Lieb-Robinson Bound in 2nd quantized description?

Question

Background

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):

$$ u_{n_1,n_2,\dots n_N} \equiv \frac{1}{\sqrt{N!}} \sum_{\text{permutations of $r_1,\dots,r_N$}}u_{n_1} (\vec r_1) \dots u_{n_N} (\vec r_N) \leftrightarrow (a_1^\dagger)^{n_1} (a_2^\dagger)^{n_2} \dots (a_N^\dagger)^{n_N} |0 \rangle$$

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.

Question

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?

Answer 1

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 (disclaimer: I'm an author).

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.

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