Let us restrict our discussion to bosons and adopt the convention First Quantised $\leftrightarrow $ Second Quantised Theory (we are following these Ashok Sen's Quantum Field Theory I of HRI institute's notes also on my google drive). Consider a single particle system's hamiltonian $\hat h$ and energy operator $\hat E$:

$$ \hat h \psi = \hat E \psi $$

with energy eigenstates $u_i$ and eigenvalues $e_n$

$$ \hat h u_n = e_n u_n$$

Now moving to an assembly of quantum mechanical Hamiltonians (many body system) $\sum_i h_i$ with an interacting potential $\hat v_{ik}$ (between two particles corresponds to a second quantised version (Page 16):

$$ \hat H_N = \sum_{i=1}^N \hat h_i + \frac{1}{2}\sum_{\substack i\neq j \\ i,j=1 }^N \hat v_{i,j} \leftrightarrow \sum_{n=1}^\infty e_n a_n^\dagger a_n + \frac{1}{2} \sum_{m,n,p,q=1}^\infty \Big(\int \int d^3 r_1 d^3 r_2 u_{m}(\vec r_1)^* u_{n}(\vec r_2)^* \hat v_{12} u_{m}(\vec r_1) u_{p}(\vec r_2) \Big) a^\dagger_m a^\dagger_n a_p a_q$$

where $e_n$ i the $n$th energy eigevalue, $u_i$ are the one-particle eigenstates and $a_i^\dagger$ is the creation operator of the $i$'th particle and they obey the energy eigenvalue system $h_i u_n = e_n u_n$. The symmetric wave function for $H_N$ corresponds as a second quantised version (Page 6):

$$ u_{n_1,n_2,\dots n_N} \equiv \frac{1}{\sqrt{N!}} \sum_{\text{Permutations of $r_1,\dots,r_N$}}u_{n_1} (\vec r_1) \dots u_{n_N} (\vec r_N) \leftrightarrow (a_{n_1}^\dagger) (a_{n_2}^\dagger) \dots (a_{n_N}^\dagger) |0 \rangle$$

Now, in QM one can derive the Lieb-Robinson bound as:

$$ || [ O_A(t), O_B(0) ] || < C e^{- \frac{L-vt}{\eta}}$$

with $|| A ||$ is the norm of the operator, $ O_A(t) = e^{ i H t} O e^{- i H t} $ where the operators $O_A$ and $O_B$ act non-trivially on the subsystems $A$ and $B$ and identity outside it.


Since the mathematical machinery is only different (QFT vs QM) and the physical system is the same. How does one derive the Lieb-Robinson (or equivalent) bound in the Second Quantized Theory?

  • $\begingroup$ For those who cannot access the original link here it on my google drive: drive.google.com/file/d/1lC-B4mTtZQm_iO-qe_vriClaw0pfRzrd/view $\endgroup$ Jan 22 at 11:09
  • $\begingroup$ I'm a bit unclear why you spend so much time explaining the formalism of 2nd quantization in your question. Your question could be considerably shortened. Also, your question stipulates there is a LR-bound for bosons. $\endgroup$ Jan 22 at 12:35
  • $\begingroup$ Cross posted: quantumcomputing.stackexchange.com/questions/15636/… $\endgroup$ Jan 22 at 12:35
  • $\begingroup$ Is this on the lattice or in the continuum? To duplicate my comment from QCSE: In QFT (->your tag), there is the a concept called "speed of light", which could be seen as an analogue of LR-bounds, if you wish. $\endgroup$ Jan 22 at 12:37
  • $\begingroup$ @NorbertSchuch It is meant for those who are not familiar with subject. Those familiar can skip that part. I would be very surprised if there isn't since I'm describing the same physical system but with different mathematical machinery. $\endgroup$ Jan 22 at 12:39

In the Lieb-Robinson bound, the velocity depends on the strength (operator norm) of the interaction. This is intuitive: Twice as strong couplings will propagate information twice as fast (effectively, you can think of this as renormalizing time).

Here comes the catch with bosonic systems: For bosons, the norm of interactions is unbounded (e.g. $a^\dagger a$ can take any value $n$). Thus, the proof of Lieb-Robinson bounds cannot be transferred to bosonic systems.

This allows to for instance construct bosonic systems where information can travel at arbitrary speed, if you just put enough energy into the system: Supersonic quantum communication. What this result tells you is that Lieb-Robinson bounds for bosonic systems will only make sense in a setup where the energy is bounded (which is indeed generally necessary to get well-behaved physics with bosons, and also physically reasonable).

On the other hand, one can prove Lieb-Robinson type bounds in certain scenarios, namely when it is about the propagation of the bosons themselves into an initially unoccupied region, rather about the propagation of information in a general bosonic system: Information propagation for interacting particle systems.

To the best of my knowledge, the general question -- whether information in a bosonic system in a general state can only travel at a finite speed, as long as the energy is suitable bounded -- is still an open question.

Note: Since the question got cross-posted to qc.se, I also cross-posted the answer.

  • $\begingroup$ Thank you for your super swift answer. I'm just going through all the links. $\endgroup$ Jan 22 at 12:54
  • 1
    $\begingroup$ If the energy/$n$ is bounded, doesn't using a boson->spin mapping recover the original Lieb-Robinson bound? $\endgroup$
    – zeldredge
    Jan 23 at 21:28
  • $\begingroup$ @zeldredge What is "energy/$n$"? And what do you mean by "boson->spin" mapping? The numbers of bosons in a mode (an this is what we are talking about - a lattice of harmonic oscillators) is unbounded. I don't see how you would map this to spins. Of course, if you bound the number of bosons per mode, then you can map back to spins and recover some LR-bound (which, unfortunately, will depend on the cutoff). $\endgroup$ Jan 23 at 21:58
  • $\begingroup$ Sorry, I misread/read too fast and see now that by "the energy" in the last paragraph you mean the total system energy rather than the the number of bosons on a single site. $\endgroup$
    – zeldredge
    Jan 24 at 12:42
  • $\begingroup$ @zeldredge I didn't mean something too specific. One would have to look up the mathematical literature on bosonic systems to see what suitable bounds look like. You could also imagine that you impose a constraint that the expectation value of the energy per site is bounded: This would be local yet not impose a cutoff. Finally, for a "good" LR-bound you would like the speed to be independent of the cutoff, which is not the case if you truncate the Fock space. $\endgroup$ Jan 24 at 12:49

Your Answer

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

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