- In special relativity, the light cone defines the set of points that can be reached by null geodesics originating from a point$^1$. It is essentially the boundary of the set of points that can be reached by timelike curves. We call a curve timelike if its tangent vector $u^\mu$ is normalized as follows$^{2}$: $u^\mu u^\nu\eta_{\mu\nu}>0$. Three basic tenets of special relativity are
Massless particles travel on null curves.
Massive particles travel on timelike curves.
Tachyons travel on spacelike curves and are unphysical.
The light cone of special relativity thus defines the set of points a real particle can occupy in the future. If the amplitude is nonzero outside of this cone, then there is a probability that the particle will propagate along a curve with a spacelike tangent vector, thus violating special relativity.
- Perhaps your quote is a little misleading. I propose the following re-write:
If the amplitude is nonzero then there will be a nonzero probability for a real particle to be found outside its forward light-cone. This is unacceptable and would spell the death of quantum theory as we've known it so far.
If you had never heard of quantum field theory, but had heard of a little inequality
$$\Delta x\Delta p\ge\frac{\hbar}{2}$$
you could reason that trajectories are a little "fuzzy" and perhaps violate special relativity.
Suppose we had heard of QFT though. Let's show the amplitude of a virtual particle is nonzero outside of the light cone. For this we have to consider the propagator of an internal line of a Feynman diagram. The canonical example here is the Feynman propagator of a real scalar field. We solve the equation
$$-(\Box+m^2)D(x)=\delta^4(x)$$
by the method of Green's functions and obtain$^3$
$$D(x)=\int\frac{d^4k}{(2\pi)^2}\frac{e^{ikx}}{k^2-m^2+i\epsilon}$$
A standard calculation by the method of residues leads to$^4$
$$D(x)=-i\int\frac{d^3k}{(2\pi)^32\omega_k}\left[e^{-i(\omega_kt-\mathbf{k}\cdot\mathbf{x})}\theta(t)+e^{i(\omega_kt-\mathbf{k}\cdot\mathbf{x})}\theta(-t)\right]$$
where $\omega_k=\sqrt{\mathbf{k}^2+m^2}$ is the on-shell energy. The physical interpretation of $D(x)$ is that it describes the amplitude for a particle to travel from the origin to the point $x$. One finds that $^5$
$$D(0,\mathbf{x})\simeq ce^{-mr}$$
where $c$ is an irrelevant constant.
So virtual particles violate special relativity! So what? They're not real and special relativity only puts restrictions on real particles. This property of virtual particles is explained away by the uncertainty principle.
So how is causality, special relativity and Lorentz invariance respected in field theory? The answer is probably important enough to be called a theorem$^6$. Let $\mathcal{H}(x)$ be the interaction Hamiltonian density. Then the $S$-matrix can be written as the Dyson series
$$S=\mathcal{T}\exp\left(-i\int d^4x\,\mathcal{H}(x)\right)$$
where $\mathcal{T}$ denotes time-ordering. Using the cluster decomposition principle, we can write the interaction Hamiltonian in terms of quantum fields.
Theorem. All quantum fields obey
$$[\psi_\ell(x),\psi_{\ell'}(y)]_\mp=[\psi_\ell(x),\psi^\dagger_{\ell'}(y)]_\mp=0$$
for $(x-y)$ spacelike. The $-$ holds for bosons and $+$ for fermions.
You can verify (laboriously) that your standard mode expansions obey this theorem.
$^1$ In general relativity, however, this is only locally true and depends furthermore on the topology of spacetime. See, e.g., Wald, General Relativity (1984).
$^2$ Here I am using the $(+--\,-)$ convention.
$^3$ See, e.g., this Phys SE post.
$^4$ See, e.g., Cahill, Physical Mathematics (2013), p. 201.
$^5$ The full calculation is found in Zee, Quantum Field Theory in a Nutshell (2010, 2nd Ed.), p. 545.
$^6$ See, e.g., Weinberg, The Quantum Theory of Fields (1995). I can't nail down a specific page because the proof is smeared out over chapters 3, 4 and 5.
