Lieb-Robinson bounds describe how effects are propagated through a system due to a local Hamiltonian. They are often described in the form $$ \left|[A,B(t)]\right|\leq Ce^{vt-l}, $$ where $A$ and $B$ are operators that are separated by a distance $l$ on a lattice where the Hamiltonian has local (e.g. nearest neighbour) interactions on that lattice, bounded by some strength $J$. The proofs of the Lieb Robinson bound typically show the existence of a velocity $v$ (that depends on $J$). This is often really useful for bounding properties in these systems. For example, there were some really nice results here regarding how long it takes to generate a GHZ state using a nearest-neighbour Hamiltonian.

The problem that I've had is that the proofs are sufficiently generic that it is difficult to get a tight value on what the velocity actually is for any given system.

To be specific, imagine a one dimensional chain of qubits coupled by a Hamiltonian $$ H=\sum_{n=1}^N\frac{B_n}{2}Z_n+\sum_{n=1}^{N-1}\frac{J_n}{2}(X_nX_{n+1}+Y_nY_{n+1}), \tag{1} $$ where $J_n\leq J$ for all $n$. Here $X_n$, $Y_n$ and $Z_n$ represent a Pauli operator being applied to a given qubit $n$, and $\mathbb{I}$ everywhere else. Can you give a good (i.e. as tight as possible) upper bound for the Lieb-Robinson velocity $v$ for the system in Eq. (1)?

This question can be asked under two different assumptions:

  • The $J_n$ and $B_n$ are all fixed in time
  • The $J_n$ and $B_n$ are allowed to vary in time.

The former is a stronger assumption which may make proofs easier, while the latter is usually included in the statement of Lieb-Robinson bounds.


Quantum computation, and more generally quantum information, comes down to making interesting quantum states. Through works such as this, we see that information takes a certain amount of time to propagate from one place to another in a quantum system undergoing evolution due to a Hamiltonian such as in Eq. (1), and that quantum states, such as GHZ states, or states with a topological order, take a certain amount of time to produce. What the result currently shows is a scaling relation, e.g. the time required is $\Omega(N)$.

So, let's say I come up with a scheme that does information transfer, or produces a GHZ state etc. in a way that scales linearly in $N$. How good is that scheme actually? If I have an explicit velocity, I can see how closely matched the scaling coefficient is in my scheme compared to the lower bound.

If I think that one day what I want to see is a protocol implemented in the lab, then I very much care about optimising these scaling coefficients, not just the broad scaling functionality, because the faster I can implement a protocol, the less chance there is for noise to come along and mess everything up.

Further Information

$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}\newcommand{\proj}[1]{\left|#1\right>\left<#1\right|}$There are some nice features of this Hamiltonian which I assume make calculation easier. In particular, the Hamiltonian has a subspace structure based on the number of 1s in the standard basis (it is said to be excitation preserving) and, even better, the Jordan-Wigner transformation shows that all the properties of higher excitation subspaces can be derived from the 1-excitation subspace. This essentially means we only have to do the maths on an $N\times N$ matrix $h$ instead of the full $2^N\times 2^N$ matrix $H$, where $$ h=\sum_{n=1}^NB_n\proj{n}+\sum_{n=1}^{N-1}J_n(\ket{n}\bra{n+1}+\ket{n+1}\bra{n}). $$ There is some evidence that the Lieb-Robinson velocity is $v=2J$, such as here and here, but these all use a close to uniformly coupled chain, which has a group velocity $2J$ (and I assume the group velocity is closely connected to the Lieb-Robinson velocity). It doesn't prove that all possible choices of coupling strength have a velocity that is so bounded.

I can add a bit further to the motivation. Consider the time evolution of a single excitation starting at one end of the chain, $\ket{1}$, and what its amplitude is for arriving at the other end of the chain $\ket{N}$, a short time $\delta t$ later. To first order in $\delta t$, this is $$ \bra{N}e^{-ih\delta t}\ket{1}=\frac{\delta t^{N-1}}{(N-1)!}\prod_{n=1}^{N-1}J_n+O(\delta t^{N}). $$ You can see the exponential functionality that you would expect being outside the 'light cone' defined by a Lieb-Robinson system, but more importantly, if you wanted to maximise that amplitude, you'd set all the $J_n=J$. So, at short times, the uniformly coupled system leads to the most rapid transfer. Trying to push this further, you can ask, as a bit of a fudge, when can $$ \frac{t^{N-1}}{(N-1)!}\prod_{n=1}^{N-1}J_n\sim 1 $$ Taking the large $N$ limit, and using Stirling's formula on the factorial leads to $$ \frac{etJ}{N-1}\sim 1, $$ which suggests a maximum velocity of about $eJ$. Close, but hardly rigorous (as the higher order terms are non-negligible)!

$$ \newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}\newcommand{\proj}[1]{\left|#1\right>\left<#1\right|} $$

  • $\begingroup$ Have you computed the best LR-bound from the proofs for that model? How does it compare to the velocity you quote? $\endgroup$ – Norbert Schuch Apr 6 '18 at 13:36
  • 1
    $\begingroup$ Ok, I concede it is a quantum computing question, at least the way I interpret it now: "What is the choice of $J_n$ and $B_n$ (subject to some constraints) which yields the maximum velocity for information/state/... transfer." --- Is this the right interpretation? $\endgroup$ – Norbert Schuch Apr 7 '18 at 8:53
  • $\begingroup$ @NorbertSchuch Not quite. I want to be able to say "I've come up with a set of couplings that achieves a protocol with a certain scaling. That protocol is known to be constrained by Lieb-Robinson bounds. How close am I to saturating that constraint?" as a measure of how fast my protocol is. $\endgroup$ – DaftWullie Apr 9 '18 at 7:11
  • $\begingroup$ @DaftWullie So - is you question: "How close am I to being optimal", or "How close am I to some kind of bound (taking the tightest possible one)"? $\endgroup$ – Norbert Schuch Apr 9 '18 at 9:00
  • 1
    $\begingroup$ @user1271772 That is correct. $B(t)=e^{-iHt}B(0)e^{iHt}$ $\endgroup$ – DaftWullie May 18 '18 at 10:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.