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.

  • $\begingroup$ 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. $\endgroup$ – stafusa Mar 17 at 2:21
  • $\begingroup$ 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? $\endgroup$ – 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.

[1] Roberts et al (2014), "Localized shocks," https://arxiv.org/abs/1409.8180


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