The Out-of-time-ordered correlator (OTOC) is defined as $\langle[W(t),V(0)]^2\rangle$, and can be considered as a new way to extract quantum chaos. However, the understanding of this special correlator is not clear. For a simple example, what the free fermion OTOC behaves? Both lattice or continuum case are welcome.

  • $\begingroup$ Since OTOC can detect the scarmbling of quantum dynamics system, the interest lies in quantum many-body interacting systems to reveal its quantum chaotic behaviour (some works on interacting fermion system, e.g. journals.aps.org/pra/abstract/10.1103/PhysRevA.95.011601, and for those work with SKY model you can find them easily) . For free fermion, this correlator will simply trivial and free of Lyapunov exponent (chaos doesn't occur in such integrable/dynamic determined system) and stay constant, which depends on what local obserables you choose and calculate with fermion commutator. $\endgroup$
    – Tom Gao
    May 25, 2017 at 14:17

1 Answer 1


First of all, your definition of the OTOC is wrong. It is supposed to be composed of the static and time-evolved versions of the same operator, so it should read $C(t) = - \langle [\hat{V}(t),\hat{V}(0)]^2 \rangle$. The minus sign has become standard in the field.

The understanding of this function is not that hard to grasp: if $\hat{V}(t)$ and $\hat{V}(0)$ commute for all times, nothing happens. Your question, regarding the free fermion case, can be generalised for all cases where $\hat{V}$ commutes with the Hamiltonian: nothing happens. This makes sense, since if you're looking for something that could be called chaos, then it shouldn't arise with such a trivial time evolution. But if your time evolution is interesting enough to provide operators that stop commuting, then your OTOC starts giving off meaningful stuff. It was postulated that a special (momentum) OTOC version growth rate should be closely tied with Lyapunov exponents, something that didn't turn out to be exactly true, but close enough in some aspects. The reason behind this postulate regarding the momentum-OTOC is that, using a test function $f$ and the static position basis,

\begin{align} [\hat{p}(t), \hat{p}(0) ]f &= -i \hbar \left[ \hat{p}(t) \frac{\partial f}{\partial q(0)} - \frac{\partial}{\partial q(0)}(\hat{p}(t) f) \right] \\ &= (i \hbar f) \frac{\partial \hat{p}(t)}{\partial q(0)} \, , \end{align}

so that

\begin{align} C(t) &= -\langle [\hat{p}(t), \hat{p}(0) ]^2 \rangle \\ &= \hbar^2 \bigg\langle \left( \frac{\partial \hat{p}(t)}{\partial q(0)} \right)^2 \bigg\rangle \\ &\approx \bigg\langle \left( \frac{\Delta{p}(t)}{\Delta q(0)} \right)^2 \bigg\rangle_{\text{phase space}} \quad (\equiv C_{cl}(t))\, , \end{align}

that is, for times smaller than Ehrenfest's time, this approximation should be quite reasonable. Now, if we're talking about a chaotic map, it means that for small deviations on position we'll have very large (exponential) deviations on momentum. This suggests that, after a little analysis,

\begin{align} C_{cl} (t) &\approx (e^{ \Lambda t})^2 \\ \Rightarrow \Lambda &= \lim_{t \to \infty} \lim_{\Delta q(0) \to 0} \frac{1}{2t} \ln \left[ \frac{C_{cl}(t+1)}{C_{cl}(1)} \right] \, . \end{align}

The classical expression for the Lyapunov exponent is given by

\begin{align} \lambda = \lim_{t \to \infty} \lim_{\delta Z(0) \to 0} \frac{1}{t} \ln \left( \frac{|\delta Z(t)|}{|\delta Z(0)|} \right) \, , \end{align}

where $\delta Z$ represents the separation between trajectories and $|.|$ the norm. Notice how $\Lambda$ and $\lambda$ are very similar. This motivates the study of OTOCs in the context of quantum chaos.

  • $\begingroup$ First thank you for answering. However I don't think the minus sign is by any means conventional or standard. For a very recent paper Phys. Rev. Lett. 120, 110603 (2018), the equation (1) is clearly without minus sign. I can see some of researchers might be more comfortable with a minus sign, such as in Phys. Rev. Lett. 118, 086801 (2017), but it's not normal. $\endgroup$ Mar 18, 2018 at 18:19
  • $\begingroup$ And I actually have a question. Is that the OTOC of a state would be interesting only with some sophisticated choice of operators? And even if I choose two sets of operators that have some non-trivial property, the two corresponding OTOC could behave differently? $\endgroup$ Mar 18, 2018 at 18:21
  • $\begingroup$ I am used to the definition with the minus sign and didn't know people used it with a plus sign too. Now, what do you mean by sophisticated operators? My example is nothing but two momenta... Quite peasant. $\endgroup$ Oct 5, 2018 at 0:42
  • $\begingroup$ Dear @QuantumBrick, could you please introduce some introductory materials (books or papers) for me on OTOC, I am recently starting to learn this but have no clue where to start, thanks a lot. $\endgroup$
    – guangcun
    Sep 23, 2022 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.