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). 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.
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?