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 ψ is written as ⟨ψ⟩=∫DψDˉψψe−S,
where the action S is bosonic (involves products even products of
ψ and
ˉψ). Therefore, unless there are source terms of the form
ˉηψ in the action, the integral over the
ψe−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