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?


2 Answers 2


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. Edit (June 15th, 2022): as Nanashi No Gombe pointed in the comments below, an example of this is provided by parastatistic field models, which need not commit to the Bose/Fermi notion of microcausality but can be Lorentz covariant nonetheless.

It is also important to point that there are non-Lorentz invariant quantum field theories that are nonetheless microcausal (e.g. some models coupled to a suitable external "ether" field). In such models, microcausality and the energy-momentum spectrum condition suffice to ensure that energy-momentum spectrum has a Lorentz-invariant shape and therefore has Lorentz-invariant dispersion laws (i.e. either of "mass-shell" of "light-cone" type), even in the absence of bona fide Lorentz invariance - this is a consequence of the Jost-Lehmann-Dyson representation of the two-point function, which does not rely on Lorentz invariance. Once more. this shows that the concepts of Lorentz invariance and microcausality in a quantum field theory are not equivalent.

  • $\begingroup$ thanks for your answer. Weinberg in fact says that we need microcausality to prove the Lorentz Invariance of the S matrix. He says that for something like the Dirac field which is not directly observable, this is the only 'justification' of the microcausality condition. The paper I linked however claims that microcausality can follow from Lorentz Invariance. However they show it holds in curved spacetime as well,in the absence of global L.I. It is quite an interesting paper actually. $\endgroup$ Jul 16, 2016 at 22:35
  • $\begingroup$ It is important to remark that the arguments in the paper you cited are not rigorous, being mostly based on formal functional integrals and perturbation theory. The result of Methée I cited, on the other hand, is completely rigorous and holds under very general assumptions (the two-point function doesn't even need to be positive or tempered). Moreover, microcausality lies at the basis of the dispersion relations for the S-matrix and its resulting high-energy behavior, which can be checked in collider experiments (Froissart bounds, etc.). $\endgroup$ Jul 17, 2016 at 3:36
  • $\begingroup$ In curved space-times, arguments based on formal functional integrals and the ensuing formal perturbation theory are even more suspicious because there is in general neither a Wick rotation nor an S-matrix to begin with - the very geometry of space-time precludes that since there may be no global isometries at all. In this case, an algebraic approach tends to be more appropriate. $\endgroup$ Jul 17, 2016 at 3:39
  • $\begingroup$ Thanks for your comments. I have one final question. If we know that for a theory the S matrix is Lorentz invariant, does microcausality automatically follow? $\endgroup$ Jul 17, 2016 at 6:29
  • $\begingroup$ @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. $\endgroup$
    – Name YYY
    Jul 28, 2016 at 20:37

Consider a Poincare-invariant scalar field theory. Assume we are guaranteed Poincare-invariance in the sense that there exists a unitary representation of the Poincare group acting on the Hilbert space, which transforms the scalar field operator $\phi(x)$ like $$U(g)\phi(x)U(g)^\dagger = \phi(g x)$$ for elements $g$ of the Poincare group.

Consider two spacelike-separated points $x,y$. We want to show $$[\phi(x),\phi(y)]\overset{?}{=}0.$$ On the other hand, we know that for any two points $x'=(0,\vec{x}')$ and $y'=(0,\vec{y}')$ at time $t=0$, we have $$[\phi(0,\vec{x}'),\phi(0,\vec{y}')] = 0,$$ which follows immediately from the wavefunctional or lattice perspectiv; e.g. in a lattice theory the operators $\phi(0,\vec{x}')$ and $\phi(0,\vec{y}')$ act on different lattice sites and thus commute.

Here's the key step: Note that for any spacelike $x,y$, there exists some Poincare transformation $g$ taking $x,y$ to $x',y'$, i.e. transforming to a frame where both point are at $t=0$. So then $$U(g)[\phi(x),\phi(y)]U(g^{-1}) =[\phi(0,\vec{x}'),\phi(0,\vec{y}')] = 0$$ and thus $[\phi(x),\phi(y)]=0$ as desired.

(In another answer Pedro Ribeiro mentions Lorentz-invariant theories that are not microcausal, and I'm not sure how these violate my assumptions.)


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