# Does Microcausality follow from Lorentz Invariance?

In a Lorentz Invariant theory, does microcausality automatically hold? In a free theory this is obvious. In an interacting theory I found some 'proof's in this paper: http://arxiv.org/abs/0709.1483

However the proofs show that for space-like separated $x$ and $y$ $$\langle 0| [\hat{\phi}(x),\hat{\phi}(y)]|0\rangle =0.$$

But for the microcausality condition to hold at an operator level we need to show that $$\langle n| [\hat{\phi}(x),\hat{\phi}(y)]|n\rangle =0\,\, \forall \,n$$

where $|n\rangle$ forms a basis. My question is, can this be shown? Or are further assumptions needed?

If all truncated $n$-point functions vanish for $n>2$ (i.e. we are dealing with a so-called generalized free field), microcausality for vacuum expectation values and at the operator level are equivalent. The former, on its turn, follows from Lorentz invariance alone in the case of scalar (but not necessarily free) fields, as shown by Pierre-Denis Methée, who was a student of de Rham (Sur les Distributions Invariantes dans le Groupe de Rotations de Lorentz, Commentarii Mathematici Helvetici 28 (1954) 225-269). If the field is interacting, this is no longer the case and then, in fact, microcausality does not follow from Lorentz invariance alone, even if positive definiteness of the $n$-point functions and the energy-momentum spectrum condition also hold. Unfortunately, though, I am not aware of any explicit example of a quantum field theory which is Lorentz-invariant but not microcausal, despite some claims in the literature that it is not difficult to build such an example. If I happen to find one, I will update my answer.
• @NirmalyaKajuri : it seems, that yes, it does. From Lorentz invariance of $S$-operator follows, that interaction hamiltonians commute on space-like intervals, which is nothing but microcausality criterion. – Name YYY Jul 28 '16 at 20:37