Can the Lieb-Robinson Bound be Intuited from the Taylor Series of an Operator?

[thatboi]

Hi all,
I was wondering if there was a way to intuit the Lieb-Robinson bound from simply looking at the taylor series for an operator $A(t) = e^{-iHt}Ae^{iHt}$ where $H$ is a k-local Hamiltonian and $A(t)$ initially starts off as a single-site operator. The generic idea is that at each order of $it$, the operator "grows" in size since it will have non-zero commutator with the local terms in $H$. My issue though, is how to see the "light-cone" that is frequently used in discussions of this topic because for any non-zero $t$, $A(t)$ will necessarily have contributions from all order of $it$, so I am not sure where to draw the boundary between operator and the rest of the system that is typically drawn in the Lieb-Robinson bound.
Any advice would be appreciated.
Thanks!

 

[Demystifier]

I never heard of Lieb-Robinson bound before, but after a brief googling my intuition is the following. The wave (quantum of classical) propagates with a finite velocity, approximately given by the group velocity
$$v_g=\frac{d\omega}{dk}$$
For example, for a free non-relativistic particle we have $\omega=k^2/2m$ (in units $\hbar=1$), so
$$v_g=k/m$$
which is finite, provided that $k$ is finite. Thus, if the wave packet has vanishing contributions from infinite $\omega$ and $k$, i.e. the wave does not have wild oscillations at arbitrarily small time and space scales, then one expects a finite speed of propagation. This is a wave analog of the fact that a classical non-relativistic particle travels with finite speed, provided that its energy and momentum are not infinite.

But that's just my intuition, I'm not certain how much is this related to the actual Lieb-Robinson bound.