Jan 29 at 21:54
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    $\begingroup$ @zooby Ink?: Lots of it. The figure is from a my book Physics of Quantum Fields, but there are many review articles by other people. It helps to be familar with Grassmann integrals to read the literature. $\endgroup$ – mike stone Jan 30 at 0:07
  • $\begingroup$ Thanks I'll have a look at your book! Yes, I tried working it out from the Grassman integral $\int exp(i\sum_i\overline{\psi}_i\gamma (\psi_i-\psi_j))\overline{\psi}_a \psi_b D\psi D\overline{\psi}$ over a lattice. I can see it would give you paths in order to get a term with a Grassman variable from each vertex. I guess the thing about anti-symmetric wave-functions is a consequence but not really needed. $\endgroup$ – zooby Jan 30 at 1:23

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