I think it is a general fact about grassmannian field, and this has nothing to do with Lorentz invariance or other symmetries (you can invent a lot of QFTs without this kind of symmetry, but the VEV of a fermionic operator will be always zero (in the absence of sources)).

In a functional integral formulation, the VEV of a grassmannian field $\psi$ is written as $$ \langle \psi \rangle= \int D\psi D\bar\psi\, \psi \,e^{-S},$$
where the action S is bosonic (involves products even products of $\psi$ and $\bar\psi$). Therefore, unless there are source terms of the form $\bar\eta\psi$ in the action, the integral over the $\psi e^{-S}$ will give zero, since we are integrating over an odd number of grassmannian fields (when the exponential is expanded).

This post imported from StackExchange Physics at 2014-04-13 11:31 (UCT), posted by SE-user Adam