# Background

The Lieb-Robinson bound is said to provide some notion of causality in non-relativistic quantum systems with bounded local interactions. Formally, it can be stated as follows.

Theorem (Lieb-Robinson): Consider a $k$-local Hamiltonian $H = \sum_{Z} h_{Z}$, where the sum runs over subsets $Z$ of qudits. Assume there exist constants $\mu, s > 0$ such that, for all qudits $i$, we can bound the interaction strengths as $\sum_{Z \ni i} \lVert h_{Z} \rVert \leq s e^{-\mu}$, where $\lVert \cdot \rVert$ is the operator norm. Now consider operators $A_{X}$ and $B_{Y}$ which are supported on subsets $X$ and $Y$ of qudits respectively. Define the time-evolved operator $A_{X}(t) = U^{\dagger}(t) A_{X} U(t)$, where $U(t) = \exp(-i H t)$. Then we have the following bound:

$$\lVert \left[ A_{X}(t) , B_{Y} \right] \rVert \leq 2 \lVert A_{X} \rVert \lVert B_{Y} \rVert \min\left\{|X|, |Y|\right\} e^{-\mu (d_{H}(X,Y) - v_{LR} t)},$$

where $d_{H}(X,Y)$ is the interaction distance between subsets $X$ and $Y$, and $v_{LR} = 2 k s / \mu$ is the Lieb-Robinson velocity.

This is said to capture the notion of causality because the appearance of the $e^{-\mu (d_{H}(X,Y) - v_{LR} t)}$ factor shows that the commutator $\left[ A_{X}(t) , B_{Y} \right]$ dies off exponentially outside the "lightcone" defined by $d_{H}(X,Y) - v_{LR} t \leq 0$.

# Motivation for question

My question is essentially about why it is natural to consider the quantity $\left[ A_{X}(t) , B_{Y} \right]$ where only one-operator is time-evolved, rather than $\left[ A_{X}(t) , B_{Y}(t) \right]$ where both operators are evaluated at the same time. Intuitively, the latter seems more natural to me, since it seems like it should capture something about the behaviour of the overlap of these operators as they evolve in time. However, it fails to do so, as the following argument shows.

Suppose the subsets $X$ and $Y$ are disjoint, so initially $\left[ A_{X}, B_{Y} \right] = 0$. Then notice that

$$\left[ A_{X}(t) , B_{Y}(t) \right] = \left[ U^{\dagger}(t) A_{X} U(t) \,,\, U^{\dagger}(t) B_{Y} U(t) \right] = U^{\dagger}(t) \left[ A_{X}, B_{Y} \right] U(t) = 0,$$

so if the operators initially commute, they also commute at later times.

Generically, for arbitrary $t > 0$ the operators $A_{X}(t)$ and $B_{Y}(t)$ will have support on the whole Hilbert space, yet the above calculation shows that if they initially act on disjoint subsets then they will still commute for all times, despite their overlap. So, clearly, $\left[ A_{X}(t) , B_{Y}(t) \right]$ is not the right quantity to consider if we're interested in notions of causality, yet the only argument I have for why is the rather unsatisfying "because it doesn't tell us anything non-trivial".

# Question

Is there a way to see that $\left[ A_{X}(t) , B_{Y} \right]$, where only one-operator is time-evolved, rather than $\left[ A_{X}(t) , B_{Y}(t) \right]$, where both operators are evaluated at the same time, is the more natural quantity to consider when dealing with notions of causality? Specifically I am looking for an answer that is not "because the latter doesn't tell us anything non-trivial".

$$\lVert \left[ A_{X}(t) , B_{Y} \right] \rVert = \lVert \left[ A_{X}(t/2) , B_{Y}(-t/2) \right] \rVert$$

due to the unitary invariance of the operator norm. In other words, evolving one operator forwards in time by $t$ is equivalent (at least for the purposes of the Lieb-Robinson bound) to evolving both operators in time by $t/2$, but one forwards in time and one backwards in time. However, I'm not sure why this might make this a more "natural" quantity to consider.

• "Intuitively, the latter seems more natural to me, since it seems like it should capture something about the behaviour of the overlap of these operators as they evolve in time." --- Why would you want to say sth. about "the overlap of operators as they evolve in time"? $A(t)$ descibes the action of an operator which has acted a time $t$ ago. So you are suggesting to look at the relation of two operators which have both acted the same time $t$ ago. Unless you break time-independence, $t$ is completely irrelevant here -- as you correctly observed. – Norbert Schuch Jul 6 '18 at 16:51

You want to see that there is a light-cone like structure: If Alice applies some operation at time $0$ and position $0$ and Bob at time $t$ and position $x$, their actions commute (i.e., are space-like separated) if Bob is outside the lightcone of Alice. (Just try to think of a Minkowski diagram.) This is formalized by the fact that Alice's time evolved observable $A(t)$ commutes with Bob's observable.
Let us do this more formally: Let's say Alice tries to pass a message to Bob by applying some unitary $U_A$ at time $0$, starting from an initial state $|\psi\rangle$. Bob then tries to infer through a measurement $M_B$ at time $t$ whether Alice has applied $U_A$ (or which $U_A$ she applied). ($M_B$ could, e.g., be a POVM element whose outcome probability would give information about $U_A$.)
The state after time $t$ is then $$|\psi(t)\rangle = e^{-iHt}U_A|\psi\rangle\ ,$$ and Bob's measurement $M_B$ will yield \begin{align*} \langle \psi(t) | B |\psi(t)\rangle &= \langle\psi|U_A^\dagger e^{iHt}M_B e^{-iHt}U_A|\psi\rangle \\ &= \langle\psi|U_A^\dagger M_B(t) U_A|\psi\rangle\ . \end{align*} As long as we are inside the Lieb-Robinson-cone, this is, Alice and Bob are spacelike separated, $U_A$ and $M_B(t)$ commute (up to an exponentially small correction, the terms arising from which can be easily bounded), and thus \begin{align*} \langle \psi(t) | B |\psi(t)\rangle &= \langle\psi|U_A^\dagger M_B(t) U_A|\psi\rangle \\ &= \langle\psi|U_A^\dagger U_A M_B(t)|\psi\rangle \\ &= \langle\psi|M_B(t)|\psi\rangle\ , \end{align*} i.e., the measurement outcome of Bob is independent of $U_A$, and thus, Alice cannot pass any information to Bob.