# Lieb-Robinson bound and spin chain

I am trying to understand the paper Localized shocks better. There is Lieb-Robinson bound on the page 6. How does formula (7) imply that:

the radius of the operator can grow no faster than linearly $$r[Z_{1}(t_{w})]<(c_{1}/c_{2})t_{w}$$

? This is not obvious for me.

• In Upper Bound on Diffusivity, Hartman et al. state "operators can spread at most linearly in time, a fact that can be deduced via repeated commutation with the Hamiltonian [1,2]", where Ref. 2 is Hasting's paper, whose Eq. 30 seems to contain an explicit linear dependence with time. But I'm out of my league here, let me know if this is wrong/useless. – stafusa Mar 17 at 2:21
• Also, couldn't one simply calculate as we do for wave propagation and, given the exponent of the Lieb-Robinson bound is $c_1 t_w − c_2 |x−y|$ and identifying $|x−y|$ with $r$, obtain the equation the OP asks about? Or this doesn't even make sense and seems to work only by accident? – stafusa Mar 17 at 2:28

From equation (7) in the cited paper [1], the Lieb-Robinson bound for the commutator of local operators $$W_x$$ and $$W_y$$, where the subscript indicates the location, is $$\big\|[W_x(t),\,W_y]\big\|\leq c_0 \|W_x\|\,\|W_y\|\exp\big(c_1 t-c_2|x-y|\big). \tag{7}$$ The preceding pages defines the "radius" of an operator $$W_x$$ to be "the region in which [other] local operators have an order-one commutator with [the given operator]." In other words, the radius of $$W_x(t)$$ is the largest value of $$|x-y|$$ for which the commutator $$[W_x(t),\,W_y]$$ is of order $$1$$. The bound (7) says that if $$|x-y|$$ is much larger than $$(c_1/c_2)t$$, then the magnitude of the commutator is exponentially suppressed (assuming $$c_2>0$$). Therefore, the radius is (softly) bounded by $$(c_1/c_2)t$$, which means that it cannot grow faster than linearly-in-time.