0
$\begingroup$

So we know when one goes from QM to QFT Lieb Robinson bounds become micro causality. But micro causality is a statement on the commutators assuming they are space-like, time-like or light-like. Naively I would guess the direct involvement of the position operator. How do Lieb-Robinson Bounds talk about locality without the position operator in QM and QFT?

$\endgroup$
4
  • 1
    $\begingroup$ Lieb-Robinson bound is about local operators in a many-body system with locality. Local operators are labeled by their positions, so here position is a parameter. You can also define position operator for a many-body system, but LB bound has nothing to say about that. $\endgroup$
    – Meng Cheng
    May 27, 2022 at 13:35
  • $\begingroup$ So let's say I have a Hamiltonian with some potential energy. I shouldn't be able to (accurately) label the position and energy at the same time since the operators don't commute, no ? $\endgroup$ May 27, 2022 at 13:39
  • $\begingroup$ I think the problem you have is that you are trying to think about LB in the context of quantum mechanics. In QM you really don't have the notion of local operators, so it does not make sense to talk about LB. You can define position and momentum operators in many-body systems, and they will reduce to usual $x$ and $p$ in QM if there is only 1 particle, so there is no contradiction. $\endgroup$
    – Meng Cheng
    May 27, 2022 at 13:43
  • $\begingroup$ @MengCheng perhaps an explicit demonstration would be a worthwhile answer? $\endgroup$ May 27, 2022 at 14:23

0

Your Answer

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